A267322 Expansion of (1 + x + x^2 + x^4 + 2*x^5)/(1 - x^3)^3.

1, 1, 1, 3, 4, 5, 6, 9, 12, 10, 16, 22, 15, 25, 35, 21, 36, 51, 28, 49, 70, 36, 64, 92, 45, 81, 117, 55, 100, 145, 66, 121, 176, 78, 144, 210, 91, 169, 247, 105, 196, 287, 120, 225, 330, 136, 256, 376, 153, 289, 425, 171, 324, 477, 190, 361, 532, 210, 400, 590, 231, 441, 651

Offset: 0

Author

Ilya Gutkovskiy, Apr 07 2016

Keywords

• nonn
• easy

Comments

Triangular numbers alternating with squares and pentagonal numbers.

Examples

			Illustration of initial terms:
==========================================================
n:    0   1   2     3     4     5       6       7       8
----------------------------------------------------------
                                                        o
                                                      o o
                                o       o   o o o   o o o
                    o   o o   o o     o o   o o o   o o o
      o   o   o   o o   o o   o o   o o o   o o o   o o o
==========================================================
      1   1   1     3     4     5       6       9      12
----------------------------------------------------------
		

Links

• Ilya Gutkovskiy, Extended illustration of initial terms
• Eric Weisstein's World of Mathematics, Triangular Number
• Eric Weisstein's World of Mathematics, Square Number
• Eric Weisstein's World of Mathematics, Pentagonal Number
• Index entries for linear recurrences with constant coefficients, signature (0,0,3,0,0,-3,0,0,1)

Crossrefs

Cf. A000217, A000290, A000326, A123596, A124093, A271391.

Programs

Mathematica

LinearRecurrence[{0, 0, 3, 0, 0, -3, 0, 0, 1}, {1, 1, 1, 3, 4, 5, 6, 9, 12}, 70]
Table[(Floor[n/3] + 1) ((n + 1) Floor[n/3] - 3 Floor[n/3]^2 + 2)/2, {n, 0, 70}] (* Bruno Berselli, Apr 08 2016 *)
CoefficientList[Series[(1+x+x^2+x^4+2x^5)/(1-x^3)^3,{x,0,70}],x] (* Harvey P. Dale, Dec 31 2023 *)

PARI

x='x+O('x^99); Vec((1+x+x^2+x^4+2*x^5)/(1-x^3)^3) \\ Altug Alkan, Apr 07 2016

Formula

G.f.: (1 + x + x^2 + x^4 + 2*x^5)/(1 - x^3)^3.

a(n) = 3*a(n-3) - 3*a(n-6) + a(n-9).

a(3k) = A000217(k+1), a(3k+1) = A000290(k+1), a(3k+2) = A000326(k+1).

Sum_{n>=0} 1/a(n) = 2 - Pi/sqrt(3) + Pi^2/6 + 3*log(3) = 5.1269715686...

a(n) = (floor(n/3) + 1)*((n+1)*floor(n/3) - 3*floor(n/3)^2 + 2)/2. - Bruno Berselli, Apr 08 2016