cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A124838 Denominators of third-order harmonic numbers (defined by Conway and Guy, 1996).

Original entry on oeis.org

1, 2, 6, 4, 20, 10, 70, 56, 504, 420, 4620, 3960, 3432, 6006, 90090, 80080, 1361360, 408408, 369512, 67184, 470288, 1293292, 29745716, 27457584, 228813200, 212469400, 5736673800, 5354228880, 155272637520, 291136195350, 273491577450
Offset: 1

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Author

Jonathan Vos Post, Nov 10 2006

Keywords

Comments

Numerators are A124837. All fractions reduced. Thanks to Jonathan Sondow for verifying these calculations. He suggests that the equivalent definition in terms of first order harmonic numbers may be computationally simpler. We are happy with the description of A027612 Numerator of 1/n + 2/(n-1) + 3/(n-2) +...+ (n-1)/2 + n, but baffled by the description of A027611.

Examples

			a(1) = 1 = denominator of 1/1.
a(2) = 2 = denominator of 1/1 + 5/2 = 7/2.
a(3) = 6 = denominator of 7/2 + 13/3 = 47/6.
a(4) = 4 = denominator of 47/6 + 77/12 = 57/4.
a(5) = 20 = denominator of 57/4 + 87/10 = 549/20.
a(6) = 10 = denominator of 549/20 + 223/20 = 341/10
a(7) = 70 = denominator of 341/10 + 481/35 = 3349/70.
a(8) = 1260 = denominator of 3349/70 + 4609/280 = 88327/1260.
a(9) = 45 = denominator of 88327/1260 + 4861/252 = 3844/45.
a(10) = 504 = denominator of 3844/45 + 55991/2520 = 54251/504, or, untelescoping:
a(10) = 504 = denominator of 1/1 + 5/2 + 13/3 + 77/12 + 87/10 + 223/20 + 481/35 + 4609/252 + 4861/252 + 55991/2520 = 54251/504.
		

References

  • J. H. Conway and R. K. Guy, The Book of Numbers, New York: Springer-Verlag, pp. 143 and 258-259, 1996.

Crossrefs

Programs

  • Haskell
    a124838 n = a213999 (n + 2) (n - 1) -- Reinhard Zumkeller, Jul 03 2012
  • Mathematica
    Table[Denominator[(n+2)!/2!/n!*Sum[1/k,{k,3,n+2}]],{n,1,40}] (* Alexander Adamchuk, Nov 11 2006 *)

Formula

A124837(n)/A124838(n) = Sum_{i=1..n} A027612(n)/A027611(n+1).
a(n) = denominator(Sum_{m=1..n} Sum_{L=1..m} Sum_{k=1..L} 1/k).
a(n) = denominator(((n+2)!/(2!*n!)) * Sum_{k=3..n+2} 1/k). - Alexander Adamchuk, Nov 11 2006
a(n) = A213999(n+2,n-1). - Reinhard Zumkeller, Jul 03 2012

Extensions

Corrected and extended by Alexander Adamchuk, Nov 11 2006