A127629 -id:A127629 - cp's OEIS frontend

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.

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

Bernard Schott, Apr 26 2020

nonn

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 (rm 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.

			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

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

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

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

More terms from Michel Marcus, Apr 26 2020

A335064 Let m = d*q + r be the Euclidean division of m by d. The terms m of this sequence satisfy that q, r, d are consecutive positive integer terms in a geometric progression with a noninteger common ratio > 1.

Original entry on oeis.org

42, 110, 156, 210, 240, 342, 420, 462, 506, 600, 702, 812, 930, 1122, 1190, 1260, 1332, 1482, 1560, 1640, 1806, 1980, 2070, 2162, 2352, 2550, 2652, 2756, 2970, 3080, 3192, 3306, 3422, 3660, 3906, 4032, 4290, 4422, 4692, 4830, 4970, 5256, 5550, 5700, 5852, 6006, 6162

Offset: 1

Bernard Schott, May 22 2020

nonn

Inspired by the problem 141 of Project Euler (see the link).
The terms of this sequence are oblong numbers m = k*(k+1) with k in A024619.
When q < r < d are consecutive terms of a geometric progression of constant b = p/s noninteger, with b>1, s>=2, p>s, it is necessary that q is a multiple of s^2, so q = q' * s^2 with q' >= 1; the Euclidean division of a term m by q becomes
p*s*q' * (1+p*s*q') = (p^2*q') * (s^2*q') + p*s*q' with k = p*s*q',
so (q, r, d) = (s^2*q', p*s*q', p^2*q') is solution. (see examples).
But, as these terms are oblong, there exists also another division where the constant ratio is the integer psq' and (q,r,d) = (1, p*s*q', (p*s*q')^2) are in geometric progression.

			Examples for 42, 110 and 156 with consecutive ratios 3/2, 5/2, 4/3:
   42 | 9         110 | 25         156 | 16
      -----           -----            -----
    6 | 4    ,     10 |  4     ,    12 |  9 ,
then with consecutive ratios 2, 10 and 12:
   42 | 12        110 | 100        156 | 144
      -----           -----            ------
    6 |  3   ,     10 |   1    ,    12 |   1.

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

Cf. A024619, A127629, A334185, A334186.

Subsequence of A002378 and of A335065.

Table[n*(n + 1), {n, Select[Range[80], PrimeNu[#] > 1 &]}] (* Amiram Eldar, May 23 2020 *)

apply(x->x*(x+1), select(x->!isprimepower(x), [2..80])) \\ Michel Marcus, May 23 2020

a(n) = A024619(n) * (1+A024619(n)).

a(n) = A002378(A024619(n)). - Michel Marcus, May 23 2020

A335065 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.

Original entry on oeis.org

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

Offset: 1

Bernard Schott, May 23 2020

nonn

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].

			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.

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

Equals A127629 Union A002378 \ {0,2}.

Subsequences: A334185, A334186, A335064.

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 *)

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

A335272 For m to be a term there must exist three Euclidean divisions of m by d, d', and d", m = dq + r = d'q' + r' = d"*q" + r", such that (r, q, d), (r', d', q'), and (q", r", d") are three geometric progressions.

Original entry on oeis.org

110, 132, 1332, 6162, 10712, 12210, 35156, 60762, 67340, 152490, 296480, 352242, 354620, 357006, 648830, 771762, 932190, 1197930, 2057790, 2803950, 3241800, 3310580, 4458432, 6454140, 7865220, 9613100, 10814232, 13976382, 16382256, 19267710, 53824232, 55138050

Offset: 1

Bernard Schott, May 30 2020

nonn

Inspired by Project Euler, Problem 141 (see link).

The terms are necessary oblong numbers >= 6.

			For 110:
   110 | 18             110 |  6               110 | 100
       -----               ------               ---------
     2 |  6       ,       2 | 18       ,        10 |   1
For 132, see A335065.
For 1332:
  1332 | 121           1332 |  11            1332 | 1296
       ------              -------                -------
     1 |  11      ,       1 | 121      ,       36 |   1   .

Giovanni Resta, Table of n, a(n) for n = 1..200
Project Euler, Problem 141: Investigating progressive numbers, n, which are also square

Intersection of A127629 and A002378.

Cf. A334185, A334186, A335064, A335065.

Select[(#^2 + #) & /@ Range[2000], (n = #; AnyTrue[ Range[1 + Sqrt@ n], #^2 == Mod[n, #] Floor[n/#] &]) &] (* Giovanni Resta, Jun 03 2020 *)

isob(n) = my(m=sqrtint(n)); m*(m+1)==n; \\ A002378
isgd(n) = {for(d=1, n, if((n\d)*(n%d)==d^2, return(1))); return(0)}; \\ A127629
isok(n) = isob(n) && isgd(n); \\ Michel Marcus, May 30 2020

More terms from Michel Marcus, May 30 2020
a(18)-a(26) from Jinyuan Wang, May 30 2020
Terms a(27) and beyond from Giovanni Resta, Jun 03 2020

Showing 1-4 of 4 results.

cp's OEIS Frontend

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.

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A335064 Let m = d*q + r be the Euclidean division of m by d. The terms m of this sequence satisfy that q, r, d are consecutive positive integer terms in a geometric progression with a noninteger common ratio > 1.

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A335065 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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A335272 For m to be a term there must exist three Euclidean divisions of m by d, d', and d", m = d*q + r = d'*q' + r' = d"*q" + r", such that (r, q, d), (r', d', q'), and (q", r", d") are three geometric progressions.

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Extensions

A335272 For m to be a term there must exist three Euclidean divisions of m by d, d', and d", m = dq + r = d'q' + r' = d"*q" + r", such that (r, q, d), (r', d', q'), and (q", r", d") are three geometric progressions.