A272398 The union of hexagonal numbers (A000384) and centered 9-gonal numbers (A060544).

1, 6, 10, 15, 28, 45, 55, 66, 91, 120, 136, 153, 190, 231, 253, 276, 325, 378, 406, 435, 496, 561, 595, 630, 703, 780, 820, 861, 946, 1035, 1081, 1128, 1225, 1326, 1378, 1431, 1540, 1653, 1711, 1770, 1891, 2016, 2080, 2145, 2278, 2415, 2485, 2556, 2701, 2850

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

Cf. A000384, A060544, A272399 (intersection).

Sequence in context: A315287 A315288 A238047 * A020159 A357529 A048017

Adjacent sequences: A272395 A272396 A272397 * A272399 A272400 A272401

Keyword

nonn

Author

Colin Barker, Apr 28 2016

Status

approved