A051574 a(n) = number of k, 1<=k<=n, such that (n*k) divides binomial(n,k).

1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 3, 1, 3, 1, 1, 1, 3, 1, 5, 3, 2, 1, 1, 1, 2, 2, 2, 4, 7, 1, 2, 2, 5, 6, 3, 1, 6, 3, 2, 1, 4, 1, 2, 1, 2, 2, 2, 1, 4, 5, 9, 5, 8, 4, 7, 3, 9, 6, 8, 2, 6, 3, 1, 4, 11, 5, 9, 4, 5, 1, 4, 1, 7, 4, 2, 4, 8, 3, 4, 1, 6, 11, 15, 3, 7, 5, 4, 5, 9, 1, 5, 3, 2, 2, 1, 1, 4, 2, 7, 7, 19, 8

Offset

1,5

Comments

k = 1 always works, so a(n) >= 1. a(n) = 1 for n = 1, 2, 3, 4, 6, 7, 8, 12, 14, 15, 16, 18, 22, 23, 24, 30, 36, 40, 42, 44, 48, 63, 70, 72, 80, 90, 95, 96, 120, 240. Are there any others? - Robert Israel, Feb 27 2024

Links

Robert Israel, Table of n, a(n) for n = 1..10000

Example

a(11)=3 since k=1, k=3, k=6 are only solutions to 11*k divides binomial(11,k).

Maple

f:= proc(n) nops(select(t -> binomial(n, t) mod (n*t) = 0, [$1..n])) end proc:
map(f, [$1..200]); # Robert Israel, Feb 27 2024

Mathematica

nkb[n_]:=Total[Table[If[Divisible[Binomial[n, k], n*k], 1, 0], {k, n}]]; Array[ nkb, 110] (* Harvey P. Dale, Jul 06 2020 *)

Prog

(PARI) a(n) = sum(k=1, n, (binomial(n, k) % (n*k)) == 0); \\ Michel Marcus, May 18 2014

Crossrefs

Cf. A051556 (similar definition).

Sequence in context: A277315 A277326 A050431 * A029386 A277325 A240837

Adjacent sequences: A051571 A051572 A051573 * A051575 A051576 A051577

Keyword

nonn

Author

Leroy Quet, Dec 11 1999

Status

approved