

A241954


Number of integers x such that the repeated application of sigma(x)>x leads to n.


2



1, 0, 1, 2, 0, 1, 3, 4, 0, 0, 0, 3, 1, 2, 5, 0, 0, 2, 0, 1, 0, 0, 0, 10, 0, 0, 0, 4, 0, 1, 2, 4, 0, 0, 0, 1, 0, 1, 3, 1, 0, 4, 0, 1, 0, 0, 0, 3, 0, 0, 0, 0, 0, 2, 0, 9, 1, 0, 0, 14, 0, 1, 5, 0, 0, 0, 0, 1, 0, 0, 0, 6, 0, 1, 0, 0, 0, 1, 0, 3, 0, 0, 0, 4, 0, 0
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OFFSET

1,4


COMMENTS

If n is A007369 (sigma(x) = n has no solution) then a(n) = 0.
Obviously a(n) >= A054973(n), number of solutions to sigma(x) = n.
The equality is obtained for terms of A007369, but not only: see a(n) for 1, 3, 6, 13, 18, 20, 30, 31, 36, 38 ...
The maxima for a(n) are : 1, 2, 3, 4, 5, 10, 14, 15 ... and are obtained for n: 1, 4, 7, 8, 15, 24, 60, 120, ...


LINKS

Jens Kruse Andersen, Table of n, a(n) for n = 1..10000
G. L. Cohen and H. J. J. te Riele, Iterating the sumofdivisors function, Experimental Mathematics, 5 (1996), pp. 93100 (see p. 97)


EXAMPLE

There is a single integer such that sigma(x) = 1 so a(1) = 1.
For n=4, we have only sigma(3) = 4 and sigma(sigma(2)) = 4, so a(4) = 2.
For n=7, we have only sigma(4) = 7, sigma(sigma(3)) = 7, and sigma(sigma(sigma(2))) = 7, so a(7) = 3.


PROG

(PARI) isok(i, n) = {j = i; while((k = sigma(j)) < n, j = k); k == n; }
a(n) = {if (n == 1, return (1)); nb = 0; for (i=2, n1, nb += isok(i, n); ); nb; }


CROSSREFS

Cf. A000203, A007369, A054973, A216200.
Sequence in context: A289229 A263097 A286011 * A049600 A318602 A004542
Adjacent sequences: A241951 A241952 A241953 * A241955 A241956 A241957


KEYWORD

nonn


AUTHOR

Michel Marcus, Aug 09 2014


STATUS

approved



