A072905 a(n) is the least k > n such that k*n is a square.

4, 8, 12, 9, 20, 24, 28, 18, 16, 40, 44, 27, 52, 56, 60, 25, 68, 32, 76, 45, 84, 88, 92, 54, 36, 104, 48, 63, 116, 120, 124, 50, 132, 136, 140, 49, 148, 152, 156, 90, 164, 168, 172, 99, 80, 184, 188, 75, 64, 72, 204, 117, 212, 96, 220, 126, 228, 232, 236, 135, 244, 248

Offset

1,1

Comments

From Peter Kagey, Jun 22 2015: (Start)

a(n) is a bijection from the positive integers to A013929 (numbers that are not squarefree). Proof:

(1) Injection: Suppose that b<c and a(b) == a(c). By definition and assumption, b < c < a(c) = a(b). Because a(c) = a(b), b, c, a(b), and a(c) must all have the same squarefree part, thus b*c must be a perfect square. However c < a(b), so a(b) must not be the minimal solution. This is a contradiction. If b<c, then a(b) != a(c) so the function is an injection.

(2) Surjection: Given some number k in A013929, a(A007913(k)*(A000188(k)-1)^2.) = k (End)

Links

Peter Kagey, Table of n, a(n) for n = 1..5000

Formula

a(n) = n + A067722(n). - Peter Kagey, Feb 05 2015

a(n) = A007913(n)*(A000188(n)+1)^2. - Peter Kagey, Feb 06 2015

Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = 1 + 2*zeta(3)/zeta(2) + Pi^2/15 = 3.11949956554216757204... . - Amiram Eldar, Feb 17 2024

Example

12 is the smallest integer > 3 such that 3*12 = 6^2 is a perfect square, hence a(3) = 12.

Maple

f:= proc(n) local F, f, x, y;
     F:= ifactors(n)[2];
     x:= mul(`if`(f[2]::odd, f[1], 1), f=F);
     y:= mul(f[1]^floor(f[2]/2), f=F);
     x*(y+1)^2
end proc:
map(f, [$1..100]); # Robert Israel, Jun 23 2015

Mathematica

a[n_] := For[k = n+1, True, k++, If[IntegerQ[Sqrt[k*n]], Return[k]]]; Array[a, 100] (* Jean-François Alcover, Jan 26 2018 *)

Prog

(PARI) a(n)=if(n<0, 0, s=n+1; while(issquare(s*n)==0, s++); s)
(PARI) a(n)=my(c=core(n)); (sqrtint(n/c)+1)^2*c \\ Charles R Greathouse IV, Jun 23 2015
(Haskell)
a072905 n = head [k | k <- [n + 1 ..], a010052 (k * n) == 1]
-- Reinhard Zumkeller, Feb 07 2015
(Ruby)
def a(n)
  k = Math.sqrt(n).to_i
  k -= 1 until n % k**2 == 0
  n + 2*n/k + n/(k**2)
end # Peter Kagey, Jul 27 2015

Crossrefs

Cf. A000188, A007913, A010052, A067722.

Cf. A182448, A253905.

Sequence in context: A251756 A357803 A158457 * A086481 A329507 A046129

Adjacent sequences: A072902 A072903 A072904 * A072906 A072907 A072908

Keyword

easy,nonn

Author

Benoit Cloitre, Aug 10 2002

Status

approved