A274682 Numbers n such that 8*n-1 is a triangular number.

2, 7, 29, 44, 88, 113, 179, 214, 302, 347, 457, 512, 644, 709, 863, 938, 1114, 1199, 1397, 1492, 1712, 1817, 2059, 2174, 2438, 2563, 2849, 2984, 3292, 3437, 3767, 3922, 4274, 4439, 4813, 4988, 5384, 5569, 5987, 6182, 6622, 6827, 7289, 7504, 7988, 8213, 8719

Offset: 1

Author

Colin Barker, Jul 02 2016

Keywords

• nonn
• easy

Examples

			2 is in the sequence since 8*2 - 1 = 15, and 15 = 1 + 2 + 3 + 4 + 5 is a triangular number. - _Michael B. Porter_, Jul 03 2016
		

Links

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

Crossrefs

Cf. A000124 (n-1), A174114 (2*n-1), A213399 (4*n-1), A069099 (7*n-1).

Programs

Mathematica

Table[(5 + 3 (-1)^n - 2 (8 + 3 (-1)^n) n + 16 n^2)/4, {n, 47}] (* or *)
Rest@ CoefficientList[Series[x (2 + 5 x + 18 x^2 + 5 x^3 + 2 x^4)/((1 - x)^3 (1 + x)^2), {x, 0, 47}], x] (* Michael De Vlieger, Jul 02 2016 *)

PARI

isok(n) = ispolygonal(8*n-1, 3)

PARI

select(n->ispolygonal(8*n-1, 3), vector(10000, n, n-1))

PARI

Vec(x*(2+5*x+18*x^2+5*x^3+2*x^4)/((1-x)^3*(1+x)^2) + O(x^100))

Formula

a(n) = (5+3*(-1)^n-2*(8+3*(-1)^n)*n+16*n^2)/4.

a(n) = (8*n^2-11*n+4)/2 for n even.

a(n) = (8*n^2-5*n+1)/2 for n odd.

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

G.f.: x*(2+5*x+18*x^2+5*x^3+2*x^4) / ((1-x)^3*(1+x)^2).