A329547 Number of natural numbers k <= n such that k^k is a square.

1, 2, 2, 3, 3, 4, 4, 5, 6, 7, 7, 8, 8, 9, 9, 10, 10, 11, 11, 12, 12, 13, 13, 14, 15, 16, 16, 17, 17, 18, 18, 19, 19, 20, 20, 21, 21, 22, 22, 23, 23, 24, 24, 25, 25, 26, 26, 27, 28, 29, 29, 30, 30, 31, 31, 32, 32, 33, 33, 34, 34, 35, 35, 36, 36, 37, 37, 38, 38, 39, 39, 40, 40

Offset

1,2

Comments

For even k, k^k is always a square. For odd k, k^k is a square if and only if k is a square.

It seems the unrepeated terms form A266304 \ {0}. - Ivan N. Ianakiev, Nov 21 2019

Indices of unrepeated terms are A081349. - Rémy Sigrist, Dec 07 2019

Links

David A. Corneth, Table of n, a(n) for n = 1..10000

Formula

a(n) = floor(n/2) + ceiling(floor(sqrt(n))/2).

Example

a(5) = 3 because among 1^1, 2^2, ..., 5^5 there are 3 squares: 1^1, 2^2, and 4^4.

Mathematica

Table[Floor[n/2] + Ceiling[Floor[Sqrt[n]]/2], {n, 1, 100}]

Prog

(PARI) a(n) = sum(k=1, n, issquare(k^k)); \\ Michel Marcus, Nov 17 2019
(PARI) first(n) = my(res=vector(n), inc); res[1] = 1; for(i=2, n, inc = (1-(i%2)) || issquare(i); res[i] = res[i-1] + inc); res \\ David A. Corneth, Dec 07 2019
(PARI) a(n) = n\2 + (sqrtint(n)+1)\2 \\ David A. Corneth, Dec 07 2019
(Python)
from math import isqrt
def A329547(n): return (n>>1)+(isqrt(n)+1>>1) # Chai Wah Wu, Sep 18 2024

Crossrefs

Cf. A081349, A176693, A266304.

Sequence in context: A007963 A137222 A077641 * A330560 A358466 A194210

Adjacent sequences: A329544 A329545 A329546 * A329548 A329549 A329550

Keyword

nonn,easy,changed

Author

Pablo Hueso Merino, Nov 16 2019

Status

approved