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a(n) = n^2*(2*n+1).
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%I #62 Aug 19 2022 13:59:55

%S 0,3,20,63,144,275,468,735,1088,1539,2100,2783,3600,4563,5684,6975,

%T 8448,10115,11988,14079,16400,18963,21780,24863,28224,31875,35828,

%U 40095,44688,49619,54900,60543,66560,72963,79764,86975,94608,102675,111188

%N a(n) = n^2*(2*n+1).

%C For a right triangle with sides of lengths 8*n^3 + 12*n^2 + 8*n + 2, 4*n^4 + 8*n^3 + 4*n^2, and 4*n^4 + 8*n^3 + 12*n^2 + 8*n + 2, dividing the area by the perimeter gives a(n). - _J. M. Bergot_, Jul 30 2013

%C This sequence is the difference between the centered icosahedral (or cuboctahedral) numbers (A005902(n)) and the centered octagonal pyramidal numbers (A000447(n+1)). - _Peter M. Chema_, Jan 09 2016

%C a(n) is the sum of the integers in the closed interval (n-1)*n to n*(n+1). - _J. M. Bergot_, Apr 19 2017

%H Vincenzo Librandi, <a href="/A099721/b099721.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).

%F G.f.: x*(3 + 8*x + x^2)/(x-1)^4.

%F a(n) = A024196(n) - A024196(n-1). - _Philippe Deléham_, May 07 2012

%F a(n) = ceiling(Sum_{i=n^2-(n-1)..n^2+(n-1)} s(i)), for n > 0 and integer i, where s(i) are the real solutions to x = i + sqrt(x), and the summation range excludes the integer solutions which occur where i is an oblong number (A002378). The fractional portion of the summation converges to 2/3 for large n. If s(i) is replaced with i, then the summation equals n^2*(2*n-1) = A015237. - _Richard R. Forberg_, Oct 15 2014

%F a(n) = A005902(n) - A000447(n+1). - _Peter M. Chema_, Jan 09 2016

%F From _Amiram Eldar_, May 17 2022: (Start)

%F Sum_{n>=1} 1/a(n) = Pi^2/6 + 4*log(2) - 4.

%F Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/12 - Pi - 2*log(2) + 4. (End)

%p A099721 := proc(n) n^2*(2*n+1) ; end proc:

%p seq(A099721(n),n=0..10) ;

%t a[n_]:=2*n^3+n^2; (* _Vladimir Joseph Stephan Orlovsky_, Dec 21 2008 *)

%t LinearRecurrence[{4,-6,4,-1},{0,3,20,63},40] (* _Harvey P. Dale_, Aug 19 2022 *)

%o (Magma) [n^2*(2*n+1): n in [0..50]]; // _Vincenzo Librandi_, May 01 2011

%o (PARI) a(n) = ceil(sum(i=n^2-(n-1), n^2+(n-1), if(!issquare(4*i+1), (2*i+1+sqrt(4*i+1))/2, 0))); \\ _Michel Marcus_, Nov 14 2014, after _Richard R. Forberg_

%Y Cf. A000578, A066023, A001093, A034262, A071568, A011379, A027444, A053698, A033431, A033562, A061317, A098547, A015237.

%Y Row n=3 of A229079.

%K nonn,easy

%O 0,2

%A Douglas Winston (douglas.winston(AT)srupc.com), Nov 07 2004