A154357 - OEIS

A154357

a(n) = 25*n^2 - 14*n + 2.

2, 13, 74, 185, 346, 557, 818, 1129, 1490, 1901, 2362, 2873, 3434, 4045, 4706, 5417, 6178, 6989, 7850, 8761, 9722, 10733, 11794, 12905, 14066, 15277, 16538, 17849, 19210, 20621, 22082, 23593, 25154, 26765, 28426, 30137, 31898, 33709, 35570, 37481, 39442, 41453, 43514

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OFFSET

0,1

COMMENTS

The identity (1250*n^2 - 700*n + 99)^2 - (25*n^2 - 14*n + 2)*(250*n - 70)^2 = 1 can be written as A154359(n)^2 - a(n)*A154361(n)^2 = 1.

Numbers of the form (4*n-1)^2 + (3*n-1)^2. - Bruno Berselli, Dec 11 2011

From Bruno Berselli, Dec 13 2011: (Start)

More generally, considering together this sequence and A154355, A154358-A154361, for

r = (1/4)*(1250*(n-1)*(n-2) + 75*(2*n-3)(-1)^n + 321) with n>=0, i.e. the interleaving of A154358 and A154359 (649, 99, 99, 649, 2049, 3699,...)

s = (5/2)*(50*n+3*(-1)^n-75), the interleaving of A154360 and A154361 (-180, -70, 70, 180, 320, 430,...)

t = (1/8)*(50*(n-1)*(n-2) + 3*(2*n-3)*(-1)^n + 13), the interleaving of A154355 and A154357 (13, 2, 2, 13, 41, 74,...)

we verify that r^2 - t*s^2 = 1.

For n even we obtain (1250*n^2 - 1800*n + 649)^2 - (25*n^2 - 36*n + 13)*(250*n - 180)^2 = 1; for n odd we have the identity shown in the first comment. (End)

sqrt(A154357(n)) for n >= 1 has the continued fraction x; [1 1 1 1 2x] where x = 5n - 2 (the part in [] being repeated). - Robert Israel, May 26 2013

For n >= 1, the continued fraction expansion of sqrt(4*a(n)) is [10n-3; {4, 1, 5n-3, 1, 4, 20n-6}] - Magus K. Chu, Sep 16 2022

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..10000

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

FORMULA

G.f.: (2 + 7*x + 41*x^2)/(1-x)^3. - R. J. Mathar, Jan 05 2011

a(n) = 3*a(n-1) -3*a(n-2) +a(n-3). - Vincenzo Librandi, Feb 08 2012

E.g.f.: (2 + 11*x + 25*x^2)*exp(x). - G. C. Greubel, Sep 14 2016

MATHEMATICA

LinearRecurrence[{3, -3, 1}, {2, 13, 74}, 40] (* Vincenzo Librandi, Feb 08 2012 *)

PROG