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A335065
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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 d, q, r are consecutive positive integer terms in a geometric progression but not necessarily in that order.
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2
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6, 9, 12, 20, 28, 30, 34, 42, 56, 58, 65, 72, 75, 90, 110, 126, 132, 156, 182, 201, 205, 210, 217, 224, 240, 246, 254, 258, 272, 294, 306, 342, 344, 380, 384, 399, 420, 436, 462, 498, 502, 506, 513, 516, 520, 552, 579, 600, 650, 657, 680, 690, 702, 730, 756, 786
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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 link).
There exist 3 possibilities to get such terms m that satisfy that d, q, r are consecutive positive integer terms in a geometric progression but not necessarily in that order:
-> the geometric progression is r < q < d (A127629).
-> the geometric progression is r < d < q (same terms of A127629).
-> the geometric progression is q < r < d (A002378 \ {0,2} = oblong numbers >= 6).
Some numbers have a geometric progression solution in the 3 cases (132, 1332, 6162, ...) [see examples].
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LINKS
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EXAMPLE
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Examples with r < q < d, r < d < q, q < r <d:
34 | 8 75 | 6 42 | 12
---- ----- -----
2 | 4 , 3 | 12 , 6 | 3
The 3 possible divisions by 132:
132 | 16 132 | 8 132 | 121
----- ------ ------
4 | 8 , 4 | 16 , 11 | 1.
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MATHEMATICA
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mx = 800; Union@ Reap[ Do[y = x+1; While[(z = y^2/x) < mx, If[ IntegerQ@ z, If[(m = z y + x) <= mx, Sow@ m]; If[(m = z x + y) <= mx, Sow@ m]]; y++], {x, mx}]][[2, 1]] (* Giovanni Resta, May 24 2020 *)
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PROG
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(PARI) isok(n) = {my(r, d); for (q=2, n-1, if (r=(n % q), d = n\q; if ((r*d == q^2) || (r*q == d^2) || (q*d == r^2), return (1)); ); ); } \\ Michel Marcus, May 25 2020
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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