A360530 -id:A360530 - cp's OEIS frontend

A362068 a(n) is the smallest positive integer k such that n can be expressed as the arithmetic mean of k squares.

1, 2, 3, 1, 2, 3, 3, 2, 1, 2, 3, 3, 2, 3, 3, 1, 2, 2, 3, 2, 4, 3, 3, 3, 1, 2, 3, 3, 2, 3, 3, 2, 3, 2, 3, 1, 2, 3, 3, 2, 2, 3, 3, 3, 2, 3, 3, 3, 1, 2, 3, 2, 2, 3, 3, 3, 3, 2, 3, 3, 2, 3, 3, 1, 2, 3, 3, 2, 4, 3, 3, 2, 2, 2, 3, 3, 4, 3, 3, 2, 1, 2, 3, 4, 2, 3, 3, 3, 2, 2, 3, 3, 4, 3, 3, 3, 2, 2, 3, 1

Offset: 1

Yifan Xie, Apr 07 2023

a(n) is the smallest number k such that n*k can be expressed as the sum of k squares.

			For n = 2, if k = 1, 2*1 = 2 is not a square; if k = 2, 2*2 = 4 = 2^2 + 0^2, so a(2) = 2.

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

Cf. A360530 (nonzero), A004215.

findsquare(k, m) = if(k == 1, issquare(m), for(j=0, m, if(j*j > m, return(0), if(findsquare(k-1, m-j*j), return(1)))));
a(n) = {for(t = 1, 3, if(findsquare(t, n*t), return(t))); return(4)};

from sympy.ntheory.primetest import is_square
from sympy import factorint
def A362068(n):
    if is_square(n):
        return 1
    if all(map(lambda x:x[0]&3<3 or x[1]&1^1, factorint(k:=n>>(m:=(~n&n-1).bit_length())).items())):
        return 2
    if m&1 or 3*k&7<7:
        return 3
    return 4 # Chai Wah Wu, Apr 27 2023

a(n) <= 4. (Lagrange)

A362110 a(n) is the smallest integer k such that n can be expressed as the arithmetic mean of k distinct nonzero squares, or 0 if no such k exists.

Original entry on oeis.org

1, 0, 0, 1, 2, 0, 3, 0, 1, 2, 5, 0, 2, 3, 3, 1, 2, 3, 5, 2, 4, 3, 3, 5, 1, 2, 3, 3, 2, 3, 3, 5, 5, 2, 3, 1, 2, 3, 3, 2, 2, 3, 3, 5, 2, 3, 3, 5, 1, 2, 3, 2, 2, 3, 3, 3, 3, 2, 4, 3, 2, 3, 3, 1, 2, 3, 3, 2, 4, 3, 3, 3, 2, 2, 3, 5, 4, 3, 3, 2, 1, 2, 3, 4, 2, 3, 3, 3, 2, 2, 3, 3, 4, 3, 3, 5, 2, 3, 3

Offset: 1

Yifan Xie, Apr 16 2023

nonn

			For n = 2, if k = 1, 2*1 = 2 is not a square; but, from the upper bound formula, (k + 1) * (2*k + 1) <= 12, so k <= 1. So, a satisfactory k does not exist; hence a(2) = 0.

Cf. A360530 (allows repeated squares).

Cf. A001422, A033461, A129210, A132777, A248509.

Upper bound: (a(n) + 1) * (2*a(n) + 1) <= 6*n. Proof: Because (Sum_{m=1..k} (i_m)^2)/k = n, n*k = Sum_{m=1..k} (i_m)^2. Since each i_m is distinct, n*k >= Sum_{m=1..k} m^2 = k * (k + 1) * (2*k + 1)/6, hence (k + 1) * (2*k + 1) <= 6*n.

a(A132777(n)) = 2. - Thomas Scheuerle, Apr 16 2023

Name qualified and other edits by Peter Munn, Apr 21 2023

cp's OEIS Frontend

A362068 a(n) is the smallest positive integer k such that n can be expressed as the arithmetic mean of k squares.

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