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a(n) = n*(3*n + 4).
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%I #59 Jan 01 2023 03:19:55

%S 0,7,20,39,64,95,132,175,224,279,340,407,480,559,644,735,832,935,1044,

%T 1159,1280,1407,1540,1679,1824,1975,2132,2295,2464,2639,2820,3007,

%U 3200,3399,3604,3815,4032,4255,4484,4719,4960,5207,5460,5719,5984,6255,6532,6815

%N a(n) = n*(3*n + 4).

%C The number of peers of a cell of an n^2 X n^2 sudoku is a(n-1). - _Neven Sajko_, Apr 20 2016

%C First differences are in A016921. - _Wesley Ivan Hurt_, Apr 21 2016

%H G. C. Greubel, <a href="/A140676/b140676.txt">Table of n, a(n) for n = 0..5000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F a(n) = 3*n^2 + 4*n.

%F a(n) = 6*n + a(n-1) + 1 for n>0, a(0)=0. - _Vincenzo Librandi_, Aug 03 2010

%F O.g.f.: x*(7 - x)/(1 - x)^3. - _Arkadiusz Wesolowski_, Dec 24 2011

%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>2. - _Harvey P. Dale_, May 04 2013

%F E.g.f.: x*(7 + 3*x)*exp(x). - _Ilya Gutkovskiy_, Apr 20 2016

%F a(n) = A000567(n+1) - 1. - _Neven Sajko_, Apr 20 2016

%F From _Amiram Eldar_, Feb 26 2022: (Start)

%F Sum_{n>=1} 1/a(n) = 15/16 - Pi/(8*sqrt(3)) - 3*log(3)/8.

%F Sum_{n>=1} (-1)^(n+1)/a(n) = 9/16 - Pi/(4*sqrt(3)). (End)

%p A140676:=n->n*(3*n+4): seq(A140676(n), n=0..100); # _Wesley Ivan Hurt_, Apr 21 2016

%t Table[n (3 n + 4), {n, 0, 50}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 7, 20}, 50] (* _Harvey P. Dale_, May 04 2013 *)

%o (PARI) a(n)=n*(3*n+4) \\ _Charles R Greathouse IV_, Oct 07 2015

%o (Magma) [n*(3*n+4) : n in [0..80]]; // _Wesley Ivan Hurt_, Apr 21 2016

%Y Cf. A000567, A016921, A033428, A045944, A067707, A067725, A140677, A140678, A140679, A140680, A140681, A140689.

%K nonn,easy

%O 0,2

%A _Omar E. Pol_, May 22 2008