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 A096617 Numerator of n*HarmonicNumber(n). 7
 1, 3, 11, 25, 137, 147, 363, 761, 7129, 7381, 83711, 86021, 1145993, 1171733, 1195757, 2436559, 42142223, 42822903, 275295799, 279175675, 56574159, 19093197, 444316699, 1347822955, 34052522467, 34395742267, 312536252003 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS 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 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 p^2 divides a(p-1) for prime p>3. - Alexander Adamchuk, Jul 16 2006 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 REFERENCES W. Feller, An Introduction to Probability Theory and Its Applications, Vol. I, 2nd Ed. 1957, p. 211, formula (3.3) LINKS Vincenzo Librandi, Table of n, a(n) for n = 1..1000 Eric Weisstein's World of Mathematics, Complete Set FORMULA a(n) = abs(Stirling1(n+1, 2))/(n-1)!. - Vladeta Jovovic, Jul 06 2004 a(n) = numerator of integral(1-(1-exp(-t/n])^n, {t, 0, infinity}). - Jean-François Alcover, Feb 17 2014 EXAMPLE 1, 3, 11/2, 25/3, 137/12, 147/10, 363/20, 761/35, 7129/280, ... MAPLE 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 MATHEMATICA Numerator[Table[(Sum[(1/k), {k, 1, n}]/Sum[(1/k), {k, 1, n-1}]), {n, 1, 20}]] (* Alexander Adamchuk, Oct 29 2004 *) Table[n*HarmonicNumber[n] // Numerator, {n, 1, 27}] (* Jean-François Alcover, Feb 17 2014 *) PROG (Magma) [Numerator(n*HarmonicNumber(n)): n in [1..40]]; // Vincenzo Librandi, Feb 19 2014 (PARI) {h(n) = sum(k=1, n, 1/k)}; for(n=1, 50, print1(numerator(n*h(n)), ", ")) \\ G. C. Greubel, Sep 01 2018 (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 CROSSREFS Cf. A027611, A001008, A002805. Differs from A025529 at 7th term. Cf. A193758. Sequence in context: A175441 A001008 A231606 * A025529 A124078 A096795 Adjacent sequences: A096614 A096615 A096616 * A096618 A096619 A096620 KEYWORD nonn,frac AUTHOR Eric W. Weisstein, Jul 01 2004 STATUS approved

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Last modified January 29 02:42 EST 2023. Contains 359915 sequences. (Running on oeis4.)