A263986 Difference between Catalan numbers and Fibonacci numbers: a(n) = C(n) - F(n).
1, 0, 1, 3, 11, 37, 124, 416, 1409, 4828, 16741, 58697, 207868, 742667, 2674063, 9694235, 35356683, 129643193, 477636116, 1767259009, 6564113655, 24466256074, 91482545929, 343059584993, 1289904100956, 4861946326427, 18367352950759, 69533550719586
Offset: 0
Keywords
Links
- Robert Israel, Table of n, a(n) for n = 0..1667
- S. B. Ekhad and M. Yang, Proofs of Linear Recurrences of Coefficients of Certain Algebraic Formal Power Series Conjectured in the On-Line Encyclopedia Of Integer Sequences, (2017)
Programs
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Maple
f:= gfun:-rectoproc({(2+4*n)*a(n)+(10+7*n)*a(1+n)+(-16-6*n)*a(n+2)+(-24-7*n)*a(n+3)+(28+6*n)*a(n+4)+(-6-n)*a(n+5), a(0) = 1, a(1) = 0, a(2) = 1, a(3) = 3, a(4) = 11}, a(n), remember): map(f, [$0..40]); # Robert Israel, Apr 26 2017 -
Mathematica
RecurrenceTable[{(2+4n) a[n] + (10+7n) a[1+n] + (-16-6n) a[n+2] + (-24-7n) a[n+3] + (28+6n) a[n+4] + (-6-n) a[n+5] == 0, a[0] == 1, a[1] == 0, a[2] == 1, a[3] == 3, a[4] == 11}, a, {n, 0, 40}] (* Jean-François Alcover, Sep 16 2022, after Robert Israel *) -
PARI
a000108(n) = binomial(2*n, n)/(n+1) a(n) = a000108(n)-fibonacci(n) \\ Felix Fröhlich, Jun 07 2016
Formula
a(n) = (2*n)!/(n!*(n+1)!) - ((1+sqrt(5))^n-(1-sqrt(5))^n)/(2^n*sqrt(5)).
G.f.: (1 - sqrt(1 - 4*x))/(2*x) - x/(1 - x - x^2).
a(n+1)/a(n) ~ 4.
Conjecture: (11*n-20)*(n-3)*(n+1)*a(n) +(-55*n^3+276*n^2-389*n+120)*a(n-1) +3*(11*n^3-64*n^2+125*n-80)*a(n-2) +2*(2*n-5)*(n-2)*(11*n-9)*a(n-3)=0. - R. J. Mathar, Jun 07 2016
Conjecture: +(n+1)*a(n) +2*(-3*n+1)*a(n-1) +(7*n-11)*a(n-2) +2*(3*n-7)*a(n-3) +(-7*n+25)*a(n-4) +2*(-2*n+9)*a(n-5)=0. - R. J. Mathar, Jun 07 2016
Conjectures confirmed since the g.f. satisfies the differential equations corresponding to these recurrences. - Robert Israel, Apr 26 2017
E.g.f.: (BesselI(0,2*x) - BesselI(1,2*x))*exp(2*x) - 2*sinh(sqrt(5)*x/2)*exp(x/2)/sqrt(5). - Ilya Gutkovskiy, Jun 07 2016