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%I #19 Sep 20 2022 02:37:21
%S 1,1,3,7,14,126,42,66,99,143,1001,273,364,476,612,3876,969,1197,1463,
%T 1771,10626,2530,2990,3510,4095,23751,5481,6293,7192,8184,46376,10472,
%U 11781,13209,14763,82251,18278,20254,22386,24682,135751,29799,32637,35673,38916
%N a(n) = binomial(n+4,4)*gcd(n,5)/5.
%C The sixth column of the triangle A107711.
%H Indranil Ghosh, <a href="/A234042/b234042.txt">Table of n, a(n) for n = 0..10000</a>
%F a(n) = A107711(n+5,5) = binomial(n+5,5)*gcd(n,5)/(n+5), with n >= 0.
%F O.g.f.: ((1+x^20) + x*(1+x^18) + 3*x^2*(1+x^16) + 7*x^3*(1+x^14) + 14*x^4*(1+x^12) + 121*x^5*(1+x^10)+37*x^6*(1+x^8) + 51*x^7*(1+x^6) + 64*x^8*(1+x^4) + 73*x^9*(1+x^2) + 381*x^10)/(1-x^5)^5. From the 5-section using n = 5*k + j, for j = 0, 1, 2, 3, 4.
%F Sum_{n>=0} 1/a(n) = 20/3 - 16*sqrt(10-22/sqrt(5))*Pi/5. - _Amiram Eldar_, Sep 20 2022
%t a[n_] := Binomial[n + 4, 4] * GCD[n, 5]/5; Table[a[n], {n, 0, 40}] (* _Amiram Eldar_, Sep 20 2022 *)
%o (PARI) a(n) = binomial(n+4,4)*gcd(n,5)/5 \\ _Charles R Greathouse IV_, Feb 16 2017
%Y Cf. A107711, A208950 (fifth column of A107711), A109009 (gcd(n,5)).
%K nonn,easy
%O 0,3
%A _Wolfdieter Lang_, Feb 24 2014
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