A277640 a(n) is the integer r with |r| < prime(n)/2 such that (T(prime(n)^2)-T(prime(n)))/prime(n)^4 == r (mod prime(n)), where T(k) denotes the central trinomial coefficient A002426(k).

-2, 1, -3, -1, 7, -1, 6, 4, -15, -15, -13, 1, -23, 1, 8, -15, -22, 13, -33, 27, 25, 11, -17, 24, -32, -53, 31, 42, -19, 18, -35, 55, -5, 38, -29, 76, 34, 44, -71, -21, -13, 16, 46, 70, 92, 70, -39, 88, -84, -118, -120, 64, 107, 111, -56, 124, -13, -23

Offset: 3

Author

Zhi-Wei Sun, Oct 25 2016

Keywords

• sign

Comments

Conjecture: (i) For any prime p > 3 and positive integer n, the number (T(p*n)-T(n))/(p*n)^2 is always a p-adic integer.

(ii) For any prime p > 3 and positive integer k, we have (T(p^k)-T(p^(k-1)))/p^(2k) == 1/6*(p^k/3)*B_{p-2}(1/3) (mod p), where (p^k/3) denotes the Legendre symbol and B_{p-2}(x) is the Bernoulli polynomial of degree p-2.

For any prime p > 3, the author has proved that (T(p*n)-T(n))/(p^2*n) is a p-adic integer for each positive integer n, and that T(p) == 1 + p^2/6*(p/3)*B_{p-2}(1/3) (mod p^3).

Examples

			a(3) = -2 since (T(prime(3)^2)-T(prime(3)))/prime(3)^4 = (T(25)-T(5))/5^4 = (82176836301-51)/5^4 = 131482938 is congruent to -2 modulo prime(3) = 5 with |-2| < 5/2.
		

Links

• Hao Pan and Zhi-Wei Sun, Supercongruences for central trinomial coefficients, arXiv:2012.05121 [math.NT], 2020.
• Zhi-Wei Sun, Congruences involving generalized central trinomial coefficients, Sci. China Math. 57(2014), no.7, 1375-1400.
• Zhi-Wei Sun, Supercongruences involving Lucas sequences, arXiv:1610.03384 [math.NT], 2016.

Crossrefs

Cf. A000040, A002426, A245089, A277860.

Programs

Mathematica

T[n_]:=T[n]=Sum[Binomial[n,2k]Binomial[2k,k],{k,0,n/2}]
rMod[m_,n_]:=Mod[Numerator[m]*PowerMod[Denominator[m],-1,n],n,-n/2]
Do[Print[n," ",rMod[(T[Prime[n]^2]-T[Prime[n]])/Prime[n]^4,Prime[n]]],{n,3,60}]