A086651 a(1)=1, a(2)=1 and for n > 2, a(n) is the smallest positive integer such that the third-order absolute difference gives the Fibonacci numbers A000045 = {1,1,2,3,5,8,...}.

1, 1, 2, 5, 5, 10, 23, 23, 2, 57, 57, 146, 379, 379, 2, 989, 989, 2586, 6767, 6767, 2, 17713, 17713, 46370, 121395, 121395, 2, 317813, 317813, 832042, 2178311, 2178311, 2, 5702889, 5702889, 14930354, 39088171, 39088171, 2, 102334157, 102334157

Offset

1,3

Comments

It appears that a(6k+3) is always 2. Is this easy to prove?

Links

Table of n, a(n) for n=1..41.

Formula

a(n)= +a(n-1) +18*a(n-6) -18*a(n-7) -a(n-12) +a(n-13), n>15. - R. J. Mathar, Sep 15 2012

G.f. x + x^2 -x^3*(2+3*x+5*x^3+13*x^4-57*x^6+x^7-x^9-x^10+3*x^12) / ( (x-1) *(x^2+x-1) *(x^2-x-1) *(x^4-x^3+2*x^2+x+1) *(x^4+x^3+2*x^2-x+1) ). - R. J. Mathar, Sep 15 2012

Crossrefs

Cf. A000045.

Sequence in context: A184443 A238655 A132295 * A074495 A081467 A194119

Adjacent sequences: A086648 A086649 A086650 * A086652 A086653 A086654

Keyword

nonn

Author

John W. Layman, Sep 11 2003

Status

approved