

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




OFFSET

1,1


COMMENTS

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
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 (sc)/(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


LINKS



PROG

(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


CROSSREFS



KEYWORD

base,easy,nonn,fini,full


AUTHOR



STATUS

approved



