A096617 Numerator of n*HarmonicNumber(n).

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

Offset: 1

Author

Eric W. Weisstein, Jul 01 2004

Keywords

• nonn
• frac

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

Examples

			1, 3, 11/2, 25/3, 137/12, 147/10, 363/20, 761/35, 7129/280, ...
		

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

Crossrefs

Cf. A027611, A001008, A002805.

Differs from A025529 at 7th term.

Cf. A193758.

Programs

Magma

[Numerator(n*HarmonicNumber(n)): n in [1..40]]; // Vincenzo Librandi, Feb 19 2014

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 *)

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

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