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A020928 Expansion of 1/(1-4*x)^(17/2). 3

%I #28 Mar 27 2022 03:54:43

%S 1,34,646,9044,104006,1040060,9360540,77558760,601080390,4407922860,

%T 30855460020,207573094680,1349225115420,8510496881880,52278766560120,

%U 313672599360720,1842826521244230,10623352887172620,60198999693978180,335847050924299320,1847158780083646260

%N Expansion of 1/(1-4*x)^(17/2).

%H Vincenzo Librandi, <a href="/A020928/b020928.txt">Table of n, a(n) for n = 0..200</a>

%F a(n) = binomial(n+8, 8)*A000984(n+8)/A000984(8), A000984: central binomial coefficients. - _Wolfdieter Lang_

%F a(n) = ((2*n+15)*(2*n+13)*(2*n+11)*(2*n+9)*(2*n+7)*(2*n+5)*(2*n+3)*(2*n+1)/2027025)*binomial(2*n, n). - _Vincenzo Librandi_, Jul 05 2013

%F Boas-Buck recurrence: a(n) = (34/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, a(0) = 1. Proof from a(n) = A046521(n+8, 8). See a comment there. - _Wolfdieter Lang_, Aug 10 2017

%F From _Amiram Eldar_, Mar 27 2022: (Start)

%F Sum_{n>=0} 1/a(n) = 39708492/1001 - 7290*sqrt(3)*Pi.

%F Sum_{n>=0} (-1)^n/a(n) = 937500*sqrt(5)*log(phi) - 3029336260/3003, where phi is the golden ratio (A001622). (End)

%t CoefficientList[Series[1/(1-4x)^(17/2), {x,0,20}], x] (* _Vincenzo Librandi_, Jul 05 2013 *)

%o (Magma) [&*[2*n+i: i in [1..15 by 2]]*Binomial(2*n, n)/2027025: n in [0..20]]; // _Vincenzo Librandi_, Jul 05 2013

%o (PARI) vector(20, n, n--; m=n+8; binomial(2*m,m)*binomial(m,8)/12870) \\ _G. C. Greubel_, Jul 21 2019

%o (Sage) [binomial(2*(n+8),n+8)*binomial(n+8,8)/12870 for n in (0..20)] # _G. C. Greubel_, Jul 21 2019

%o (GAP) List([0..20], n-> Binomial(2*(n+8),n+8)*Binomial(n+8,8)/12870); # _G. C. Greubel_, Jul 21 2019

%Y Cf. A000984, A001622, A020926, A046521 (ninth column).

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_

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Last modified March 29 08:53 EDT 2024. Contains 371268 sequences. (Running on oeis4.)