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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
2, 4, 6, 8, 18, 72 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
A000027(n)*A007953(n)/(A000027(n)+A007953(n))= c, c positive integer.
No further term < 10000000. - Michel Marcus, Jun 02 2013
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 (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
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
Sequence in context: A344902 A104001 A048784 * A355557 A329738 A133296
KEYWORD
base,easy,nonn,fini,full
AUTHOR
Ctibor O. Zizka, Dec 06 2008
STATUS
approved

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Last modified March 3 07:26 EST 2024. Contains 370499 sequences. (Running on oeis4.)