A127629 Numbers m such that a divisor, together with its quotient and remainder, are consecutive terms (in that order) in a geometric sequence.

9, 28, 34, 58, 65, 75, 110, 126, 132, 201, 205, 217, 224, 246, 254, 258, 294, 344, 384, 399, 436, 498, 502, 513, 516, 520, 579, 657, 680, 690, 730, 786, 810, 866, 880, 978, 979, 1001, 1008, 1028, 1038, 1105, 1128, 1164, 1330, 1332, 1365, 1370, 1374, 1388

Offset

1,1

Comments

The sequence misses the primes.

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

Links

Amiram Eldar, Table of n, a(n) for n = 1..10000

C. Hughes, Geometric Division

Wikipedia, Euclidean division

Example

58 is in the sequence because 58 = 9*6 + 4, where 9, 6 and 4 are consecutive terms in a geometric sequence.
For a(4) = 58 with noninteger ratio = 3/2:
     58 | 9          58 | 6
        ------          ------
      4 | 6           4 | 9
For a(16) = 258 with integer ratio = 4:
    258 | 32         258 |  8
        ------           -------
      2 |  8           2 | 32

Mathematica

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 *)

Prog

(PARI) is(n)={for(d=1, n, if((n\d)*(n%d)==d^2, return(1))); return(0)}

Crossrefs

Disjoint union of A334185 and A334186.

Subsequence: A001093 \ {0, 1, 2} (for remainder = 1).

Sequence in context: A031454 A044999 A155473 * A334185 A267686 A024670

Adjacent sequences: A127626 A127627 A127628 * A127630 A127631 A127632

Keyword

easy,nonn

Author

Nick Hobson, Jan 20 2007

Status

approved