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a(n) = n-th concentric 12-gonal number.
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%I #62 Jan 16 2023 08:53:11

%S 0,1,12,25,48,73,108,145,192,241,300,361,432,505,588,673,768,865,972,

%T 1081,1200,1321,1452,1585,1728,1873,2028,2185,2352,2521,2700,2881,

%U 3072,3265,3468,3673,3888,4105,4332,4561,4800,5041,5292,5545,5808,6073,6348

%N a(n) = n-th concentric 12-gonal number.

%C Concentric dodecagonal numbers. [corrected by _Ivan Panchenko_, Nov 09 2013]

%C Sequence found by reading the line from 0, in the direction 0, 12,..., and the same line from 1, in the direction 1, 25,..., in the square spiral whose vertices are the generalized octagonal numbers A001082. Main axis, perpendicular to A028896 in the same spiral.

%C Partial sums of A091998. - _Reinhard Zumkeller_, Jan 07 2012

%C Column 12 of A195040. - _Omar E. Pol_, Sep 28 2011

%H Vincenzo Librandi, <a href="/A195143/b195143.txt">Table of n, a(n) for n = 0..10000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,-2,1).

%F From _Vincenzo Librandi_, Sep 27 2011: (Start)

%F a(n) = 3*n^2+(-1)^n-1.

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

%F G.f.: -x*(1+10*x+x^2) / ( (1+x)*(x-1)^3 ). - _R. J. Mathar_, Sep 18 2011

%F a(n) = Sum_{k=1..n} (2*(-1)^(n-k+1) + 3*(2*k-1)), n>0, a(0) = 0. - _L. Edson Jeffery_, Sep 14 2014

%F Sum_{n>=1} 1/a(n) = Pi^2/72 + tan(Pi/sqrt(6))*Pi/(4*sqrt(6)). - _Amiram Eldar_, Jan 16 2023

%t Table[Sum[2*(-1)^(n - k + 1) + 6*k - 3, {k, n}], {n, 0, 47}] (* _L. Edson Jeffery_, Sep 14 2014 *)

%o (Magma) [(3*n^2+(-1)^n-1): n in [0..50]]; // _Vincenzo Librandi_, Sep 27 2011

%o (Haskell)

%o a195143 n = a195143_list !! n

%o a195143_list = scanl (+) 0 a091998_list

%o -- _Reinhard Zumkeller_, Jan 07 2012

%Y A135453 and A069190 interleaved.

%Y Cf. A001082, A032527, A032528, A077221, A195043, A195045, A195040, A195142, A195145, A195146, A195147, A195148, A195149.

%Y Cf. A016921 (6n+1), A016969 (6n+5), A091998 (positive integers of the form 12*k +- 1), A092242 (positive integers of the form 12*k +- 5).

%K nonn,easy

%O 0,3

%A _Omar E. Pol_, Sep 17 2011