OFFSET
1,2
COMMENTS
The construction of the g.f. works basically as follows every third entry of A000384 equals every second entry of A060544, A000384(3n+1) = A060544(2n+1) = (3*n+1)*(6*n+1), which is an immediate consequence of their polynomial representations. So the sequence is the union of A000384 and the bisection 10, 55, 136, 253,... of A060544. Following Section 4.3 of Riordan's book "Combinatorial identities", subsampling and "aering" are done by replacing the independent variable of the g.f. by roots of the independent variable. So this sequence has rational g.f. because it is derived by regular interlacing of the two original sequences which also have rational g.f.'s. - R. J. Mathar, Jul 15 2016
LINKS
Colin Barker, Table of n, a(n) for n = 1..1000
FORMULA
a(4*n-3) = A272399(n).
Conjectures:
a(n) = (-1+(-1)^n-6*((-i)^n+i^n)*n+18*n^2)/16 where i is the imaginary unit.
a(n) = 2*a(n-1)-2*a(n-2)+2*a(n-3)-2*a(n-5)+2*a(n-6)-2*a(n-7)+a(n-8) for n>8.
G.f.: x*(1+4*x+5*x^3+6*x^4+x^5+x^6) / ((1-x)^3*(1+x)*(1+x^2)^2).
CROSSREFS
KEYWORD
nonn
AUTHOR
Colin Barker, Apr 28 2016
STATUS
approved