A379857 Number of values of k for which n can be written as a sum of k distinct positive squares.

1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 2, 2, 0, 0, 2, 2, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 2, 1, 0, 0, 2, 2, 0, 0, 2, 3, 1, 1, 2, 2, 1, 1, 1, 1, 1, 0, 2, 2, 1, 1, 3, 3, 0, 1, 1, 2, 1, 0, 1, 3, 3, 0, 1, 2, 2, 1, 3, 2, 1, 2

Offset: 0

Author

Luke E. Holland, Jan 04 2025

Keywords

• nonn
• changed

Comments

It appears that a(n) is not bounded, but grows very slowly.

First differs from A033461 at n=62 which can be written as A033461(62) = 3 sums of squares, but among them only a(62) = 2 different numbers of squares.

Examples

			a(0) = 1, as it is the sum of 0 squares.
a(25) = 2, as 25 = 5^2 = 4^2 + 3^2, so k can be 1 or 2.
a(90) = 4, as 90 = 9^2 + 3^2 = 8^2 + 5^2 + 1^2 = 8^2 + 4^2 + 3^2 + 1^2 = 6^2 + 5^2 + 4^2 + 3^2 + 2^2, so k can be 2, 3, 4 or 5.
		

Links

• Luke E. Holland, Table of n, a(n) for n = 0..10000
• Paul T. Bateman, Adolf J. Hildebrand, and George B. Purdy, Sums of distinct squares, Acta Arithmetica 67 (1994), pp. 349-380.

Crossrefs

Cf. A001422 (indices of 0's), A003995 (indices of not 0).

Cf. A379831 (indices of records).

Programs

Python

MAXSQUARE = 500
possibleSums = {i: [[], []] for i in range(MAXSQUARE ** 2 + 1)}
possibleSums[0] = [[0],[0]]
for val in range(MAXSQUARE ** 2):
    for posSquare in range(len(possibleSums[val][0])):
        newSum = possibleSums[val][0][posSquare] + 1
        curr = possibleSums[val][1][posSquare] + 1
        while val + curr ** 2 <= MAXSQUARE ** 2:
            nVal = val + curr ** 2
            if newSum not in possibleSums[nVal][0]:
                possibleSums[nVal][0].append(newSum)
                possibleSums[nVal][1].append(curr)
            else:
                index = possibleSums[nVal][0].index(newSum)
                if curr < possibleSums[nVal][1][index]:
                    possibleSums[nVal][1][index] = curr
            curr += 1
posKVals = tuple([len(possibleSums[i][0]) for i in range(MAXSQUARE ** 2 + 1)])

Python

from itertools import count
from sympy.solvers.diophantine.diophantine import power_representation
def A379857(n):
    if n == 0: return 0
    c = 0
    for i in count(1):
        if i*(i+1)*((i<<1)+1)>6*n:
            break
        if any(len(set(t))==i for t in power_representation(n,2,i)):
            c += 1
    return c # Chai Wah Wu, Jan 28 2025

Formula

a(n) < (3n)^(1/3) for n > 0. I conjecture that a(n) ~ (3n)^(1/3). - Charles R Greathouse IV, Feb 05 2025