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%I #9 Apr 23 2023 12:19:56
%S 1,2,8,44,256,1502,8912,53510,324352,1980332,12160008,75015162,
%T 464566144,2886488906,17985045464,112333392044,703119387648,
%U 4409231140086,27696141476336,174229516043630,1097501783152256,6921721148337452,43701895245221848
%N a(n) = [x^n] (F(x)/F(-x))^n where F(x) = (1 + x)*(1 + x^3).
%C Compare with A228960(n) = [x^n] F(x)^n.
%C Let k and m be positive integers and let f(x) be a finite product of cyclotomic polynomials. Define b(n) = [x^(k*n)] (f(x)/f(-x))^(m*n). Then we conjecture that the supercongruences a(p) == a(1) (mod p^3) and, for n >= 2, a(n*p) == a(n) (mod p^2) hold for all primes p, with a finite number of exceptions depending on f(x).
%C The present sequence is the case k = m = 1 and f(x) = (1 + x)*(1 + x^3) = C(2,x)^2 * C(6,x), where C(n,x) denotes the n-th cyclotomic polynomial. See A002003 for the case k = m = 1 and f(x) = (1 + x).
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Cyclotomic_polynomial">Cyclotomic polynomial</a>
%F Conjectures: 1) the supercongruence a(p) == 2 (mod p^3) holds for all primes p >= 5 (checked up to p = 47).
%F 2) for n >= 2, the supercongruence a(n*p) == a(n) (mod p^2) holds for all primes p >= 5.
%p F(x) := (1 + x)*(1 + x^3): G(x) := taylor(F(x)/F(-x),x = 0, 50); seq(coeftayl(G(x)^n, x = 0, n), n = 0..50);
%Y Cf. A002003, A228960, A333579.
%K nonn,easy
%O 0,2
%A _Peter Bala_, Apr 18 2023
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