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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

If p is prime, binomial(2*p,p) == 2^p (mod p).

a(17) > 10^9.

From Gabriel Guedes and Ricardo Machado, Nov 16 2023: (Start)

Theorem. Let j = (2^k)*p, where p is an odd prime and k is in N; then binomial(2*j,j) == 2^j (mod j) if and only if p satisfies the following conditions:

a) p divides binomial(2^(k+1),2^k) - 2^(2^k);

b) p has at least k 1's in its binary expansion.

Theorem. If m is an even perfect number then j = 2m satisfies the congruence binomial(2*j,j) == 2^j (mod j). See A000396.

Theorem. Let j = p^2 with p a prime number. Then p is a Wieferich prime if and only if binomial(2*j,j) == 2^j (mod j). See A001220. (End)

Contains 17179738112 and 274877382656 (from Guedes-Machado paper). - Michael De Vlieger, Nov 22 2023

Contains 3386741824, 750984028672, 33029195197184, 1145067923695616, 422612863956511744. - Ricardo Machado, Nov 23 2023

Contains 84385517065596416, 62648180117928433664, 273984397779878971648, 36506097537257040703232. - Max Alekseyev, Dec 07 2023