A246132 Binomial(2n, n) - 2 mod n^2.

0, 0, 0, 4, 0, 22, 0, 4, 18, 54, 0, 122, 0, 102, 43, 68, 0, 274, 0, 18, 361, 246, 0, 538, 250, 342, 504, 166, 0, 722, 0, 580, 865, 582, 5, 50, 0, 726, 18, 818, 0, 1510, 0, 310, 493, 1062, 0, 538, 1029, 2254, 2041, 406, 0, 922, 855, 1206, 379, 1686, 0, 3454, 0, 1926, 3538, 580, 3123, 922, 0, 4114, 547, 1298, 0, 4930, 0, 2742, 2518, 790, 3309, 2950, 0

Offset: 1

Author

Stanislav Sykora, Aug 16 2014

Keywords

• nonn

Comments

When e=2, the numbers binomial(2n, n)-2 mod n^e are 0 whenever n is a prime (see A246130 for introductory comments). This follows from Wolstenholme's theorem or, in a simpler way, from the identity binomial(2n, n)-2 = sum_{k=1..(n-1)} binomial(n,k)^2, in which every RHS term is divisible by n^2 whenever n is a prime. No composite number n for which a(n)=0 was found up to n=431500; nevertheless, the existence of such a composite is likely (personal opinion, based on the combinatorial nature of the problem).

Examples

			a(7) = (binomial(14,7)-2) mod 7^2 = (3432-2) mod 49 = 70*49 mod 49 = 0.
		

Links

• Stanislav Sykora, Table of n, a(n) for n = 1..10000
• Wikipedia, Wolstenholme's theorem

Crossrefs

Cf. A000984, A246130 (e=1), A246133 (e=3), A246134 (e=4).

Programs

Mathematica

Table[Mod[Binomial[2n,n]-2,n^2],{n,80}] (* Harvey P. Dale, Apr 13 2025 *)

PARI

a(n) = (binomial(2*n,n)-2)%n^2

Formula

For any prime p, a(p)=0.