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Denominator of L(n) = (Sum_{k=1..n} k^n)/(Sum_{k=1..n-1} k^n).
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%I #22 May 14 2022 11:29:39

%S 1,1,49,52,20515,7689,1976849,769072,196573677,1176564625,

%T 2252928456427,915495729492,116920050750711,202297407264253,

%U 1206847874699507489,1507470694179701824,6945343389873635897155

%N Denominator of L(n) = (Sum_{k=1..n} k^n)/(Sum_{k=1..n-1} k^n).

%C L(n) has the amazing asymptotic development L(n) = e + c(1)/n + c(2)/n^2 + c(3)/n^3 + ... with c(1) = e*(e+1)/(2*(e-1)), c(2) = e*(11*e^3 + 3*e^2 - 51*e - 11)/(24*(e-1)^3), etc., where e = exp(1).

%D "A sequence convergent to Napier's Constant" by Alexandru Lupas from the University "Lucian Blaga" of Sibiu / e-mail: lupas(AT)jupiter.sibiu.ro

%H Amiram Eldar, <a href="/A043300/b043300.txt">Table of n, a(n) for n = 2..388</a>

%H Alexandru Lupas, <a href="https://web.archive.org/web/20170710020154/http://eureka.vu.edu.au/~rgmia/memberspapers/E-NEW.ps">A sequence convergent to Napier's Constant</a>.

%t a[n_] := Denominator[Sum[k^n, {k, 1, n}]/Sum[k^n, {k, 1, n - 1}]]; Array[a, 17, 2] (* _Amiram Eldar_, May 14 2022 *)

%o (PARI) a(n) = denominator(sum(k = 1, n, k^n)/sum(k = 1, n-1, k^n)); \\ _Michel Marcus_, Nov 21 2013

%Y Cf. A001113, A043299 (numerators).

%K easy,frac,nonn

%O 2,3

%A _Benoit Cloitre_, Apr 04 2002