Dear SeqFan members,
I would like to share an observation I recently made while experimenting with Pascal's triangle and divisibility by integers.
I have not found this particular construction in the literature, so I would be very interested in hearing the opinion of the SeqFan community and, in particular, whether this is already known in some form.
The construction is as follows.
For a fixed positive integer n, consider the rows preceding row n of Pascal's triangle. For the row n-d, I modify each internal coefficient C(n-d,k) by adding an alternating correction
(-1)^(k+1) C(k+d-1,d-1).
Thus, for the row immediately preceding row n (d=1), the corrections are
+1, -1, +1, -1, ...
For the row n-2 (d=2), they are
+2, -3, +4, -5, ...
For the row n-3 (d=3), they are
+3, -6, +10, -15, ...
For the row n-4 (d=4), they are
+4, -10, +20, -35, ...
and so on.
The positive numbers appearing in these corrections are precisely the successive figurate-number sequences: 1's, natural numbers starting at 2, triangular numbers starting at 3, tetrahedral numbers starting at 4, etc.
For prime n, I found that all the corrected internal coefficients in every preceding row are divisible by n. This follows from the congruence
C(p-d,k) ≡ (-1)^k C(k+d-1,d-1) (mod p),
for p prime.
What particularly caught my attention is that the same construction can also be investigated for composite n. I have computationally examined the behavior for n up to 100 and found a structured pattern: composite numbers can sometimes have individual preceding rows for which all the corrected internal values are divisible by n, although they do not satisfy the complete prime pattern.
For reference, below I give the first 11 corrected triangles. Here "first 11" means n = 1, 2, ..., 11, not the first 11 prime cases. The uncorrected rows are left unchanged; for each n, the rows preceding n are corrected according to the rule above. The endpoints 1 are not included in the divisibility condition.
n = 1
1
1 1
n = 2
1
1 1
1 2 1
n = 3
1
1 1
1 3 1
1 3 3 1
n = 4
1
1 1
1 4 1
1 4 2 1
1 4 6 4 1
n = 5
1
1 1
1 5 1
1 5 0 1
1 5 5 5 1
1 5 10 10 5 1
n = 6
1
1 1
1 6 1
1 6 -3 1
1 6 3 8 1
1 6 9 11 4 1
1 6 15 20 15 6 1
n = 7
1
1 1
1 7 1
1 7 -7 1
1 7 0 14 1
1 7 7 14 0 1
1 7 14 21 14 7 1
1 7 21 35 35 21 7 1
n = 8
1
1 1
1 8 1
1 8 -12 1
1 8 -4 24 1
1 8 4 20 -10 1
1 8 12 24 10 12 1
1 8 20 36 34 22 6 1
1 8 28 56 70 56 28 8 1
n = 9
1
1 1
1 9 1
1 9 -18 1
1 9 -9 39 1
1 9 0 30 -30 1
1 9 9 30 0 27 1
1 9 18 39 30 27 0 1
1 9 27 57 69 57 27 9 1
1 9 36 84 126 126 84 36 9 1
n = 10
1
1 1
1 10 1
1 10 -25 1
1 10 -15 60 1
1 10 -5 45 -65 1
1 10 5 40 -20 62 1
1 10 15 45 20 42 -21 1
1 10 25 60 65 62 21 16 1
1 10 35 85 125 127 83 37 8 1
1 10 45 120 210 252 210 120 45 10 1
n = 11
1
1 1
1 11 1
1 11 -33 1
1 11 -22 88 1
1 11 -11 66 -121 1
1 11 0 55 -55 132 1
1 11 11 55 0 77 -77 1
1 11 22 66 55 77 0 44 1
1 11 33 88 121 132 77 44 0 1
1 11 44 121 209 253 209 121 44 11 1
1 11 55 165 330 462 462 330 165 55 11 1
One aspect I find particularly interesting is the appearance of zeros in some corrected rows for both prime and composite values of n. For example, n=5 has a zero in the corrected row 3, while n=7 has zeros in corrected rows 4 and 5. Among composite values, n=9 has a completely zero-modulo corrected row at row 3, and n=25 has several consecutive corrected rows with this property.
I am currently investigating whether there is a deeper interpretation of these patterns, perhaps in terms of binomial congruences, figurate numbers, finite differences, or known properties of Pascal's triangle.
I would therefore be very grateful for your expert opinion:
Is this construction or this particular family of congruences already known?
Is there a standard name for the corrected arrays or for this type of transformation of Pascal's triangle?
Are there known results describing the composite cases?
Does the general congruence above lead to any interesting identities or sequences?
Any references, comments, or suggestions for further investigation would be greatly appreciated.
Best regards,
Davide Rotondo
Independent researcher
Italy