cp's OEIS Frontend

A350407 a(n) = [x^n] 1/(1 + x + x^2 + x^3 + x^4)^n.

%I A350407 #26 Apr 17 2023 11:01:55
%S A350407 1,-1,1,-1,1,4,-35,146,-447,1133,-2464,4355,-4355,-9296,77078,-314636,
%T A350407 1006145,-2738565,6398155,-12175809,14621376,14828330,-179236815,
%U A350407 786257460,-2625667395,7412386254,-17983703880,36057065030,-49553122730,-14585596720
%N A350407 a(n) = [x^n] 1/(1 + x + x^2 + x^3 + x^4)^n.
%H A350407 Seiichi Manyama, <a href="/A350407/b350407.txt">Table of n, a(n) for n = 0..1000</a>
%F A350407 a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n-1+k,k) * binomial(n,5*k).
%F A350407 Recurrence: 125*(n-3)*(n-2)*(n-1)*n*(2*n - 7)*(3*n - 11)*(3*n - 8)*(6*n - 23)*(6*n - 17)*(6*n - 11)*a(n) = -50*(n-3)*(n-2)*(n-1)*(3*n - 11)*(6*n - 23)*(6*n - 17)*(576*n^4 - 4896*n^3 + 14402*n^2 - 16875*n + 6250)*a(n-1) - 30*(n-3)*(n-2)*(6*n - 23)*(6*n - 5)*(7884*n^6 - 123516*n^5 + 791601*n^4 - 2652565*n^3 + 4894096*n^2 - 4707500*n + 1842500)*a(n-2) - 2*(n-3)*(3*n - 5)*(6*n - 11)*(6*n - 5)*(60192*n^6 - 1113552*n^5 + 8528546*n^4 - 34608379*n^3 + 78470893*n^2 - 94255700*n + 46855500)*a(n-3) - 5*(2*n - 5)*(3*n - 8)*(3*n - 5)*(5*n - 19)*(5*n - 18)*(5*n - 17)*(5*n - 16)*(6*n - 17)*(6*n - 11)*(6*n - 5)*a(n-4). - _Vaclav Kotesovec_, Mar 18 2023
%F A350407 From _Peter Bala_, Apr 16 2023: (Start)
%F A350407 a(n) = (-1)^n*hypergeom([-n/5, 1/5 - n/5, 2/5 - n/5, 3/5 - n/5, 4/5 - n/5, n], [1/5, 2/5, 3/5, 4/5, 1], 1).
%F A350407 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 >= 3. (End)
%p A350407 a := n -> (-1)^n*hypergeom([-n/5, 1/5 - n/5, 2/5 - n/5, 3/5 - n/5, 4/5 - n/5, n], [1/5, 2/5, 3/5, 4/5, 1], 1): seq(simplify(a(n)), n = 0..30); # _Peter Bala_, Apr 16 2023
%t A350407 a[n_] := Coefficient[Series[1/(1 + x + x^2 + x^3 + x^4)^n, {x, 0, n}], x, n]; Array[a, 30, 0] (* _Amiram Eldar_, Dec 29 2021 *)
%o A350407 (PARI) a(n) = sum(k=0, n, (-1)^(n-k)*binomial(n-1+k, k)*binomial(n, 5*k));
%Y A350407 Cf. A187925, A228960, A350383, A350406.
%K A350407 sign,easy
%O A350407 0,6
%A A350407 _Seiichi Manyama_, Dec 29 2021