A260636 a(n) = binomial(3n, n) mod n.

0, 1, 0, 3, 3, 0, 3, 7, 3, 5, 3, 0, 3, 8, 9, 15, 3, 15, 3, 15, 0, 4, 3, 12, 3, 2, 3, 12, 3, 24, 3, 15, 18, 15, 0, 9, 3, 34, 6, 31, 3, 21, 3, 0, 15, 38, 3, 36, 3, 40, 33, 40, 3, 42, 0, 16, 27, 44, 3, 0, 3, 46, 45, 47, 39, 51, 3, 53, 15, 0, 3, 45, 3, 15, 9, 20, 76, 0, 3, 7, 3

Offset

1,4

Comments

Motivated by A080469: C(3n,n)=3^n (mod n), A109641, A109642 and other sequences.

See A059288 for the "2n" analog. Sequence A260640 yields the indices of zeros (analog to A014847 for 2n).

Links

Chai Wah Wu, Table of n, a(n) for n = 1..10000

M. Alekseyev, PARI/GP Scripts for Miscellaneous Math Problems, sect. III: Binomial coefficients modulo integers, binomod.gp (Vers. 1.4, 11/2015).

Example

n=1: C(3,1) = 3 = 0 (mod 1).
n=2: C(3*2,2) = 15 = 1 (mod 2).
n=3: C(3*3,3) = 84 = 0 (mod 3).
n=4: C(3*4,4) = 495 = 3 (mod 4).

Maple

A260636:=n->binomial(3*n, n) mod n: seq(A260636(n), n=1..100); # Wesley Ivan Hurt, Nov 12 2015

Mathematica

Array[Mod[Binomial[3 #, #], #] &, 112] (* Michael De Vlieger, Nov 12 2015 *)

Prog

(PARI) a(n)=binomial(3*n, n)%n
(PARI) A260636(n)=lift(binomod(3*n, n, n)) \\ using binomod.gp by M. Alekseyev, cf. Links.
(Magma) [Binomial(3*n, n) mod n : n in [1..100]]; // Wesley Ivan Hurt, Nov 12 2015
(Python)
from __future__ import division
A260636_list, b = [], 3
for n in range(1, 10001):
    A260636_list.append(b % n)
    b = b*3*(3*n+2)*(3*n+1)//((2*n+2)*(2*n+1)) # Chai Wah Wu, Jan 27 2016

Crossrefs

Cf. A080469, A109641, A109642; A260640 (indices of zeros); A059288, A014847 (analogs for 2n).

Sequence in context: A330013 A128164 A339702 * A245256 A347149 A140686

Adjacent sequences: A260633 A260634 A260635 * A260637 A260638 A260639

Keyword

nonn,easy

Author

M. F. Hasler, Nov 11 2015

Status

approved