A241954 Number of integers x such that the repeated application of sigma(x)->x leads to n.

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

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 sum-of-divisors function, Experimental Mathematics, 5 (1996), pp. 93-100 (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, n-1, nb += isok(i, n); ); nb; }

Crossrefs

Cf. A000203, A007369, A054973, A216200.

Sequence in context: A289229 A263097 A286011 * A307047 A049600 A318602

Adjacent sequences: A241951 A241952 A241953 * A241955 A241956 A241957

Keyword

nonn

Author

Michel Marcus, Aug 09 2014

Status

approved