A330987 Alternatively add and half-multiply pairs of the nonnegative integers.

1, 3, 9, 21, 17, 55, 25, 105, 33, 171, 41, 253, 49, 351, 57, 465, 65, 595, 73, 741, 81, 903, 89, 1081, 97, 1275, 105, 1485, 113, 1711, 121, 1953, 129, 2211, 137, 2485, 145, 2775, 153, 3081, 161, 3403, 169, 3741, 177, 4095, 185, 4465, 193, 4851, 201, 5253, 209

Offset: 1

Author

George E. Antoniou, Jan 05 2020

Keywords

• nonn
• easy

Comments

In groups of two, add and half-multiply the integers: 0+1, (2*3)/2, 4+5, (6*7)/2, ....

From Bernard Schott, Jan 06 2020: (Start)

The bisection of this sequence gives:

For n odd = 2*k+1, k >= 0: a(2*k+1) = 8*k+1 = A017077(k),

For n even = 2*k, k >= 1: a(2*k) = T(4*k-2) = A000217(4*k-2) = (2*k-1)*(4*k-1) = A033567(k) where T(j) is the j-th triangular number. (End)

Links

• Colin Barker, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

Crossrefs

Cf. A330983.

Interspersion of A017077 and A033567 (excluding first term). - Michel Marcus, Jan 06 2020

Programs

Mathematica

a[n_]:=If[OddQ[n],4n-3,(n-1)(2n-1)]; Array[a,53] (* Stefano Spezia, Jan 05 2020 *)

PARI

Vec(x*(1 + 3*x + 6*x^2 + 12*x^3 - 7*x^4 + x^5) / ((1 - x)^3*(1 + x)^3) + O(x^50)) \\ Colin Barker, Jan 06 2020

Formula

From Colin Barker, Jan 05 2020: (Start)

G.f.: x*(1 + 3*x + 6*x^2 + 12*x^3 - 7*x^4 + x^5) / ((1 - x)^3*(1 + x)^3).

a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6) for n>6.

a(n) = -1 + 2*(-1)^n - (1/2)*(-1+7*(-1)^n)*n + (1+(-1)^n)*n^2.

(End)

E.g.f.: (1 + 4*x + 2*x^2)*cosh(x) - (3 + x)*sinh(x) - 1. - Stefano Spezia, Jan 05 2020 after Colin Barker