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A358464 a(n) is the greatest m such that Sum_{k = 1..m} 1/(1 + n*k) <= 1.

%I A358464 #39 Apr 05 2024 10:06:41
%S A358464 2,6,16,42,110,288,761,2020,5388,14417,38681,103994,280032,755031,
%T A358464 2037848,5504884,14880978,40250609,108926101,294902398,798703663,
%U A358464 2163878141
%N A358464 a(n) is the greatest m such that Sum_{k = 1..m} 1/(1 + n*k) <= 1.
%C A358464 This sequence coincidences with 2*Fibonacci(2*n) (A025169) for the first 6 terms.
%H A358464 David Borwein and Jonathan M. Borwein, <a href="https://carmamaths.org/resources/db90/pdfs/db90-119.00.pdf">Some Remarkable Properties of Sinc and Related Integrals</a>, The Ramanujan Journal, 5 (2001) 73 - 89.
%H A358464 Hanspeter Schmid, <a href="http://schmid-werren.ch/hanspeter/publications/2014elemath.pdf">Two curious integrals and a graphic proof</a>, Elemente der Mathematik, 69 (2014) 11 - 17.
%F A358464 ceiling(digamma(a(n)+(1/n)+1) - digamma((1/n)+1)) = n.
%F A358464 Integral_{x=0..oo} Product_{k=0..m} sinc(x/(n*k+1)) dx = Pi for 0 <= m <= a(n). See links Schmid and Borwein.
%F A358464 ceiling(Sum_{m = 0..oo} ( 1/(m+1) * Sum_{k = 0..m} (-1)^k*binomial(m, k)*log( (a(n)+(1/n)+1+k) / ((1/n)+1+k) ) )) = n.
%F A358464 a(n) ~ floor(exp(n + digamma(1+(1/n))) - (1/2) - (1/n)). This appears to be accurate for at least n < 22.
%e A358464 a(2) = 6 because Sum_{m = 1..a(2)} 1/(1+2*m) = 43024/45045 < 1, but Sum_{m = 1..a(2)+1} 1/(1+2*m) = 46027/45045 > 1.
%o A358464 (PARI) a(n) = {my(k = 2*fibonacci(2*n)-1);my(b = (psi(k+(1/n))-psi(1+(1/n)))/n); until(b > 1, b = b+(1/(1+n*k)); k=k+1 );k-2}
%Y A358464 Cf. A002387, A025169, A064169.
%K A358464 nonn,more
%O A358464 1,1
%A A358464 _Thomas Scheuerle_, Nov 18 2022