login
A350406
a(n) = [x^n] 1/(1 + x + x^2 + x^3)^n.
4
1, -1, 1, -1, 5, -26, 91, -246, 597, -1540, 4576, -14521, 44915, -132328, 380290, -1102076, 3268437, -9838428, 29616364, -88538500, 263489380, -785026110, 2348923875, -7053379710, 21204016275, -63716916276, 191394838116, -575200476046, 1730575897202
OFFSET
0,5
LINKS
FORMULA
a(n) = (-1)^n * Sum_{k=0..n} binomial(n-1+k,k) * binomial(n,4*k).
From Peter Bala, Mar 17 2023: (Start)
a(n) = Sum_{k = 0..floor(n/2)} (-1)^(n-k)*binomial(n+k-1,k)*binomial(2*n-2*k-1,n-2*k).
16*n*(n+1)*(n-1)*(5*n-4)*(5*n-8)*(5*n-9)*a(n) = - ( 4*n*(n-1)*(5*n-9)* (550*n^3-1045*n^2+164*n+223)*a(n-1) + (n-1)*(5*n+1)*(5075*n^4-22330*n^3 +33771*n^2-20232*n+4032)*a(n-2) + 8*(2*n-3)*(4*n-5)*(4*n-7)*(5*n+1)*(5*n-3)*(5*n-4)*a(n-3) ).
Conjecture: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(2*r)) holds for positive integers n and r and all primes p >= 5. (End)
a(n) = (-1)^n*hypergeom([1/4 - n/4, 1/2 - n/4, 3/4 - n/4, -n/4, n], [1/4, 1/2, 3/4, 1], 1). - Peter Luschny, Mar 19 2023
MAPLE
a := proc (n) option remember; if n = 0 then 1 elif n = 1 then -1 elif n = 2 then 1 else -(4*n*(n-1)*(5*n-9)*(550*n^3-1045*n^2+164*n+223)*a(n-1) + (n-1)*(5*n+1)*(5075*n^4-22330*n^3+33771*n^2-20232*n+4032)*a(n-2) + (16*n-24)*(4*n-5)*(4*n-7)*(5*n+1)*(5*n-3)*(5*n-4)*a(n-3))/(16*n*(n+1)*(n-1)*(5*n-4)*(5*n-8)*(5*n-9)) end if; end:
seq(a(n), n = 0..30); # Peter Bala, Mar 18 2023
a := n -> (-1)^n*hypergeom([1/4 - n/4, 1/2 - n/4, 3/4 - n/4, -n/4, n], [1/4, 1/2, 3/4, 1], 1): seq(simplify(a(n)), n = 0..28); # Peter Luschny, Mar 19 2023
MATHEMATICA
a[n_] := Coefficient[Series[1/(1 + x + x^2 + x^3)^n, {x, 0, n}], x, n]; Array[a, 30, 0] (* Amiram Eldar, Dec 29 2021 *)
PROG
(PARI) a(n) = (-1)^n*sum(k=0, n, binomial(n-1+k, k)*binomial(n, 4*k));
CROSSREFS
KEYWORD
sign,easy
AUTHOR
Seiichi Manyama, Dec 29 2021
STATUS
approved