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A349593 Square array read by downward diagonals: for n >= 0, k >= 1, T(n,k) is the period of {binomial(N,n) mod k: N in Z}.

%I A349593 #80 Jul 07 2025 10:20:39
%S A349593 1,1,1,1,2,1,1,3,4,1,1,4,3,4,1,1,5,8,9,8,1,1,6,5,8,9,8,1,1,7,12,5,16,
%T A349593 9,8,1,1,8,7,36,5,16,9,8,1,1,9,16,7,72,25,16,9,16,1,1,10,9,16,7,72,25,
%U A349593 16,9,16,1,1,11,20,27,32,7,72,25,32,27,16,1
%N A349593 Square array read by downward diagonals: for n >= 0, k >= 1, T(n,k) is the period of {binomial(N,n) mod k: N in Z}.
%C A349593 Since binomial(N,n) is defined for all integers N, there is no need to assume that N >= n.
%C A349593 Let Q(N) = 1 if k | binomial(N,n), 0 otherwise. Then T(n,k) is also the period of {Q(N): N in Z}.
%C A349593 By the formula given below, the n-th row is identical to the (n-1)th row if and only if n is not a power of a prime, i.e., n is in A024619. - _Jianing Song_, Jul 03 2025
%H A349593 Jianing Song, <a href="/A349593/b349593.txt">Table of n, a(n) for antidiagonals 1..100</a> (T(n,k) occurs at the ((n+k)*(n+k-1)/2+n)-th place)
%H A349593 Andrew Granville, <a href="http://www.cecm.sfu.ca/organics/papers/granville/paper/binomial/html/binomial.html">Arithmetic Properties of Binomial Coefficients I: Binomial Coefficients modulo prime powers</a>
%H A349593 Jianing Song, <a href="/A349593/a349593.pdf">Proof for my formula for A349593</a>
%H A349593 Jianing Song, <a href="/A349593/a349593_1.pdf">A more simple proof of the formula for A349593</a>
%H A349593 Wikipedia, <a href="https://en.m.wikipedia.org/wiki/Kummer%27s_theorem">Kummer's_theorem</a>
%F A349593 The n-th row is multiplicative with T(n,p^e) = 1 if n = 0, p^(e+floor(log(n)/log(p))) otherwise. In other words, for n > 0, T(n,k) = k * Product_{prime p|k} p^(floor(log(n)/log(p))). See my pdf file for a proof.
%e A349593 Rows 0..10:
%e A349593   1,  1,  1,  1,  1,   1,  1,  1,  1,   1, ...
%e A349593   1,  2,  3,  4,  5,   6,  7,  8,  9,  10, ...
%e A349593   1,  4,  3,  8,  5,  12,  7, 16,  9,  20, ...
%e A349593   1,  4,  9,  8,  5,  36,  7, 16, 27,  20, ...
%e A349593   1,  8,  9, 16,  5,  72,  7, 32, 27,  40, ...
%e A349593   1,  8,  9, 16, 25,  72,  7, 32, 27, 200, ...
%e A349593   1,  8,  9, 16, 25,  72,  7, 32, 27, 200, ...
%e A349593   1,  8,  9, 16, 25,  72, 49, 32, 27, 200, ...
%e A349593   1, 16,  9, 32, 25, 144, 49, 64, 27, 400, ...
%e A349593   1, 16, 27, 32, 25, 432, 49, 64, 81, 400, ...
%e A349593   1, 16, 27, 32, 25, 432, 49, 64, 81, 400, ...
%e A349593 Example showing that T(4,4) = 16: for N == 0, 1, ..., 15 (mod 16), binomial(N,4) == {0, 0, 0, 0, 1, 1, 3, 3, 2, 2, 2, 2, 3, 3, 1, 1} (mod 4).
%e A349593 Example showing that T(3,10) = 20: for N == 0, 1, ..., 19 (mod 20), binomial(N,3) == {0, 0, 0, 1, 4, 0, 0, 5, 6, 4, 0, 5, 0, 6, 4, 5, 0, 0, 6, 9} (mod 10).
%t A349593 A349593[n_, k_] := If[n == 0 || k == 1, 1, k*Product[p^Floor[Log[p, n]], {p, FactorInteger[k][[All, 1]]}]];
%t A349593 Table[A349593[k - 1, n - k + 2], {n, 0, 15}, {k, n + 1}] (* _Paolo Xausa_, Jul 07 2025 *)
%o A349593 (PARI) T(n,k) = if(n==0, 1, my(r=1, f=factor(k)); for(j=1, #f[, 1], my(p=f[j, 1], e=f[j, 2]); r *= p^(logint(n,p)+e)); return(r))
%Y A349593 Cf. A022998 (row n = 2), A385555 (row n = 3), A385556 (row n = 4), A385557 (rows n = 5 and 6), A385558 (row n = 7), A385559 (row n = 8), A385560 (rows n = 9 and 10).
%Y A349593 Cf. A062383 (2nd column), A064235 (3rd column if offset 0), A385552 (5th column), A385553 (6th column), A385554 (10th column).
%Y A349593 Cf. A349221.
%K A349593 nonn,tabl,easy
%O A349593 0,5
%A A349593 _Jianing Song_, Nov 27 2021