A Bunch of Examples of Sequence Numbers for Combinatoric Functions

10/05/2015

Sequence Numbers: Combinatorial Functions

By David Brooks

C (n,2): The number of ways to choose two items from a group of n items.  (This sequence is also called the Triangular numbers, and the third diagonal in Pascal’s Triangle.  At the end of this post I will show how this sequence is related the 3, 3, 1 Tribonacci sequence.) The Sequence Number is: 999,997,000,002,999,999 1/999997000002999999 = 0. 000000  000000  000001  000003  000006  000010  000015  000021  000028  000036  000045  000055  000066  000078  … Terms are written in six digit strings. Compare with OEIS sequence A000217.

C (n, 3): The number of ways to choose three items from a group of n items.  (This sequence is also called the Tetrahedral numbers and the fourth diagonal in Pascal’s Triangle.) The Sequence Number is: 999,996,000,005,999,996,000, 001 1/999996000005999996000 001 = 0. 000000  000000  000000  000001  000004  000010  000020  000035  000056  000084  000120  000165  000220  000286  … Terms are written in six digit strings. Compare with OEIS sequence A000292.

C (n, 4): The number of ways to choose four items from a group of n items. The Sequence Number is: 999,995,000,009,999,990,000, 004,999,999 1/9999950000099999900000 04999999 = 0. 000000  000000  000000  000000  000001  000005  000015  000035  000070  000126  000210  000330  000495  000715  001001  001365  001820  … Terms are written in six digit strings. These terms are also the numbers found in the fifth diagonal of Pascal’s Triangle. Compare with OEIS sequence A000332.

C (n, 5): The number of ways to choose five items from a group of n items. The Sequence Number is: 999,994,000,014,999,980,000, 014,999,994,000,001 1/9999940000149999800000 14999994000001 = 0. 000000  000000  000000  000000  000000  000001  000006  000021  000056  000126  000252  000462  000792  001287  002002  003003  004368  … Terms are written in six digit strings. These terms are also the numbers found in the sixth diagonal of Pascal’s Triangle. Compare with OEIS sequence A000389.

C (n, 6): The number of ways to choose six items from a group of n items. The Sequence Number is: 999993000020999965000034 999979000006999999 1/9999930000209999650000 34999979000006999999 = 0. 000000  000000  000000  000000  000000  000000  000001  000007  000028  000084  000210  000462  000924  001716  Terms are written in six digit strings. These terms are also the numbers found in the seventh diagonal of Pascal’s Triangle. Compare with OEIS sequence A000579.

C (n, 7):  The number of ways to choose seven items from a group of n items. The Sequence Number is: 9999920000279999440000699 99944000027999992000001 1/99999200002799994400006 9999944000027999992000001= 0. 000000  000000  000000  000000  000000  000000  000000  000001  000008  000036  000120  000330  000792  001716  … Terms are written in six digit strings. These terms are also the numbers found in the eighth diagonal of Pascal’s Triangle. Compare with OEIS sequence A000580.

C (n, 8): The number of ways to choose eight items from a group of n items. The Sequence number is: 999991000035999916000125 999874000083999964000008 999999 1/99999100003599991600012 599987400008399996400000 8999999 = 0, 000000  000000  000000  000000  000000  000000  000000  000000  000001  000009  000045  000165  000495  001287  003003  006435  012870  024310  043758  075582  … Terms are written in six digit strings. These terms are also the numbers found in the ninth diagonal of Pascal’s Triangle. Compare with OEIS sequence A000581.

C (n, 9): The number of ways to choose nine items from a group of n items. The Sequence Number is:

If you recall, 999,997,000,002,999,999 was the Sequence Number we used to produce the terms of C (n,2).  But this sequence number looks like a Sequence number that we would use to produce a tribonacci like sequence. If we interpret 999,997,000,002,999,999 as a Sequence Number for a Tribonacci like sequence it translates into a sequence defined as: a(0) = a(1) = 0, a(2) = 1, and when n>1 then a(n) = 3*a(n-1) – 3*a(n-2) + a(n-3).  This would be called the 3, -3, 1 Tribonacci sequence. I set up a spreadsheet to calculate the terms of this sequence, and printed the results below.  I calculated the terms using the recursive formula at the end of the paragraph above.  You can compare the results with the results from the Sequence Number for C (n, 3). Index Term a(0) 0 a(1) 0 a(2) 1 3 3 4 6 5 10 6 15 7 21 8 28 9 36 10 45 11 55 12 66 13 78 14 91 15 105 So C (n, 2), Triangular Numbers, the third diagonal of Pascal’s Triangle, and the 3, -3, 1 Tribonacci sequence are mathematically related. I think it is amazing how many branches of mathematics have a relationship to Pascal’s Triangle.  I guess I need to do a series of posts in the near future about Pascal’s Triangle.

David