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

Author

Michel Marcus, Aug 09 2014

Keywords

• nonn

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, ...

Examples

			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.
		

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)

Crossrefs

Cf. A000203, A007369, A054973, A216200.

Programs

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;}