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

1,8

Comments

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.

Links

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

Formula

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

Example

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.

Maple

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

Mathematica

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 *)

Prog

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

Crossrefs

Cf. A005117, A007913, A013929, A072905.

Sequence in context: A300723 A263788 A355335 * A300324 A298368 A072069

Adjacent sequences: A292141 A292142 A292143 * A292145 A292146 A292147

Keyword

nonn

Author

Peter Kagey, Sep 09 2017

Status

approved