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%I #20 May 12 2017 19:40:53
%S 0,1,2,7,20,52,144,389,1058,2876,7817,21250,57763,157017,426817,
%T 1160207,3153770,8572836,23303385,63345169,172190019,468061001,
%U 1272321714,3458528995,9401256521,25555264765,69466411833,188829284972
%N For k >= 1, let b(k) = ceiling( Sum_{i=1..k} 1/i ); a(n) = number of b(k) that are equal to n.
%F a(n+1)/a(n) approaches e = exp(1) = 2.71828...
%F First differences of A002387. - _Vladeta Jovovic_, Jul 30 2004
%e The ceilings of the first several partial sums of the reciprocal of the positive integers are 1 2 2 3 3 3 3 3 3 3 4 4 and the series is monotonically increasing, so a(0) = 0 (there being no zero), a(1) = 1 (there being but one 1) and a(3) = 7 (there being seven 3s).
%t fh[0] = 0; fh[1] = 1; fh[k_] := Module[{tmp}, If[ Floor[tmp = Log[k + 1/2] + EulerGamma] == Floor[tmp + 1/(24k^2)], Floor[tmp], UNKNOWN]]; a[0] = 1; a[1] = 2; a[n_] := Module[{val}, val = Round[Exp[n - EulerGamma]]; If[fh[val] == n && fh[val - 1] == n - 1, val, UNKNOWN]]; Table[ a[n + 1] - a[n], {n, 0, 27}] (* _Robert G. Wilson v_, Aug 05 2004 *)
%o (C)
%o int A096005(int k)
%o { if(k<3) return k;
%o double sum = 0, n = 1; int ceiling = 2, cnt = 0;
%o for(;;) {
%o sum += 1/n++;
%o if(sum < ceiling) { cnt++; continue; }
%o if(ceiling++ == k) return cnt; else cnt = 1; }
%o } /* _Oskar Wieland_, May 01 2014 */
%Y Cf. A002387, A096143.
%K nonn
%O 0,3
%A _W. Neville Holmes_, Jul 29 2004
%E More terms from _Robert G. Wilson v_, Aug 05 2004
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