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%I #13 May 30 2013 16:57:29
%S 4,8,13,17,22,26,31,36,40,45,49,54,58,63,68,72,77,81,86,90,95,99,104,
%T 109,113,118,122,127,131,136,141,145,150,154,159,163,168,173,177,182,
%U 186,191,195,200
%N a(n) = n-th positive integer such that only one integer lies between exp(s(m)) and exp(s(m+1)), where s(m) = 1 + 1/2 + 1/3 + . . . + 1/m.
%C Equivalently, this is a list of integers m for which there is exactly one integer k satisfying 1+1/2+...+1/m < log(k) < 1+1/2+...+1/(m+1). - _Clark Kimberling_ May 30 2013.
%H Clark Kimberling, <a href="/A004081/b004081.txt">Table of n, a(n) for n = 1..400</a>
%F Conjecturally, a(n) = floor(n/(2 - exp(g)) - 1/2 + exp(g)/(24n)), where g is the Euler-Mascheroni constant. - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Aug 11 2007
%t Flatten[Position[Differences[Table[Floor[E^Sum[1/k, {k, 1, m}]], {m, 1, 500}]], 1]] (* _Clark Kimberling_, May 30 2013 *)
%o (PARI) lista(n) = {old = 1; expo = exp(old); for (i=2, n, new = old + 1/i; expn = exp(new); if (floor(expn)==ceil(expo), print1(i-1, ", ")); old = new; expo = expn;);} \\ _Michel Marcus_, Mar 21 2013
%K nonn
%O 1,1
%A _Clark Kimberling_
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