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

Author

Paul Curtz, Jun 15 2007

Keywords

• nonn

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.

Examples

			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.
		

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)

Extensions

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