Finite(?) Sequence (involving divisors)

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From: "Vladeta Jovovic" <vladeta at EUnet.yu>
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Subject: A conjecture
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> %I A131883
> %S A131883 1,2,2,2,2,4,4,4,4,4,4,6,6,6,6,6,6,8,8,8
> %N A131883 a(n) = the minimum value from among 
> (phi(n+1),phi(n+2),phi(n+3),...,phi(2n)), where phi(m) is the number of 
> positive integers which are coprime to m and are <= m.
> %C A131883 Conjecture: After omitting multiple occurrences we get A036912.
> %O A131883 1
> %K A131883 ,nonn,
> %A A131883 Vladeta Jovovic (vladeta at Eunet.yu), Oct 31 2007

V. 

Yesterday I wrote:
:Leroy Quet <q1qq2qqq3qqqq at yahoo.com> wrote:
::Here is a seemingly dumb question.
::
::Is sequence A134320 finite?
::
::Here is the info on this sequence, for those too lazy
::to check the EIS:
::
::%I A134320
::%S A134320 2,4,6,12,20,30,42,90
::%N A134320 Positive integers with more non-isolated
::divisors than isolated divisors. A divisor, k, of n is
::non-isolated if (k-1) and/or (k+1) also divides n. A
::divisor, k, of n is isolated if neither (k-1) nor
::(k+1) divides n.
::%C A134320 Is this sequence finite?
::
::With the exception of a(2) = 4, all terms of this
::sequence are of the form m*(m+1). (Oblong numbers:
[...]
:A quick program to check found no other example up to 3e6, which
:certainly suggests it is not just finite but complete.

Partial proof: if adjacent integers k, k+1 both divide n then since
they are coprime we also have that k(k+1) divides n, so k < sqrt(n).
Ie the largest non-isolated factor a number can have is ceiling(sqrt(n)).

Since the divisors are symmetrically disposed around the square root, we
with all divisors below the square root non-isolated; if n is square,
So the only square entry is n = 4.

It remains to prove that there is no oblong number greater than 9 x 10
that avoids isolated divisors below the square root.

:Using the same program to find more terms for A134321 "positive integers
:with the same number of non-isolated divisors as isolated divisors" found
:them much denser:
:8 10 14 18 22 24 26 34 38 40 46 56 58 60 62 72 74 82 84 86 94
:106 110 118 122 132 134 142 146 156 158 166 178 182 194 202 206 210 214 218
:220 226 254 262 274 278 298 302 314 326 334 346 358 362 380 382 386 394 398
:422 446 454 458 466 478 482 502 506 514 526 538 542 554 562 566 586 614 622
:626 634 662 674 694 698 706 718 734 746 758 766 778 794 802 818 838 842 862
:866 878 886 898 914 922 926 934 958 974 982 998 [...]
:
:Almost all these entries are of the form 2p or 2pq where q = 2p +/- 1
:(and so p is in A005383 or A005384). The exceptions are:
:  8 18 24 40 56 60 72 84 132 156 210 220 380
:.. with no other up to 2e6, suggesting the exception list is also finite
:and complete.

Similarly for A134321, we have: elements cannot be perfect squares since
those have an odd number of divisors. Thus they must either be oblong
numbers with one isolated divisor below the square root (such as the
isolated 5 for 110) or non-oblong numbers with all divisors below the

I expect that proving this sequence consists only of the two general classes
and the finite, complete list of exceptions describe above is also possible
and would use a similar approach to the first case.

Hugo