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a(n) = floor(H(n) + exp(H(n))*log(H(n))) - n, where H(n) = Sum_{k=1..n} 1/k.
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%I #31 May 18 2022 10:09:19

%S 0,1,2,3,5,6,8,9,11,13,14,16,18,19,21,23,24,26,28,30,32,34,35,37,39,

%T 41,43,45,47,49,51,53,55,57,59,61,63,65,67,69,71,73,75,77,79,81,83,85,

%U 87,89,91,93,96,98,100,102,104,106,108,110,113,115,117,119,121,123,126,128

%N a(n) = floor(H(n) + exp(H(n))*log(H(n))) - n, where H(n) = Sum_{k=1..n} 1/k.

%C About Lagarias's theorem and the Riemann hypothesis the graph of A057640 vs. A000203 is essentially equivalent to the graph of this sequence vs. A001065 (see Plot 2 in the Links section and A057640, A057641).

%H J. C. Lagarias, <a href="https://arxiv.org/abs/math/0008177">An elementary problem equivalent to the Riemann hypothesis</a>, arXiv:math/0008177 [math.NT], 2000-2001; Am. Math. Monthly 109 (#6, 2002), 534-543.

%H OEIS Plot 2, <a href="https://oeis.org/plot2a?name1=A353130&amp;name2=A001065&amp;tform1=untransformed&amp;tform2=untransformed&amp;shift=0&amp;radiop1=matp&amp;drawlines=true">A353130 vs A001065</a>

%F a(n) = A057640(n) - n.

%F a(n) = A057641(n) + A001065(n).

%t a[n_] := Module[{h = HarmonicNumber[n]}, Floor[h + Exp[h]*Log[h]] - n]; Array[a, 100] (* _Amiram Eldar_, Apr 26 2022 *)

%o (PARI) H(n) = sum(k=1, n, 1/k)

%o a(n) = floor(H(n) + exp(H(n))*log(H(n))) - n \\ _Felix Fröhlich_, Apr 26 2022

%Y Cf. A000203, A057640, A057641, A001065.

%K nonn

%O 1,3

%A _Omar E. Pol_, Apr 24 2022