A379831 Positions of records in A379857.

0, 25, 50, 90, 146, 169, 260, 289, 425, 529, 625, 900, 1156, 1521, 1681, 2025, 2500, 2704, 3434, 3600, 4225, 4624, 4900, 5625, 7146, 7225, 8281, 9409, 10404, 11236, 11881, 13225, 14400, 15129, 16900, 18769, 19600, 21316, 23409, 25281, 26896, 28561, 30625, 32400, 34969, 36100, 40000, 41209, 44944, 47524

Offset: 1

Author

Luke E. Holland, Jan 03 2025

Keywords

• nonn

Comments

This sequence is infinite in length.

Not all terms are squares, meaning that sometimes k=1 is not among the k values which set a new record.

Elements of this sequence are often, but not always, records of A033461.

Examples

			25 is a member of this sequence, as 25 = 5^2 = 3^2 + 4^2, meaning that it can be expressed as a sum of k squares for two values of k, in this case 1 and 2, which is more than any smaller value.
50 is a member of this sequence, as 50 = 7^2 + 1^2 = 5^2 + 4^2 + 3^2 = 6^2 + 3^2 + 2^2 + 1^2, giving 3 possible values of k, being 2, 3 and 4, more than any smaller value.
0 begins this sequence, as it is the sum of zero squares, so it has one possible value of k, being k = 0.
		

Links

• Index entries for sequences related to sums of squares

Crossrefs

Cf. A033461, A003995, A001156, A379857.

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
best = 0
for i in range(MAXSQUARE ** 2 + 1):
    if len(possibleSums[i][0]) > best:
        print(i)
        best = len(possibleSums[i][0])
records = ()
best = 0
for i in range(MAXSQUARE ** 2 + 1):
    sums = len(possibleSums[i][0])
    if sums > best:
        records += (i, )
        best = sums

Formula

a(n) >> n^3. - Charles R Greathouse IV, Jan 06 2025