A020932 Expansion of 1/(1-4*x)^(21/2).
1, 42, 966, 16100, 217350, 2521260, 26053020, 245642760, 2149374150, 17672631900, 137846528820, 1027583214840, 7364346373020, 50983936428600, 342320716020600, 2236495344667920, 14257657822257990, 88900689950549820, 543281994142248900, 3259691964853493400
Offset: 0
Links
- Harvey P. Dale, Table of n, a(n) for n = 0..1000
Programs
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GAP
List([0..20], n-> Binomial(2*(n+10),n+10)*Binomial(n+10, 10)/Binomial(20,10)); # G. C. Greubel, Jul 21 2019
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Magma
[&*[2*n+i: i in [1..19 by 2]]*Binomial(2*n, n)/654729075: n in [0..20]]; // Vincenzo Librandi, Jul 05 2013
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Mathematica
CoefficientList[Series[1/(1-4x)^(21/2),{x,0,30}],x] (* Harvey P. Dale, Oct 10 2011 *) -
PARI
vector(20, n, n--; m=n+10; binomial(2*m,m)*binomial(m, 10)/binomial(20,10) ) \\ G. C. Greubel, Jul 21 2019
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Sage
[binomial(2*(n+10), n+10)*binomial(n+10, 10)/binomial(20,10) for n in (0..20)] # G. C. Greubel, Jul 21 2019
Formula
a(n) = binomial(n+10, 10)*A000984(n+10)/A000984(10), where A000984 are the central binomial coefficients. - Wolfdieter Lang
a(n) = ((2*n+19)*(2*n+17)*(2*n+15)*(2*n+13)*(2*n+11)*(2*n+9)*(2*n+7)*(2*n+5)*(2*n+3)*(2*n+1)/654729075)*Binomial(2*n, n). - Vincenzo Librandi, Jul 05 2013
Boas-Buck recurrence: a(n) = (42/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, a(0) = 1. Proof from a(n) = A046521(n+10, 10). See a comment there. - Wolfdieter Lang, Aug 10 2017
From Amiram Eldar, Mar 27 2022: (Start)
Sum_{n>=0} 1/a(n) = 38476615836/85085 - 83106*sqrt(3)*Pi.
Sum_{n>=0} (-1)^n/a(n) = 29687500*sqrt(5)*log(phi) - 4892382624460/153153, where phi is the golden ratio (A001622). (End)