|
|
A051556
|
|
a(n) = number of 0<=k<=n such that n+k divides binomial(n,k).
|
|
2
|
|
|
1, 0, 0, 1, 0, 0, 0, 1, 2, 1, 1, 1, 0, 0, 0, 6, 3, 1, 2, 2, 1, 3, 2, 3, 6, 2, 3, 5, 5, 2, 3, 6, 4, 6, 5, 10, 12, 6, 4, 10, 8, 3, 8, 5, 8, 5, 3, 5, 11, 4, 4, 10, 7, 8, 9, 19, 21, 12, 8, 7, 8, 4, 3, 17, 13, 19, 23, 10, 11, 13, 16, 18, 14, 9, 11, 19, 13, 9, 16, 16, 25, 21, 17, 15, 18, 13, 16, 17, 19
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,9
|
|
LINKS
|
|
|
EXAMPLE
|
For n=9, 0<=k<=n, n+k divides C(n,k) only when k=3 and k=5, so a(9)=2.
|
|
MAPLE
|
f:= proc(n)
nops(select(t -> binomial(n, t) mod (n+t) = 0, [$0..n]))
end proc:
|
|
MATHEMATICA
|
A051556[n_] := Count[Divisible[Binomial[n, Range[0, n]], Range[n, 2*n]], True];
|
|
PROG
|
(PARI) a(n) = sum(k=0, n, (binomial(n, k) % (n+k)) == 0); \\ Michel Marcus, May 18 2014
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|