A260640 Numbers n such that binomial(3*n,n) == 0 (mod n).

1, 3, 6, 12, 21, 35, 44, 55, 60, 70, 78, 88, 90, 99, 102, 110, 117, 119, 120, 133, 156, 171, 176, 180, 184, 204, 207, 220, 225, 230, 231, 234, 238, 240, 247, 252, 255, 285, 286, 300, 312, 341, 342, 348, 360, 368, 372, 391, 403, 408, 414, 425, 434, 460, 462, 465, 468, 481, 483, 494, 495, 504, 506, 510, 550, 555, 561, 572, 574, 585, 600

Offset: 1

Author

M. F. Hasler, Nov 11 2015

Keywords

• nonn

Comments

See A014847 for the analog 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 (v.1.4, 11/2015).

Crossrefs

Cf. A260636, A080469, A109641, A109642, A014847.

Programs

Magma

[n: n in [1..600] |Binomial(3*n,n) mod n eq 0]; // Vincenzo Librandi, Jan 29 2016

Mathematica

Select[Range@ 600, Mod[Binomial[3 #, #], #] == 0 &] (* Michael De Vlieger, Nov 12 2015 *)

PARI

for(n=1,999,binomod(3*n,n,n)==0&&print1(n",")) \\ Using binomod.gp by M. Alekseyev, cf. links.

Python

from _future_ import division
A260640_list, b = [], 3
for n in range(1,10**3):
    if not b % n:
        A260640_list.append(n)
    b = b*3*(3*n+2)*(3*n+1)//((2*n+2)*(2*n+1)) # Chai Wah Wu, Jan 27 2016