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a(n) = (3 + 2*n + 6*n^2 + 4*n^3)/3.
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%I #33 Jul 27 2024 04:05:25

%S 1,5,21,57,121,221,365,561,817,1141,1541,2025,2601,3277,4061,4961,

%T 5985,7141,8437,9881,11481,13245,15181,17297,19601,22101,24805,27721,

%U 30857,34221,37821,41665,45761,50117,54741,59641,64825,70301,76077,82161,88561,95285,102341,109737,117481,125581

%N a(n) = (3 + 2*n + 6*n^2 + 4*n^3)/3.

%C Atomic number of first transition metal of period 2n (n>3) or of the element after n-th alkaline earth metal. This can be calculated by finding the sum of the first n even squares plus 1. - _Natan Arie Consigli_, Jul 03 2016

%D JANET,Charles, La structure du Noyau de l'atome,consideree dans la Classification periodique,des elements chimiques,1927 (Novembre),N. 2,BEAUVAIS,67 pages,3 leaflets.

%H G. C. Greubel, <a href="/A166464/b166464.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 a(n) - a(n-1) = 4*(n+1)^2 = A016742(n+1).

%F a(n) - 2*a(n-1) + a(n-2) = -4 + 8*n = A017113(n+1).

%F a(n) - 3*a(n-1) + 3*a(n-2) - a(n-3) = 8 = A010731(n).

%F a(n) - 4*a(n-1) + 6*a(n-2) - 4*a(n-3) + a(n-4) = 0.

%F Binomial transform of quasi-finite sequence 1,4,12,8,0,(0 continued).

%F G.f.: (1+x+7*x^2-x^3)/(1-x)^4. - _R. J. Mathar_, Feb 15 2010

%F From _Natan Arie Consigli_, Jul 03 2016: (Start)

%F a(n) = A018227(2*n) + 3.

%F a(n) = A002492(n) + 1. (End)

%F E.g.f.: (1/3)*(3 + 12*x + 18*x^2 + 4*x^3)*exp(x). - _G. C. Greubel_, Jul 27 2024

%t Table[(3+2*n+6*n^2+4*n^3)/3, {n,0,60}] (* _G. C. Greubel_, May 15 2016 *)

%o (PARI) a(n)=(3+2*n+6*n^2+4*n^3)/3 \\ _Charles R Greathouse IV_, Oct 07 2015

%o (Magma) [(3+2*n+6*n^2+4*n^3)/3: n in [0..60]]; // _G. C. Greubel_, Jul 27 2024

%o (SageMath) [(3+2*n+6*n^2+4*n^3)//3 for n in range(61)] # _G. C. Greubel_, Jul 27 2024

%Y Cf. A002492, A010731, A016742, A017113, A018227.

%K nonn,easy

%O 0,2

%A _Paul Curtz_, Oct 14 2009

%E Edited by _N. J. A. Sloane_, Oct 17 2009

%E More terms a(11)-a(35) from _Vincenzo Librandi_, Oct 17 2009