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%I #21 Apr 23 2018 04:05:19
%S 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,
%T 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,
%U 0,0,0,6,0,1,0,0,0,1,0,3,0,0,0,4,0,0
%N Number of integers x such that the repeated application of sigma(x)->x leads to n.
%C If n is A007369 (sigma(x) = n has no solution) then a(n) = 0.
%C Obviously a(n) >= A054973(n), number of solutions to sigma(x) = n.
%C 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 ...
%C 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, ...
%H Jens Kruse Andersen, <a href="/A241954/b241954.txt">Table of n, a(n) for n = 1..10000</a>
%H G. L. Cohen and H. J. J. te Riele, <a href="http://projecteuclid.org/euclid.em/1047565640">Iterating the sum-of-divisors function</a>, Experimental Mathematics, 5 (1996), pp. 93-100 (see p. 97)
%e There is a single integer such that sigma(x) = 1 so a(1) = 1.
%e For n=4, we have only sigma(3) = 4 and sigma(sigma(2)) = 4, so a(4) = 2.
%e For n=7, we have only sigma(4) = 7, sigma(sigma(3)) = 7, and sigma(sigma(sigma(2))) = 7, so a(7) = 3.
%o (PARI) isok(i, n) = {j = i; while((k = sigma(j)) < n, j = k); k == n;}
%o a(n) = {if (n == 1, return (1)); nb = 0; for (i=2, n-1, nb += isok(i, n);); nb;}
%Y Cf. A000203, A007369, A054973, A216200.
%K nonn
%O 1,4
%A _Michel Marcus_, Aug 09 2014
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