A334186 - OEIS

A334186

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.

58, 201, 224, 254, 384, 498, 516, 690, 786, 880, 1008, 1038, 1105, 1370, 1388, 1462, 1518, 1545, 1740, 1755, 1968, 2032, 2094, 2262, 2585, 2666, 2674, 2752, 2932, 3009, 3108, 3402, 3488, 3633, 3670, 4002, 4016, 4134, 4370, 4398, 4410, 4548, 4845, 5152, 5340, 5440

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

Inspired by the problem 141 of Project Euler (see the link).

If b is the fractional common ratio, then b = p/s irreducible > 1 and r > 0.

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.

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).

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.

LINKS

Table of n, a(n) for n=1..46.

Project Euler, Problem 141: Investigating progressive numbers, n, which are also square

Wikipedia, Euclidean division

EXAMPLE

a(4) = 254 = 25 * 10 + 4 with (4, 10, 25) and ratio = 5/2;

a(6) = 498 = 27 * 18 + 12 with (12, 18, 27) and ratio = 3/2;

a(19) = 1740 = 49 * 35 + 25 with (25, 35, 49) and ratio = 7/5;

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:

1755 | 48 1755 | 36

------ ------

27 | 36 27 | 48