A106521 -id:A106521 - cp's OEIS frontend

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

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

A094196 Indices of the start of a string of 24 consecutive squares whose sum is a square.

Original entry on oeis.org

1, 9, 20, 25, 44, 76, 121, 197, 304, 353, 540, 856, 1301, 2053, 3112, 3597, 5448, 8576, 12981, 20425, 30908, 35709, 54032, 84996, 128601, 202289, 306060, 353585, 534964, 841476, 1273121, 2002557, 3029784, 3500233, 5295700, 8329856, 12602701

Offset: 1

Lekraj Beedassy, May 25 2004

The sequence could also include -11, -8 and -4; and if N is in the sequence, then so is -23-N.
Equivalently, 24*a(n)^2 + 552*a(n) + 4324 is a square.
All sequences of this type (i.e., sequences with fixed offset k, and a discernible pattern: k=0...23 for this sequence, k=0...22 for A269447, k=0..1 for A001652) can be extended using a formula such as x(n) = a*x(n-p) - x(n-2p) + b, where a and b are various constants, and p is the period of the series. Alternatively, 'p' can be considered the number of concurrent series. - Daniel Mondot, Aug 05 2016

Daniel Mondot, Table of n, a(n) for n = 1..106
K. S. Brown, Sum of Consecutive Nth Powers Equals an Nth Power
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,10,-10,0,0,0,0,-1,1).

Cf. A001032, A001652, A106521, A269447.

LinearRecurrence[{1,0,0,0,0,10,-10,0,0,0,0,-1,1},{1,9,20,25,44,76,121,197,304,353,540,856,1301},60] (* Harvey P. Dale, Oct 10 2011 *)

for(n=1,15000000,if(issquare(sum(j=n,n+23,j^2)),print1(n,","))) \\ Klaus Brockhaus, Jun 01 2004

Recurrence: a(n+12) = 10a(n+6) - a(n) + 92.
O.g.f.: x*(-1-8*x-11*x^2-5*x^3-19*x^4-32*x^5-35*x^6+4*x^7+3*x^8+x^9+3*x^10+4*x^11+4*x^12) / ((-1+x) * (1-10*x^6+x^12)). - R. J. Mathar, Dec 02 2007
a(0)=1, a(1)=9, a(2)=20, a(3)=25, a(4)=44, a(5)=76, a(6)=121, a(7)=197, a(8)=304, a(9)=353, a(10)=540, a(11)=856, a(12)=1301; thereafter a(n) = a(n-1)+10*a(n-6)-10*a(n-7)-a(n-12)+a(n-13). - Harvey P. Dale, Oct 10 2011
a(1)=1, a(2)=9, a(3)=20, a(4)=25, a(5)=44, a(6)=76, a(7)=121, a(8)=197, a(9)=304, a(10)=353, a(11)=540, a(12)=856; a(n)=10*a(n-6)-a(n-12) + 92 for n>12. - Daniel Mondot, Aug 05 2016

More terms from Don Reble (djr(AT)hotmail.com) and Klaus Brockhaus, Jun 01 2004

A269447 The first of 23 consecutive positive integers the sum of the squares of which is a square.

Original entry on oeis.org

7, 17, 881, 1351, 42787, 65337, 2053401, 3135331, 98520967, 150431057, 4726953521, 7217555911, 226795248547, 346292253177, 10881444977241, 16614810597091, 522082563659527, 797164616407697, 25049081610680561, 38247286776972871, 1201833834749007907

Offset: 1

Colin Barker, Feb 27 2016

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

All sequences of this type (i.e. sequences with fixed offset k, and a discernible pattern: k=0...22 for this sequence, k=0..1 for A001652, k=0...10 for A106521) can be continued using a formula such as x(n) = a*x(n-p) - x(n-2p) + b, where a and b are various constants, and p is the period of the series. Alternatively 'p' can be considered the number of concurrent series. - Daniel Mondot, Aug 05 2016

			7 is in the sequence because sum(k=7, 29, k^2) = 8464 = 92^2.

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

Cf. A001032, A001652, A094196, A106521, A257761, A269448, A269449, A269451.

LinearRecurrence[{1,48,-48,-1,1},{7,17,881,1351,42787},30] (* Harvey P. Dale, May 21 2024 *)

Vec(x*(7+10*x+528*x^2-10*x^3-29*x^4)/((1-x)*(1-48*x^2+x^4)) + O(x^30))

a(n) = a(n-1)+48*a(n-2)-48*a(n-3)-a(n-4)+a(n-5) for n>5.
G.f.: x*(7+10*x+528*x^2-10*x^3-29*x^4) / ((1-x)*(1-48*x^2+x^4)).
a(1)=7, a(2)=17, a(3)=881, a(4)=1351, a(n) = 48*a(n-2)-a(n-4)+506. - Daniel Mondot, Aug 05 2016

A269448 The first of 26 consecutive positive integers the sum of the squares of which is a square.

Original entry on oeis.org

25, 301, 454, 3850, 31966, 47569, 393925, 3261481, 4852834, 40177750, 332640346, 494942749, 4097737825, 33926055061, 50479308814, 417929081650, 3460124977126, 5148394557529, 42624668591725, 352898821613041, 525085765560394, 4347298267275550

Offset: 1

Colin Barker, Feb 27 2016

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

All sequences of this type (i.e. sequences with fixed offset k, and a discernible pattern: k=0...25 for this sequence, k=0...22 for A269447, k=0..1 for A001652) can be continued using a formula such as x(n) = a*x(n-p) - x(n-2p) + b, where a and b are various constants, and p is the period of the series. Alternatively 'p' can be considered the number of concurrent series. - Daniel Mondot, Aug 05 2016

			25 is in the sequence because sum(k=25, 50, k^2) = 38025 = 195^2.

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

Cf. A001032, A001652, A094196, A106521, A257765, A269447, A269449, A269451.

Rest@ CoefficientList[Series[x (25 + 276 x + 153 x^2 + 846 x^3 - 36 x^4 - 3 x^5 - 11 x^6)/((1 - x) (1 - 102 x^3 + x^6)), {x, 0, 22}], x] (* Michael De Vlieger, Aug 07 2016 *)

Vec(x*(25+276*x+153*x^2+846*x^3-36*x^4-3*x^5-11*x^6)/((1-x)*(1-102*x^3+x^6)) + O(x^30))

G.f.: x*(25+276*x+153*x^2+846*x^3-36*x^4-3*x^5-11*x^6) / ((1-x)*(1-102*x^3+x^6)).

a(1)=25, a(2)=301, a(3)=454, a(4)=3850, a(5)=31966, a(6)=47569, a(n)=102*a(n-3) - a(n-6) + 1250. - Daniel Mondot, Aug 05 2016

A269449 The first of 33 consecutive positive integers the sum of the squares of which is a square.

Original entry on oeis.org

7, 27, 60, 181, 227, 612, 1085, 1985, 3492, 9047, 11161, 28860, 50607, 91987, 161276, 416685, 513883, 1327652, 2327541, 4230121, 7415908, 19159167, 23628161, 61043836, 107016983, 194494283, 340971196, 880905701, 1086382227, 2806689508, 4920454381, 8942507601

Offset: 1

Colin Barker, Feb 27 2016

Positive integers y in the solutions to 2*x^2-66*y^2-2112*y-22880 = 0.
All sequences of this type (i.e. sequences with fixed offset k, and a discernible pattern: k=0...32 for this sequence, k=0..1 for A001652, k=0...10 for A106521) can be extended using a formula such as x(n) = a*x(n-p) - x(n-2p) + b, where a and b are various constants, and p is the period of the series. Alternatively 'p' can be considered the number of concurrent series. - Daniel Mondot, Aug 08 2016
Numbers x such that 11440+33*x*(32+x)is a square. - Harvey P. Dale, Oct 18 2020

			7 is in the sequence because sum(k=7, 39, k^2) = 20449 = 143^2.

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

Cf. A001032, A001652, A094196, A106521, A257767, A269447, A269448, A269451.

Rest@ CoefficientList[Series[x (7 + 20 x + 33 x^2 + 121 x^3 + 46 x^4 + 385 x^5 + 151 x^6 - 20 x^7 - 11 x^8 - 11 x^9 - 2 x^10 - 11 x^11 - 4 x^12)/((1 - x) (1 - 46 x^6 + x^12)), {x, 0, 32}], x] (* Michael De Vlieger, Aug 08 2016 *)
LinearRecurrence[{1,0,0,0,0,46,-46,0,0,0,0,-1,1},{7,27,60,181,227,612,1085,1985,3492,9047,11161,28860,50607},50] (* Harvey P. Dale, Oct 18 2020 *)

Vec(x*(7 +20*x +33*x^2 +121*x^3 +46*x^4 +385*x^5 +151*x^6 -20*x^7 -11*x^8 -11*x^9 -2*x^10 -11*x^11 -4*x^12) / ((1 -x)*(1 -46*x^6 +x^12)) + O(x^40))

G.f.: x*(7 +20*x +33*x^2 +121*x^3 +46*x^4 +385*x^5 +151*x^6 -20*x^7 -11*x^8 -11*x^9 -2*x^10 -11*x^11 -4*x^12) / ((1 -x)*(1 -46*x^6 +x^12)).

a(1)=7, a(2)=27, a(3)=60, a(4)=181, a(5)=227, a(6)=612, a(7)=1085, a(8)=1985, a(9)=3492, a(10)=9047, a(11)=11161, a(12)=28860, a(n)=46*a(n-6)-a(n-12)+704. - Daniel Mondot, Aug 08 2016

A269451 The first of 50 consecutive positive integers the sum of the squares of which is a square.

Original entry on oeis.org

7, 28, 44, 67, 87, 124, 168, 287, 379, 512, 628, 843, 1099, 1792, 2328, 3103, 3779, 5032, 6524, 10563, 13687, 18204, 22144, 29447, 38143, 61684, 79892, 106219, 129183, 171748, 222432, 359639, 465763, 619208, 753052, 1001139, 1296547, 2096248, 2714784

Offset: 1

Colin Barker, Feb 27 2016

Positive integers y in the solutions to 2*x^2-100*y^2-4900*y-80850 = 0.

Numbers n such that 40425 + 2450*n + 50*n^2 is a square. - Harvey P. Dale, Oct 22 2016

			7 is in the sequence because sum(k=7, 56, k^2) = 60025 = 245^2.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,6,-6,0,0,0,0,-1,1).

Cf. A001032, A001652, A094196, A106521, A257781, A269447, A269448, A269449.

Select[Range[3*10^6],IntegerQ[Sqrt[40425+2450#+50#^2]]&] (* or *) LinearRecurrence[ {1,0,0,0,0,6,-6,0,0,0,0,-1,1},{7,28,44,67,87,124,168,287,379,512,628,843,1099},40] (* Harvey P. Dale, Oct 22 2016 *)

Vec(x*(7+21*x+16*x^2+23*x^3+20*x^4+37*x^5+2*x^6-7*x^7-4*x^8-5*x^9-4*x^10-7*x^11-x^12) / ((1-x)*(1+2*x^3-x^6)*(1-2*x^3-x^6)) + O(x^40))

G.f.: x*(7+21*x+16*x^2+23*x^3+20*x^4+37*x^5+2*x^6-7*x^7-4*x^8-5*x^9-4*x^10-7*x^11-x^12) / ((1-x)*(1+2*x^3-x^6)*(1-2*x^3-x^6)).

A350886 Let X,Y,Z be positive integer solutions to X^2 = Sum_{j=0..Y-1} (1+Z*j)^2, where solutions for Y or Z < 1 are excluded. This sequence lists the sorted values for X.

Original entry on oeis.org

54, 70, 1618, 2344, 2541, 27597, 48486, 73795, 184162, 320739, 648009, 766669, 990983, 1452962, 3816551, 4456264, 6287116, 23251921, 37396339, 43540374, 51136014, 53005618, 63668661, 147115419, 205943541, 236317895, 253970684, 275914803, 386480829, 629467300

Offset: 1

Thomas Scheuerle, Feb 25 2022

nonn

A generalization of the cannonball problem for pyramids with a slope of 1/A350888(n). In the cannonball problem, a square pyramid of stacked balls shall contain a square number of balls in total. Each layer of such a pyramid consists of a square number of balls, in the classic case the top layer has one ball, the layers below contain adjacent square numbers of balls. For this sequence we ignore the fact that if adjacent layers do not alternate between odd and even squares the stacking will become difficult at least for sphere-like objects. We start in the top layer always with one ball, but will skip a constant count of square numbers between each layer. This results in pyramids which slope <= 1.

a(n) may be interpreted as the length of a Pythagorean vector with gcd = 1 (over all coordinates) and no duplicate coordinate values. Such vectors may have applications in the theory of lattices.

			a(1) = 54 and A350887(1) = 4, A350888(1) = 14:
   54^2 = 1^2 + 15^2 + 29^2 + 43^2.
a(2) = 70 and  A350887(2) = 24, A350888(2) = 1:
   70^2 = 1^2 + 2^2 + 3^2 + ... + 23^2 + 24^2. This is the classic solution for the cannonball problem.

Thomas Scheuerle, Some solutions to this problem sorted by A350887.
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.
Thomas Scheuerle, Recursive solution formulas.
Index entries for sequences related to sums of squares.

Cf. A350887 (number of layers), A350888 (denominator of slope).
Cf. A000330, A000447, A024215, A024381 (square pyramidal numbers for slope 1, 1/2, 1/3, 1/4).
Cf. A076215, A001032, A134419, A106521. Some related problems.

sqtest(n,c)={q=1; for(t=2, c, t+=n; q+=(t*t); if(issquare(q), break)); q}
z=500000; a=[];for(n=0,z,r=sqtest(n,z); if(issquare(r), a=concat(a, sqrtint(r)))); a=vecsort(a) \\ Last valid value for z=500000 is 990983.

a(n)^2 = A350888(n)^2*binomial(2*A350887(n), 3)/4 + 2*A350888(n)*binomial(A350887(n), 2) + A350887(n).
a(n)^2 = c*((b*(4*b*c^2-(6*c-2)*b + 12*(c-1)) + 12)/12), with c = A350887(n) and b = A350888(n). Expanded to see factors more clearly.
a(n)^2 = c*b^2*( ((1/b) + (c-1)/2)^2 + (c^2-1)/12 ). Shorter form of above.
(12*a(n)^2) mod A350887(n) = 0.
((12*a(n)^2/A350887(n)) - 12) mod A350888(n) = 0.
Choose n such that A350887(n) = 4 and a(n) = 54 and A350888(n) = 14, then we may find further solutions recursively for all A350887(m) = 4 with
  x = -A350888(n) = -14; y = -a(n) = -54 and also x = A350888(n) = 14; y = a(n) = 54. Recursive solutions:
  x_(n+1) = 15*x_n + 4*y_n + 6
  y_(n+1) = 56*x_n + 15*y_n + 24 and also:
  x_(n+1) = 15*x_n - 4*y_n + 6
  y_(n+1) = -56*x_n + 15*y_n - 24.
Choose n such that A350887(n) = 9 and a(n) = 27597 and A350888(n) = 1932, then we may find further solutions recursively for all A350887(m) = 9 with x = -A350888(n) = -1932; y = -a(n) = -27597 and also x = A350888(n) = 1932; y = a(n) = 27597. Recursive solutions:
  x_(n+1) = 4999*x_n + 350*y_n + 882
  y_(n+1) = 71400*x_n + 4999*y_n + 12600 and also:
  x_(n+1) = 4999*x_n - 350*y_n + 6
  y_(n+1) = -71400*x_n + 4999*y_n - 12600.
Further recursive solution formulas for other values of A350887(n) will be provided in a link as for some values the coefficients become very large sometimes with several hundred digits.
a(n) != a(m) if n != m.

A350887 Let X,Y,Z be positive integer solutions to X^2 = Sum_{j=0..Y-1} (1+Z*j)^2, where solutions for Y or Z < 1 are excluded. This sequence lists the values for Y sorted by X.

Original entry on oeis.org

4, 24, 4, 64, 49, 9, 4, 50, 484, 3249, 81, 361, 49, 4, 289, 64, 16, 3938, 5041, 4, 36, 568, 441, 121, 4761, 33, 1936, 9, 49, 25872, 4, 64, 8257, 24, 361, 12024

Offset: 1

Thomas Scheuerle, Feb 25 2022

A generalization of the cannonball problem for pyramids with a slope of 1/A350888(n). In the cannonball problem, a square pyramid of stacked balls shall contain a square number of balls in total. Each layer of such a pyramid consists of a square number of balls, in the classic case the top layer has one ball, the layers below contain adjacent square numbers of balls. For this sequence we ignore the fact that if adjacent layers do not alternate between odd and even squares the stacking will become difficult at least for sphere-like objects. We start in the top layer always with one ball, but will skip a constant count of square numbers between each layer. This results in pyramids which slope <= 1. a(n) is the number of layers needed to stack such a pyramid. As this sequence is only a list of solutions, n has no known relation to its definition.

This sequence contains only numbers which appear in A186699, too. A number which is in A186699, does not appear in this sequence if it is of the form: 2^m*p where p is a prime of the form 12*k+1 or 12*k-1. There are also some perfect squares like 25 excluded, as these would only provide valid solutions in the case of A350888(n) = 0, which is not part of the sequence definition.

			a(1) = 4 and A350886(1) = 54, A350888(1) = 14: 54^2 = 1^2 + 15^2 + 29^2 + 43^2.
a(2) = 24 and A350886(2) = 70, A350888(2) = 1: 70^2 = 1^2 + 2^2 + 3^2 + ... + 23^2 + 24^2. This is the classic solution for the cannonball problem.

Thomas Scheuerle, Some solutions to this problem sorted by a(n).
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.
Thomas Scheuerle, Recursive solution formulas.
Index entries for sequences related to sums of squares

Cf. A350886 (squareroot of pyramid), A350888 (denominator of slope).
Cf. A000330, A000447, A024215, A024381 (square pyramidal numbers for slope 1, 1/2, 1/3, 1/4).
Cf. A076215, A001032, A134419, A106521. Some related problems.
Cf. A186699.

sqtest(n, c)={q=1; r=1; for(t=2, c, t+=n; r+=1; q+=(t*t); if(issquare(q), break)); [q, r]}
z=500000; b=[;]; for(n=0,z,r=sqtest(n,z); if(issquare(r[1]), b=concat(b,[sqrtint(r[1]); r[2]]))); b=vecsort(b,1); vector(#b, k, b[2, k]) \\ Last valid value for z=500000 is 49.

A350886(n)^2 = A350888(n)^2*binomial(2*a(n), 3)/4 + 2*A350888(n)*binomial(a(n), 2) + a(n).
A350886(n)^2 = c*((b*(4*b*c^2-(6*c-2)*b + 12*(c-1)) + 12)/12), with c = a(n) and b = A350888(n). Expanded to see factors more clearly.
A350886(n)^2 = c*b^2*( ((1/b) + (c-1)/2)^2 + (c^2-1)/12 ). Shorter form of above.

A350888 Let X,Y,Z be positive integer solutions to X^2 = Sum_{j=0..Y-1} (1+Z*j)^2, where solutions for Y or Z < 1 are excluded. This sequence lists the values for Z sorted by X.

Original entry on oeis.org

14, 1, 432, 8, 13, 1932, 12958, 367, 30, 3, 1554, 194, 5082, 388320, 1349, 15254, 178542, 163, 181, 11636654, 418782, 6791, 11928, 192638, 1086, 2209447, 5166, 19317900, 1981979, 262, 348711312, 4799102, 7379, 60240793

Offset: 1

Thomas Scheuerle, Feb 25 2022

This is a generalization of the cannonball problem for pyramids with a slope of 1/a(n). In the cannonball problem, a square pyramid of stacked balls shall contain a square number of balls in total. Each layer of such a pyramid consists of a square number of balls, in the classic case the top layer has one ball, the layers below contain adjacent square numbers of balls. For this sequence we ignore the fact that if adjacent layers do not alternate between odd and even squares the stacking will become difficult at least for sphere-like objects. We start in the top layer always with one ball, but will skip a constant count of square numbers between each layer. This results in pyramids which slope <= 1. A350887(n) is the number of layers needed to stack such a pyramid. As this sequence is only a list of solutions, n has no known relation to its definition.

For each slope 1/a(n) there exists exactly one or no such pyramid with a square number of balls.

			a(1) = 14 and A350886(1) = 54, A350887(1) = 4:
  54^2 = 1^2 + 15^2 + 29^2 + 43^2.
a(2) = 1 and A350886(2) = 70, A350887(2) = 24:
  70^2 = 1^2 + 2^2 + 3^2 + ... + 23^2 + 24^2. This is the classic solution for the cannonball problem.

Thomas Scheuerle, Some solutions to this problem sorted by A350887.
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.
Thomas Scheuerle, Recursive solution formulas.
Index entries for sequences related to sums of squares.

Cf. A350886 (square root of pyramid), A350887 (number of layers).
Cf. A000330, A000447, A024215, A024381 (square pyramidal numbers for slope 1, 1/2, 1/3, 1/4).
Cf. A076215, A001032, A134419, A106521. Some related problems.
Cf. A186699.

sqtest(n, c)={q=1; for(t=2, c, t+=n; q+=(t*t); if(issquare(q), break)); q}
z=500000; b=[; ]; for(n=0, z, r=sqtest(n, z); if(issquare(r), b=concat(b, [sqrtint(r); n+1]))); b=vecsort(b, 1); vector(#b, k, b[2,k]) \\ Last valid value for z=500000 is 5082.

A350886(n)^2 = a(n)^2*binomial(2*A350887(n), 3)/4 + 2*a(n)*binomial(A350887(n), 2) + A350887(n).
A350886(n)^2 = c*((b*(4*b*c^2-(6*c-2)*b + 12*(c-1)) + 12)/12), with c = A350887(n) and b = a(n). Expanded to see factors more clearly.
A350886(n)^2 = c*b^2*( ((1/b) + (c-1)/2)^2 + (c^2-1)/12 ). Shorter form of above.
a(n) != a(m) if n != m.
Let s(n) be the sequence of numbers such that A350887(s(n)) = 4 and s(n) is sorted in ascending order, then a(s(n)) has the ordinary generating function (2*(-7 + z))/(-1 + 31*z - 31*z^2 + z^3).
Let s(n) be the sequence of numbers such that A350887(s(n)) = 49 and s(n) is sorted in ascending order, then a(s(n)) has the ordinary generating function (-13 + z)/(-1 + 391*z - 391*z^2 + z^3).

cp's OEIS Frontend

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A269448 The first of 26 consecutive positive integers the sum of the squares of which is a square.

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A269449 The first of 33 consecutive positive integers the sum of the squares of which is a square.

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A269451 The first of 50 consecutive positive integers the sum of the squares of which is a square.

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A350886 Let X,Y,Z be positive integer solutions to X^2 = Sum_{j=0..Y-1} (1+Z*j)^2, where solutions for Y or Z < 1 are excluded. This sequence lists the sorted values for X.

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