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

%I A053611 #42 Feb 16 2025 08:32:42
%S A053611 1,5,6,85
%N A053611 Numbers k such that 1 + 4 + 9 + ... + k^2 = 1 + 2 + 3 + ... + m for some m.
%C A053611 These are the only possibilities for a sum of the first n squares to equal a triangular number.
%C A053611 From _Seiichi Manyama_, Aug 25 2019: (Start)
%C A053611 The complete list of solutions to k*(k+1)*(2*k+1)/6 = m*(m+1)/2 is as follows.
%C A053611 (k,m) = (-1, 0), (0, 0), (1, 1), (5, 10), (6, 13), (85, 645),
%C A053611         (-1,-1), (0,-1), (1,-2), (5,-11), (6,-14), (85,-646). (End)
%D A053611 E. T. Avanesov, The Diophantine equation 3y(y+1) = x(x+1)(2x+1), Volz. Mat. Sb. Vyp., 8 (1971), 3-6.
%D A053611 R. K. Guy, Unsolved Problems in Number Theory, Section D3.
%D A053611 Joe Roberts, Lure of the Integers, page 245 (entry for 645).
%H A053611 R. Finkelstein, H. London, <a href="http://dx.doi.org/10.1016/0022-314X(72)90036-4">On triangular numbers which are sums of consecutive squares</a>, J. Number Theory 4 (1972), 455-462.
%H A053611 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/SquarePyramidalNumber.html">Square Pyramidal Number</a>
%e A053611 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1 + 2 + 3 + ... + 10, so 5 is in the sequence.
%p A053611 istriangular:=proc(n) local t1; t1:=floor(sqrt(2*n)); if n = t1*(t1+1)/2 then RETURN(true) else RETURN(false); fi; end;
%p A053611 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_
%p A053611 # second Maple program:
%p A053611 q:= n-> issqr(8*sum(j^2, j=1..n)+1):
%p A053611 select(q, [$1..100])[];  # _Alois P. Heinz_, Oct 10 2024
%t A053611 Select[Range[90], IntegerQ[(Sqrt[(4/3) * (# + 3 * #^2 + 2 * #^3) + 1] - 1)/2] &] (* _Harvey P. Dale_, Sep 22 2014 *)
%Y A053611 Cf. A039596, A053612 (values of m).
%Y A053611 Cf. A000217, A000330.
%K A053611 fini,full,nonn
%O A053611 1,2
%A A053611 _Jud McCranie_, Mar 19 2000
%E A053611 Edited by _N. J. A. Sloane_, May 25 2008