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A096617
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Numerator of n*HarmonicNumber(n).
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7
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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
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OFFSET
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1,2
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COMMENTS
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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
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
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REFERENCES
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W. Feller, An Introduction to Probability Theory and Its Applications, Vol. I, 2nd Ed. 1957, p. 211, formula (3.3)
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LINKS
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FORMULA
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EXAMPLE
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1, 3, 11/2, 25/3, 137/12, 147/10, 363/20, 761/35, 7129/280, ...
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MAPLE
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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
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MATHEMATICA
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Numerator[Table[(Sum[(1/k), {k, 1, n}]/Sum[(1/k), {k, 1, n-1}]), {n, 1, 20}]] (* Alexander Adamchuk, Oct 29 2004 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,frac
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AUTHOR
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STATUS
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approved
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