OFFSET
3,1
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).
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.
EXAMPLE
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.
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}]
CROSSREFS
KEYWORD
sign
AUTHOR
Zhi-Wei Sun, Oct 25 2016
STATUS
approved