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

Offset: 1

Author

Bernard Schott, Apr 18 2020

Keywords

• nonn

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).

Examples

			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
     +-----                 +-----
   4 |  20                4 | 100
		

Links

• Project Euler, Problem 141: Investigating progressive numbers, n, which are also square
• Wikipedia, Euclidean division

Crossrefs

Cf. A334186 (similar, with b is an irreducible fraction).

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

Programs

Mathematica

Select[Range[2000], Length @ Solve[r * (1 + r*b^3) == # && r >=1 && b >= 2, {r, b}, Integers] > 0 &] (* Amiram Eldar, Apr 18 2020 *)

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 19 2020

Extensions

Name improved by Michel Marcus, Apr 19 2020

More terms from Michel Marcus, Apr 19 2020