A180442 -id:A180442 - cp's OEIS frontend

A001032 Numbers k such that sum of squares of k consecutive integers >= 1 is a square.

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

N. J. A. Sloane

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

			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

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).

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.
Laurent Beeckmans, Squares expressible as sum of consecutive squares, Amer. Math. Monthly, 101 (1994), 437-442.
Kevin S. Brown, Sum of Consecutive Nth Powers Equals an Nth Power
Anji Dong, Katerina Saettone, Kendra Song, and Alexandru Zaharescu, An Equidistribution Result for Differences Associated with Square Pyramidal Numbers, arXiv:2412.10097 [math.NT], 2024. See p. 1.
Moshe Laub, Squares Expressible as a Sum of n Consecutive Squares, Advanced Problem 6552, Amer. Math. Monthly 97 (1990), 622-625.
Stanton 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

Cf. A007475, A076215, A151557, A274469.

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

(* 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 *)

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

Corrected by T. D. Noe, Aug 25 2004
Offset changed to 1 by N. J. A. Sloane, Jun 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

A184763 Irregular triangle in which row n has the values of k>n such that Sum_{i=n..k} i^2 is a square.

Original entry on oeis.org

24, 4, 580, 963, 29, 39, 56, 190, 2215, 32, 552057, 22908, 108, 111, 326, 39, 5345, 28, 21, 43, 308, 1221044, 116, 80, 6910, 48, 50, 73, 578, 624, 3625, 21624, 59, 364, 77, 123, 198, 609, 4087, 61281, 148856, 48, 96, 349, 686, 11918, 67, 93, 171, 15674, 147, 389, 3533, 2132, 92, 3238, 305, 282, 928, 116, 8516, 194, 22873, 36554, 99, 276, 26003, 136, 332

Offset: 1

T. D. Noe, Jan 21 2011

Sequence A184762 gives the length of row n. A180442 lists the nonempty rows.

			The triangle is
(row 1)  24
(row 3)  4, 580, 963
(row 7)  29, 39, 56, 190, 2215
(row 9)  32, 552057
(row 11) 22908
(row 13) 108
(row 15) 111, 326

See A180442.

Jon E. Schoenfield, List of all integer solutions (i.e., triples (n, k, m) where k > n and Sum_{i=n..k} i^2 = m^2) for n in 1..460 (includes a derivation of the elliptic curves and Magma code used to find the solutions)

Cf. A184885 (number of squares in each sum), A184886 (sqrt of sums).

A075404 Smallest m > n such that Sum_{i=n..m} i^2 is a square, or 0 if no such m exists.

Original entry on oeis.org

24, 0, 4, 0, 0, 0, 29, 0, 32, 0, 22908, 0, 108, 0, 111, 0, 39, 28, 0, 21, 116, 80, 0, 0, 48, 0, 59, 77, 0, 198, 0, 609, 0, 0, 0, 0, 0, 48, 0, 0, 0, 0, 0, 67, 0, 0, 0, 0, 0, 171, 0, 147, 0, 0, 3533, 0, 0, 2132, 0, 92, 0, 0, 0, 305, 282, 0, 116, 0, 0, 0, 0, 0, 194, 36554, 0, 99, 0, 0, 0, 0, 0, 0, 276, 0, 0, 0, 136, 0, 0, 0, 332, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 1

Zak Seidov, Sep 13 2002

nonn

For a(1) see A000330.
The corresponding squares are in A075405, the numbers of terms in the sum = a(n)-n+1 are in A075406.
All terms were verified by solving elliptic curves. If a(n)>0, then there may be additional values of m that produce squares. See A184763 for more information.

			a(1) = 24 because 1^2+...+24^2 = 70^2, a(7) = 29 because 7^2+...+29^2 = 92^2.

See A180442.

Cf. A000330, A075405, A075406, A180442 (n such that a(n) > 0).

s[n_,k_]:=Module[{m=n+k-1},(m(m+1)(2m+1)-n(n-1)(2n-1))/6]; mx=40000; Table[k=2; While[k

Corrected and extended by Lior Manor, Sep 19 2002

Corrected and edited by T. D. Noe, Jan 21 2011

A184762 The number of numbers k > n such that Sum_{i=n..k} i^2 is a square.

Original entry on oeis.org

1, 0, 3, 0, 0, 0, 5, 0, 2, 0, 1, 0, 1, 0, 2, 0, 2, 1, 0, 4, 1, 2, 0, 0, 7, 0, 2, 2, 0, 1, 0, 4, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 2, 0, 2, 0, 0, 1, 0, 0, 1, 0, 2, 0, 0, 0, 1, 2, 0, 2, 0, 0, 0, 0, 0, 2, 1, 0, 1, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 1

T. D. Noe, Jan 21 2011

It is an old result (see Watson) that for n=1 the only k>n is k=24. Bremner, Stroeker, and Tzanakis compute the k for n <= 100 by solving elliptic curves. This sequence lists the number of k for each n; the values of k are in A184763. Sequence A180442 lists the n for which a(n) is nonzero.

A. Bremner, R. J. Stroeker, N. Tzanakis, On Sums of Consecutive Squares, J. Number Theory 62 (1997), 39-70.
G. N. Watson, The problem of the square pyramid, Messenger of Mathematics 48 (1918), pp. 1-22.

Cf. A075404, A075405, A075406.

A180259 Squares which are the sum of consecutive squares starting with 25^2.

Original entry on oeis.org

625, 33124, 38025, 127449, 64529089, 81180100, 15884821225, 3370675683600

Offset: 1

Zhining Yang, Jan 17 2011

That is, terms are squares of the form sum_{i=25..m} i^2 = (m-24) *(2*m^2+51*m+1225) / 6 for some m. Known solutions refer to m = 25, 48, 50, 73, 578, 624, 3625 and 21624, and no further in the range m <= 70000000.

This sequence is complete. See A180442 and A184763.

			38025 is in the sequence because 38025 = 195^2 = 25^2 + 26^2 + ... + 50^2.

Cf. A001032, A094196, A151557, A180273.

Select[Accumulate[Range[25, 22000]^2], IntegerQ[Sqrt[#]] &] (* Harvey P. Dale, Aug 10 2023 *)

for(n=26,9999999,t=n*(n+1)*(2*n+1)/6-4900;if(issquare(t),print1(t,",")))

A180273 Squares which are a sum of consecutive squares starting with 7^2.

Original entry on oeis.org

49, 8464, 20449, 60025, 2304324, 3624882849

Offset: 1

Zhining Yang, Jan 17 2011

That is, terms are squares of the form sum_{i=7..m} i^2 for some m. This sequence is complete. See A180442 and A184763.

			a(2) = 8464 = 92^2 = 7^2+8^2+9^2+...+28^2+29^2.

Cf. A001032, A094196, A151557, A180259.

Select[Accumulate[Range[7,3000]^2],IntegerQ[Sqrt[#]]&] (* Harvey P. Dale, Jul 16 2014 *)

for(n=8,9999999,t=n*(n+1)*(2*n+1)/6-91;if(issquare(t),print1(t,",")))

A184885 Irregular triangle in which row n has the values of k>1 such that sum_{i=n..n+k-1} i^2 is a square.

Original entry on oeis.org

24, 2, 578, 961, 23, 33, 50, 184, 2209, 24, 552049, 22898, 96, 97, 312, 23, 5329, 11, 2, 24, 289, 1221025, 96, 59, 6889, 24, 26, 49, 554, 600, 3601, 21600, 33, 338, 50, 96, 169, 578, 4056, 61250, 148825, 11, 59, 312, 649, 11881, 24, 50, 122, 15625, 96, 338, 3479, 2075, 33, 3179, 242, 218, 864, 50, 8450, 122, 22801, 36481, 24, 194, 25921, 50, 242

Offset: 1

T. D. Noe, Jan 24 2011

That is, this sequence gives the number of squares in the sums described in A184763. Sequence A184762 gives the length of row n. A180442 lists the nonempty rows. All these numbers must appear in A001032.

			The triangle is
(row 1)  24
(row 3)  2, 578, 961
(row 7)  23, 33, 50, 184, 2209
(row 9)  24, 552049
(row 11) 22898
(row 13) 96
(row 15) 97, 312

Row n = row n of A184763 minus n-1.

A184886 Irregular triangle of the square root of the sums of squares mentioned in A184763.

Original entry on oeis.org

70, 5, 8075, 17267, 92, 143, 245, 1518, 60207, 106, 236818619, 2001863, 652, 679, 3406, 138, 225643, 77, 29, 158, 3128, 778998480, 724, 413, 331668, 182, 195, 357, 8033, 9010, 126035, 1835940, 253, 4017, 385, 788, 1612, 8687, 150878, 8758575, 33158210, 143, 531, 3770, 10384, 751228, 274, 495, 1281, 1133000, 1012, 4433, 121268, 56855, 440, 106403, 3069, 2725, 16332, 655, 453765, 1525, 1997277, 4035066, 430, 2619, 2420957, 795, 3465

Offset: 1

T. D. Noe, Jan 24 2011

Sequence A184762 gives the length of row n. A180442 lists the nonempty rows. These numbers are in Table 3 of the paper by Bremner, Stroeker, and Tzanakis.

			The triangle is
(row 1)  70
(row 3)  5, 8075, 17267
(row 7)  92, 143, 245, 1518, 60207
(row 9)  106, 236818619
(row 11) 2001863
(row 13) 652
(row 15) 679, 3406

A. Bremner, R. J. Stroeker, N. Tzanakis, On Sums of Consecutive Squares, J. Number Theory 62 (1997), 39-70.

A180465 Squares which are a sum of consecutive squares starting with 38^2.

Original entry on oeis.org

1444, 20449, 281961, 14212900, 107827456, 564343507984

Offset: 1

Zhining Yang, Jan 19 2011

This sequence is complete. See A180442 and A184763.

Cf. A001032, A094196, A151557, A180259, A180273

for(n=38, 9999999, t=n*(n+1)*(2*n+1)/6-17575; if(issquare(t), print1(t, ", ")))

cp's OEIS Frontend

A001032 Numbers k such that sum of squares of k consecutive integers >= 1 is a square.

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A184763 Irregular triangle in which row n has the values of k>n such that Sum_{i=n..k} i^2 is a square.

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A075404 Smallest m > n such that Sum_{i=n..m} i^2 is a square, or 0 if no such m exists.

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A184762 The number of numbers k > n such that Sum_{i=n..k} i^2 is a square.

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A180259 Squares which are the sum of consecutive squares starting with 25^2.

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A180273 Squares which are a sum of consecutive squares starting with 7^2.

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A184885 Irregular triangle in which row n has the values of k>1 such that sum_{i=n..n+k-1} i^2 is a square.

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A184886 Irregular triangle of the square root of the sums of squares mentioned in A184763.

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A180465 Squares which are a sum of consecutive squares starting with 38^2.

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