OFFSET
0,2
LINKS
Robert Israel, Table of n, a(n) for n = 0..850
Eric Weisstein's World of Mathematics, Quintic Equation.
Wikipedia, Lagrange inversion theorem.
FORMULA
25000*(2*n+7)*(4*n-1)*(4*n+9)*(n+1)*a(n) + (n+5)*(n+4)*(n+3)*(n+2)*a(n+5) = 0.
a(5*k) = Pochhammer(-1/10, 2*k)*Pochhammer(2/5, 2*k)*(-50000)^k/(Pochhammer(4/5, k)*Pochhammer(3/5, k)*Pochhammer(2/5, k)*k!).
a(5*k+1) = -5*Pochhammer(4/5, 2*k)*Pochhammer(3/10, 2*k)*(-50000)^k/(Pochhammer(6/5, k)*Pochhammer(4/5, k)*Pochhammer(3/5, k)*k!).
a(5*k+2) = -25*Pochhammer(6/5, 2*k)*Pochhammer(7/10, 2*k)*(-50000)^k/(Pochhammer(7/5, k)*Pochhammer(6/5, k)*Pochhammer(4/5, k)*k!).
a(5*k+3) = -125*Pochhammer(8/5, 2*k)*Pochhammer(11/10, 2*k)*(-50000)^k/(Pochhammer(8/5, k)*Pochhammer(7/5, k)*Pochhammer(6/5, k)*k!).
a(5*k+4) = 0.
From Seiichi Manyama, Jun 21 2025: (Start)
a(n) = (-25)^n * binomial(n/5+1/5,n)/(n+1).
G.f. A(x) satisfies A(x) = 1/A(-x/A(x)^3).
G.f.: (1/x) * Series_Reversion(x/(1-25*x)^(1/5)). (End)
MAPLE
f:= gfun:-rectoproc({(25000*(2*n+7))*(4*n-1)*(4*n+9)*(n+1)*a(n)+(n+5)*(n+4)*(n+3)*(n+2)*a(n+5), a(0)=1, a(1)=-5, a(2)=-25, a(3)=-125, a(4)=0}, a(n), remember):
map(f, [$0..40]);
MATHEMATICA
CoefficientList[Root[#^5 + 25*x*# - 1&, 1] + O[x]^40, x] (* Jean-François Alcover, Aug 27 2022 *)
PROG
(PARI) a(n) = (-25)^n*binomial(n/5+1/5, n)/(n+1); \\ Seiichi Manyama, Jun 21 2025
(PARI) my(N=30, x='x+O('x^N)); Vec(serreverse(x/(1-25*x)^(1/5))/x) \\ Seiichi Manyama, Jun 21 2025
CROSSREFS
KEYWORD
sign
AUTHOR
Robert Israel, Feb 22 2018
STATUS
approved