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A285317
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Squarefree numbers n for which A019565(n) < n.
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4
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33, 65, 66, 129, 130, 131, 257, 258, 259, 514, 515, 517, 518, 521, 1027, 1030, 1031, 1033, 1034, 1041, 1042, 1057, 2049, 2051, 2053, 2054, 2055, 2059, 2065, 2066, 2081, 2082, 2113, 2114, 2177, 2305, 2561, 3073, 4097, 4098, 4099, 4101, 4102, 4103, 4105, 4106, 4109, 4115, 4129, 4130, 4161, 4162, 4226, 4353, 4354, 4609, 4610, 5122
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OFFSET
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1,1
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COMMENTS
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Any finite cycle in A019565, if such cycles exist at all, must have at least one member that occurs somewhere in this sequence, although certainly not all terms of this sequence could occur in a finite cycle. Specifically, such a number n must occur also in subsequence A285319, and in general, it should satisfy A019565(n) < n and that A048675^{k}(n) is squarefree for all k = 0 .. ∞.
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LINKS
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FORMULA
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MATHEMATICA
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a019565[n_]:=Times @@ Prime@ Flatten@ Position[#, 1] &@ Reverse@ IntegerDigits[n, 2] ; Select[Range[5200], SquareFreeQ[#] && a019565[#]<# &] (* Indranil Ghosh, Apr 18 2017, after Michael De Vlieger *)
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PROG
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(PARI)
A019565(n) = {my(j, v); factorback(Mat(vector(if(n, #n=vecextract(binary(n), "-1..1")), j, [prime(j), n[j]])~))}; \\ This function from M. F. Hasler
isA285317(n) = (issquarefree(n) & (A019565(n) < n));
n=0; k=1; while(k <= 10000, n=n+1; if(isA285317(n), write("b285317.txt", k, " ", n); k=k+1));
(Python)
from operator import mul
from functools import reduce
from sympy import prime
from sympy.ntheory.factor_ import core
def a019565(n): return reduce(mul, (prime(i+1) for i, v in enumerate(bin(n)[:1:-1]) if v == '1')) if n > 0 else 1
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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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