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%I #42 Aug 04 2022 16:09:56
%S 12,29,49,69,108,120,203,243,285,382,404,453,592,645,677,788,848,996,
%T 1140,1149,1241,1365,1779,1796,1797,1857,2032,2236,2649,2704,2812,
%U 2870,3143,3188,3388,3443,3525,3831,4372,4379,4592,4799,4911,5204,5364,5520,5814
%N Numbers k such that the arithmetic differential equation m'' - m'm + k = 0 has exactly one positive solution in m with two prime factors (counted with multiplicity).
%C This is a second-order nonlinear ADE. It is known that many linear second-order ADEs have infinitely many solutions (A334261), but nonlinear cases haven't been studied.
%H Nathan Mabey, <a href="https://drive.google.com/file/d/1XLmndd9TBYff15kszUTgIvlFqoInLrSo/view?usp=sharing">C Script</a>
%e k = 12 is in the sequence, since for m = 4, we have m' = m'' = 4, so m'm - m'' = 16 - 4 = 12 = k.
%o (C) See Link
%o (MATLAB)
%o function a = A353872( max_pow_2 )
%o a = [];
%o maxad2 = ad(ad(2^max_pow_2));
%o for m = 1:2^max_pow_2
%o if length(factor(m)) == 2
%o d = ad(m); b = ad(d); c = d*m;
%o k(m) = b - c;
%o end
%o end
%o for n = 1:length(k)
%o if k(n) > -maxad2;
%o if isempty(find(a == k(n),1))
%o if 1 == length(find(k == k(n)))
%o a = [a k(n)];
%o end
%o end
%o end
%o end
%o a = sort(-a);
%o end
%o function y = ad( x )
%o y = 0;
%o if(x > 1)
%o p = factor(x); pu = unique(p);
%o for n = 1:length(pu);
%o y = y + (x*length(find(p == pu(n))))/pu(n);
%o end
%o end
%o end % _Thomas Scheuerle_, Jun 15 2022
%Y Cf. A003415 (n'), A068346 (n''), A334261 (2m'' - m' - 4 = 0).
%K nonn
%O 1,1
%A _Nathan Mabey_, May 08 2022
%E More terms from _Jinyuan Wang_, Jun 15 2022
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