A092055 a(n) = binomial(2 + 2^n,3).
1, 4, 20, 120, 816, 5984, 45760, 357760, 2829056, 22500864, 179481600, 1433753600, 11461636096, 91659526144, 733141975040, 5864598896640, 46914643623936, 375308558925824, 3002434111406080, 24019335451770880, 192154133857304576, 1537230871833083904, 12297838178567454720
Offset: 0
Examples
a(5) = C(2+2^5,3) = C(34,3) = 5984.
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (14,-56,64).
Programs
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Magma
[Binomial(2^n+2, 3): n in [0..30]]; // G. C. Greubel, Dec 27 2024
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Maple
seq(binomial(2+2^n, 3), n=0..25); # Zerinvary Lajos, Feb 22 2008
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Mathematica
nn=20;Table[Coefficient[Series[1/(1-x)^(2^n),{x,0,nn}],x^3],{n,0,nn}] (* Geoffrey Critzer, Jul 10 2013 *) Binomial[2+2^Range[0,30], 3] (* G. C. Greubel, Dec 27 2024 *)
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PARI
Vec((1-10*x+20*x^2)/((1-2*x)*(1-4*x)*(1-8*x)) + O(x^100)) \\ Colin Barker, Sep 13 2014
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Python
def A092055(n): return binomial(pow(2,n)+2,3) print([A092055(n) for n in range(41)]) # G. C. Greubel, Dec 27 2024
Formula
a(n) = (2^(3*n-1) +3*2^(2*n-1) + 2^n)/3 = A092056(3, n).
a(n) = 2*a(n-1) + 4^(n-1) + 8^(n-1).
a(n) = [x^3] 1/(1-x)^(2^n). - Geoffrey Critzer, Jul 11 2013
a(n) = 14*a(n-1) - 56*a(n-2) + 64*a(n-3). - Colin Barker, Sep 13 2014
G.f.: (1-10*x+20*x^2) / ((1-2*x)*(1-4*x)*(1-8*x)). - Colin Barker, Sep 13 2014
E.g.f.: (1/6)*(2*exp(2*x) + 3*exp(4*x) + exp(8*x)). - G. C. Greubel, Dec 27 2024
Comments