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_{t=0..oo} 1-(1-exp(-t/n))^n dt. - 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
KEYWORD
nonn,frac
AUTHOR
Eric W. Weisstein, Jul 01 2004
STATUS
approved