[seqfan] Re: Two Sequences

[M. F. Hasler]

On Wed, Jul 3, 2019 at 5:17 PM C Boyd  wrote:

> For values > 29, the indices given match A006093 a(n) = prime(n) - 1.
>

Indeed, it appears that A1008(p-1) is divisible by p^2, at least for the
first values.
I'm just working out drafts related to factorization of A001008 : bigomega,
list of factors with/without repetition, smallest & largest prime factors...
(cf. A308967 .. A308971)
(The latter seem to coincide with A120299 = Largest prime factor of
Stirling numbers of first kind A000254,
but those are the numerators without reduction to lowest terms, so maybe(?)
we can expect that the gpf() will eventually differ...).

(I'm quite ignorant about known results concerning this topic, please feel
free to add comments & references.)

- Maximilian

In fact, ignoring the odd terms 7 & 29, it appears to be a subsequence of
> A006093.
>
> Does this similarity extend to your larger terms?
>
> CB
>
> On 03/07/2019, M. F. Hasler <seqfan at hasler.fr> wrote:
> > PS: it appears that indices of non-squarefree terms of
> > A001008 Numerators of harmonic numbers H(n) = Sum_{i=1..n} 1/i.
> > are:
> > 4,6,7,10,12,16,18,22,28,29,30,36,40,42,46,52,58,60,66,70,72,78,82,88,96,
> > 100,102,106,108,112,126,130,136,138,148,150,
> > 156,162,166,172,178,180,190,192,196,198,
> > 210,222,226,228,232,238,240,250,256,262,268,270,276,...
> > (sequence not in the OEIS -- could it be worth adding ?)
> >
> > It appears that
> > A1008(848) = 11^3 * 1871 * C359
> > is the first term to have a cubic factor.
> >
> > (I say "it appears" because I did not factor large cofactors.)
> >
> > Could anyone confirm and / or know whether this is studied anywhere in
> the
> > literature ?
> >
> > - Maximilian
> >
> >
> > On Wed, Jul 3, 2019 at 2:35 PM M. F. Hasler <seqfan at hasler.fr> wrote:
> >
> >> On Wed, Jul 3, 2019 at 4:19 AM Ali Sada via SeqFan <
> seqfan at list.seqfan.eu>
> >> wrote:
> >>
> >>> The first sequence is 1, 66, 6, 12, 8220, 20, 420,213080, 17965080,
> >>> 153720, 210951720, 14109480, 31766925960,....
> >>> A(n) is the least integer that when multiplied by the harmonic number
> sum
> >>> of n generates a square number (M*M.)
> >>> The second sequence is M itself.
> >>> 1, 3, 11, 5, 137, 7, 33, 761, 7129, 671, 83771, 6617,....
> >>> I searched OEIS and found that this sequence is related to A120299
> >>> (Largest prime factor of Stirling numbers of first kind s(n,2)
> A000254.)
> >>> A120299 also represents the largest prime factor of M, except for M(1.)
> >>>
> >>
> >> A(n,H=sum(k=1,n,1/k))=core(numerator(H))*denominator(H)}
> >> /*=A007913(A001008(n))*A002805(n)*/
> >> A_vec(13)
> >> %21 = [1, 6, 66, 12, 8220, 20, 420, 213080, 17965080, 153720,
> 2320468920,
> >> 14109480, 412970037480]
> >> Then, M = sqrt(H*A) = sqrt(N*core(N)) = A019554(N) (= "outer square
> root")
> >> of N = numerator(H),
> >> this does not depend at all on the denominator, and it could be simpler
> to
> >> use  A(n) = M(n)^2/H(n) .)
> >>
> >> M_vec(Nmax, H=0)=vector(Nmax, k, H+=1/k;
> sqrtint(core(N=numerator(H))*N))
> >>  = [1, 3, 11, 5, 137, 7, 33, 761, 7129, 671, 83711, 6617, 1145993]
> >>
> >> (So far this yields the same result as the radical of N, A007947, but
> this
> >> will no more be the case as soon as N has a cube as factor. However,
> this
> >> isn't the case soon - you'll have to go beyond 150 terms or more!)
> >>
> >>
> >
> > --
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> >
>
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> Seqfan Mailing list - http://list.seqfan.eu/
>


-- 
Maximilian