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A166464 a(n) = (3+2n+6n^2+4n^3)/3. 10

%I #27 Jan 11 2020 15:57:46

%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

%N a(n) = (3+2n+6n^2+4n^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+8n+4n^2 or a(n)-a(n-1)=4*(n+1)^2 = A016742(n+1).

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

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

%F a(n) = 4a(n-1)-6a(n-2)+4a(n-3)-a(n-4) or a(n)-4a(n-1)+6a(n-2)-4a(n-3)+a(n-4)=0 = A000004(n).

%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 a(n) = A018227(2n) + 3; a(n) = A002492(n) + 1. - _Natan Arie Consigli_, Jul 03 2016

%t Table[(3 + 2*n + 6*n^2 + 4*n^3)/3, {n, 0, 100}](* _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

%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

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Last modified March 28 21:57 EDT 2024. Contains 371254 sequences. (Running on oeis4.)