

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).


7



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
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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 padic integer.
(ii) For any prime p > 3 and positive integer k, we have (T(p^k)T(p^(k1)))/p^(2k) == 1/6*(p^k/3)*B_{p2}(1/3) (mod p), where (p^k/3) denotes the Legendre symbol and B_{p2}(x) is the Bernoulli polynomial of degree p2.
For any prime p > 3, the author has proved that (T(p*n)T(n))/(p^2*n) is a padic integer for each positive integer n, and that T(p) == 1 + p^2/6*(p/3)*B_{p2}(1/3) (mod p^3).


LINKS



EXAMPLE

a(3) = 2 since (T(prime(3)^2)T(prime(3)))/prime(3)^4 = (T(25)T(5))/5^4 = (8217683630151)/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



STATUS

approved



