OFFSET
1,3
COMMENTS
It is interesting that, for k > 0, it appears that a(2k) is the square of A005251(k+2). (This has since been proved by Andrew Weimholt; see link.)
If we denote by d2 the second difference of {a(n)}, it appears that d2(2k) is the square of A005314(k).
LINKS
Colin Barker, Table of n, a(n) for n = 1..1000
Proof by Andrew Weimholt, SeqFan Jan 2010
Index entries for linear recurrences with constant coefficients, signature (2,0,-1,-1,3,-1,0,1,-1).
FORMULA
Andrew Weimholt has shown that a(2*n) = A005251(n+2) ^ 2, and a(2*n+1) = A005251(n+2) * A005251(n+3). (See the link.)
G.f.: x*(1 - x + x^3 + 2*x^4 - x^8) / ((1 - 2*x + x^2 - x^3)*(1 + x - x^3)*(1 - x + x^3)). - Colin Barker, Feb 15 2016
MATHEMATICA
LinearRecurrence[{2, 0, -1, -1, 3, -1, 0, 1, -1}, {1, 1, 2, 4, 8, 16, 28, 49, 84}, 32] (* Jean-François Alcover, Feb 15 2016 *)
PROG
(PARI) Vec(x*(1-x+x^3+2*x^4-x^8)/((1-2*x+x^2-x^3)*(1+x-x^3)*(1-x+x^3)) + O(x^50)) \\ Colin Barker, Feb 15 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
John W. Layman, Jan 22 2010
STATUS
approved