A302906 a(0) = 0; for n > 0, a(n) = a(n-1) + 5*n + 4.

0, 9, 23, 42, 66, 95, 129, 168, 212, 261, 315, 374, 438, 507, 581, 660, 744, 833, 927, 1026, 1130, 1239, 1353, 1472, 1596, 1725, 1859, 1998, 2142, 2291, 2445, 2604, 2768, 2937, 3111, 3290, 3474, 3663, 3857, 4056, 4260, 4469, 4683, 4902, 5126, 5355, 5589

Offset: 0

Author

J. Stauduhar, Apr 15 2018

Keywords

• nonn
• easy

Comments

In A302717, if we count the terms added from each 4-tuple during each iteration we find that either two or three terms are added: 2, 2, 2, 3, 2, 2, 2, 3, 2, 3, 2, 2, 2, 3, ... where the set of three twos (2, 2, 2) appears with decreasing frequency. a(n) is the index into A302717 where each such set begins.

40*a(n) + 169 is a square, whose square root belongs to A063226 and A063206. - Bruno Berselli, Apr 17 2018

Examples

			In A302717:
with a=9  and b=10, a(20,21,22) are appended;
with a=10 and b=11, a(23,24) are appended;
with a=11 and b=12, a(25,26) are appended;
with a=12 and b=13, a(27,28) are appended;
with a=13 and b=14, a(29,30,31) are appended, so a(2) = 23, because A302717(23) is the start of three consecutively added pairs of terms.
		

Links

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

Crossrefs

Cf. A302717.

Programs

Magma

[n*(5*n+13)/2: n in [0..60]]; // Vincenzo Librandi, Apr 17 2018

Mathematica

LinearRecurrence[{3, -3, 1}, {0, 9, 23}, 50] (* Vincenzo Librandi, Apr 17 2018 *)

PARI

concat(0, Vec(x*(9 - 4*x) / (1 - x)^3 + O(x^50))) \\ Colin Barker, Apr 16 2018

Formula

From Colin Barker, Apr 16 2018: (Start)

G.f.: x*(9 - 4*x)/(1 - x)^3.

a(n) = n*(5*n + 13)/2.

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>2. (End)