A001032 Numbers k such that sum of squares of k consecutive integers >= 1 is a square. (Formerly M1996 N0787)

1, 2, 11, 23, 24, 26, 33, 47, 49, 50, 59, 73, 74, 88, 96, 97, 107, 121, 122, 146, 169, 177, 184, 191, 193, 194, 218, 239, 241, 242, 249, 289, 297, 299, 311, 312, 313, 337, 338, 347, 352, 361, 362, 376, 383, 393, 407, 409, 431, 443, 457, 458, 479, 481, 491, 506

Offset

1,2

Comments

It was shown by Watson (and again by Ljunggren) that if 0^2 + 1^2 + ... + r^2 is a square then r = 0, 1 or 24.

The terms up to 1391 are == 0, 1, 2, 9, 11, 16, 23 (mod 24). Start number is in A007475(n). Square root of sum is in A076215(n). - Ralf Stephan, Nov 04 2002

The solutions in the case n=2 are in A001652 or A082291.

For k > 5 and k == 1 or 5 (mod 6), it appears that all k^2 are here. When n is not a square, the solution to problem 6552 shows that there are an infinite number of sums of n consecutive squares that equal a square. There are only a finite number when n is a square. For example, the only sum having 49 terms is 25^2 + ... + 73^2 = 357^2. - T. D. Noe, Jan 20 2011

In the previous comment, "it appears" can be removed because the k^2 squares beginning at (k^2+1)(k^2-25)/48 sum to a square. - Thomas Andrews, Feb 14 2011

See A180442 for the complementary problem of finding numbers n such that there are consecutive squares beginning with n^2 that sum to a square.

From Thomas Andrews, Feb 22 2011: (Start)

Elementary necessary conditions for n to be in this sequence:

1. If n=s^2b where b is squarefree, then:

a. If s is divisible by 3 then b is divisible by 3.

b. If s is divisible by 2, then b is divisible by 2.

c. If b is divisible by 3, then b = 6 (mod 9)

d. b only has prime factors p where 3 is a square, modulo p. (So, p=2, p=3, or p=12k+-1)

2.

a. If n+1 is divisible by 3, then (n+1)/3 is the sum of two perfect squares.

b. If n+1 is not divisible by 3, then n+1 is the sum of two perfect squares

The smallest number which satisfies these conditions which is not in this sequence is 842.

These conditions can be used to establish the conjecture of Ralf Stephan, above, that all the terms are == 0, 1, 2, 9, 11, 16, or 23 (mod 24). (End)

The numbers satisfying the above conditions but which are not in this sequence can be found in A274469. - Christopher E. Thompson, Jun 28 2016

References

S. Dinh, The Hard Mathematical Olympiad Problems And Their Solutions, AuthorHouse, 2011, Problem 6 of the Irish Mathematical Olympiad 1990 (in fact, it is 1991), page 96.

W. Ljunggren, New solution of a problem proposed by E. Lucas, Norsk Mat. Tid. 34 (1952), 65-72.

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Christopher E. Thompson, Table of n, a(n) for n = 1..10438 (up to 250000, extends first 128 terms computed by T. D. Noe).

U. Alfred, Consecutive integers whose sum of squares is a perfect square, Math. Mag., 37 (1964), 19-32.

L. Beeckmans, Squares expressible as sum of consecutive squares, Amer. Math. Monthly, 101 (1994), 437-442.

K. S. Brown, Sum of Consecutive Nth Powers Equals an Nth Power

M. Laub, Squares Expressible as a Sum of n Consecutive Squares, Advanced Problem 6552, Amer. Math. Monthly 97 (1990), 622-625.

S. Philipp, Note on consecutive integers whose sum of squares is a perfect square, Math. Mag., 37 (1964), 218-220.

Vladimir Pletser, Congruence conditions on the number of terms in sums of consecutive squared integers equal to squared integers, arXiv:1409.7969 [math.NT], 2014.

Vladimir Pletser, Finding all squared integers expressible as the sum of consecutive squared integers using generalized Pell equation solutions with Chebyshev polynomials, arXiv preprint arXiv:1409.7972 [math.NT], 2014.

John Scholes, 4th Irish Mathematical Olympiad 1991, Problem B1.

G. N. Watson, The problem of the square pyramid, Messenger of Mathematics 48 (1918), pp. 1-22.

Eric Weisstein's World of Mathematics, Cannonball Problem

Index to sequences related to Olympiads.

Index entries for sequences related to sums of squares

Example

3^2 + 4^2 = 5^2, with two consecutive terms, so 2 is in the sequence.
Sum_{m=18..28} m^2 = 77^2, with eleven consecutive terms, so 11 is in the sequence and A007475(3) = 18. - Bernard Schott, Jan 03 2022

Mathematica

(* An empirical recomputation, assuming Ralf Stephan's conjecture *) nmax = 600; min[_](* minimum start number *) = 1; max[_](* maximum start number *) = 10^5; min[457(* the first not-so-easy term *)] = 10^7; min[577] = 10^5; min[587] = 10^7; max[457] = max[577] = max[587] = Infinity; okQ[n_ /; ! MemberQ[{0, 1, 2, 9, 11, 16, 23}, Mod[n, 24]]] = False; okQ[n_] := For[m = min[n], m < max[n], m++, If[IntegerQ[ r = Sqrt[1/6*n*(1 + 6*m^2 + 6*m*(n - 1) - 3*n + 2*n^2)]], Return[True]]]; nmr = Reap[k = 1; Do[If[okQ[n] === True, Print["a(", k, ") = ", n, ", start nb = A007475(", k, ") = ", m, ", sqrt(sum) = A076215(", k, ") = ", r]; k++; Sow[{n, m, r}]], {n, 1, nmax}]][[2, 1]]; A001032 = nmr[[All, 1]]; A007475 = nmr[[All, 2]]; A076215 = nmr[[All, 3]] (* Jean-François Alcover, Sep 09 2013 *)

Prog

(PARI) is(n, L=max(999, n^5\2e5), s=norml2([1..n-1]))={bittest(8456711, n%24) && for(x=n, L, issquare(s+=(2*x-n)*n)&&return(x))} \\ Returns the smallest "ending number" x (such that (x-n+1)^2+...+x^2 is a square) if n is in the sequence, otherwise zero. - M. F. Hasler, Feb 02 2016

Crossrefs

Cf. A007475, A076215, A151557, A274469.

Cf. A097812 (n^2 is the sum of two or more consecutive squares).

Sequence in context: A004642 A346497 A185545 * A045386 A084354 A066725

Adjacent sequences: A001029 A001030 A001031 * A001033 A001034 A001035

Keyword

nonn,easy,nice

Author

N. J. A. Sloane

Extensions

Corrected by T. D. Noe, Aug 25 2004

Offset changed to 1 by N. J. A. Sloane, June 2008

Additional terms up to 30000 added to b-file by Christopher E. Thompson, Jun 10 2016

Additional terms up to 250000 added to b-file by Christopher E. Thompson, Feb 20 2018

Status

approved