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A336630 a(n) = 2*F(2*n+1) + 4*F(n+1)*F(n-1) for n > 0, with a(0) = 0 and F(n) = A000045(n). 0

%I #14 Sep 11 2020 12:04:52

%S 0,4,18,38,108,274,726,1892,4962,12982,33996,88994,232998,609988,

%T 1596978,4180934,10945836,28656562,75023862,196415012,514221186,

%U 1346248534,3524524428,9227324738,24157449798,63245024644,165577624146,433487847782

%N a(n) = 2*F(2*n+1) + 4*F(n+1)*F(n-1) for n > 0, with a(0) = 0 and F(n) = A000045(n).

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,2,-1).

%F a(n) = (F(4n+1) - F(n+1)^4)/F(n)^2 for n > 0 and a(0) = 0, where F(n) = A000045(n).

%F a(n) = 2*A001519(n+1) + 4*A059929(n-1) for n > 0.

%F From _Stefano Spezia_, Jul 28 2020: (Start)

%F O.g.f.: 2*x*(2 + 5*x - 3*x^2)/(1 - 2*x - 2*x^2 + x^3).

%F a(n) = 2*a(n-1) + 2*a(n-2) - a(n-2) for n > 3. (End)

%Y Cf. A000045, A001519, A059929.

%K nonn,easy

%O 0,2

%A _Michael Tulskikh_, Jul 28 2020

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Last modified April 25 03:15 EDT 2024. Contains 371964 sequences. (Running on oeis4.)