A057780 Multiples of 3 that are one less than a perfect square.
0, 3, 15, 24, 48, 63, 99, 120, 168, 195, 255, 288, 360, 399, 483, 528, 624, 675, 783, 840, 960, 1023, 1155, 1224, 1368, 1443, 1599, 1680, 1848, 1935, 2115, 2208, 2400, 2499, 2703, 2808, 3024, 3135, 3363, 3480, 3720, 3843, 4095, 4224, 4488, 4623, 4899, 5040
Offset: 1
Links
- Jason Kimberley, Table of n, a(n) for n = 1..2001
- Robert J. Lemke Oliver, Eta quotients and theta functions, Advances in Mathematics, Vol. 241, Jul. 2013, pp. 1-17.
- Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).
Crossrefs
Numbers of the form 9n^2+kn, for integer n: A016766 (k=0), A132355 (k=2), A185039 (k=4), this sequence (k=6), A218864 (k=8). - Jason Kimberley, Nov 08 2012
Programs
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Magma
a:=func
;[0]cat[a(n*m):m in[-1, 1],n in[1..24]]; // Jason Kimberley, Nov 09 2012 -
Mathematica
Select[3*Range[0,2000],IntegerQ[Sqrt[#+1]]&] (* or *) LinearRecurrence[ {1,2,-2,-1,1},{0,3,15,24,48},50] (* Harvey P. Dale, Sep 10 2019 *)
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PARI
concat(0, Vec(3*x^2*(1+4*x+x^2)/((1-x)^3*(1+x)^2) + O(x^100))) \\ Colin Barker, Dec 26 2015
Formula
G.f.: 3*x^2*(1+4*x+x^2) / ((1-x)^3*(1+x)^2). - Colin Barker, Nov 24 2012
From Colin Barker, Dec 26 2015: (Start)
a(n) = 3/8*(6*n^2-2*((-1)^n+3)*n+(-1)^n-1).
a(n) = a(n-1)+2*a(n-2)-2*a(n-3)-a(n-4)+a(n-5) for n>5. (End)
Extensions
Since this is a list, offset corrected to 1 by Jason Kimberley, Nov 09 2012
Comments