OFFSET
1,2
COMMENTS
As n increases, this sequence is approximately geometric with common ratio r = lim_{n->infinity} a(n)/a(n-1) = (4+sqrt(15))^2 = 31 + 8*sqrt(15). - Ant King, Dec 15 2011
LINKS
Colin Barker, Table of n, a(n) for n = 1..558
Eric Weisstein's World of Mathematics, Heptagonal Pentagonal Number.
Index entries for linear recurrences with constant coefficients, signature (63,-63,1).
FORMULA
From Ant King, Dec 15 2011: (Start)
a(n) = 63*a(n-1) - 63*a(n-2) + a(n-3).
a(n) = 62*a(n-1) - a(n-2) - 10.
a(n) = (1/60)*((3*sqrt(15)-5)*(4+sqrt(15))^(2*n-1) - (3*sqrt(15)+5)*(4-sqrt(15))^(2*n-1)+10).
a(n) = ceiling((1/60)*(3*sqrt(15)-5)*(4+sqrt(15))^(2*n-1)).
G.f.: x*(1-9*x-2*x^2)/((1-x)*(1-62*x+x^2)).
(End)
MATHEMATICA
LinearRecurrence[{63, -63, 1}, {1, 54, 3337}, 14] (* Ant King, Dec 15 2011 *)
CoefficientList[Series[x (1-9x-2x^2)/((1-x)(1-62x+x^2)), {x, 0, 20}], x] (* Harvey P. Dale, Jul 09 2021 *)
PROG
(PARI) Vec(x*(2*x^2+9*x-1)/((x-1)*(x^2-62*x+1)) + O(x^30)) \\ Colin Barker, Jun 23 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved