%N Let m = d*q + r be the Euclidean division of m by d. The terms m of this sequence satisfy that r, q, d are consecutive positive integer terms in a geometric progression with a noninteger common ratio > 1.
%C Inspired by the problem 141 of Project Euler (see the link).
%C If b is the fractional common ratio, then b = p/s irreducible > 1 and r > 0.
%C To get r, d, q as integers, it is necessary that r is a multiple of s^2; in this case, if r = s^2 *r' with r' >= 1, q = p*s*r' and d = p^2*r', then every m = s*r' * (s+p^3*r') with p/s>1 is a term, and the Euclidean division becomes : s*r' * (s+p^3*r') = (p^2*r') * (p*s*r') + s^2*r'. The integers (s^2*r', p*s*r', p^2*r') are in geometric progression.
%C When (r<q<d) is solution with m = d * q + r, then, with d' = q and q' = d, m = d' * q' + r and (r<d'<q') is also a solution with another order between remainder, divisor and quotient (see last example).
%C m is a term iff m = s*r' * (s+p^3*r') with r' >= 1 and p > s, p no multiple of s. For every irreducible ratio b = p/s, there are infinitely many terms.
%H Project Euler, <a href="https://projecteuler.net/problem=141">Problem 141: Investigating progressive numbers, n, which are also square</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Euclidean_division">Euclidean division</a>
%e a(4) = 254 = 25 * 10 + 4 with (4, 10, 25) and ratio = 5/2;
%e a(6) = 498 = 27 * 18 + 12 with (12, 18, 27) and ratio = 3/2;
%e a(19) = 1740 = 49 * 35 + 25 with (25, 35, 49) and ratio = 7/5;
%e a(20) = 1755 = 48 * 36 + 27 with (r=27, q=36, d=48) but also 1755 = 36 * 48 + 27 with (r=27, d'=36, q'=48) both with ratio = 4/3:
%e 1755 | 48 1755 | 36
%e ------ ------
%e 27 | 36 27 | 48
%o (PARI) isok(m) = for (d=1, m, if (m % d, q = m\d; r = m % d; if ((d % q) && (d/q == q/r), return (1)); ); ) \\ _Michel Marcus_, Apr 26 2020
%Y Cf. A334185 (similar with integer ratio), A127629 (similar with integer and noninteger ratio).
%A _Bernard Schott_, Apr 26 2020
%E More terms from _Michel Marcus_, Apr 26 2020