A134419 -id:A134419 - cp's OEIS frontend

A274471 Numbers missing from A134419 despite satisfying the necessary congruence conditions (see comments).

564, 842, 1284, 2306, 2308, 2402, 2459, 3602, 3650, 3803, 6242, 6338, 6779, 7044, 7058, 7319, 7643, 8088, 8354, 8363, 8402, 8543, 8628, 9122, 9168, 9412, 10607, 10826, 10852, 11257, 11378, 11447, 12203, 12436, 12458, 12722, 12984, 13682, 14162, 14388, 14424, 14639

Offset: 1

Christopher E. Thompson, Jun 24 2016

nonn

A134419 consists of those n where x^2 - n*y^2 = n(n-1)(n+1)/3 has integer solutions for x and y. There are easily verified necessary congruence conditions for that to occur:
  (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.
However, these conditions are not sufficient. This sequence consists of the numbers n satisfying the congruence conditions but for which the Pellian equation has no integer solutions.
If n = k^2*m where m is squarefree, then a necessary (but not sufficient) condition for n to occur in this sequence is that the narrow class group of quadratic forms of discriminant 4*m has more than one class per genus, or equivalently that the narrow class group is not an elementary 2-group.

Christopher E. Thompson, Table of n, a(n) for n = 1..799 [values up to 250000]

Cf. A134419, A274469, A274470.

A001033 Numbers n such that the sum of the squares of n consecutive positive odd numbers x^2 + (x+2)^2 + ... + (x+2n-2)^2 = k^2 for some integer k. The least values of x and k for each n are in A056131 and A056132, respectively.

Original entry on oeis.org

1, 16, 25, 33, 49, 52, 64, 73, 97, 100, 121, 148, 169, 177, 193, 196, 241, 244, 249, 256, 276, 289, 292, 297, 313, 337, 361, 388, 393, 400, 409, 457, 481, 484, 528, 529, 537, 577, 592, 625, 628, 649, 673, 676, 708, 724, 753, 772, 784, 793, 832, 841, 852, 897, 913, 961, 964, 976, 996

Offset: 1

N. J. A. Sloane

Papers by Sollfrey, Hunter and Makowski correct and extend the work of Alfred. However, they do not consider n = 97, 241, 244, 276, 528 and 832, which are in this sequence. I have verified that there are no other n < 1000. - T. D. Noe, Oct 24 2007
A134419 shows how A001032 and this sequence are related. - T. D. Noe, Nov 04 2007
The number 4 is not in this sequence due to the requirement that the odd integers be positive, otherwise 6^2 = (-1)^2 + 1^2 + 3^2 + 5^2.

			a(1) = 1 from 1^2.
a(2) = 16 from 27^2 + 29^2 + ... + 55^2 + 57^2 = 172^2.
a(4) = 33 from 91^2 + 93^2 + ... + 153^2 + 155^2 = 715^2.

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..7103 (up to 250000, extending first 100 terms computed by T. D. Noe).
U. Alfred, Sums of squares of consecutive odd integers, Math. Mag., 40 (1967), 194-199.
J. A. H. Hunter, A note on sums of squares of consecutive odd numbers, Math. Mag. 42 (1969), 145.
Andrzej Makowski, Remark on the paper "Sums of squares of consecutive odd numbers", Math. Mag. 43 (1970), 212-213.
William Sollfrey, Note on sums of squares of consecutive odd integers, Math. Mag. 41 (1968), 255-258.

Cf. A056131, A056132, A274470.

r[1] = {True, {1, 1}}; r[n_] := (rn = Reduce[x > 0 && k > 0 && Sum[(x + 2*j)^2, {j, 0, n - 1}] == k^2, {x, k}, Integers]; srn = Simplify[(rn /. C[1] -> 0) || (rn /. C[1] -> 1) || (rn /. C[1] -> 2)]; rnOdd = Which[rn === False, False, srn[[0]] === And, srn, True, Select[srn, OddQ[x /. ToRules[#1]] & ]]; If[ rnOdd === False, {False, {0, 0}}, {True, {x, k} /. Flatten[{ToRules[rnOdd]}]}]); A001033 = Reap[Do[rn = r[n]; {x0, k0} = rn[[2]]; If[rn[[1]] && OddQ[x0], Print[{n, x0, k0}]; Sow[n]], {n, 1, 1000}]][[2, 1]] (* Jean-François Alcover, Mar 14 2012 *)

We must solve m*(3*x^2 + 6*m*x - 6*x + 4*m^2 - 6*m + 2)/3 = k^2 in integers (x, m, k). - N. J. A. Sloane

For a given n, we must determine whether the generalized Pell equation 4n*y^2 + 4y*n^2 + n(4n^2-1)/3 = k^2 has any integer solutions with y >= 0. Note that x = 2y+1 will be the first odd number being squared. If there are solutions then n is in this sequence. - T. D. Noe, Oct 24 2007

More terms from Robert G. Wilson v
Corrected and extended by T. D. Noe, Oct 24 2007
1024 was missing from b-file. - Christopher E. Thompson, Feb 05 2016

A186699 Numbers n such that there are n numbers in arithmetic progression whose squares sum to a perfect square.

Original entry on oeis.org

1, 2, 4, 9, 11, 16, 23, 24, 25, 26, 33, 36, 47, 49, 50, 52, 59, 64, 73, 74, 81, 88, 96, 97, 100, 107, 121, 122, 144, 146, 148, 169, 177, 184, 191, 193, 194, 196, 218, 225, 239, 241, 242, 244, 249, 256, 276, 289, 292, 297, 299, 311, 312, 313, 324, 337, 338

Offset: 1

Thomas Andrews, Feb 25 2011, Mar 12 2011

nonn

A positive integer n is in this sequence if and only if there is a solution to the Pell-like equation x^2-ny^2=d^2(n-1)n(n+1)/3 for some x,y,d integers.
A positive integer n is in this sequence if and only if it can be written in the form: (u^2-3w^2)/(v^2+3w^2), with u,v,w integers and gcd(v,w)=1. This can also be written as a  n(v^2) + 3(n+1)(w^2) = z^2.
If n is in this sequence, then we can find an arithmetic progression of *positive* integers which satisfy this equation. (The description above does not require the sequence to be positive.)
By using the method of Legendre to find whether there exists rational numbers r,s on the curve nr^2 + 3(n+1)s^2 = 1, we get the following necessary and sufficient conditions on n:
  A. Factor n=a^2b, with b squarefree, then
    1. If 3 does not divide b(n+1), then b ≅ 1 (mod 3)
    2. If b is divisible by 3, then b ≅ 6 (mod 9)
    3. 3 is a square (mod b.)
  B.
    1. If n+1 is divisible by 3, then (n+1)/3 is the sum of two perfect squares
    2. If n+1 is not divisible by 3, then n+1 is the sum of two perfect squares
When n is a perfect square, we can use the arithmetic sequence starting at m=(3n+2)(sqrt(n)-1)/2 + 6 and common difference 6.

			For n=4, (13,19,25,31) is an arithmetic progression of length 4, and 13^2+19^2+25^2+31^2 = 46^2, so 4 is in the sequence.

Thomas Andrews, Article about this and related problems
Thomas Andrews, Initial Terms
Index entries for sequences related to sums of squares

Cf. A134419 is a subsequence.

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.

A274470 Numbers missing from A001033 despite satisfying the necessary congruence conditions (see comments).

Original entry on oeis.org

4, 564, 1284, 2308, 3601, 7044, 7057, 7513, 8628, 9168, 9412, 10561, 10852, 11257, 12436, 13897, 14113, 14388, 14425, 16144, 16692, 16753, 17124, 17257, 17737, 18064, 18433, 18708, 19408, 19428, 20628, 20688, 20752, 20788, 20977, 21073, 23668, 25153, 27193, 28212, 28228

Offset: 1

Christopher E. Thompson, Jun 24 2016

nonn

A001033 consists of those n for which there is a sequence of n consecutive positive odd 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 even;
  if n is odd, then n=1 (mod 8).
However, these conditions are not sufficient. This sequence consists of the numbers n that satisfy all of the congruence conditions but for which there is no sequence of n consecutive positive odd squares whose sum is a square.
The term 4 is present despite the Pellian equation having a solution with the correct parity, because it leads only to (-1)^2 + 1^2 + 3^2 + 5^2 = 6^2, and the specification of A001033 disallows squares of negative numbers. 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..325 [values up to 250000]

Cf. A001033, A134419, A274471.

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.

A185545 Numbers n such that there exists a sequence of n consecutive perfect squares that add up to a perfect square.

Original entry on oeis.org

1, 2, 11, 23, 24, 25, 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

Thomas Andrews, Feb 13 2011

nonn

This is A001032 with the addition of 25, because in this sequence the perfect squares may include 0. Subsequence of A134419.

A248194 Positive integers n such that the equation x^2 - ny^2 = n(n+1)/2 has integer solutions.

Original entry on oeis.org

1, 3, 8, 9, 17, 19, 24, 25, 33, 49, 51, 57, 67, 72, 73, 81, 88, 89, 96, 97, 99, 121, 129, 136, 147, 152, 163, 169, 177, 179, 193, 201, 211, 225, 233, 241, 243, 249, 264, 288, 289, 297, 313, 337, 339, 352, 361, 369, 387, 393, 408, 409, 441, 449, 451, 456, 457

Offset: 1

Colin Barker, Oct 03 2014

nonn

All odd squares are in this sequence. Proof: Set n = (2k + 1)^2, then we have x^2 - (2k + 1)^2 * y^2 = (2k + 1)^2 * (2k^2 + 2k + 1). Rearranging gives x^2 = (2k + 1)^2 * (y^2 + 2k^2 + 2k + 1). As 2k^2 + 2k + 1 is odd, a careful selection of y makes the RHS square. So [(2k+1) * (k(k + 1) + 1), k(k + 1)]. E.g., if k=2, then (5*7)^2 - 25*6^2 = 1225 - 900 = 325 = 25*26/2. - Jon Perry, Nov 07 2014
No even squares are in the sequence. Proof: Rearrange the equation to read x^2 = n(n + 1 + 2y^2)/2, with n = 4k^2. n + 1 + 2y^2 is always odd and so the RHS contains an odd exponent of 2, and therefore cannot be square. - Jon Perry, Nov 15 2014
From Jon Perry, Nov 15 2014: (Start)
Odd squares + 8 are always in this sequence. Proof: Let m = 4k^2 + 4k + 9 and let n = (m+1)/2 = 2k^2 + 2k + 5.
Rearranging the equation x^2 - m*y^2 = m(m + 1)/2, we get x^2  = m(m + 1 + 2y^2)/2, and so x^2 = m(n + y^2) = (4k^2 + 4k + 9)(2k^2 + 2k + 5 + y^2).
We aim to find a y such that the last bracket on the RHS is z^2 * (4k^2 + 4k + 9), so that x equals z*m. We claim that if we let Y = ((n-3)/2)^2*m - n, then Y is a square, and letting Y = y^2, we have y^2 + n = Y + n = z^2 * m as required, with z = (n-3)/2 = k^2 + k + 1.
To prove that Y is a square, Y = [(n^2 - 6n + 9)*(2n - 1) - 4n]/4 = [2n^3 - 13n^2 + 20n - 9]/4 = [(n-1)^2*(2n-9)]/4, and with n as it is, 2n - 9 = 4k^2 + 4k + 1 = (2k + 1)^2, and so we arrive at Y = [(n-1)^2*(2k+1)^2]/4 = [(n-1)(2k+1)/2]^2 = [(k^2 + k + 2)(2k + 1)]^2, a square as required, with y = (k^2 + k + 2)(2k + 1). Also GCD(n-3,2n-1)=1 as required.
This gives a solution as [(k^2 + k + 1)*(4k^2 + 4k + 9), (k^2 + k + 2)*(2k + 1)]. E.g., if k=4, n=45 and a solution is [21*89, 22*9] = [1869, 198]. To validate, 1869^2 - 89*198^2 = 3493161 - 3489156 = 4005 = 89*45.
(End)

			3 is in the sequence because x^2 - 3*y^2 = 6 has integer solutions, including (x, y) = (3, 1) and (9, 5).

Lars Blomberg, Table of n, a(n) for n = 1..10000

Cf. A134419, A245226.

More terms from Lars Blomberg, Nov 02 2014

"Positive" added to definition by N. J. A. Sloane, Nov 02 2014

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

A274471 Numbers missing from A134419 despite satisfying the necessary congruence conditions (see comments).

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A001033 Numbers n such that the sum of the squares of n consecutive positive odd numbers x^2 + (x+2)^2 + ... + (x+2n-2)^2 = k^2 for some integer k. The least values of x and k for each n are in A056131 and A056132, respectively.

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A186699 Numbers n such that there are n numbers in arithmetic progression whose squares sum to a perfect square.

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

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A274470 Numbers missing from A001033 despite satisfying the necessary congruence conditions (see comments).

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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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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.

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A185545 Numbers n such that there exists a sequence of n consecutive perfect squares that add up to a perfect square.

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A248194 Positive integers n such that the equation x^2 - n*y^2 = n*(n+1)/2 has integer solutions.

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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.

A248194 Positive integers n such that the equation x^2 - ny^2 = n(n+1)/2 has integer solutions.