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

Author

M. F. Hasler, Nov 11 2015

Keywords

• nonn
• easy

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).

Examples

			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).
		

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).

Crossrefs

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

Programs

Magma

[Binomial(3*n,n) mod n : n in [1..100]]; // Wesley Ivan Hurt, Nov 12 2015

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 *)

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.

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