login
This site is supported by donations to The OEIS Foundation.

 

Logo

Annual Appeal: Today, Nov 11 2014, is the 4th anniversary of the launch of the new OEIS web site. 70,000 sequences have been added in these four years, all edited by volunteers. Please make a donation (tax deductible in the US) to help keep the OEIS running.

Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A001032 Numbers n such that sum of squares of n consecutive integers >= 1 is a square.
(Formerly M1996 N0787)
18
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 (list; graph; refs; listen; history; text; internal format)
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.

Contribution 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 square-free, 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 show the conjecture by Ralf Stephan, above, that all the numbers are == 0, 1, 2, 9, 11, 16, or 23 (mod 24.) (End)

REFERENCES

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

T. D. Noe, Table of n, a(n) for n=1..128

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.

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 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.

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-Fran├žois Alcover, Sep 09 2013 *)

CROSSREFS

Cf. A007475, A076215, A151577.

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

Sequence in context: A018351 A004642 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

STATUS

approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Transforms | Superseeker | Recent | More pages
The OEIS Community | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement .

Last modified November 26 03:41 EST 2014. Contains 250017 sequences.