

A001032


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


35



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
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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
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^225)/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.
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)


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), 6572.
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



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 notsoeasy 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]] (* JeanFrançois Alcover, Sep 09 2013 *)


PROG

(PARI) is(n, L=max(999, n^5\2e5), s=norml2([1..n1]))={bittest(8456711, n%24) && for(x=n, L, issquare(s+=(2*xn)*n)&&return(x))} \\ Returns the smallest "ending number" x (such that (xn+1)^2+...+x^2 is a square) if n is in the sequence, otherwise zero.  M. F. Hasler, Feb 02 2016


CROSSREFS

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


KEYWORD

nonn,easy,nice


AUTHOR



EXTENSIONS



STATUS

approved



