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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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.

Original entry on oeis.org

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
Offset: 1

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Author

Bernard Schott, Apr 26 2020

Keywords

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
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.

Examples

			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
		

Crossrefs

Cf. A334185 (similar, with integer ratio), A127629 (similar, with integer and noninteger ratio).

Programs

  • 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

Extensions

More terms from Michel Marcus, Apr 26 2020