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A294656
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Size of the orbit of n under iteration of the map A125256: x -> smallest odd prime divisor of n^2+1.
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4
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3, 3, 4, 2, 4, 3, 3, 6, 5, 7, 3, 2, 4, 4, 4, 3, 3, 6, 5, 3, 3, 3, 4, 4, 4, 3, 3, 4, 4, 3, 3, 3, 3, 4, 4, 3, 3, 5, 5, 5, 3, 3, 3, 4, 5, 3, 3, 4, 6, 5, 3, 3, 4, 4, 4, 3, 3, 6, 3, 6, 3, 3, 4, 4, 4, 3, 3, 7, 3, 5, 3, 3, 4, 5, 4, 3, 3, 6, 4, 4, 3, 3, 4, 4, 3, 3, 3, 4, 8, 6, 3
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OFFSET
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2,1
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COMMENTS
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The orbit or trajectory under A125256 appears to end in the cycle 5 -> 13 -> 5 -> etc. for any initial value n.
Sequence A294658 gives the number of steps to reach either 5 or 13, i.e. an element of this terminating cycle. Therefore a(n) (which counts these two elements as well as the initial value) is 2 more than A294658(n) for all n. This is confirmed by careful examination of special cases - assuming, of course, that all trajectories end in the cycle (5, 13).
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LINKS
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FORMULA
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EXAMPLE
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For n = 1 the map A125256 is not defined.
a(2) = 3 = # { 2, 5, 13 }, because under A125256, 2 -> 2^2+1 = 5 (= its smallest odd prime factor), 5 -> least odd prime factor(5^2+1 = 26) = 13, 13 -> least odd prime factor(13^2 + 1 = 170 = 2*5*17) = 5, etc.
a(3) = 3 = # { 3, 5, 13 }, because under A125256, 3 -> smallest odd prime factor(3^2+1 = 10) = 5, 5 -> 13, 13 -> 5 etc.
a(4) = 4 = # { 4, 17, 5, 13 }, because under A125256, 4 -> 4^2+1 = 17 (= its smallest odd prime factor), 17 -> smallest odd prime factor(17^2+1 = 290 = 2*5*29) = 5, 5 -> 13, 13 -> 5 etc.
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PROG
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(PARI) A294656(n, f=A125256, S=[n])={while(#S<#S=setunion(S, [n=f(n)]), ); #S} \\ Does not assume the terminating cycle is (5, 13): also works correctly in case there are other terminating cycles.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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