A166911 - OEIS

A166911

a(n) = (9 + 14*n + 12*n^2 + 4*n^3)/3.

3, 13, 39, 89, 171, 293, 463, 689, 979, 1341, 1783, 2313, 2939, 3669, 4511, 5473, 6563, 7789, 9159, 10681, 12363, 14213, 16239, 18449, 20851, 23453, 26263, 29289, 32539, 36021, 39743, 43713, 47939, 52429, 57191, 62233, 67563, 73189, 79119, 85361, 91923

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OFFSET

0,1

COMMENTS

The inverse binomial transform yields the quasi-finite sequence 3,10,16,8,0,.. (0 continued).

These are the bottom-left numbers in the blocks (each with 2 rows) shown in A172002, the

atomic number of the leftmost element in the 2nd, 4th, 6th etc. row of the Janet table.

REFERENCES

Charles Janet, La structure du noyau de l'atome .., Nov 1927, page 15.

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..10000

Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

FORMULA

First differences: a(n)-a(n-1) = 2+4*n+4*n^2 = 1+(1+2n)^2 = 1 + A016754(n+1) = A069894(n+1).

Second differences: a(n) - 2*a(n-1) + a(n-2) = 8*n = A008590(n+2).

Third differences: a(n) - 3*a(n-1) + 3*a(n-2) - a(n-3) = 8.

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

G.f.: (3 + x + 5*x^2 - x^3)/(1-x)^4.

a(n) = A166464(n) + 2*(n+1)^2 = A166464(n) + A001105(n+1).

E.g.f.: (1/3)*(9 + 30*x + 24*x^2 + 4*x^3)*exp(x). - G. C. Greubel, May 28 2016

MATHEMATICA

LinearRecurrence[{4, -6, 4, -1}, {3, 13, 39, 89}, 100] (* G. C. Greubel, May 28 2016 *)

PROG

(Magma) [(9+14*n+12*n^2+4*n^3)/3: n in [0..40] ]; // Vincenzo Librandi, Aug 06 2011

(PARI) a(n)=n*(4*n^2+12*n+14)/3+3 \\ Charles R Greathouse IV, Dec 21 2011

CROSSREFS

Sequence in context: A072790 A323009 A328703 * A103657 A320661 A122504

Adjacent sequences: A166908 A166909 A166910 * A166912 A166913 A166914

KEYWORD

nonn,easy

AUTHOR

Paul Curtz, Oct 23 2009

EXTENSIONS

Edited and extended by R. J. Mathar, Mar 02 2010

STATUS

approved