A334185 - OEIS

A334185

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 common integer ratio.

9, 28, 34, 65, 75, 110, 126, 132, 205, 217, 246, 258, 294, 344, 399, 436, 502, 513, 520, 579, 657, 680, 730, 810, 866, 978, 979, 1001, 1028, 1128, 1164, 1330, 1332, 1365, 1374, 1582, 1605, 1729, 1736, 1815, 1947, 2004, 2050, 2064, 2196, 2198, 2310, 2329, 2610, 2710

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OFFSET

1,1

COMMENTS

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

If b is the common ratio, then b is an integer >= 2.

So, when b >= 2 and r >= 1, q=r*b, d=r*b^2, then every m = r * (1+r*b^3) is a term, and the division becomes: r*(1+r*b^3) = (r*b^2) * (r*b) + r. The integers (r, r*b, r*b^2) 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 if m = r * (1+r*b^3) with r >= 1 and b >= 2; so, when r = 1, A001093(n) for n > 1 are terms (see 1st example).

LINKS

Table of n, a(n) for n=1..50.

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

Wikipedia, Euclidean division

EXAMPLE

a(2) = 28 = 9*3 + 1 with (1,3,9) and ratio = 3;

a(5) = 75 = 12*6 + 3 with (3,6,12) and ratio = 2;

a(12) = 258 = 32*8 + 2 with (2,8,32) and ratio = 4;

a(42) = 2004 = 100*20 + 4 with (r=4, q=20, d=100) but also 2004 = 20*100 + 4 with (r=4, d'=20, q'=100) both with ratio = 5:

2004 | 100 2004 | 20

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