A301926 a(n+3) = a(n) + 24*n + 32, a(0)=0, a(1)=3, a(2)=13.

0, 3, 13, 32, 59, 93, 136, 187, 245, 312, 387, 469, 560, 659, 765, 880, 1003, 1133, 1272, 1419, 1573, 1736, 1907, 2085, 2272, 2467, 2669, 2880, 3099, 3325, 3560, 3803, 4053, 4312, 4579, 4853, 5136, 5427, 5725

Offset: 0

Author

Paul Curtz, Jun 20 2018

Keywords

• nonn
• easy

Comments

Difference table:

0, 3, 13, 32, 59, 93, 136, 187, ...

3, 10, 19, 27, 34, 43, 51, ... = b(n)

7, 9, 8, 7, 9, 8, ... .

The sequence of last decimal digits of a(n) has period 15 and contain no 1's, 4's or 8's.

a(n) is e(n), hexasection, in A262397(n-1).

b(n) mod 9 is of period 9: 3, 1, 1, 0, 7, 7, 6, 4, 4.

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-2,1).

Crossrefs

Cf. A262997, A262397. A000290, A240438, A016754, A262523 (hexasections). Cf. A130518.

Programs

Mathematica

CoefficientList[ Series[ -x (5^3 +9x^2 +7x +3)/(x -1)^3 (x^2 +x +1), {x, 0, 40}], x] (* or *)LinearRecurrence[{2, -1, 1, -2, 1}, {0, 3, 13, 32, 59, 93}, 41] (* Robert G. Wilson v, Jun 20 2018 *)

PARI

concat(0, Vec(x*(1 + x)*(3 + 4*x + 5*x^2) / ((1 - x)^3*(1 + x + x^2)) + O(x^40))) \\ Colin Barker, Jun 20 2018

Formula

a(-n) = A262997(n).

a(n) = 2*a(n-1) - a(n-2) + a(n-3) - 2*a(n-4) + a(n-5).

Trisections: a(3n) = 4*n*(9*n-1), a(3n+1) = 3 + 20*n + 36*n^2, a(3n+2) = 13 + 44*n + 36*n^2.

a(n+15) = a(n) + 40*(22+3*n).

G.f.: x*(1 + x)*(3 + 4*x + 5*x^2) / ((1 - x)^3*(1 + x + x^2)). - Colin Barker, Jun 20 2018