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A285319
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Squarefree numbers n for which A019565(n) < n and A048675(n) is also squarefree.
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9
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66, 129, 130, 258, 514, 1034, 1041, 1042, 2049, 2054, 2055, 2066, 2082, 2114, 4098, 4101, 4102, 4130, 4161, 4162, 4226, 4353, 4354, 4610, 5122, 8193, 8198, 8202, 8205, 8206, 8210, 8211, 8229, 8259, 8706, 9218, 9219, 12291
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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. Furthermore, such a number n should satisfy A019565(n) < n and that A048675(n)^k is squarefree for all k >= 0.
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LINKS
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MATHEMATICA
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lim = 4000;
A019565 = Table[Times @@ Prime@Flatten@Position[#, 1] &@
Reverse@IntegerDigits[n, 2], {n, 1, lim}]; (* From Michael De Vlieger in A019565 *)
A048675 = Table[Total[#[[2]]*2^(PrimePi[#[[1]]] - 1) & /@ FactorInteger[n]], {n, 1, lim}]; (* From Jean-François Alcover in A048675 *)
Select[Range[lim], A019565[[#]] < # && SquareFreeQ[#] &&
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PROG
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(PARI)
allocatemem(2^30);
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
A048675(n) = my(f = factor(n)); sum(k=1, #f~, f[k, 2]*2^primepi(f[k, 1]))/2; \\ Michel Marcus, Oct 10 2016
isA285319(n) = (issquarefree(n) & (A019565(n) < n) && issquarefree(A048675(n)));
n=0; k=0; while(k <= 60, n=n+1; if(isA285319(n), print1(n, ", "); k=k+1));
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CROSSREFS
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KEYWORD
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nonn,hard,more
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AUTHOR
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STATUS
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approved
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