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%I #42 Apr 18 2022 22:55:15
%S 14,1,432,8,13,1932,12958,367,30,3,1554,194,5082,388320,1349,15254,
%T 178542,163,181,11636654,418782,6791,11928,192638,1086,2209447,5166,
%U 19317900,1981979,262,348711312,4799102,7379,60240793
%N 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.
%C 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.
%C For each slope 1/a(n) there exists exactly one or no such pyramid with a square number of balls.
%H Thomas Scheuerle, <a href="/A350887/a350887.txt">Some solutions to this problem sorted by A350887</a>.
%H Thomas Scheuerle, <a href="/A350886/a350886.txt">Recursive solution formulas</a>.
%H <a href="/index/Su#ssq">Index entries for sequences related to sums of squares</a>.
%F A350886(n)^2 = a(n)^2*binomial(2*A350887(n), 3)/4 + 2*a(n)*binomial(A350887(n), 2) + A350887(n).
%F 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.
%F A350886(n)^2 = c*b^2*( ((1/b) + (c-1)/2)^2 + (c^2-1)/12 ). Shorter form of above.
%F a(n) != a(m) if n != m.
%F 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).
%F 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).
%e a(1) = 14 and A350886(1) = 54, A350887(1) = 4:
%e 54^2 = 1^2 + 15^2 + 29^2 + 43^2.
%e a(2) = 1 and A350886(2) = 70, A350887(2) = 24:
%e 70^2 = 1^2 + 2^2 + 3^2 + ... + 23^2 + 24^2. This is the classic solution for the cannonball problem.
%o (PARI)
%o sqtest(n, c)={q=1; for(t=2, c, t+=n; q+=(t*t); if(issquare(q), break)); q}
%o 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.
%Y Cf. A350886 (square root of pyramid), A350887 (number of layers).
%Y Cf. A000330, A000447, A024215, A024381 (square pyramidal numbers for slope 1, 1/2, 1/3, 1/4).
%Y Cf. A076215, A001032, A134419, A106521. Some related problems.
%Y Cf. A186699.
%K nonn,more
%O 1,1
%A _Thomas Scheuerle_, Feb 25 2022
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