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%I #11 Aug 27 2015 08:17:24
%S 1,4,20257,43910,563101,18895033,119847146,305478634708,7461770367940,
%T 29820549118362
%N Solutions of the equation (n+2)' = (n+1)' + n', where n' is the arithmetic derivative of n.
%C a(9) > 5*10^11. - _Giovanni Resta_, Apr 18 2014
%C a(11) > 5*10^13. - _Hiroaki Yamanouchi_, Aug 27 2015
%e The arithmetic derivative of 43910+2 is 69948, of 43910+1 is 39201, of 43910 is 30747 and 69948 = 39201 + 30747.
%p with(numtheory); P:= proc(q) local a,b,c,n,p;
%p for n from 1 to q do
%p a:=(n+2)*add(op(2,p)/op(1,p),p=ifactors(n+2)[2]);
%p b:=(n+1)*add(op(2,p)/op(1,p),p=ifactors(n+1)[2]);
%p c:=n*add(op(2,p)/op(1,p),p=ifactors(n)[2]);
%p if a=b+c then print(n); fi; od; end: P(10^9);
%Y Cf. A003415.
%K nonn,more
%O 1,2
%A _Paolo P. Lava_, Apr 17 2014
%E a(7)-a(8) from _Giovanni Resta_, Apr 18 2014
%E a(9)-a(10) from _Hiroaki Yamanouchi_, Aug 27 2015
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