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

1, 5, 21, 57, 121, 221, 365, 561, 817, 1141, 1541, 2025, 2601, 3277, 4061, 4961, 5985, 7141, 8437, 9881, 11481, 13245, 15181, 17297, 19601, 22101, 24805, 27721, 30857, 34221, 37821, 41665, 45761, 50117, 54741, 59641, 64825, 70301, 76077, 82161, 88561, 95285, 102341, 109737, 117481, 125581

Offset: 0

Author

Paul Curtz, Oct 14 2009

Keywords

• nonn
• easy

Comments

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

References

• 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.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

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

Programs

Magma

[(3+2*n+6*n^2+4*n^3)/3: n in [0..60]]; // G. C. Greubel, Jul 27 2024

Mathematica

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

PARI

a(n)=(3+2*n+6*n^2+4*n^3)/3 \\ Charles R Greathouse IV, Oct 07 2015

SageMath

[(3+2*n+6*n^2+4*n^3)//3 for n in range(61)] # G. C. Greubel, Jul 27 2024

Formula

a(n) - a(n-1) = 4*(n+1)^2 = A016742(n+1).

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

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

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

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

G.f.: (1+x+7*x^2-x^3)/(1-x)^4. - R. J. Mathar, Feb 15 2010

From Natan Arie Consigli, Jul 03 2016: (Start)

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

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

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

Extensions

Edited by N. J. A. Sloane, Oct 17 2009

More terms a(11)-a(35) from Vincenzo Librandi, Oct 17 2009