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A152491 Numbers n such that 1/c = 1/n + 1/S(n). c, n positive integers (A000027(n)), S(n) sum of digits of n (A007953(n)). 1

%I #15 Nov 16 2017 15:52:24

%S 2,4,6,8,18,72

%N Numbers n such that 1/c = 1/n + 1/S(n). c, n positive integers (A000027(n)), S(n) sum of digits of n (A007953(n)).

%C A000027(n)*A007953(n)/(A000027(n)+A007953(n))= c, c positive integer.

%C No further term < 10000000. - _Michel Marcus_, Jun 02 2013

%C For a given n let x be the minimal natural number such that n*x/(n+x)=c. I conjecture: from a certain n onward, x>S(n) for all n. Thus there is no other solution bigger than 72, and this sequence is finite. - _Ctibor O. Zizka_, Sep 13 2015

%C Sequence is complete. To prove it, let s denote the sum of digits of n and observe that 1/c - 1/s = 1/n is equivalent to (s-c)/(c*s) = 1/n. Hence we must have s > c and c*s >= n, otherwise the denominators cannot match. But if n is greater than, say, 1000, it is easy to see that s^2 < n and this implies c*s < n, since c < s. - _Giovanni Resta_, Sep 13 2015

%o (PARI) lista(nn) = {for (n=1, nn, digs = Vec(Str(n)); sn = sum(i=1, #digs, eval(digs[i])); if (type(1/(1/n+1/sn)) == "t_INT", print1(n, ", ")););} \\ _Michel Marcus_, Jun 02 2013

%Y Cf. A007953, A000027, A146567, A146564, A146747,

%K base,easy,nonn,fini,full

%O 1,1

%A _Ctibor O. Zizka_, Dec 06 2008

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