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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. 12
77, 847, 9317, 102487, 596778, 1127357, 1193556, 6161805, 12323610, 12400927 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

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

Harmonic numbers are defined as H(n) = Sum[ 1/k, {k,1,n} ] = A001008(n)/A002805(n).

Alternating harmonic numbers are defined as H'(n) = Sum[ (-1)^(k+1)*1/k, {k,1,n} ] = 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.

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 that completes the proof.

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

LINKS

Table of n, a(n) for n=1..10.

Tanya Khovanova, Non Recursions

Eric Weisstein, Link to a section of The World of Mathematics. Harmonic Number.

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}]

CROSSREFS

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

Sequence in context: A222728 A223649 A205434 * A176632 A093277 A217643

Adjacent sequences:  A125578 A125579 A125580 * A125582 A125583 A125584

KEYWORD

hard,more,nonn

AUTHOR

Alexander Adamchuk, Jan 03 2007

EXTENSIONS

More terms from Max Alekseyev, Mar 11 2007

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

STATUS

approved

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Last modified September 26 10:46 EDT 2017. Contains 292518 sequences.