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A334186
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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.
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5
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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
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OFFSET
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1,1
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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
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PROG
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(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
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CROSSREFS
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Cf. A334185 (similar, with integer ratio), A127629 (similar, with integer and noninteger ratio).
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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