A260708 a(2n) = n*(2*n+1), a(2n+7) = a(2n+1) + 12*n + 28, with a(1)=1, a(3)=6, a(5)=16.

0, 1, 3, 6, 10, 16, 21, 29, 36, 46, 55, 68, 78, 93, 105, 122, 136, 156, 171, 193, 210, 234, 253, 280, 300, 329, 351, 382, 406, 440, 465, 501, 528, 566, 595, 636, 666, 709, 741, 786, 820, 868, 903, 953, 990, 1042, 1081, 1136, 1176, 1233, 1275, 1334, 1378

Offset

0,3

Comments

Conjecture: this sequence is 0 followed by A264041.

After 3, if a(n) is prime then n == 1 (mod 6).

a(n) is a square for n = 0, 1, 5, 8, 145, 288, 1777, 6533, 9800, 168097, 332928, 2051425, 7539845, ...

Links

Colin Barker, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (1,1,-1,0,0,1,-1,-1,1).

Formula

From Colin Barker, Nov 17 2015: (Start)

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

a(n) = a(n-1) + a(n-2) - a(n-3) + a(n-6) - a(n-7) - a(n-8) + a(n-9) for n>8. (End)

a(2*k) = A000217(2*k) by definition; for odd indices:

a(6*k+1) = 18*k^2 + 10*k + 1,

a(6*k+3) = 2*(9*k^2 + 11*k + 3),

a(6*k+5) = 2*(k + 1)*(9*k + 8), that is A178574.

a(n) = A260699(n) + A008615(n).

a(n) = n*(n + 1)/2 + (1 - (-1)^n)*floor(n/6 + 1/3)/2. [Bruno Berselli, Nov 18 2015]

Example

a(0) = 0*1 = 0,
a(1) = 1,
a(2) = 1*3 = 3,
a(3) = 6,
a(4) = 2*5 = 10,
a(5) = 16,
a(6) = 3*7 = 21,
a(7) = a(1) +12*0 +28 = 29, etc.

Mathematica

LinearRecurrence[{1, 1, -1, 0, 0, 1, -1, -1, 1}, {0, 1, 3, 6, 10, 16, 21, 29, 36}, 50] (* Bruno Berselli, Nov 18 2015 *)

Prog

(PARI) concat(0, Vec(-x*(x^6+x^5+3*x^4+2*x^3+2*x^2+2*x+1)/((x-1)^3*(x+1)^2*(x^2-x+1)*(x^2+x+1)) + O(x^100))) \\ Colin Barker, Nov 18 2015
(Sage) [n*(n+1)/2+(1-(-1)^n)*floor(n/6+1/3)/2 for n in (0..60)] # Bruno Berselli, Nov 18 2015

Crossrefs

Cf. A000217, A008615, A014105, A260160, A260699, A264041.

Sequence in context: A310082 A153453 A264041 * A310083 A310084 A025215

Adjacent sequences: A260705 A260706 A260707 * A260709 A260710 A260711

Keyword

nonn,easy

Author

Paul Curtz, Nov 17 2015

Extensions

Edited by Bruno Berselli, Nov 18 2015

Status

approved