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%I #7 Jan 30 2021 01:52:19
%S 1,1,2,5,5,10,23,23,2,57,57,146,379,379,2,989,989,2586,6767,6767,2,
%T 17713,17713,46370,121395,121395,2,317813,317813,832042,2178311,
%U 2178311,2,5702889,5702889,14930354,39088171,39088171,2,102334157,102334157
%N 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,...}.
%C It appears that a(6k+3) is always 2. Is this easy to prove?
%F 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
%F 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
%Y Cf. A000045.
%K nonn
%O 1,3
%A _John W. Layman_, Sep 11 2003
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