A125581 Numbers n such that n does not divide the denominator of the n-th harmonic number nor the denominator of the n-th alternating harmonic number.

77, 847, 9317, 102487, 596778, 1127357, 1193556, 6161805, 12323610, 12400927

Offset: 1

Author

Alexander Adamchuk, Jan 03 2007

Keywords

• hard
• more
• nonn

Comments

Note that a(1) = 7*11, a(2) = 7*11^2, and a(3) = 7*11^3.

Harmonic numbers are defined as H(n) = Sum_{k=1..n} 1/k = A001008(n)/A002805(n).

Alternating harmonic numbers are defined as H'(n) = Sum_{k=1..n} (-1)^(k+1)*1/k = A058313(n)/A058312(n).

Numbers n such that n does not divide the denominator of the n-th harmonic number are listed in A074791. Numbers n such that n does not divide the denominator of the n-th alternating harmonic number are listed in A121594.

This sequence is the intersection of A074791 and A121594.

Comments from Max Alekseyev, Mar 07 2007: (Start)

While A125581 indeed contains the geometric progression 7*11^n as a subsequence, it also contains other geometric progressions such as: 546*1093^n, 1092*1093^n, 1755*3511^n, 3510*3511^n and 4896*5557^n (see A126196 and A126197). It may also contain some "isolated" terms (i.e. not participating in the geometric progressions) but such terms are harder to find and at the moment I have no proof that they exist.

This is a sketch of my proof that geometric progression 7*11^n and the others mentioned above belong to A125581.

Lemma 1. H'(n) = H(n) - H([n/2]).

Lemma 2. For prime p and integer n >= p, valuation(H(n),p) >= valuation(H([n/p]),p) - 1.

Theorem. For an integer b > 1 and a prime number p such that p divides the numerators of both H(b) and H([b/2]), the geometric progression b*p^n belongs to A125581.

Proof. It is enough to show that valuation(H(b*p^n),p) > -n and valuation(H'(b*p^n), p) > -n. By Lemma 2 we have valuation(H(b*p^n), p) >= valuation(H(b), p) - n >= 1 - n > -n.

From this inequality and Lemma 1, we have valuation(H'(b*p^n), p) >= min{ valuation(H(b*p^n), p), valuation(H([b*p^n/2]), p) } >= min{ 1 - n, valuation(H([b*p^n/2]), p) }. It remains to show that valuation(H([b*p^n/2]), p) >= 1 - n.

Again by Lemma 2, we have valuation(H([b*p^n/2]), p) >= valuation(H([b/2]), p) - n >= 1 - n, which completes the proof.

It is easy to check that this Theorem holds for the aforementioned geometric progressions. (End)

Links

• Tanya Khovanova, Non Recursions.
• Eric Weisstein's World of Mathematics, Harmonic Number.

Crossrefs

Cf. A001008, A002805, A003599, A058312, A058313, A074791, A119955, A121594.

Programs

Mathematica

f=0; g=0; Do[g=g+1/n; f=f+(-1)^(n+1)/n; If[ !IntegerQ[Denominator[g]/n]&&!IntegerQ[Denominator[f]/n], Print[n]], {n, 1, 10000}]

Extensions

More terms from Max Alekseyev, Mar 11 2007

a(8)-a(10) from Max Alekseyev, Mar 19 2007