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%I #43 Sep 08 2022 08:45:14
%S 1,3,11,25,137,147,363,761,7129,7381,83711,86021,1145993,1171733,
%T 1195757,2436559,42142223,42822903,275295799,279175675,56574159,
%U 19093197,444316699,1347822955,34052522467,34395742267,312536252003
%N Numerator of n*HarmonicNumber(n).
%C a(1) = 1, a(n) = Numerator( H(n) / H(n-1) ), where H(n) = HarmonicNumber(n) = A001008(n)/A002805(n). - _Alexander Adamchuk_, Oct 29 2004
%C Sampling a population of n distinct elements with replacement, n HarmonicNumber(n) is the expectation of the sample size for the acquisition of all n distinct elements. - _Franz Vrabec_, Oct 30 2004
%C p^2 divides a(p-1) for prime p>3. - _Alexander Adamchuk_, Jul 16 2006
%C It appears that a(n) = b(n) defined by b(n+1) = b(n)*(n+1)/g(n) + f(n), f(n) = n*f(n-1)/g(n) and g(n) = gcd(b(n)*(n+1), n*f(n-1)), b(1) = f(1) = g(1) = 1, i.e., the recurrent formula of A000254(n) where both terms are divided by their GCD at each step (and remain divided by this factor in the sequel). Is this easy to prove? - _M. F. Hasler_, Jul 04 2019
%D W. Feller, An Introduction to Probability Theory and Its Applications, Vol. I, 2nd Ed. 1957, p. 211, formula (3.3)
%H Vincenzo Librandi, <a href="/A096617/b096617.txt">Table of n, a(n) for n = 1..1000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/CompleteSet.html">Complete Set</a>
%F a(n) = abs(Stirling1(n+1, 2))/(n-1)!. - _Vladeta Jovovic_, Jul 06 2004
%F a(n) = numerator of integral(1-(1-exp(-t/n])^n, {t, 0, infinity}). - _Jean-François Alcover_, Feb 17 2014
%e 1, 3, 11/2, 25/3, 137/12, 147/10, 363/20, 761/35, 7129/280, ...
%p ZL:=n->sum(sum(1/i, i=1..n), j=1..n): a:=n->floor(numer(ZL(n))): seq(a(n), n=1..27); # _Zerinvary Lajos_, Jun 14 2007
%t Numerator[Table[(Sum[(1/k), {k, 1, n}]/Sum[(1/k), {k, 1, n-1}]), {n, 1, 20}]] (* _Alexander Adamchuk_, Oct 29 2004 *)
%t Table[n*HarmonicNumber[n] // Numerator, {n, 1, 27}] (* _Jean-François Alcover_, Feb 17 2014 *)
%o (Magma) [Numerator(n*HarmonicNumber(n)): n in [1..40]]; // _Vincenzo Librandi_, Feb 19 2014
%o (PARI) {h(n) = sum(k=1,n,1/k)};
%o for(n=1,50, print1(numerator(n*h(n)), ", ")) \\ _G. C. Greubel_, Sep 01 2018
%o (PARI) A=List(f=1); for(k=1,999, t=[A[k]*(k+1),f*=k]); t/=gcd(t); listput(A,t[1]+f=t[2])) \\ Illustrate conjectured equality. - _M. F. Hasler_, Jul 04 2019
%Y Cf. A027611, A001008, A002805.
%Y Differs from A025529 at 7th term.
%Y Cf. A193758.
%K nonn,frac
%O 1,2
%A _Eric W. Weisstein_, Jul 01 2004
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