A001032 - OEIS

%I M1996 N0787 #110 Dec 23 2024 11:37:59

%S 1,2,11,23,24,26,33,47,49,50,59,73,74,88,96,97,107,121,122,146,169,

%T 177,184,191,193,194,218,239,241,242,249,289,297,299,311,312,313,337,

%U 338,347,352,361,362,376,383,393,407,409,431,443,457,458,479,481,491,506

%N Numbers k such that sum of squares of k consecutive integers >= 1 is a square.

%C 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.

%C 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

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

%C 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

%C 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

%C 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.

%C From _Thomas Andrews_, Feb 22 2011: (Start)

%C Elementary necessary conditions for n to be in this sequence:

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

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

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

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

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

%C   2.

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

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

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

%C 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)

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

%D 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.

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

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

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

%H Christopher E. Thompson, <a href="/A001032/b001032.txt">Table of n, a(n) for n = 1..10438</a> (up to 250000, extends first 128 terms computed by T. D. Noe).

%H U. Alfred, <a href="http://www.jstor.org/stable/2688243">Consecutive integers whose sum of squares is a perfect square</a>, Math. Mag., 37 (1964), 19-32.

%H Laurent Beeckmans, <a href="http://www.jstor.org/stable/2974904">Squares expressible as sum of consecutive squares</a>, Amer. Math. Monthly, 101 (1994), 437-442.

%H Kevin S. Brown, <a href="http://www.mathpages.com/home/kmath147.htm">Sum of Consecutive Nth Powers Equals an Nth Power</a>

%H Anji Dong, Katerina Saettone, Kendra Song, and Alexandru Zaharescu, <a href="https://arxiv.org/abs/2412.10097">An Equidistribution Result for Differences Associated with Square Pyramidal Numbers</a>, arXiv:2412.10097 [math.NT], 2024. See p. 1.

%H Moshe Laub, <a href="http://www.jstor.org/stable/2324647">Squares Expressible as a Sum of n Consecutive Squares, Advanced Problem 6552</a>, Amer. Math. Monthly 97 (1990), 622-625.

%H Stanton Philipp, <a href="http://www.jstor.org/stable/2688590">Note on consecutive integers whose sum of squares is a perfect square</a>, Math. Mag., 37 (1964), 218-220.

%H Vladimir Pletser, <a href="http://arxiv.org/abs/1409.7969">Congruence conditions on the number of terms in sums of consecutive squared integers equal to squared integers</a>, arXiv:1409.7969 [math.NT], 2014.

%H Vladimir Pletser, <a href="http://arxiv.org/abs/1409.7972">Finding all squared integers expressible as the sum of consecutive squared integers using generalized Pell equation solutions with Chebyshev polynomials</a>, arXiv preprint arXiv:1409.7972 [math.NT], 2014.

%H John Scholes, <a href="https://webee.technion.ac.il/people/aditya/www.kalva.demon.co.uk/irish/irish91.html">4th Irish Mathematical Olympiad 1991, Problem B1</a>.

%H G. N. Watson, <a href="http://archive.org/stream/messengerofmathe4849cambuoft#page/n9/mode/2up">The problem of the square pyramid</a>, Messenger of Mathematics 48 (1918), pp. 1-22.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/CannonballProblem.html">Cannonball Problem</a>

%H <a href="/index/O#Olympiads">Index to sequences related to Olympiads</a>.

%H <a href="/index/Su#ssq">Index entries for sequences related to sums of squares</a>

%e 3^2 + 4^2 = 5^2, with two consecutive terms, so 2 is in the sequence.

%e 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

%t (* 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 *)

%o (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

%Y Cf. A007475, A076215, A151557, A274469.

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

%K nonn,easy,nice

%O 1,2

%A _N. J. A. Sloane_

%E Corrected by _T. D. Noe_, Aug 25 2004

%E Offset changed to 1 by _N. J. A. Sloane_, June 2008

%E Additional terms up to 30000 added to b-file by _Christopher E. Thompson_, Jun 10 2016

%E Additional terms up to 250000 added to b-file by _Christopher E. Thompson_, Feb 20 2018