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%I #24 Sep 08 2022 08:46:17
%S 1,5,41,61,145,181,313,365,545,613,841,925,1201,1301,1625,1741,2113,
%T 2245,2665,2813,3281,3445,3961,4141,4705,4901,5513,5725,6385,6613,
%U 7321,7565,8321,8581,9385,9661,10513,10805,11705,12013,12961,13285,14281,14621,15665
%N Subsequence of centered square numbers obtained by adding four triangles from A276914 and a central element, a(n) = 4*A276914(n) + 1.
%C All terms of this sequence are centered square numbers. Graphically, each term of the sequence is made of four squares, eight triangles and a central element.
%C a(A220185(n+1)) = A008844(2n) = A079291(4n+1), which is a square of a Pell number.
%H Daniel Poveda Parrilla, <a href="/A276916/b276916.txt">Table of n, a(n) for n = 0..10000</a>
%H Daniel Poveda Parrilla, <a href="/A276916/a276916.gif">Illustration of initial terms</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,2,-2,-1,1).
%F a(n) = 4*n*(2*n + (-1)^n) + 1.
%F a(n) = 4*n*(2*n + 1) + 1 for n even.
%F a(n) = 4*n*(2*n - 1) + 1 for n odd.
%F a(n) is sum of two squares; a(n) = k^2 + (k+1)^2 where k = 2n-(n mod 2). - _David A. Corneth_, Sep 27 2016
%F From _Colin Barker_, Sep 27 2016: (Start)
%F a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n > 4.
%F G.f.: (1+4*x+34*x^2+12*x^3+13*x^4) / ((1-x)^3*(1+x)^2). (End)
%F E.g.f.: (1+8*x+8*x^2)*exp(x) - 4*x*exp(-x). - _G. C. Greubel_, Aug 19 2022
%p A276916:=n->4*n*(2*n+(-1)^n)+1: seq(A276916(n), n=0..60); # _Wesley Ivan Hurt_, Sep 27 2016
%t Table[4 n (2 n + (-1)^n) + 1, {n, 0, 44}] (* or *)
%t CoefficientList[Series[(1 +4x +34x^2 +12x^3 +13x^4)/((1-x)^3*(1+x)^2), {x, 0, 44}], x] (* _Michael De Vlieger_, Sep 28 2016 *)
%o (PARI) Vec((1+4*x+34*x^2+12*x^3+13*x^4)/((1-x)^3*(1+x)^2) + O(x^50)) \\ _Colin Barker_, Sep 27 2016
%o (Magma) [4*n*(2*n+(-1)^n)+1 : n in [0..60]]; // _Wesley Ivan Hurt_, Sep 27 2016
%o (SageMath) [4*n*(2*n+(-1)^n) +1 for n in (0..60)] # _G. C. Greubel_, Aug 19 2022
%Y Cf. A008844, A079291, A220185, A276914.
%K nonn,easy
%O 0,2
%A _Daniel Poveda Parrilla_, Sep 27 2016
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