A001032 -id:A001032 - cp's OEIS frontend

A076215 Square root of sum defined in A007475(n) and A001032(n).

1, 5, 77, 92, 70, 195, 143, 3854, 357, 245, 413, 4088, 2257, 2222, 652, 679, 278949, 3366, 1281, 67963, 1612, 8555, 1518, 63412, 1159158, 2619, 2725, 13862, 60973, 3069, 10790, 3128, 4620, 5083, 42918, 3406

Offset: 1

Ralf Stephan, Nov 03 2002

nonn

6a(n)^2 is divisible by A001032(n). Proof: Let s = A007475(n), n = A001032(n), then a(n)^2 = sum(k=s, s+n-1, k^2) = n/6*(2n^2+(6s-3)n+6s^2-6s+1).

			A001032(3)=11, A007475(3)=18, so 18^2+19^2+...+28^2 (11 terms) = 77^2.

Christopher E. Thompson, Table of n, a(n) for n = 1..10438

Cf. A001032, A007475.

Offset corrected to 1 (to match A001032) by M. F. Hasler, Feb 02 2016

A007475 a(n) is the smallest positive number such that the sum of A001032(n) consecutive squares starting with a(n)^2 is a square.

Original entry on oeis.org

1, 3, 18, 7, 1, 25, 7, 539, 25, 7, 22, 442, 225, 192, 13, 15, 26914, 244, 50, 5552, 30, 553, 7, 4493, 83342, 83, 65, 775, 3807, 64, 556, 20, 106, 132, 2277, 15, 1788, 5063, 27, 11320, 280, 358, 1805, 210, 9985, 802, 183, 71752, 10123, 16806, 94707486, 1081

Offset: 1

N. J. A. Sloane, Mira Bernstein

nonn

			a(3)=18 because A001032(3)=11 and the sum of 11 squares 18^2 + 19^2 + ... + 28^2 = 77^2.

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
Laurent Beeckmans, Squares Expressible as Sum of Consecutive Squares, Am. Math. Monthly, Volume 101, Number 5, pp. 437-442, May 1994.
Index entries for sequences related to sums of squares

Cf. A001032.

Square roots of sum are in A076215.

Better description and more terms from Ralf Stephan, Nov 03 2002
Corrected by T. D. Noe, Aug 25 2004
Offset corrected to 1 by M. F. Hasler, Feb 02 2016

A274469 Numbers missing from A001032 despite satisfying the necessary congruence conditions (see comments).

Original entry on oeis.org

25, 842, 2306, 2402, 2459, 3602, 3650, 3803, 6081, 6242, 6338, 6779, 7058, 7319, 7643, 8088, 8354, 8363, 8402, 8543, 8761, 9122, 10607, 10826, 11257, 11378, 11447, 12203, 12458, 12722, 12984, 13273, 13682, 14162, 14424, 14639, 14738, 15362, 15626, 15698, 16475, 16634

Offset: 1

Christopher E. Thompson, Jun 24 2016

nonn

A001032 consists of those n for which there is a sequence of n consecutive positive squares whose sum is a square. For the associated Pellian equation, see A134419. The necessary congruence conditions described in A274471 apply here:
  (defining x||y to mean x|y and x and y/x are coprime)
  if 3^e||n with e>0, then e is odd and (n/3^e)=2 (mod 3);
  if p^e||n with p=5 or 7 (mod 12), then e is even;
  if 3^e||(n+1) with e>0, then e is odd;
  if p^e||(n+1) with p=3 (mod 4) and p>3, then e is even.
In addition, in order that the Pellian equation has solutions of the correct parity, one must have:
  if 2^e||n with e>0, then e is odd.
However, these conditions are not sufficient. This sequence consists of the numbers n that satisfy all the congruence conditions but for which there is no sequence of n consecutive positive squares whose sum is a square.
The term 25 is present despite the Pellian equation having a solution with the correct parity, because it leads only to 0^2 + 1^2 + ... + 24^2 = 70^2 and the specification of A001032 requires the squares to be strictly positive. (One may wonder whether this is quite natural, compare A185545.) In every other case the Pellian equation lacks solutions with the right parity. Note however that it may still have solutions with the opposite parity (this can happen only if n=1 mod 8) and so this sequence is not a subsequence of A274471.

Christopher E. Thompson, Table of n, a(n) for n = 1..727 [values up to 250000]
H. T. Freitag, G. M. Phillips, On the sum of consecutive squares, Proc. 6th Intl Res. Conf. Fib. Numbers, Pullman, July 18-22, 1994 (1996), p 137.

Cf. A001032, A134419, A274471.

A180274 Integers whose squares are the sums of 24 consecutive squares.

Original entry on oeis.org

70, 106, 158, 182, 274, 430, 650, 1022, 1546, 1786, 2702, 4250, 6430, 10114, 15302, 17678, 26746, 42070, 63650, 100118, 151474, 174994, 264758, 416450, 630070, 991066, 1499438, 1732262, 2620834, 4122430, 6237050, 9810542, 14842906, 17147626, 25943582

Offset: 1

Zhining Yang, Jan 17 2011

The corresponding starts of 24 consecutive squares to be summed are A094196.

Colin Barker, Table of n, a(n) for n = 1..1000
K. S. Brown, Sum of Consecutive Nth Powers Equals an Nth Power
Fun With Num3ers, Sets of 24 consecutive squares whose sum is a square, July 27 2016.
V. 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. See Table 2 p. 8.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,10,0,0,0,0,0,-1).

Cf. A094196.

Cf. A001032 (24 is a term of that sequence).

A094196 := proc(n) if n <= 12 then op(n,[1, 9, 20, 25, 44, 76, 121, 197, 304, 353, 540, 856]) ; else 10*procname(n-6)-procname(n-12)+92 ; end if ; end proc:
A180274 := proc(n) local a96 ; a96 := A094196(n) ; 24*a96^2+552*a96+4324 ; sqrt(%) ; end proc:
seq(A180274(n),n=1..30) ; # R. J. Mathar, Jan 20 2011

Select[Sqrt[#]&/@(Total[#]&/@Partition[Range[900000]^2, 24, 1]), IntegerQ] (* Harvey P. Dale, Jan 21 2011 *)
t={70, 106, 158, 182, 274, 430, 650, 1022, 1546, 1786, 2702, 4250}; Do[AppendTo[t, 10*t[[-6]] - t[[-12]]], {n, 13, 100}]; t

{ for(n=1,999999,t=((n+23)*(n+24)*(2*n+47)-n*(n-1)*(2*n-1))/6;if(issquare(t),print1(ceil(sqrt(t)),","))) }

Vec(-2*x*(25*x^11+19*x^10+17*x^9+17*x^8+19*x^7+25*x^6-215*x^5-137*x^4-91*x^3-79*x^2-53*x-35) / (x^12-10*x^6+1) + O(x^100)) \\ Colin Barker, May 09 2015

a(n) = +10*a(n-6) -a(n-12). G.f. ( 70+106*x+158*x^2+182*x^3+274*x^4+430*x^5-50*x^6-38*x^7-34*x^8-34*x^9-38*x^10-50*x^11 ) / ( 1-10*x^6+x^12 ). - Joerg Arndt, Jan 17 2011

a(n) = sqrt( 24*(A094196(n))^2 +552*A094196(n)+4324) . - R. J. Mathar, Jan 20 2011

A218395 Numbers whose square is the sum of the squares of 11 consecutive integers.

Original entry on oeis.org

11, 77, 143, 1529, 2849, 30503, 56837, 608531, 1133891, 12140117, 22620983, 242193809, 451285769, 4831736063, 9003094397, 96392527451, 179610602171, 1923018812957, 3583208949023, 38363983731689, 71484568378289, 765356655820823, 1426108158616757

Offset: 0

Paul Weisenhorn, Oct 28 2012

a(n)^2 = Sum_{j=0..10} (x(n)+j)^2 = 11*(x(n)+5)^2 + 110 and b(n) = x(n)+5 give the Pell equation a(n)^2 - 11*b(n)^2 = 110 with the 2 fundamental solutions (11; 1) and (77; 23) and the solution (10; 3) for the unit form. A198949(n+1) = b(n); A106521(n) = x(n) and x(0) = -4.

General: If the sum of the squares of c neighboring numbers is a square with c = 3*k^2-1 and 1 <= k, then a(n)^2 = Sum_{j=0..c-1} (x(n)+j)^2 and b(n) = 2*x(n)+c-1 give the Pell equation a(n)^2 - c*(b(n)/2)^2 = binomial(c+1,3)/2. a(n) = 2*e1*a(n-k) - a(n-2*k); b(n) = 2*e1*b(n-k) - b(n-2*k); a(n) = e1*a(n-k) + c*e2*b(n-k); b(n) = e2*a(n-k) + e1*b(n-k) with the solution (e1; e2) for the unit form.

			For n=6, Sum_{z=17132..17142} z^2 = 3230444569;
a(6) = sqrt(3230444569) = 56837;
b(6) = sqrt((a(6)^2-110)/11) = 17137; x(6) = b(6)-5 = 17132.

V. 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. See Table 1 p. 7.
Index entries for linear recurrences with constant coefficients, signature (0,20,0,-1).

c=2: A001653(n+1) = a(n); A002315(n) = b(n); A001652(n) = x(n).

Cf. A001032 (11 is a term of that sequence), A198947.

s:=0: n:=-1:
for j from -5 to 5 do s:=s+j^2: end do:
for z from -4 to 100000 do
  s:=s-(z-1)^2+(z+10)^2: r:=sqrt(s):
  if (r=floor(r)) then
    n:=n+1: a(n):=r: x(n):=z:
    b(n):=sqrt((s-110)/11):
    print(n,a(n),b(n),x(n)):
  end if:
end do:

LinearRecurrence[{0,20,0,-1},{11,77,143,1529},30] (* Harvey P. Dale, Aug 15 2022 *)

a(n) = 20*a(n-2) - a(n-4); b(n) = 20*b(n-2) - b(n-4);
a(n) = 10*a(n-2) + 33*b(n-2); b(n) = 3*a(n-2) + 10*b(n-2).
a(n) = a(n-1) + 20*a(n-2) - 20*a(n-3) - a(n-4) + a(n-5).
G.f.: 11 * (1-x)*(1+8*x+x^2) / (1 - 20*x^2 + x^4).
With r=sqrt(11); s=10+3*r; t=10-3*r:
a(2*n) = ((11+r)*s^n + (11-r)*t^n)/2.
a(2*n+1) = ((77+23*r) * s^n + (77-23*r)*t^n)/2.
a(n) = 11 * A198947(n+1). - Bill McEachen, Dec 01 2022

A134419 Numbers n for which the generalized Pell equation x^2 - n*y^2 = n(n-1)(n+1)/3 has an integer solution for x and y.

Original entry on oeis.org

1, 2, 4, 11, 16, 23, 24, 25, 26, 33, 47, 49, 50, 52, 59, 64, 73, 74, 88, 96, 97, 100, 107, 121, 122, 146, 148, 169, 177, 184, 191, 193, 194, 196, 218, 239, 241, 242, 244, 249, 256, 276, 289, 292, 297, 299, 311, 312, 313, 337, 338, 347, 352, 361, 362, 376, 383

Offset: 1

T. D. Noe, Oct 25 2007

nonn

This generalized Pell equation appears in the solution of problems posed in A001032 (and A001033): numbers n such that the sum of squares of n consecutive (odd) positive integers is a square. This sequence is the union of A001032, A001033 and the number 4, which is not a solution to either problem. When n is a square > 1 and not divisible by 3, then the equation has only a finite number of solutions; otherwise it has an infinite number of solutions.
For an n in this sequence, consider solutions with x>0 and y>n. (For n=4, there will be no such solutions.) If y-n+1 is even, then n is in A001032, the n consecutive positive integers begin with (y-n+1)/2 and the sum of the squares is x/2. If y-n+1 is odd, then the n is in A001033, the n consecutive odd positive integers begin with y-n+1 and the sum of the squares is x. For some n, such as 33, there are solutions y1 and y2 such that y1-n+1 is even and y2-n+1 is odd. In this case, n is in both A001032 and A001033.
The reason that 4 is not in A001033 is that there is no sequence of 4 consecutive positive odd squares that add to a square.  However, there is a sequence of 4 consecutive odd integers whose squares add up to a square:  (-1)^2 + 1^2 + 3^2 + 5^2 = 6^2. - Thomas Andrews, Feb 22 2011

Christopher E. Thompson, Table of n, a(n) for n = 1..13437 (up to 250000, extends first 200 terms computed by T. D. Noe).

Cf. A001032, A001033, A274471.

t={}; n=0; While[Length[t]<200, n++; If[Reduce[x^2-n*y^2==n(n^2-1)/3, {x,y}, Integers] =!= False, AppendTo[t,n]]]; t

A257761 Positive integers whose square is the sum of 23 consecutive squares.

Original entry on oeis.org

92, 138, 4278, 6532, 205252, 313398, 9847818, 15036572, 472490012, 721442058, 22669672758, 34614182212, 1087671802372, 1660759304118, 52185576841098, 79681832415452, 2503820016570332, 3823067196637578, 120131175218534838, 183427543606188292

Offset: 1

Colin Barker, May 07 2015

Positive integers x in the solutions to 2*x^2-46*y^2-1012*y-7590 = 0.

			92 is in the sequence because 92^2 = 8464 = 7^2+8^2+...+29^2.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,48,0,-1).

Cf. A001032, A001653, A180274, A218395, A257765, A257767, A257780, A257781, A257823-A257828.

I:=[92,138,4278,6532]; [n le 4 select I[n] else 48*Self(n-2)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, May 11 2015

LinearRecurrence[{0, 48, 0, -1}, {92, 138, 4278, 6532}, 30] (* Vincenzo Librandi, May 11 2015 *)

Vec(-46*x*(x-1)*(x+2)*(2*x+1)/(x^4-48*x^2+1) + O(x^100))

a(n) = 48*a(n-2)-a(n-4).

G.f.: -46*x*(x-1)*(x+2)*(2*x+1) / (x^4-48*x^2+1).

A257765 Positive integers whose square is the sum of 26 consecutive squares.

Original entry on oeis.org

195, 1599, 2379, 19695, 163059, 242619, 2008695, 16630419, 24744759, 204867195, 1696139679, 2523722799, 20894445195, 172989616839, 257394980739, 2131028542695, 17643244777899, 26251764312579, 217344016909695, 1799437977728859, 2677422564902319

Offset: 1

Colin Barker, May 07 2015

Positive integers x in the solutions to 2*x^2-52*y^2-1300*y-11050 = 0.

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

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,102,0,0,-1).

Cf. A001032, A001653, A180274, A218395, A257761, A257767, A257780, A257781, A257823-A257828.

I:=[195,1599,2379,19695,163059,242619 ]; [n le 6 select I[n] else 102*Self(n-3)-Self(n-6): n in [1..30]]; // Vincenzo Librandi, May 11 2015

LinearRecurrence[{0, 0, 102, 0, 0, -1}, {195, 1599, 2379, 19695, 163059, 242619}, 30] (* Vincenzo Librandi, May 11 2015 *)
Select[Sqrt[#]&/@Total/@Partition[Range[10^6]^2,26,1],IntegerQ] (* The program generates the first 7 terms of the sequence. *) (* Harvey P. Dale, Mar 10 2024 *)

Vec(-39*x*(x^5+x^4+5*x^3-61*x^2-41*x-5) / (x^6-102*x^3+1) + O(x^100))

a(n) = 102*a(n-3)-a(n-6).

G.f.: -39*x*(x^5+x^4+5*x^3-61*x^2-41*x-5) / (x^6-102*x^3+1).

A257767 Positive integers whose square is the sum of 33 consecutive squares.

Original entry on oeis.org

143, 253, 440, 1133, 1397, 3608, 6325, 11495, 20152, 52063, 64207, 165880, 290807, 528517, 926552, 2393765, 2952125, 7626872, 13370797, 24300287, 42601240, 110061127, 135733543, 350670232, 614765855, 1117284685, 1958730488, 5060418077, 6240790853

Offset: 1

Colin Barker, May 07 2015

Positive integers x in the solutions to 2*x^2-66*y^2-2112*y-22880 = 0.

			143 is in the sequence because 143^2 = 20449 = 7^2+8^2+...+39^2.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,46,0,0,0,0,0,-1).

Cf. A001032, A001653, A180274, A218395, A257761, A257765, A257780, A257781, A257823-A257828.

I:=[143,253,440,1133,1397,3608,6325,11495,20152, 52063,64207,165880]; [n le 12 select I[n] else 46*Self(n-6)-Self(n-12): n in [1..30]]; // Vincenzo Librandi, May 11 2015

LinearRecurrence[{0, 0, 0, 0, 0, 46, 0, 0, 0, 0, 0, -1}, {143, 253, 440, 1133, 1397, 3608, 6325, 11495, 20152, 52063, 64207, 165880}, 50] (* Vincenzo Librandi, May 08 2015 *)

Vec(-11*x*(8*x^11+5*x^10+5*x^9+8*x^8+13*x^7+23*x^6-328*x^5-127*x^4-103*x^3-40*x^2-23*x-13) / (x^12-46*x^6+1) + O(x^100))

a(n) = 46*a(n-6)-a(n-12).

G.f.: -11*x*(8*x^11+5*x^10+5*x^9+8*x^8+13*x^7+23*x^6-328*x^5-127*x^4-103*x^3-40*x^2-23*x-13) / (x^12-46*x^6+1).

A097812 Numbers n such that n^2 is the sum of two or more consecutive positive squares.

Original entry on oeis.org

5, 29, 70, 77, 92, 106, 138, 143, 158, 169, 182, 195, 245, 253, 274, 357, 385, 413, 430, 440, 495, 531, 650, 652, 655, 679, 724, 788, 795, 985, 1012, 1022, 1055, 1133, 1281, 1365, 1397, 1518, 1525, 1529, 1546, 1599, 1612, 1786, 1828, 2205, 2222, 2257, 2372

Offset: 1

T. D. Noe, Aug 25 2004

nonn

These numbers were found by exhaustive search. The sums are not unique; for n = 143, there are two representations. The Mathematica code prints n, the range of squares in the sum and the number of squares in the sum. Because the search included sums of all squares up to 2000, this sequence is complete up to 2828.

			29 is in this sequence because 20^2 + 21^2 = 29^2.
Contribution from _Donovan Johnson_, Feb 19 2011: (Start)
For seven terms < (10^15)^(1/2), the square is a sum in two different ways:
143^2 = 7^2 + ... + 39^2 = 38^2 + ... + 48^2.
2849^2 = 294^2 + ... + 367^2 = 854^2 + ... + 864^2.
208395^2 = 2175^2 + ... + 5199^2 = 29447^2 + ... + 29496^2.
2259257^2 = 9401^2 + ... + 25273^2 = 26181^2 + ... + 32158^2.
6555549^2 = 41794^2 + ... + 58667^2 = 87466^2 + ... + 92756^2.
11818136^2 = 10898^2 + ... + 74906^2 = 29929^2 + ... + 76392^2.
19751043^2 = 39301^2 + ... + 107173^2 = 249217^2 + ... + 255345^2. (End)

Donovan Johnson, Table of n, a(n) for n = 1..5077
K. S. Brown, Sum of Consecutive Nth Powers Equals an Nth Power
Index entries for sequences related to sums of squares

Cf. A097811 (n^3 is the sum of consecutive cubes).

Cf. A001032, A151557.

g[m0_, m1_] := (1 - m0 + m1)(-m0 + 2m0^2 + m1 + 2m0 m1 + 2m1^2)/6; A097812 = {}; Do[n = g[m0, m1]^(1/2); If[IntegerQ[n], Print[{n, m0, m1, m1 - m0 + 1}]; AppendTo[A097812, n]], {m1, 2, 2000}, {m0, m1 - 1, 1, -1}]; Union[A097812]

Name edited by Altug Alkan, Dec 07 2015

cp's OEIS Frontend

A076215 Square root of sum defined in A007475(n) and A001032(n).

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A007475 a(n) is the smallest positive number such that the sum of A001032(n) consecutive squares starting with a(n)^2 is a square.

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A274469 Numbers missing from A001032 despite satisfying the necessary congruence conditions (see comments).

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A180274 Integers whose squares are the sums of 24 consecutive squares.

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A218395 Numbers whose square is the sum of the squares of 11 consecutive integers.

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A134419 Numbers n for which the generalized Pell equation x^2 - n*y^2 = n(n-1)(n+1)/3 has an integer solution for x and y.

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A257761 Positive integers whose square is the sum of 23 consecutive squares.

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A257765 Positive integers whose square is the sum of 26 consecutive squares.

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