A333579 a(n) = [x^n] ( (1 + x + x^2)/(1 - x + x^2) )^n.

1, 2, 8, 32, 128, 502, 1904, 6862, 22784, 64832, 120008, -223606, -4311424, -33271366, -205802344, -1142307968, -5919738880, -29159028386, -137718099760, -626077804826, -2740865583872, -11523690799904, -46214332516520, -174358991625134, -601230820510720

Offset: 0

Author

Peter Bala, Mar 29 2020

Keywords

• sign
• easy

Comments

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.

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.

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.

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).

Examples

			Examples of congruences a(p) - a(1) == 0 ( mod p^3 ):
a(11) - a(1) = -223606 - 2 = -(2^3)*3*7*11^3 == 0 ( mod 11^3 )
a(19) - a(1) = -626077804826 - 2 = -(2^2)*7*(19^3)*151*21589 == 0 ( mod 19^3 )
		

Links

• Wikipedia, Cyclotomic polynomial

Crossrefs

Cf. A002003, A002426, A187925, A318114.

Programs

Maple

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);
#alternative program
G := x -> (1 + x + x^2)/(1 - x + x^2):
H := (x,n) -> series(G(x)^n, x, n+1):
a:= n -> coeff(H(x, n), x, n):
seq(a(n), n = 0..25);

Mathematica

a[n_]:=SeriesCoefficient[((1 + x + x^2)/(1 - x + x^2))^n,{x,0,n}]; Array[a,25,0] (* Stefano Spezia, Apr 30 2024 *)

PARI

a(n) = polcoeff(((1 + x + x^2)/(1 - x + x^2))^n+ O(x^(n+1)), n, x); \\ Michel Marcus, Mar 31 2020

Formula

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).