A112524 a(n) = a(n-1) + 2*n^2 with a(1) = 1.
1, 9, 27, 59, 109, 181, 279, 407, 569, 769, 1011, 1299, 1637, 2029, 2479, 2991, 3569, 4217, 4939, 5739, 6621, 7589, 8647, 9799, 11049, 12401, 13859, 15427, 17109, 18909, 20831, 22879, 25057, 27369, 29819, 32411, 35149, 38037, 41079, 44279, 47641
Offset: 1
Links
- Harvey P. Dale, Table of n, a(n) for n = 1..1000
- Eric Weisstein's World of Mathematics, Condensation
- Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
Crossrefs
Cf. A006331.
Programs
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Magma
[n*(n+1)*(2*n+1)/3 - 1: n in [1..40]]; // G. C. Greubel, Jan 12 2022
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Maple
a[1]:=1: for n from 2 to 50 do a[n]:=a[n-1]+2*n^2 od: seq(a[n],n=1..50); # Emeric Deutsch, Feb 13 2006 a:=n->sum(k^2, k=1..n):seq(a(n)+sum(k^2, k=2..n), n=1...40); # Zerinvary Lajos, Jun 11 2008
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Mathematica
Table[n*(n+1)*(2n+1)/3 - 1, {n, 50}] (* Stefan Steinerberger, Mar 11 2006 *) 2*Accumulate[Range[50]^2]-1 (* or *) LinearRecurrence[{4,-6,4,-1},{1,9,27,59},50] (* Harvey P. Dale, Dec 03 2012 *)
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Sage
[n*(n+1)*(2*n+1)/3 - 1 for n in (1..40)] # G. C. Greubel, Jan 12 2022
Formula
Twice the sum of the first n square numbers - 1 = n*(n + 1)*(2n + 1)/3 - 1. - Stefan Steinerberger, Mar 11 2006
From R. J. Mathar, Sep 09 2008: (Start)
G.f.: x*(1 +5*x -3*x^2 +x^3)/(1-x)^4.
a(n) = A006331(n) - 1. (End)
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), a(1)=1, a(2)=9, a(3)=27, a(4)=59. - Harvey P. Dale, Dec 03 2012
E.g.f.: ( 3 + (-3 + 6*x + 9*x^2 + 2*x^3)*exp(x) )/3. - G. C. Greubel, Jan 12 2022
Extensions
Definition corrected by Alexandre Wajnberg, Jan 02 2006
More terms from Emeric Deutsch, Feb 13 2006
More terms from Stefan Steinerberger, Mar 11 2006
Comments