This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A333579 #20 Apr 30 2024 15:56:33
%S A333579 1,2,8,32,128,502,1904,6862,22784,64832,120008,-223606,-4311424,
%T A333579 -33271366,-205802344,-1142307968,-5919738880,-29159028386,
%U A333579 -137718099760,-626077804826,-2740865583872,-11523690799904,-46214332516520,-174358991625134,-601230820510720
%N A333579 a(n) = [x^n] ( (1 + x + x^2)/(1 - x + x^2) )^n.
%C A333579 If F(x) = 1 + f(1)*x + f(2)*x^2 + ... is a formal power series with integer coefficients then the sequence u(n) := [x^n] F(x)^n is an integer sequence satisfying the Gauss congruences: u(n*p^k) == u(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k. In particular u(p^k) == u(p^(k-1)) ( mod p^k ) for all prime p and positive integer k. For certain power series F(x) we may get stronger congruences.
%C A333579 According to Zhi-Wei Sun (see his comment in A002426 posted Nov 30, 2016), the central trinomial coefficients A002426(n) = [x^n] (1 + x + x^2)^n, satisfy the congruences A002426(p) == 1 ( mod p^2 ) for prime p >= 5. More generally, calculation suggests that the congruences A002426(p^k) == A002426(p^(k-1)) ( mod p^(2*k) ) hold for prime p >= 5 and any positive integer k.
%C A333579 We conjecture that the present sequence satisfies the congruences a(p) == 2 ( mod p^3 ) for prime p >= 5 (checked up to p = 499). Calculation suggests that a(p^k) == a(p^(k-1)) ( mod p^(2*k) ) for prime p >= 5 and k > 1.
%C A333579 More generally, if c(m,x) denotes the m-th cyclotomic polynomial then the sequences a(m,n) := [x^n] c(m,x)^n and b(m,n) := [x^n] ( c(m,x)/c(m,-x) )^n may satisfy the congruences a(m,p) == a(m,1) ( mod p^2 ) and b(m,p) == b(m,1) ( mod p^3 ), both for prime p >= 5, p not a divisor of m. The present sequence is b(3,n). Note that b(2,n) = A002003(n).
%H A333579 Wikipedia, <a href="https://en.wikipedia.org/wiki/Cyclotomic_polynomial">Cyclotomic polynomial</a>
%F A333579 a(n) = Sum_{0 <= i,j,k <= n} (-1)^(n-k-i-j)*C(n,k)*C(k,i)*C(n+j-1,j)*C(j,n-k-i-j).
%e A333579 Examples of congruences a(p) - a(1) == 0 ( mod p^3 ):
%e A333579 a(11) - a(1) = -223606 - 2 = -(2^3)*3*7*11^3 == 0 ( mod 11^3 )
%e A333579 a(19) - a(1) = -626077804826 - 2 = -(2^2)*7*(19^3)*151*21589 == 0 ( mod 19^3 )
%p A333579 seq(add(add(add((-1)^(n-k-i-j)*binomial(n, k)*binomial(k, i)*binomial(n+j-1, j)*binomial(j, n-k-i-j), j = 0..n-k-i), i = 0..n-k), k = 0..n), n = 0..25);
%p A333579 #alternative program
%p A333579 G := x -> (1 + x + x^2)/(1 - x + x^2):
%p A333579 H := (x,n) -> series(G(x)^n, x, n+1):
%p A333579 a:= n -> coeff(H(x, n), x, n):
%p A333579 seq(a(n), n = 0..25);
%t A333579 a[n_]:=SeriesCoefficient[((1 + x + x^2)/(1 - x + x^2))^n,{x,0,n}]; Array[a,25,0] (* _Stefano Spezia_, Apr 30 2024 *)
%o A333579 (PARI) a(n) = polcoeff(((1 + x + x^2)/(1 - x + x^2))^n+ O(x^(n+1)), n, x); \\ _Michel Marcus_, Mar 31 2020
%Y A333579 Cf. A002003, A002426, A187925, A318114.
%K A333579 sign,easy
%O A333579 0,2
%A A333579 _Peter Bala_, Mar 29 2020