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A130002
Alternating sum along antidiagonals of the array of A129952 and its iterated differences.
0
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.
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
KEYWORD
nonn
AUTHOR
Paul Curtz, Jun 15 2007
EXTENSIONS
Edited and extended by R. J. Mathar, Jun 30 2008
STATUS
approved