OFFSET
1,2
COMMENTS
3 divides a(n) for n>1. For n>1 a(n) = 3*A001108(n-1), where A001108(k) = {0, 1, 8, 49, 288, 1681, ...}, A001108(k)-th triangular number is a square. - Alexander Adamchuk, Jan 19 2007
Disregarding the term 1, numbers k such that A071910(k) is a nonzero square; i.e., numbers k such that A000096(k) = k*(k+3)/2 is a nonzero square. - Rick L. Shepherd, Jul 13 2012
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (7, -7, 1).
FORMULA
For n>1, a(n+2) = 6*a(n+1) - a(n) + 6.
For n>1, a(n) = ((3+2*sqrt(2))^(n-1) + (3-2*sqrt(2))^(n-1))*3/4 - 3/2.
a(n) = 3*A001108(n-1) for n>1. - Alexander Adamchuk, Jan 19 2007
For n>1, a(2)=3, a(3)=24, a(4)=147, a(n)=7*a(n-1)-7*a(n-2)+a(n-3) [From Harvey P. Dale, May 15 2011]
G.f.: (-1+x(4+(-10+x)x))/((-1+x)(1+(-6+x)x)) [From Harvey P. Dale, May 15 2011]
MATHEMATICA
Join[{1}, LinearRecurrence[{7, -7, 1}, {3, 24, 147}, 35]] (* or *) CoefficientList[Series[(-1+x(4+(-10+x)x))/((-1+x)(1+(-6+x) x)), {x, 0, 35}], x] (* Harvey P. Dale, May 15 2011 *)
PROG
(Magma) I:=[1, 3, 24, 147]; [n le 4 select I[n] else 7*Self(n-1)-7*Self(n-2)+Self(n-3): n in [1..30]]; Vincenzo Librandi, May 21 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Alexander Adamchuk, Nov 29 2006
EXTENSIONS
Edited by Max Alekseyev, Jan 11 2007
STATUS
approved