A145607 Numbers k such that (3*(2*k + 1)^2 + 2)/5 is a square.

0, 4, 35, 279, 2200, 17324, 136395, 1073839, 8454320, 66560724, 524031475, 4125691079, 32481497160, 255726286204, 2013328792475, 15850904053599, 124793903636320, 982500325036964, 7735208696659395, 60899169248238199

Offset: 1

Author

Richard Choulet, Oct 14 2008

Keywords

• nonn
• easy

Comments

Square roots of (3*(2*k+1)^2+2)/5 are listed in A070997, therefore (3*(2*a(n) + 1)^2 + 2)/5 = A070997(n-1)^2.

Links

• Index entries for linear recurrences with constant coefficients, signature (9,-9,1).

Crossrefs

Cf. A070997, A131751.

Cf. A001091 (first differences).

Formula

a(n+2) = 8*a(n+1) - a(n) + 3.

From R. J. Mathar, Oct 24 2008: (Start)

G.f.: x^2*(4 - x)/((1 - x)*(1 - 8*x + x^2)).

a(n) = (A057080(n-1)-1)/2. (End)

Extensions

a(4) corrected, extended, definition corrected by R. J. Mathar, Oct 24 2008

Offset changed by Bruno Berselli, Apr 06 2018