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%I #34 Sep 08 2022 08:45:58
%S 1,15,53,115,201,311,445,603,785,991,1221,1475,1753,2055,2381,2731,
%T 3105,3503,3925,4371,4841,5335,5853,6395,6961,7551,8165,8803,9465,
%U 10151,10861,11595,12353,13135,13941,14771,15625,16503,17405,18331,19281
%N a(n) = 12*n^2 + 2*n + 1.
%C A142241 gives the first differences.
%C Inverse binomial transform of this sequence: 1, 14, 24, 0, 0 (0 continued).
%C a(n)*a(n-1)-11 is a square, precisely 4*A051866(n)^2.
%C Sequence found by reading the line from 1, in the direction 1, 15, ..., in the square spiral whose vertices are the generalized octagonal numbers A001082. - _Omar E. Pol_, Jul 18 2012
%H Bruno Berselli, <a href="/A194454/b194454.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F G.f.: (1+x)*(1+11*x)/(1-x)^3.
%F a(n) = A154106(-n-1).
%F a(n) = 2*A049453(n) + 1.
%F a(n) = A051866(n) + A051866(n+1). - _Charlie Marion_, Nov 15 2019
%F E.g.f.: exp(x)*(1 + 14*x + 12*x^2). - _Stefano Spezia_, Nov 15 2019
%e Using these numbers we can write:
%e 1, 15, 53, 115, 201, 311, 445, 603, 785, 991, 1221, ...
%e 0, 0, 1, 15, 53, 115, 201, 311, 445, 603, 785, ...
%e 0, 0, 0, 0, 1, 15, 53, 115, 201, 311, 445, ...
%e 0, 0, 0, 0, 0, 0, 1, 15, 53, 115, 201, ...
%e 0, 0, 0, 0, 0, 0, 0, 0, 1, 15, 53, ...
%e 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, ...
%e ======================================================
%e The sums of the columns give the sequence A172073 (after 0):
%e 1, 15, 54, 130, 255, 441, 700, 1044, 1485, 2035, 2706, ...
%t Table[12 n^2 + 2 n + 1, {n, 0, 50}] (* _Vincenzo Librandi_, Mar 26 2013 *)
%o (Magma) [12*n^2+2*n+1: n in [0..40]];
%o (PARI) for(n=0, 40, print1(12*n^2+2*n+1", "));
%Y Cf. A154106, A172073, A049453.
%K nonn,easy
%O 0,2
%A _Bruno Berselli_, Aug 24 2011
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