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%I #30 Feb 11 2024 12:40:00
%S 9,28,34,58,65,75,110,126,132,201,205,217,224,246,254,258,294,344,384,
%T 399,436,498,502,513,516,520,579,657,680,690,730,786,810,866,880,978,
%U 979,1001,1008,1028,1038,1105,1128,1164,1330,1332,1365,1370,1374,1388
%N Numbers m such that a divisor, together with its quotient and remainder, are consecutive terms (in that order) in a geometric sequence.
%C The sequence misses the primes.
%C When m is a term, then m = d*q + r and r<q<d are in geometric progression; but also, in this case, m = d'*q' + r with r<d'<q' that are in geometric progression with d'=q and q'=d (see examples). - _Bernard Schott_, May 15 2020
%H Amiram Eldar, <a href="/A127629/b127629.txt">Table of n, a(n) for n = 1..10000</a>
%H C. Hughes, <a href="http://mathschallenge.net/index.php?section=problems&show=true&titleid=geometric_division">Geometric Division</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Euclidean_division">Euclidean division</a>
%e 58 is in the sequence because 58 = 9*6 + 4, where 9, 6 and 4 are consecutive terms in a geometric sequence.
%e For a(4) = 58 with noninteger ratio = 3/2:
%e 58 | 9 58 | 6
%e ------ ------
%e 4 | 6 4 | 9
%e For a(16) = 258 with integer ratio = 4:
%e 258 | 32 258 | 8
%e ------ -------
%e 2 | 8 2 | 32
%t mx = 1388; m = Ceiling @ Sqrt[mx]; s={}; Do[r = Select[Divisors[k^2], #<k &]; q = k^2/r; v = q * k + r; s = Join[s, v], {k, 1, m}]; Select[Union[s], # <= mx &] (* _Amiram Eldar_, Aug 28 2019 *)
%o (PARI) is(n)={for(d=1, n, if((n\d)*(n%d)==d^2, return(1))); return(0)}
%Y Disjoint union of A334185 and A334186.
%Y Subsequence: A001093 \ {0, 1, 2} (for remainder = 1).
%K easy,nonn
%O 1,1
%A _Nick Hobson_, Jan 20 2007
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