|
| |
|
|
A099325
|
|
Expansion of (sqrt(1+2x) + sqrt(1-2x))/(2*(1-2x)^(3/2)).
|
|
7
|
|
|
|
1, 3, 7, 16, 35, 76, 162, 344, 723, 1516, 3158, 6568, 13598, 28120, 57956, 119344, 245123, 503116, 1030542, 2109704, 4311786, 8808328, 17969372, 36644176, 74640430, 151985016, 309170332, 628741264, 1277540828, 2595198256
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,2
|
|
|
COMMENTS
|
The g.f. is transformed to 1/(1-x)^3 under the Chebyshev transformation A(x)->1/(1+x^2)A(x/(1+x^2)). Second binomial transform of the sequence with g.f. 1/c(-x), where c(x) is the g.f. of the Catalan numbers A000108.
Image of 2n+1 under the Riordan array (1/sqrt(1-4x^2),xc(x^2)). Hankel transform is (n+1)*(-1)^n. - Paul Barry, Oct 06 2007
a(n) is the minimum possible value for the apex of a triangle of numbers whose base consists of a permutation of the numbers 1 to n+1, and each number in a higher row is the sum of the two numbers directly below it. - Nathaniel Johnston, Apr 20 2011
|
|
|
LINKS
|
|
|
|
FORMULA
|
a(n) = Sum_{k=0..n} (k+1)*binomial(n, (n-k)/2)*binomial(k+2, 2)*(1+(-1)^(n-k))/(n+k+2).
a(n) = 2^n + Sum_{k=0..floor((n-1)/2)} (2*n-4*k-1)*binomial(n, k). - Nathaniel Johnston, Apr 20 2011
a(n) = M^n*V topmost term. M = an infinite tridiagonal matrix with all 1's in the super and subdiagonals, [1,0,0,0,...] as the main diagonal; and the rest zeros. V = the vector [1,2,3,...]. - Gary W. Adamson, Jan 30 2012
D-finite with recurrence n*a(n) + 2*(-n-1)*a(n-1) + 4*(-n+3)*a(n-2) + 8*(n-2)*a(n-3) = 0. - R. J. Mathar, Nov 30 2012
a(n) ~ 2^(n+1/2)*sqrt(n)/sqrt(Pi) * (1 + sqrt(2*Pi/n)/4). - Vaclav Kotesovec, Feb 13 2014
|
|
|
MAPLE
|
a:=proc(n)return 2^n+add((2*n-4*k-1)*binomial(n, k), k=0..floor((n-1)/2)): end:
|
|
|
MATHEMATICA
|
CoefficientList[Series[(Sqrt[1+2*x]+Sqrt[1-2*x])/(2*(1-2*x)^(3/2)), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 13 2014 *)
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
easy,nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
