A072905 -id:A072905 - cp's OEIS frontend

A292144 a(n) is the greatest k < n such that k*n is square.

0, 0, 0, 1, 0, 0, 0, 2, 4, 0, 0, 3, 0, 0, 0, 9, 0, 8, 0, 5, 0, 0, 0, 6, 16, 0, 12, 7, 0, 0, 0, 18, 0, 0, 0, 25, 0, 0, 0, 10, 0, 0, 0, 11, 20, 0, 0, 27, 36, 32, 0, 13, 0, 24, 0, 14, 0, 0, 0, 15, 0, 0, 28, 49, 0, 0, 0, 17, 0, 0, 0, 50, 0, 0, 48, 19, 0, 0, 0, 45

Offset: 1

Peter Kagey, Sep 09 2017

nonn

a(n) = 0 if and only if n is squarefree: a(A005117(n)) = 0 for all n, and a(A013929(n)) > 0 for all n.
A072905 is the right inverse of a: a(A072905(n)) = n.
If a(n) = a(m) != 0, then n = m.
Proof: Without loss of generality, assume a(n) = a(m) < n < m. Then n*a(n)*m*a(m) is square and a(n)*a(m) is square, which implies that n*m is square. Notice that n > a(m), so a(m) is not the greatest integer k such that k*m is square. This is a contradiction.

			For n = 63, a(63) = 28 because 28*63 = (7*4)*(7*9) = (7*2*3)^2 = 42^2, and there is no integer 28 < k < 63 such that 63*k is square.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A005117, A007913, A013929, A072905.

f:= proc(n) local F,r;
   F:= ifactors(n)[2];
   r:= mul(t[1], t = select(t -> t[2]::odd, F));
   r*(ceil(sqrt(n/r))-1)^2;
end proc: # Robert Israel, Sep 10 2017

a[n_] := If[SquareFreeQ[n], 0, For[k = n-1, k > 0, k--, If[IntegerQ[ Sqrt[ k*n] ], Return[k]]]]; Array[a, 80] (* Jean-François Alcover, Sep 11 2017 *)

forstep (k=n-1, 1, -1, if (issquare(k*n), return (k))); return (0); \\ Michel Marcus, Sep 10 2017

a(n) = A007913(n)*(ceiling(sqrt(n/A007913(n))-1)^2). - Robert Israel and Michel Marcus, Sep 11 2017

A299117 Sorted terms of A277781.

Original entry on oeis.org

4, 8, 9, 16, 18, 24, 25, 27, 32, 36, 40, 48, 49, 50, 54, 56, 64, 72, 75, 80, 81, 88, 96, 98, 100, 104, 108, 112, 120, 121, 125, 128, 135, 136, 144, 147, 152, 160, 162, 168, 169, 176, 180, 184, 189, 192, 196, 200, 208, 216, 224, 225, 232, 240, 242, 243, 245

Offset: 1

Peter Kagey, Feb 02 2018

nonn

It appears that all of these numbers are nonsquarefree (i.e., this sequence is a subsequence of A013929).

This sequence is to A277781 what A013929 is to A072905. That is, A277781 is a bijection from the positive integers to this sequence.

Peter Kagey, Table of n, a(n) for n = 1..1000

Cf. A013929, A072905, A277781.

With[{nn = 57}, Take[#, nn] &@ Sort@ Table[SelectFirst[n + Range[7 + n^2], AnyTrue[Power[#, 1/3] & /@ {n #, n #^2}, IntegerQ] &], {n, 8 nn}]] (* Michael De Vlieger, Feb 03 2018 *)

A379705 a(n) is the least integer k > n such that integers p, q exist for which n, p, k are in arithmetic and n, q, k are in geometric progression.

Original entry on oeis.org

9, 8, 27, 16, 45, 24, 63, 18, 25, 40, 99, 48, 117, 56, 135, 36, 153, 32, 171, 80, 189, 88, 207, 54, 49, 104, 75, 112, 261, 120, 279, 50, 297, 136, 315, 64, 333, 152, 351, 90, 369, 168, 387, 176, 125, 184, 423, 108, 81, 72, 459, 208, 477, 96, 495, 126, 513, 232

Offset: 1

Felix Huber, Jan 07 2025

nonn

			a(9) = 25 because 9, 17, 25 are in arithmetic progression (common difference = 8) and 9, +-15, 25 are in geometric progression (common ratio = +-5/3) and there is no other integer k with 9 < k < 25 such that integers p and q exist for which 9, p, k are in arithmetic and 9, q, k are in geometric progression.

Felix Huber, Table of n, a(n) for n = 1..10000
Wikipedia, Arithmetic Progression
Wikipedia, Geometric Progression

Cf. A000188, A008833, A057918, A072905, A379663.

A379705:=proc(n)
   local d;
   d:=expand(NumberTheory:-LargestNthPower(n,2));
   if is(n*(1+(d+1)^2/d^2),even) then
      n*(d+1)^2/d^2
   else
      n*(d+2)^2/d^2
   fi;
end proc;
seq(A379705(n),n=1..58);

a(n) = n/A008833(n)*(A000188(n) + k)^2, where k = 1 if n*(1+(A000188(n)+1)^2/A008833(n)) is even or k = 2 else.

a(n) = A072905(n) if n*(1+(A000188(n)+1)^2/A008833(n)) is even.

A260896 a(n) gives the number of integers m such that there exist k and h with 2n^2 < mk^2 < 2(n+1)^2 and 2n^2 < 2mh^2 < 2(n+1)^2.

Original entry on oeis.org

0, 1, 0, 1, 1, 1, 3, 3, 2, 3, 2, 2, 3, 3, 0, 3, 1, 4, 2, 3, 3, 1, 6, 3, 4, 4, 5, 3, 2, 5, 4, 8, 4, 4, 5, 1, 5, 6, 4, 5, 3, 6, 2, 5, 7, 5, 8, 4, 7, 4, 7, 7, 7, 10

Offset: 0

Peter Kagey, Aug 03 2015

nonn

A072905(2n^2) > A006255(2n^2) and A066400(2n^2) > 2 for all n such that a(n) > 0.

Conjecture: a(n) > 0 for all n > 14.

			For n=12 the a(12)=3 solutions are 3, 6, and 37:
  (1) (a) 2 * 12^2 <      3 * 10^2 < 2 * 13^2
      (b) 2 * 12^2 < 2 *  3 *  7^2 < 2 * 13^2
  (2) (a) 2 * 12^2 <      6 *  7^2 < 2 * 13^2
      (b) 2 * 12^2 < 2 *  6 *  5^2 < 2 * 13^2
  (3) (a) 2 * 12^2 <     37 *  3^2 < 2 * 13^2
      (b) 2 * 12^2 < 2 * 37 *  2^2 < 2 * 13^2

Peter Kagey, Table of n, a(n) for n = 0..10000

Cf. A006255, A066400, A072905.

A305709 Least k such that there exists a three-term sequence n = b_1 < b_2 < b_3 = k such that b_1 * b_2 * b_3 is square.

Original entry on oeis.org

8, 6, 8, 16, 10, 12, 14, 18, 25, 20, 22, 20, 26, 24, 27, 32, 34, 27, 38, 30, 28, 33, 46, 32, 48, 52, 40, 45, 58, 42, 62, 45, 48, 54, 56, 64, 74, 57, 52, 50, 82, 56, 86, 55, 60, 69, 94, 54, 72, 63, 75, 78, 106, 75, 90, 72, 76, 96, 118, 80, 122, 96, 84, 98, 104

Offset: 1

Peter Kagey, Jun 08 2018

nonn

a(n) >= A006255(n), and a(n) = A006255(n) if and only if A066400(n) = 3.

Conjecture: a(n) < A072905(n) with finitely many nonsquare exceptions.

			For n = 3 the sequence is 3, 6, 8; so a(3) = 8;
for n = 4 the sequence is 4, 9, 16; so a(4) = 16;
for n = 5 the sequence is 5, 8, 10; so a(5) = 10.

Peter Kagey, Table of n, a(n) for n = 1..10000

Cf. A006255, A066400, A072905.

A343881 Table read by antidiagonals upward: T(n,k) is the least integer m > k such that k^x * m^y = c^n for some positive integers c, x, and y where x < n and y < n; n >= 2, k >= 1.

Original entry on oeis.org

4, 8, 8, 4, 4, 12, 32, 4, 9, 9, 4, 4, 9, 16, 20, 128, 4, 9, 8, 25, 24, 4, 4, 9, 8, 20, 36, 28, 8, 4, 9, 8, 25, 24, 49, 18, 4, 4, 9, 8, 20, 36, 28, 27, 16, 2048, 4, 9, 8, 25, 24, 49, 18, 24, 40, 4, 4, 9, 8, 20, 36, 28, 16, 12, 80, 44, 8192, 4, 9, 8, 25, 24, 49

Offset: 2

Peter Kagey, May 02 2021

For prime p, the p-th row consists of distinct integers.

Conjecture: T(p,k) = A064549(k) for fixed k > 1 and sufficiently large p.

			Table begins:
  n\k|    1  2   3   4   5   6   7   8   9   10
-----+-----------------------------------------
   2 |    4, 8, 12,  9, 20, 24, 28, 18, 16,  40
   3 |    8, 4,  9, 16, 25, 36, 49, 27, 24,  80
   4 |    4, 4,  9,  8, 20, 24, 28, 18, 12,  40
   5 |   32, 4,  9,  8, 25, 36, 49, 16, 27, 100
   6 |    4, 4,  9,  8, 20, 24, 28,  9, 16,  40
   7 |  128, 4,  9,  8, 25, 36, 49, 16, 27, 100
   8 |    4, 4,  9,  8, 20, 24, 28, 16, 12,  40
   9 |    8, 4,  9,  8, 25, 36, 49, 16, 24,  80
  10 |    4, 4,  9,  8, 20, 24, 28, 16, 16,  40
  11 | 2048, 4,  9,  8, 25, 36, 49, 16, 27, 100
T(2, 3) = 12 with  3   * 12   =  6^2.
T(3,10) = 80 with 10^2 * 80   = 20^3.
T(4, 5) = 20 with  5^2 * 20^2 = 10^4.
T(5, 1) = 32 with  1   * 32   =  2^5.
T(6, 8) =  9 with  8^2 *  9^3 =  6^6.

Rows: A072905 (n=2), A277781 (n=3).

Cf. A064549, A343825.

T(n,1) = 2^A020639(n).

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A292144 a(n) is the greatest k < n such that k*n is square.

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A299117 Sorted terms of A277781.

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A379705 a(n) is the least integer k > n such that integers p, q exist for which n, p, k are in arithmetic and n, q, k are in geometric progression.

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A260896 a(n) gives the number of integers m such that there exist k and h with 2n^2 < mk^2 < 2(n+1)^2 and 2n^2 < 2mh^2 < 2(n+1)^2.

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A305709 Least k such that there exists a three-term sequence n = b_1 < b_2 < b_3 = k such that b_1 * b_2 * b_3 is square.

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A343881 Table read by antidiagonals upward: T(n,k) is the least integer m > k such that k^x * m^y = c^n for some positive integers c, x, and y where x < n and y < n; n >= 2, k >= 1.

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