A388659 Numbers k for which there exists m such that the sum from 1 to m and the sum from m + 1 to k are both perfect squares.

4, 9, 16, 25, 40, 89, 144, 148, 233, 289, 520, 800, 841, 865, 936, 1360, 3033, 4513, 4900, 4904, 5044, 5329, 7929, 9801, 10953, 11024, 13000, 14841, 17680, 18793, 19024, 26280, 28561, 28585, 29008, 29401, 29800, 31040, 46216, 46696, 47353, 57460, 62164

Offset

1,1

Comments

Given the set {1, 2, 3,..., k}, we want to separate it into the disjoint sets {1, 2, 3,..., m} and {m + 1, m + 2,..., k}, so that the sum of the elements of each set is a square.

We want the sum from 1 to m to be a square, that is, m*(m + 1)/2 to be a square. Therefore, m is in A001108. Furthermore, (m + 1) + (m + 2) + ... + k = u^2, where u is an integer. Then, k*(k + 1)/2 - m*(m + 1)/2 = u^2, which can be rewritten as (2*k + 1)^2 - 8*u^2 = (2*m + 1)^2, which is a generalized Pell equation.

Links

David Consiglio, Jr., Table of n, a(n) for n = 1..100

David Consiglio, Jr., Python

Example

k = 16 is in this sequence since for m = 8, 1 + 2 + 3 +...+ 8 = 6^2 and also 9 + 10 + 11 +...+ 16 = 10^2.

Maple

N:= 10^5: # for terms <= N
mf:= proc(i) option remember; 6*procname(i-1) - procname(i-2) + 2 end proc: mf(0):= 0: mf(1):= 1:
R:= {}:
for i from 1 do
  m:= mf(i);
  if m >= N then break fi;
  eq:= K^2 = 8*z^2 + (2*m+1)^2;
  S:= [isolve(eq)];
  for s in S do
    if subs(expand(eval(s, _Z1=1)), [K, z])::list(posint) then
      for zi from 0 do
        v:= (expand(subs(_Z1=zi, subs(s, K)))-1)/2;
        if v > N then break fi;
        if v > m then R:= R union {v} fi;
od fi od od:
sort(convert(R, list)); # Robert Israel, Jan 08 2026

Crossrefs

Cf. A000290, A001108, A374087.

Sequence in context: A008056 A352048 A206920 * A108612 A065741 A188061

Adjacent sequences: A388656 A388657 A388658 * A388660 A388661 A388662

Keyword

nonn

Author

Gonzalo Martínez, Dec 16 2025

Status

approved