A082138 A transform of C(n,3).
1, 4, 20, 80, 280, 896, 2688, 7680, 21120, 56320, 146432, 372736, 931840, 2293760, 5570560, 13369344, 31752192, 74711040, 174325760, 403701760, 928514048, 2122317824, 4823449600, 10905190400, 24536678400, 54962159616, 122607894528
Offset: 0
Examples
a(0) = (2^(-1) + 0^0/2)*C(3,0) = 2*(1/2) = 1 (using 0^0=1).
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..400
- Index entries for linear recurrences with constant coefficients, signature (8,-24,32,-16).
Programs
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GAP
a:=[4,20,80,280];; for n in [5..30] do a[n]:=8*a[n-1]-24*a[n-2] +32*a[n-3]-16*a[n-4]; od; Concatenation([1], a);
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Magma
[(Ceiling(Binomial(n+3, 3)*2^(n-1))) : n in [0..30]]; // Vincenzo Librandi, Sep 22 2011
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Maple
[seq (ceil(binomial(n+3,3)*2^(n-1)),n=0..30)]; # Zerinvary Lajos, Nov 01 2006
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Mathematica
Join[{1}, LinearRecurrence[{8,-24,32,-16}, {4,20,80,280}, 30]] (* G. C. Greubel, Jul 23 2019 *)
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PARI
my(x='x+O('x^30)); Vec((1-4*x+12*x^2-16*x^3 + 8*x^4)/(1-2*x)^4) \\ G. C. Greubel, Jul 23 2019
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Sage
((1-4*x+12*x^2-16*x^3+8*x^4)/(1-2*x)^4).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jul 23 2019
Formula
a(n) = (2^(n-1) + 0^n/2)*C(n+3, n).
a(n) = Sum_{j=0..n} C(n+3, j+3)*C(j+3, 3)*(1 + (-1)^j)/2.
G.f.: (1 - 4*x + 12*x^2 - 16*x^3 + 8*x^4)/(1-2*x)^4.
E.g.f.: (x^3/3!)*exp(x)*cosh(x) (preceded by 3 zeros).
a(n) = ceiling(binomial(n+3,3)*2^(n-1)). - Zerinvary Lajos, Nov 01 2006
From Amiram Eldar, Jan 07 2022: (Start)
Sum_{n>=0} 1/a(n) = 12*log(2) - 7.
Sum_{n>=0} (-1)^n/a(n) = 108*log(3/2) - 43. (End)
Comments