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

Author

Nick Hobson, Jan 20 2007

Keywords

• easy
• nonn

Comments

The sequence misses the primes.

When m is a term, then m = d*q + r and rBernard Schott, May 15 2020

Examples

			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
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• C. Hughes, Geometric Division
• Wikipedia, Euclidean division

Crossrefs

Disjoint union of A334185 and A334186.

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

Programs

Mathematica

mx = 1388; m = Ceiling @ Sqrt[mx]; s={}; Do[r = Select[Divisors[k^2], #Amiram Eldar, Aug 28 2019 *)

PARI

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