A368590 Numbers k such that all of k, k+1 and k+2 are the sums of consecutive squares.

728, 1013, 2813, 3309, 4323, 4899, 12438, 21259, 23113, 31394, 35719, 37812, 38023, 111894, 143449, 194053, 418613, 418614, 487368, 535309, 2232593, 2452644, 2490669, 9226854, 17367998, 19637644, 20341453, 28553671, 33406839, 174398434, 468936719, 1468970139, 2136314464

Offset: 1

Author

David A. Corneth, Dec 31 2023

Keywords

• nonn

Comments

418613 is the smallest k such that k through k + 3 are the sums of consecutive squares.

After an idea by Allan C. Wechsler.

a(30)-a(33) were calculated using the b-file at A368570.

Examples

			728 is in the sequence via 728 = 7^2 + 8^2 + 9^2 + 10^2 + 11^2 + 12^2 + 13^2, 729 = 27^2 and 730 = 10^2 + 11^2 + 12^2 + 13^2 + 14^2.
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..94 (terms 1..74 from Frank A. Stevenson)
• David A. Corneth, PARI program

Crossrefs

Subsequence of A034705 and of A368570.

Programs

PARI

\\ See PARI program

Python

import heapq
from itertools import islice
def agen(): # generator of terms
    m = 1; h = [(m, 1, 1)]; nextcount = 2
    v1 = v2 = -1
    while True:
        (v, s, l) = heapq.heappop(h)
        if v != v1:
            if v2 + 2 == v1 + 1 == v: yield v2
            v2, v1 = v1, v
        if v >= m:
            m += nextcount*nextcount
            heapq.heappush(h, (m, 1, nextcount))
            nextcount += 1
        v -= s*s; s += 1; l += 1; v += l*l
        heapq.heappush(h, (v, s, l))
print(list(islice(agen(), 33))) # Michael S. Branicky, Jan 01 2024