OFFSET
1,2
COMMENTS
From Ant King, Oct 19 2011: (Start)
lim(n->Infinity,a(2n+1)/a(2n))=1/2(47+21*sqrt(5)).
lim(n->Infinity,a(2n)/a(2n-1))=1/2(7+3*sqrt(5)).
(End)
LINKS
Colin Barker, Table of n, a(n) for n = 1..798
Eric Weisstein's World of Mathematics, Heptagonal Triangular Number
Index entries for linear recurrences with constant coefficients, signature (1,322,-322,-1,1).
FORMULA
G.f.: (-2x^4-9x^3+161x^2+9x+1)/[(1-x)(1-18x+x^2)(1+18x+x^2)].
a(n+2) = 322*a(n+1)-a(n)+160 a(n+1) = 161*a(n)+80+36*(20*a(n)^2+20*a(n)+9)^0.5 - Richard Choulet, Sep 29 2007
From Ant King, Oct 19 2011: (Start)
a(n) = a(n-1)+322a(n-2)-322a(n-3)-a(n-4)+a(n-5).
a(n) = 1/20*sqrt(5)*(( sqrt(5)-(-1)^n)*(2+ sqrt(5))^(2n-1)+( sqrt(5)+(-1)^n)*(2- sqrt(5))^(2n-1)-2* sqrt(5)).
a(n) = floor(1/20* sqrt(5)*(sqrt(5)-(-1)^n)*(2+ sqrt(5))^(2n-1))(End)
MATHEMATICA
LinearRecurrence[{1, 322, -322, -1, 1}, {1, 10, 493, 3382, 158905}, 16] (* Ant King, Oct 19 2011 *)
PROG
(PARI) Vec((-2*x^4-9*x^3+161*x^2+9*x+1)/((1-x)*(1-18*x+x^2)*(1+18*x+x^2))+O(x^99))
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved