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Partial sums of A002522, starting at n=1.
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%I #27 Oct 30 2024 21:09:47

%S 2,7,17,34,60,97,147,212,294,395,517,662,832,1029,1255,1512,1802,2127,

%T 2489,2890,3332,3817,4347,4924,5550,6227,6957,7742,8584,9485,10447,

%U 11472,12562,13719,14945,16242,17612,19057,20579,22180,23862,25627

%N Partial sums of A002522, starting at n=1.

%H G. C. Greubel, <a href="/A145066/b145066.txt">Table of n, a(n) for n = 1..5000</a>

%F a(1) = 2; a(n) = a(n-1) + n^2 + 1 for n > 1.

%F From Christoph Pacher (christoph.pacher(AT)ait.ac.at), Jul 23 2010: (Start)

%F a(n) = Sum_{k=1..n} (k^2 + 1).

%F a(n) = A000330(n) + n.

%F a(n) = n*(n+1)*(2*n+1)/6 + n. (End)

%F G.f.: x*(2-x+x^2)/(1-x)^4. - _Colin Barker_, Apr 04 2012

%F E.g.f.: (1/6)*x*(12 + 9*x + 2*x^2)*exp(x). - _G. C. Greubel_, Jul 22 2017

%F a(n) = A081489(n+1) - 1. - _Jianing Song_, Oct 10 2021

%e a(2) = a(1) + 2^2 + 1 = 2 + 4 + 1 = 7; a(3) = a(2) + 3^2 + 1 = 7 + 9 + 1 = 17.

%t Accumulate[Range[50]^2+1] (* _Harvey P. Dale_, Dec 07 2016 *)

%o (PARI) {a=0; for(n=1, 42, print1(a=a+n^2+1, ","))}

%o (Python)

%o def A145066(n): return (n*(n*(2*n + 3) + 1))//6 + n # _Chai Wah Wu_, Oct 30 2024

%Y Cf. A002522 (n^2 + 1), A005563 ((n+1)^2 - 1), A051925 (zero followed by partial sums of A005563), A081489 (partial sums of A002522 starting at n=0).

%K nonn,easy,changed

%O 1,1

%A _Vladimir Joseph Stephan Orlovsky_, Sep 30 2008

%E Edited by _Klaus Brockhaus_, Oct 17 2008