|
|
A166911
|
|
a(n) = (9 + 14*n + 12*n^2 + 4*n^3)/3.
|
|
6
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
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
|
|
|
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.
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
|
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|