OFFSET
0,5
COMMENTS
Since binomial(N,n) is defined for all integers N, there is no need to assume that N >= n.
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}.
LINKS
Jianing Song, Table of n, a(n) for antidiagonals 1..100 (T(n,k) occurs at the ((n+k)*(n+k-1)/2+n)-th place)
Andrew Granville, Arithmetic Properties of Binomial Coefficients I: Binomial Coefficients modulo prime powers
Jianing Song, Proof for my formula for A349593
Wikipedia, Kummer's_theorem
FORMULA
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.
EXAMPLE
Rows 0..10:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ...
1, 4, 3, 8, 5, 12, 7, 16, 9, 20, ...
1, 4, 9, 8, 5, 36, 7, 16, 27, 20, ...
1, 8, 9, 16, 5, 72, 7, 32, 27, 40, ...
1, 8, 9, 16, 25, 72, 7, 32, 27, 200, ...
1, 8, 9, 16, 25, 72, 7, 32, 27, 200, ...
1, 8, 9, 16, 25, 72, 49, 32, 27, 200, ...
1, 16, 9, 32, 25, 144, 49, 64, 27, 400, ...
1, 16, 27, 32, 25, 432, 49, 64, 81, 400, ...
1, 16, 27, 32, 25, 432, 49, 64, 81, 400, ...
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).
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).
PROG
(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))
CROSSREFS
KEYWORD
AUTHOR
Jianing Song, Nov 27 2021
STATUS
approved