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A098106
Hankel transform of sequence (b(n)) where b(n) = Sum_{i=0..n} binomial(2*i,i).
0
1, 3, 6, -8, -72, -144, 64, 960, 1920, -512, -10752, -21504, 4096, 110592, 221184, -32768, -1081344, -2162688, 262144, 10223616, 20447232, -2097152, -94371840, -188743680, 16777216, 855638016, 1711276032, -134217728, -7650410496, -15300820992, 1073741824, 67645734912, 135291469824
OFFSET
0,2
COMMENTS
Barker's g.f. is a simplification of the sum of the three g.f.s for a(3*n), a(3*n+1) and a(3*n+2): +(-1/(-8*x^3-1)) + (-3*x*(8*x^3-1)/(2*x+1)^2/(4*x^2-2*x+1)^2) + (-(-48*x^5+6*x^2)/(-64*x^6-16*x^3-1)). - Georg Fischer, Nov 16 2022
LINKS
J. W. Layman, The Hankel Transform and Some of its Properties, J. Integer Sequences, 4 (2001), #01.1.5.
FORMULA
a(3*n) = (-8)^n; a(3*n+1) = 3*(-8)^n*(2*n+1); a(3*n+2) = 6*(-8)^n*(2*n+1).
G.f.: -(12*x^3-6*x^2+x-1) / (4*x^2-2*x+1)^2. - Colin Barker, Jun 27 2013
D-finite with recurrence: (2*n-3)^2*a(n) - (8*(n-2)^2+16*n-18)*a(n-1) + 4*(2*n-1)^2*a(n-2) = 0. - Georg Fischer, Nov 16 2022
MATHEMATICA
CoefficientList[Series[-(12*x^3-6*x^2+x-1) / (4*x^2-2*x+1)^2, {x, 0, 35}], x] (* Georg Fischer, Nov 16 2022 *)
PROG
(PARI) a(n)=if(n<0, 0, if(n%3, if(n%3-1, 6*(-8)^floor(n/3)*(2*floor(n/3)+1), 3*(-8)^floor(n/3)*(2*floor(n/3)+1)), (-8)^(n/3)))
CROSSREFS
Cf. A006134.
Sequence in context: A306157 A262613 A137129 * A059361 A103268 A114157
KEYWORD
sign,easy
AUTHOR
Benoit Cloitre, Sep 22 2004
STATUS
approved