Matra Meru in the Meru Prastaar

By Chandrahas M. Halai

This is in continuation of my earlier articles:

https://chandrahasblogs.wordpress.com/2020/10/04/meru-prastaar-%E0%A4%AE%E0%A5%87%E0%A4%B0%E0%A5%81-%E0%A4%AA%E0%A5%8D%E0%A4%B0%E0%A4%B8%E0%A5%8D%E0%A4%A4%E0%A4%BE%E0%A4%B0%E0%A4%83/

https://www.indica.today/quick-reads/virahanka-numbers-part-iv/

https://chandrahasblogs.wordpress.com/2022/09/05/virahanka-sankhyanks-in-the-meru-prastaar/

Figure 1

Here Figure 1 gives us the Meru Prastaar.

Figure 2

Here Figure 2 gives us the Matra Meru.

Here Table 1 gives us the Virahanka Sankhayas in the slant diagonals of the Meru Prastaar.

From the above tables it is clear that the Matra Meru is embedded within the Meru Prastaar. Refer Figure 3, here each row of the Matra Meru is coded in the same colour.

Figure 3

Do you want to Bell the CAT?

Author: Chandrahas M. Halai

The 288^th term of the series a, b, b, c, c, c, d, d, d, d, e, e, e, e, e, … is

There are 8436 steel balls, each with a radius of 1cm, stacked in a pile, with 1 ball on top, 3 balls in the second layer, 6 balls in the third, 10 in the fourth and so on. Find the number of horizontal layers in the pile?

Interesting and challenging problems, aren’t they?
The above two problems were asked in CAT, the entrance exams for admission into India’s most prestigious Business schools “The IIMs”. In these exams you are expected to solve each problem in less than two minutes. These questions are multiple choice questions. But, you get negative marks for the wrong choice.

So, how do you solve such problems not only accurately but also quickly?

Let us begin by solving the first problem.

We have one ‘a‘, two ‘b‘s, three ‘c‘s, four ‘d‘s, and so on. These can be arranged in the shape of a triangle. Hence, it is obvious we are dealing with triangular numbers. For more on triangular numbers refer chapter 11 “How Many Spheres?” from my new book “Meru Prastaar“. Available at https://garudabooks.com/meru-prastaar

Triangular Numbers

We have to find the row number n for the 288^th term. Hence, we have

Triangular numbers are given by the formula:

Thus, we have

and

We have,

Thus the 288^th term lies in the 24^th row. And the 24^th character in the English alphabets is ‘x‘. Hence, the 288^th term in the given series will be ‘x‘.

Now, let us solve the second problem.

We have 1 ball on top, 3 balls in the second layer, 6 balls in the third, 10 in the fourth and so on. Now 1, 3, 6 and 10 are triangular numbers. Hence, here we have a stack of triangles. And a stack of triangles is a tetrahedron. It is now obvious that 8436 is a tetrahedral number. To understand tetrahedral numbers read chapter 12 “Aryabhata’s Sum of Sums” from my new book “Meru Prastaar“.

 

Tetrahedral Numbers

The nth tetrahedral number is given by:

Hence, we have

We have

Hence, we have n = 36

Thus, we will have 36 horizontal layers of triangles.

“Meru Prastaar can Bell the CAT“.

 

 

Virahanka Sankhyanks in the Meru Prastaar

By Chandrahas M. Halai

This is in continuation of my earlier articles:

Meru Prastaar (मेरु प्रस्तारः)

 

Virahanka Numbers – Part I

 

https://www.indica.today/tag/virahanka-series/

What we popularly call as Fibonacci numbers are predated by Virahanka sankhyas.

In this article we derive a direct combinatorial formula to calculate the Virahanka sankhya.

 

Refer Table 1. This is left justified Meru Prastaar. If we add up the numbers in the shallow diagonals (represented by the lines in Table 1) we get Virahanka Sankhyanks.

The shallow diagonal that starts from row 0, gives us the 0^th Virahanka sankhya, which is 1. The shallow diagonal that starts from row 1, gives us the 1^st Virahanka sankhya, which is 1.

The shallow diagonal that starts from row 2, gives us the 2^nd Virahanka sankhya. This is = 1 + 1 = 2.

The shallow diagonal that starts from row 3, gives us the 3^rd Virahanka sankhya. This is = 1 + 2 = 3.

The shallow diagonal that starts from row 4, gives us the 4^th Virahanka sankhya. This is = 1 + 3 + 1 = 5.

The shallow diagonal that starts from row 5, gives us the 5^th Virahanka sankhya. This is = 1 + 4 + 3 = 8.

The shallow diagonal that starts from row 6, gives us the 6^th Virahanka sankhya. This is = 1 + 5 + 6 +1 = 13.

And so on.

Refer Table 2. This gives the combinatorial formula for the binomial coefficients. Using Table 2, we can calculate the Virahanka sankhyas using the combinatorial formulas. Thus we have:

Do we discern a pattern here? Can we have a formula to calculate the Virahanka sankhyank for a quarter having n matras?

Let us consider a quarter having 5 matras. The metrical forms in a 5 matra prastaar will only have odd numbers of Laghus, i.e. 1, 3 and 5. To put it another way, the metrical forms in a 5 matra prastaar will have 0, 1 and 2 Gurus. The metrical forms in a 5 matra prastaar will have at least 1 Laghu syllable.

The total metrical forms in a 5 matra prastaar are

If there is 1 Guru in the metrical forms then there will be 5 – 1 = 4 syllables in the quarter. If there are 2 Gurus in the metrical forms then there will be 5 – 2 = 3 syllables in the quarter.

In general, for a n matra prastaar, if there are r Gurus in the metrical forms then there will be n – r syllables in the quarter.

Let us now consider a quarter having 6 matras. The metrical forms in a 6 matra prastaar will only have even numbers of Laghus, i.e. 0, 2, 4 and 6. To put it another way, the metrical forms in a 6 matra prastaar will have 0, 1, 2 and 3 Gurus.

The total metrical forms in a 6 matra prastaar are

Let us now consider a quarter having 7 matras. The metrical forms in a 7 matra prastaar will only have odd numbers of Laghus, i.e. 1, 3, 5 and 7. To put it another way, the metrical forms in a 7 matra prastaar will have 0, 1, 2 and 3 Gurus. The metrical forms in a 7 matra prastaar will have at least 1 Laghu syllable.

The total metrical forms in a 7 matra prastaar are

Let us now consider a quarter having 8 matras. The metrical forms in a 8 matra prastaar will only have even numbers of Laghus, i.e. 0, 2, 4, 6 and 8. To put it another way, the metrical forms in a 8 matra prastaar will have 0, 1, 2, 3 and 4 Gurus.

The total metrical forms in a 8 matra prastaar are

From above, we observe that:

In general in a n matra vrutta prastaar, if n is even the metrical forms will have only even number of Laghus. And if n is odd the metrical forms will have only odd number of Laghus.

In general in a n matra vrutta prastaar, if n is even the metrical forms can have maximum n/2 number of Gurus. And if n is odd the metrical forms can have maximum (n – 1)/2 number of Gurus.

Therefore, the total metrical forms in a n matra prastaar are:

1. If n is even
   

   

   
   

   The number of terms that are added in this formula will be n/2 + 1.
   

    

2. If n is odd
   

   

The number of terms that are added in this formula will be (n + 1)/2.

Thus we now have a direct combinatorial formula for calculating the Virahanka sankhyas.

Rectangular Numbers

By Chandrahas M. Halai

In my earlier article

https://chandrahasblogs.wordpress.com/2022/03/18/sum-of-odd-numbers/

I had shown that the sum of first n positive odd numbers gives us the square of n.

You must be wondering what would be the sum of first n positive even numbers? Let us find out.

We have, the first positive even number is 2 = 1 x 2

Next, we have, 2 + 4 = 6 = 2 x 3

Next, we have, 2 + 4 + 6 = 12 = 3 x 4

Next, we have, 2 + 4 + 6 + 8 = 20 = 4 x 5

Next, we have, 2 + 4 + 6 + 8 + 10 = 30 = 5 x 6

Next, we have, 2 + 4 + 6 + 8 + 10 + 12 = 42 = 6 x 7

From this discussion can we say that the sum of first n positive even numbers is

Let us prove it.

Let 

The nth even number is given by 2n.

We have

Let us now add (1) and (2), we have

Thus it is proved.

The n(n+1) number of things can be arranged in a rectangle as shown in Figure 1. Hence, these numbers are called Rectangular numbers or Oblong numbers. In figure 1, the numbers in the first column are rectangular numbers.

Figure 1

The numbers which are a product of two consecutive numbers are also called Pronic numbers.

Pronic numbers are twice the triangular numbers. Refer my earlier article on Triangular numbers.

https://chandrahasblogs.wordpress.com/2022/03/09/triangular-numbers/

Sum of Odd Numbers

By Chandrahas M. Halai

We have,

Here, n is any positive integer.

Also, 2*n will always be even. Hence, 2n+1 will always be odd.

Let us begin with ,

We have,

From the above discussion, can we say that the sum of first n positive odd numbers gives . Refer Figure 1. The numbers in the top row in this figure are all square numbers.

Figure 1

Let us prove it.

The nth odd number is given by 2n-1. Here, n is any positive integer.

Adding (1) and (2), we get

Hence, we have

Thus, it is proved.

Tetrahedral Numbers, Part 2

By Chandrahas M. Halai

This is in continuation of my earlier article:

https://chandrahasblogs.wordpress.com/2022/03/11/tetrahedral-numbers/

In this article we will take an alternate view of addition of two consecutive tetrahedral numbers.

Before adding the fifth and the sixth tetrahedral numbers let us have a relook at some basics.

We have

This can be rewritten as

Now let us prove that adding two consecutive tetrahedral numbers gives us a square pyramidal number.

We have

This can be rewritten as

Hence it is proven.

Tetrahedral Numbers

By Chandrahas M. Halai

This is in continuation of my earlier articles:

https://chandrahasblogs.wordpress.com/2021/07/30/how-many-laddoos/

https://chandrahasblogs.wordpress.com/2021/08/02/aryabhatts-sum-of-sums/

https://chandrahasblogs.wordpress.com/2022/03/09/triangular-numbers/

In my earlier article I had shown how adding two consecutive triangular numbers gives you a square number. Now let us explore what happens when we add two consecutive tetrahedral numbers. Refer Figure 1.

Figure 1

Tetrahedral numbers are given by

Let us add the fourth and fifth triangular numbers:

Now, let us add the fifth and sixth triangular numbers:

Now, 55 and 91 are the fifth and sixth square pyramidal numbers. Refer Figure 2.

Figure 2

Square pyramidal numbers are given by

Let us calculate the 5th square pyramidal number:

Let us now calculate the 6th square pyramidal number:

Can we say that the sum of two consecutive tetrahedral numbers gives us a square pyramidal number? Let us prove it.

We have,

Thus, it is proved.

The sum of two consecutive tetrahedral numbers gives us a square pyramidal number.

Triangular Numbers

By Chandrahas M. Halai

This is in continuation of my earlier article:

https://chandrahasblogs.wordpress.com/2021/07/30/how-many-laddoos/

In the earlier article I had described the Triangular Numbers. The 5th triangular number is given by:

This is shown in Figure 1.

Figure 1

The spheres can be rearranged as shown in Figure 2.

Figure 2

Now, the 6th triangular number is

This is as shown in figure 3.

Figure 3

Now, let us add the 5th and 6th triangular numbers, we get:

This is as shown in figure 4.

Figure 4

Let us now add the 7th and 8th triangular numbers:

Now let us add the 10th and 11th triangular numbers:

From the above discussion we can say that adding two consecutive triangular numbers gives a square number. That is

Now, let us prove this.

We have

Hence, we have

Iterative Search Algorithms

Iterative Search Algorithms for unique numbers that satisfy given conditions

– Chandrahas M. Halai

Find the unique four-digit number, where all the digits are not the same, such that when the number that you get by arranging the digits in ascending order is subtracted from the number that you get by arranging them in descending order you get the same four-digit number.

Let us choose any four-digit number and test whether it satisfies the given condition. Let the number be 1947 the year our country got its independence.

Arranging the digits in 1947 in descending order we get 9741 and arranging them in ascending order we get 1479. Now, subtracting 1479 from 9741 we get:

9741 – 1479 = 8262

Now, let us do the same operations on 8262 as the new input. We get:

8622 – 2268 = 6354

Now, let us do the above operations on every new output in iteration. We get:

6543 – 3456 = 3087

8730 – 0378 = 8352

8532 – 2358 = 6174

7641 – 1467 = 6174

Since, we are getting the same output for 6174, we have found the number we were looking out for.

Let us try the same procedure on one more four-digit number. Let that number be the current year 2021.

2210 – 0122 = 2088

8820 – 0288 = 8532

8532 – 2358 = 6174

7641 – 1467 = 6174

Again, we get the same number 6174.

If you do these operations on any four-digit number in which all the digits are not the same (a four-digit number consisting at least two different digits), then you will get the result as 6174 in not more than seven steps. This amazing discovery was made by Indian mathematician Shri Dattatreya Ramchandra Kaprekar (1905 – 1986) in the year 1949. In honour of Shri Kaprekar the number 6174 is called Kaprekar’s constant and the above procedure is called Kaprekar’s operations.

Let us define Kaprekar’s operations as a function k(x). The iterations in Kaprekar’s operations stop when k(a) = a. Here, we now know that a = 6174. That is k(6174) = 6174.

Let us say we begin Kaprekar’s operation with an initial four-digit number x₁. The operations will stop when you reach the final value x_n, where x_n = 6174. Before reaching the final value x_n, we get all the intermediate values, x₂, x₃, …, x_n-1. All these values form what is known as the Kaprekar’s sequence. 6174, is the limiting value of these sequences.

From the above two examples we have the Kaprekar’s sequences as:

1) 1947, 8262, 6354, 3087, 8352, 6174

2) 2021, 2088, 8532, 6174

Let us generate a few more Kaprekar’s sequences:

3) 3355, 2178, 7443, 3996, 6264, 4176, 6174

4) 4434, 0999, 8991, 8082, 8532, 6174

5) 3579, 6174

6) 1357, 6174

7) 0246, 6174

8) 2468, 6174

Now, let us look at one more interesting number puzzle.

After a few hours of getting a new mobile phone number I forgot it. In India, we have ten digit mobile phone numbers. Fortunately, I had paid premium amount to get a special number. In this special number, the first digit (from left to right) is equal to number of zeroes in that number, the second digit is equal to the number of ones in that number, third digit is equal to the number of twos in that number and so on and so forth. For example, if first digit of number is 4 then there are four zeroes in that number. There is only one such ten-digit number. So, what is my mobile phone number?

The desired number is actually a self-descriptive number in base 10 (decimal number) number system.

A self-descriptive number is an integer m that in a given base b is b digits long in which each digit d at position n (the most significant digit being at position 0 and the least significant at position b−1) counts how many instances of digit n are in m.

Solution:

There are ten-billion ten-digit numbers from 0000000000 to 9999999999. Finding that one unique number amongst ten-billion is like finding a needle in a haystack.

I have developed an algorithm that constructs the number using the specified conditions.

We can begin solving the problem with a clean slate, i.e.

00000 00000

(I am breaking the number into two groups of five digits for better readability.)

Here we have ten zeroes. Hence, as per given conditions we should put a 10 in the first digit. Firstly 10 is not a digit and secondly putting any digit other than zero in the place of first digit will alter our count of zeroes in the number. The number of zeroes will be decreased by one, that is, now the number of zeroes will now be 9. Let us now put 9 in the first digit.

Hence, now our number will be

90000 00000

Our algorithm will work on this number to arrive at the required number. We begin by checking every digit in the number to ensure that it contains the right count of the corresponding digits. If not, then the right count is put in that particular place.

First pass or First Iteration:

(You will understand later why I am calling this step the First pass / iteration)

Checking the First digit:

This digit should have the count of zeroes in the number.

We see that it contains 9 and when we count the number of zeroes we see that it is also 9. Hence no change is made to the number. The number will be:

90000 00000

The digits from second to the ninth all contain zeroes and they are correct.

Checking the Tenth or the Last digit:

This digit should have the count of 9s in the number.

We see that there is one 9 in the number. Hence, we put a 1 in that place. Thus, our number becomes:

90000 00001

Have we got the desired ten-digit number? Let us check every digit of the number once again.

Second pass or Iteration:

    Checking the First digit:

At present this place contains a 9. Let us count the number of zeroes in the number. We see that there are eight zeroes instead of 9. Hence, we will replace the 9 with 8. Thus, the number becomes:

80000 00001

Checking the Second digit:

This digit should have the count of 1s in the number. At present it contains a 0. Let us count the number of 1s in the number. We see that there is one 1 (in the last digit). Hence, we will put a 1 in that place. Thus, the number becomes:

81000 00001

The digits from third to the Eighth all contain zeroes and they are correct.

Checking the Ninth digit:

This digit should have the count of 8s in the number. At present it contains a 0. Let us count the number of 8s in the number. We see that there is one 8 (in the first digit). Hence, we will put a 1 in that place. Thus, the number becomes:

81000 00011

Checking the Tenth or the Last digit:

This digit should have the count of 9s in the number.

At present that place contains a 1. Let us count the number of 9s in the number. We find that there are no 9s in the number. Hence, we put a 0 in that place. Thus, our number becomes:

81000 00010

Have we got the desired ten-digit number? Let us check every digit of the number once again. We are going to repeat the same steps (procedure) once again. We are going into a loop or iteration (iteration means repetition). When we find that all the digits contain the right count, we come out of the loop.

Third pass or Iteration:

    Checking the First digit:

At present this place contains an 8. Let us count the number of zeroes in the number. We see that there are seven zeroes instead of 8. Hence, we will replace the 8 with 7. Thus, the number becomes:

71000 00010

Checking the Second digit:

At present this place contains a 1. Let us count the number of 1s in the number. We see that there are two 1s instead of one. Hence, we will replace the 1 with 2. Thus, the number becomes:

72000 00010

Checking the Third digit:

At present it contains a 0. Let us count the number of 2s in the number. We see that there is one 2. Hence, we will put a 1 in that place. Thus, the number becomes:

72100 00010

The digits from fourth to the seventh all contain zeroes and they are correct.

Checking the Eighth digit:

At present it contains a 0. Let us count the number of 7s in the number. We see that there is one 7 (in the first digit). Hence, we will put a 1 in that place. Thus, the number becomes:

72100 00110

Checking the Ninth digit:

At present that place contains a 1. Let us count the number of 8s in the number. We find that there are no 8s in the number. Hence, we put a 0 in that place. Thus, our number becomes:

72100 00100

The last digit contains a zero and that is correct. Hence there will be no change in the number.

Have we got the desired ten-digit number? Let us check every digit of the number once again.

Fourth pass or Iteration:

    Checking the First digit:

At present this place contains a 7. Let us count the number of zeroes in the number. We see that there are six zeroes instead of 7. Hence, we will replace the 7 with 6. Thus, the number becomes:

62100 00100

The second and third digits contain a 2 and a 1 respectively and both are found to be correct.

The digits from fourth to the sixth all contain zeroes and they are correct.

Checking the Seventh digit:

At present this place contains a 0. Let us count the number of 6s in the number. We see that there is one 6 (in the first digit). Hence, we will put a 1 in that place. Thus, the number becomes:

62100 01100

Checking the Eighth digit:

At present that place contains a 1. Let us count the number of 7s in the number. We find that there are no 7s in the number. Hence, we put a 0 in that place. Thus, our number becomes:

62100 01000

The ninth and the last digits both contain a zero and that is correct. Hence there will be no change in the number.

Have we got the desired ten-digit number? Let us check every digit of the number once again.

Fifth pass or Iteration:

In this pass we find that all the digits are correct. No changes are made to the number in this pass. This means that we have got our desired mobile phone number.

Thus the mobile phone number is 62100 01000.

To get the mobile number we had applied the algorithm on the number 90000 00000. Instead of applying the algorithm on this, we can use any random ten-digit number and still get the correct answer.

I have written a program in Python to carry out the above operations on any ten-digit number. This can be found at:

https://chandrahasblogs.wordpress.com/2020/08/16/using-python-to-find-the-needle-in-the-haystack/

Let us try the above operations on another number, 9988776654. The results that we get after every iterations are as listed below:

Your 10 digit mobile number: 9988776654

[0, 0, 0, 0, 1, 1, 2, 0, 0, 0]

[7, 2, 2, 0, 0, 0, 0, 1, 0, 0]

[6, 1, 1, 0, 0, 0, 1, 0, 0, 0]

[6, 3, 0, 1, 0, 0, 1, 0, 0, 0]

[6, 2, 1, 0, 0, 0, 1, 0, 0, 0]

[6, 2, 1, 0, 0, 0, 1, 0, 0, 0]

Let me call these operations as the Chandrahas’ search operations. We can define these operations as a function c(x). The iterations in the search operations stop when c(a) = a. Here, we now know that a = 6210001000. That is c(6210001000) = 6210001000.

Let us say we begin the search operation with an initial ten-digit number x₁. The operations will stop when you reach the final value x_n, where x_n = 6210001000. Before reaching the final value x_n, we get all the intermediate values, x₂, x₃, …, x_n-1. All these values form what is known as the Chandrahas’ search sequence. 6210001000, is the limiting value of these sequences.

From the above two examples we have the Chandrahas’ search sequences as:

1) 9000000000, 9000000001, 8100000010, 7210000100, 6210001000

2) 9988776654, 0000112000, 7220000100, 6110001000, 6301001000, 6210001000

Let us generate a few more sequences:

3) 0123456789, 1221111111, 0710000100, 7210000100, 6210001000

4) 5461532497, 0111100100, 5500020000, 7010000100, 7210000100, 6210001000

The algorithms / operations to find the solutions for both the above problems are similar. They are iterative search algorithms. You begin with any random number as the initial value and perform operations on it. The output is used as the new input. The operations are performed in iterations till the desired value is reached. The above algorithms can be grouped in to a class of Iterative Search Algorithms.

There will be similar mathematical problems which can be solved using iterative search algorithms.

Naryan Pandit’s: Sum of Sums of Sums of Sums…

– Chandrahas M. Halai

This is in continuation of my earlier articles

https://chandrahasblogs.wordpress.com/2020/10/04/meru-prastaar-%e0%a4%ae%e0%a5%87%e0%a4%b0%e0%a5%81-%e0%a4%aa%e0%a5%8d%e0%a4%b0%e0%a4%b8%e0%a5%8d%e0%a4%a4%e0%a4%be%e0%a4%b0%e0%a4%83/

https://chandrahasblogs.wordpress.com/2021/07/30/how-many-laddoos/

https://chandrahasblogs.wordpress.com/2021/08/02/aryabhatts-sum-of-sums/

In the previous article I had discussed about the sum of sums. Let me start this article with the sum of sums of sums. That is, we want to find:

We get:

These are called the Pentatope numbers. They are the sum of Tetrahedral numbers. The fifth pentatope number is the sum of first five tetrahedral numbers. That is:

The above formula (i) and a general formula for sums of higher orders were derived by Naryan Pandit in his work Ganitakaumudi (1356 CE).

How about adding pentatope numbers? Sums of pentatope numbers are called 5-simplex numbers. Sums of sequences of natural numbers are in general called the Simplex numbers. Are we going to find formula for 5-simplex numbers and higher order sums recursively, like the way we have found the formulas for tetrahedral and pentatope numbers? Well, that will be a very cumbersome way to do it. Let us have a look at an alternative way of doing it.

Let me begin by making a table as shown table 1. First let us fill up column 0 with 1s. Now, put in cell of row1 and column 1, the sum of the cells above and above-left of it. That is the sum of columns 1 and 0 of row 0. Now, in the cell of row 2 and column 1, put the sum of the cells above and above-left of it. That is the sum of columns 1 and 0 of row 1 which 1 + 1 = 2. Now, in the cell of row 3 and column 1, put the sum of the cells above and above-left of it. That is the sum of columns 1 and 0 of row 2 which 1 + 2 = 3. In the same way, we fill up all the cells of column 1. And we see that column 1 contains the sequence of natural numbers. The nth number of column 1 is the sum of n 1s of column 0.

	Col 0	Col 1	Col 2	Col 3	Col 4	Col 5	Col 6	Col 7	Col 8	Col 9
Row 0	1
Row 1	1	1
Row 2	1	2	1
Row 3	1	3	3	1
Row 4	1	4	6	4	1
Row 5	1	5	10	10	5	1
Row 6	1	6	15	20	15	6	1
Row 7	1	7	21	35	35	21	7	1
Row 8	1	8	28	56	70	56	28	8	1
Row 9	1	9	36	84	126	126	84	36	9	1

Table 1

The number in cell of row 5 and column 3, is the sum of numbers in columns 3 and 2 of row 4, which is 4 + 6 = 10. You can fill the rest of the cells of the remaining columns in the same way. That is each cell is filled with the sum of numbers in the cells above and above-left.

Table 1 is actually left justified Pingala’s Meru Prastaar

(मेरु प्रस्तार). Refer my earlier article:

https://chandrahasblogs.wordpress.com/2020/10/04/meru-prastaar-%e0%a4%ae%e0%a5%87%e0%a4%b0%e0%a5%81-%e0%a4%aa%e0%a5%8d%e0%a4%b0%e0%a4%b8%e0%a5%8d%e0%a4%a4%e0%a4%be%e0%a4%b0%e0%a4%83/

The column 2 contains the sequence of triangular numbers. The nth number of column 2 is the sum of first n numbers of column 1, that is, the sum of first n natural numbers. The 5th number of column 2 is the sum of first 5 numbers (first 5 natural numbers) of column 1, which is 1 + 2 + 3 + 4 + 5 = 15.

The column 3 contains the sequence of tetrahedral numbers. The nth number of column 3 is the sum of first n numbers of column 2, that is, the sum of first n triangular numbers. The 4th number of column 3 is the sum of first 4 numbers (first 4 triangular numbers) of column 2, which is 1 + 3 + 6 + 10 = 20.

The column 4 contains the sequence of pentatope numbers. The nth number of column 4 is the sum of first n numbers of column 3, that is, the sum of first n tetrahedral numbers. The 5th number of column 4 is the sum of first 5 numbers (first 5 tetrahedral numbers) of column 3, which is 1 + 4 + 10 + 20 +35 = 70.

All these sums of sequence of natural numbers are in general called simplex numbers. The numbers in column 1, that is, the sequence of natural numbers are called 1-simplex numbers. The numbers in column 2, that is, the sequence of triangular numbers are called 2-simplex numbers. The numbers in column 3, that is, the sequence of tetrahedral numbers are called 3-simplex numbers. The numbers in column 4, that is, the sequence of pentatope numbers are called 4-simplex numbers. And so on. The numbers in column r are the r-simplex numbers.

The r-simplex number is denoted by S_r. The nth r-simplex number is denoted by S_r(n).

The fifth pentatope number is S₄(5) = 70.

S_r(n) is the number in the cell of rth column and (r + n – 1)th row.

The numbers in the cells of Meru Prastaar are binomial coefficients. Refer table 2.

Table 2

The number in the cell of row n and column r is given by

Hence, we have

This generic formula for sums of higher orders was derived by Naryan Pandit in his work Ganitakaumudi (1356 CE).

Thus the fifth pentatope number is given by: