A130002 Alternating sum along antidiagonals of the array of A129952 and its iterated differences.

1, 1, 2, 3, 10, 23, 60, 139, 326, 735, 1648, 3635, 7962, 17287, 37316, 80091, 171118, 364079, 771864, 1631107, 3436994, 7223511, 15146092, 31690283, 66176790, 137945983, 287076800, 596523219, 1237785706, 2565049895, 5309056788, 10976027515, 22667882942

Offset

0,3

Comments

Define the square array T of A129952 and its iterated differences: T(0,n)=A129952(n), T(d,n)=T(d-1,n+1)-T(d-1,n), d>0. Then a(n) = sum_{d=0..n} (-1)^d*T(d,n-d), the sum along the antidiagonals of T(d,n), alternating signs.

Links

Table of n, a(n) for n=0..32.

Formula

From Chai Wah Wu, Jan 30 2018: (Start)

a(n) = 4*a(n-1) - 2*a(n-2) - 8*a(n-3) + 7*a(n-4) + 4*a(n-5) - 4*a(n-6) for n > 6.

G.f.: (-2*x^6 - 6*x^5 + 3*x^4 + 5*x^3 - 3*x + 1)/((x - 1)^2*(x + 1)^2*(2*x - 1)^2). (End)

Example

The original series and first, 2nd etc. differences are the rows of
1..1..2...6..16..40.. <- A129952 = T(0,n)
0..1..4..10..24..56.. <- A129953 = T(1,n)
1..3..6..14..32..72.. <- A129954 = T(2,n)
2..3..8..18..40..88.. <- A129955 = T(3,n)
1..5.10..22..48......
...
a(2) = 2-1+1 = 2. a(3) = 6-4+3-2 = 3. a(4) = 16-10+6-3+1 = 10.

Crossrefs

Sequence in context: A148059 A089880 A271740 * A320812 A162034 A105286

Adjacent sequences: A129999 A130000 A130001 * A130003 A130004 A130005

Keyword

nonn

Author

Paul Curtz, Jun 15 2007

Extensions

Edited and extended by R. J. Mathar, Jun 30 2008

Status

approved