A053611 Numbers k such that 1 + 4 + 9 + ... + k^2 = 1 + 2 + 3 + ... + m for some m.

1, 5, 6, 85

Offset: 1

Author

Jud McCranie, Mar 19 2000

Keywords

• fini
• full
• nonn

Comments

These are the only possibilities for a sum of the first n squares to equal a triangular number.

From Seiichi Manyama, Aug 25 2019: (Start)

The complete list of solutions to k*(k+1)*(2*k+1)/6 = m*(m+1)/2 is as follows.

(k,m) = (-1, 0), (0, 0), (1, 1), (5, 10), (6, 13), (85, 645),

(-1,-1), (0,-1), (1,-2), (5,-11), (6,-14), (85,-646). (End)

Examples

			1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1 + 2 + 3 + ... + 10, so 5 is in the sequence.
		

References

• E. T. Avanesov, The Diophantine equation 3y(y+1) = x(x+1)(2x+1), Volz. Mat. Sb. Vyp., 8 (1971), 3-6.
• R. K. Guy, Unsolved Problems in Number Theory, Section D3.
• Joe Roberts, Lure of the Integers, page 245 (entry for 645).

Links

• R. Finkelstein, H. London, On triangular numbers which are sums of consecutive squares, J. Number Theory 4 (1972), 455-462.
• Eric Weisstein's World of Mathematics, Square Pyramidal Number

Crossrefs

Cf. A039596, A053612 (values of m).

Cf. A000217, A000330.

Programs

Maple

istriangular:=proc(n) local t1; t1:=floor(sqrt(2*n)); if n = t1*(t1+1)/2 then RETURN(true) else RETURN(false); fi; end;
M:=1000; for n from 1 to M do if istriangular(n*(n+1)*(2*n+1)/6) then lprint(n,n*(n+1)*(2*n+1)/6); fi; od: # N. J. A. Sloane
# second Maple program:
q:= n-> issqr(8*sum(j^2, j=1..n)+1):
select(q, [$1..100])[];  # Alois P. Heinz, Oct 10 2024

Mathematica

Select[Range[90], IntegerQ[(Sqrt[(4/3) * (# + 3 * #^2 + 2 * #^3) + 1] - 1)/2] &] (* Harvey P. Dale, Sep 22 2014 *)

Extensions

Edited by N. J. A. Sloane, May 25 2008