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1, 2, 4, 2, 8, 4, 6, 2, 16, 8, 18, 4, 12, 6, 10, 2, 32, 16, 54, 8, 36, 18, 50, 4, 24, 12, 30, 6, 20, 10, 14, 2, 64, 32, 162, 16, 108, 54, 250, 8, 72, 36, 150, 18, 100, 50, 98, 4, 48, 24, 90, 12, 60, 30, 70, 6, 40, 20, 42, 10, 28, 14, 22, 2, 128, 64, 486, 32, 324, 162, 1250, 16, 216, 108, 750, 54, 500, 250, 686, 8, 144
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OFFSET
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0,2
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COMMENTS
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As the underlying sequence A163511 can be represented as a binary tree, so can this be also:
1
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...................2...................
4 2
8......../ \........4 6......../ \........2
/ \ / \ / \ / \
/ \ / \ / \ / \
/ \ / \ / \ / \
16 8 18 4 12 6 10 2
32 16 54 8 36 18 50 4 24 12 30 6 20 10 14 2
etc.
Each rightward leaning branch stays constant, because a(2n+1) = a(n).
Conjecture: Mersenne primes (A000668) gives all such odd numbers k for which a(k) = A348717(k). If true, then it immediately implies that map n -> A163511(n) [or equally: map n -> A243071(n)] has no other fixed points than those given by A007283. But see also A364959. - Edited Sep 03 2023
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LINKS
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FORMULA
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a(0) = 1, a(1) = 2, a(2n) = A163511(2n) = 2*A163511(n), and for n > 0, a(2n+1) = a(n).
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PROG
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(PARI)
A163511(n) = if(!n, 1, my(p=2, t=1); while(n>1, if(!(n%2), (t*=p), p=nextprime(1+p)); n >>= 1); (t*p));
A348717(n) = if(1==n, 1, my(f = factor(n), k = primepi(f[1, 1])-1); for (i=1, #f~, f[i, 1] = prime(primepi(f[i, 1])-k)); factorback(f));
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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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