A153644 - OEIS

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A153644

a(n) = 4*n^2 + 28*n + 10.

42, 82, 130, 186, 250, 322, 402, 490, 586, 690, 802, 922, 1050, 1186, 1330, 1482, 1642, 1810, 1986, 2170, 2362, 2562, 2770, 2986, 3210, 3442, 3682, 3930, 4186, 4450, 4722, 5002, 5290, 5586, 5890, 6202, 6522, 6850, 7186, 7530, 7882, 8242, 8610, 8986, 9370

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OFFSET

1,1

COMMENTS

Sequence gives values of x such that x^3 + 39x^2 = y^2 since a(n)^3 + 39*a(n)^2 = (8n^3 + 84n^2 + 216n + 70)^2.

a(n) = 2*(seventh diagonal to A153238).

About the first comment, naturally, the complete list of nonnegative values of x in x^3 + 39*x^2 = y^2 is given by x = m^2-39 with m>6. - Bruno Berselli, Jan 25 2012

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 1..10000

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

FORMULA

From Colin Barker, Jan 24 2012: (Start)

a(1)=42, a(2)=82, a(3)=130, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).

G.f.: 2*x*((3-x)*(7-5*x))/(1-x)^3. (End)

E.g.f.: 2*(-5 + (5 + 16*x + 2*x^2)*exp(x)). - G. C. Greubel, Aug 23 2016

Sum_{n>=1} 1/a(n) = 62/1995 + tan(sqrt(39)*Pi/2)*Pi/(4*sqrt(39)). - Amiram Eldar, Mar 02 2023

MATHEMATICA

LinearRecurrence[{3, -3, 1}, {42, 82, 130}, 25] (* G. C. Greubel, Aug 23 2016 *)

Table[4n^2+28n+10, {n, 70}] (* Harvey P. Dale, Jan 15 2023 *)