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.

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

1,1

Comments

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.

Links

Table of n, a(n) for n=1..30.

Thomas Scheuerle, Some solutions to this problem sorted by A350887.

Thomas Scheuerle, Recursive solution formulas.

Index entries for sequences related to sums of squares.

Formula

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.

Example

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.

Prog

(PARI)
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.

Crossrefs

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.

Sequence in context: A025323 A157934 A335035 * A281920 A005129 A039532

Adjacent sequences: A350883 A350884 A350885 * A350887 A350888 A350889

Keyword

nonn

Author

Thomas Scheuerle, Feb 25 2022

Status

approved