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A257670
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Minimum term in the sigma(x) -> x subtree whose root is n.
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6
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1, 2, 2, 2, 5, 5, 2, 2, 9, 10, 11, 5, 9, 9, 2, 16, 17, 10, 19, 19, 21, 22, 23, 2, 25, 26, 27, 5, 29, 29, 16, 16, 33, 34, 35, 22, 37, 37, 10, 27, 41, 19, 43, 43, 45, 46, 47, 33, 49, 50, 51, 52, 53, 34, 55, 5, 49, 58, 59, 2, 61, 61, 16, 64, 65, 66, 67, 67, 69
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OFFSET
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1,2
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LINKS
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FORMULA
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a(m) = m for m in A007369: sigma(x) = m has no solution. [Corrected by M. F. Hasler, Nov 19 2019]
a(A007497(n)) = 2; a(A051572(n)) = 5; a(A257349(n)) = 16. (These sequences being the trajectory of 2, 5 resp. 16 under iterations of sigma = A000203.)
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EXAMPLE
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We have the following trees (a <- b means sigma(a) = b):
2 <-- 3 <-- 4 <-- 7 <-- 8 <-- 15 <-- 24 <-- 60 <-- ...
9 <-- 13 <-- 14 <-’
5 <-- 6 <-- 12 <-- 28 <-- 56 <-- 120 <-- ...
11 <-’ /
10 <-- 18 <-- 39 <-’
The number 1 has strictly speaking an arrow to itself, so it is not part of a tree. (For all n > 1, sigma(n) > n, so no other fixed point or longer "cycle" can exist.) But actually we rather consider connected components, and let a(1) = 1 as the smallest element of this connected component.
a(2) = 2, since there is no smaller x such that sigma(x) = 2: the subtree with root 2 is reduced to a single node: 2. Similarly, a(m) = m for all m in A007369.
For n=3, since sigma(2) = 3, the tree whose root is 3 has 2 nodes: 2 and 3, and the smallest one is 2, hence a(3) = 2.
Similarly, although 24 occurs directly first at sigma(14), it is also reached from 15 which is in turn reached, via intermediate steps, from 2. Thus, the subtree with root 24 has as 2 as smallest element, whence a(24) = 2.
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PROG
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(PARI) lista(nn) = {my(v = vector(nn)); v[1] = 1; for (i=2, nn, my(s = i); while (s <= nn, if (v[s] == 0, v[s] = i); s = sigma(s); ); ); for (i=1, nn, if (v[i] == 0, v[i] = i); ); v; } \\ Michel Marcus, Nov 19 2019
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CROSSREFS
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Cf. A216200 (number of disjoint trees), A257348 (minimal node of all trees).
Cf. A257669 (number of terms in current tree).
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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