A080469 Composite n such that binomial(3*n,n)==3^n (mod n).

36, 57, 121, 132, 552, 8397, 7000713, 9692541, 36294723, 564033861

Offset: 1

Author

Benoit Cloitre, Oct 15 2003

Keywords

• nonn
• more
• hard

Comments

If p is prime, binomial(3*p,p)==3^p (mod p)

No other terms below 10^9.

A subsequence of A109641. The terms a(n) with n=2, 6, 7, 8, 9, 10 are of the form 3^k*p where p is prime and k=1, 3, 2, 5, 6, 7, respectively. It is tempting to conjecture that there are (infinitely many?) more terms of that form. - M. F. Hasler, Nov 11 2015

Examples

			57 is a term because binomial(3*57, 57) = 12039059761216294940321619222324879408784636200 mod 57 = 27 == 3^57 mod 57.
		

Links

• Max Alekseyev, PARI scripts for various problems

Crossrefs

Cf. A109641, A109642; A109760, A109769.

Programs

Mathematica

Do[If[ !PrimeQ[n], k = Binomial[3*n, n]; m = 3^n; If[Mod[k, n] == Mod[m, n], Print[n]]], {n, 1, 70000}] (* Ryan Propper, Aug 12 2005 *)

PARI

forcomposite(n=1,1e9, binomod(3*n,n,n)==Mod(3,n)^n && print1(n",")) \\ Cf. Alekseyev link. - M. F. Hasler, Nov 14 2015

Extensions

One more term a(6) from Ryan Propper, Aug 12 2005

Four new terms a(7)-a(10) added by Max Alekseyev, Nov 05 2009