OFFSET
1,1
COMMENTS
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 consecutively nested subsequences A285317, A285319, ..., and in general, it should satisfy A019565(n) < n and that A048675^{k}(n) is squarefree for all k = 0 .. ∞.
LINKS
Antti Karttunen, Table of n, a(n) for n = 1..10000
MATHEMATICA
a019565[n_]:=Times @@ Prime@ Flatten@ Position[#, 1] &@ Reverse@ IntegerDigits[n, 2] ; Select[Range[3000], a019565[#]<# &] (* Indranil Ghosh, Apr 18 2017, after Michael De Vlieger *)
PROG
(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
isA285315(n) = (A019565(n) < n);
n=0; k=1; while(k <= 10000, n=n+1; if(isA285315(n), write("b285315.txt", k, " ", n); k=k+1));
(Scheme, with Antti Karttunen's IntSeq-library)
(Python)
from sympy import prime, prod
def a019565(n): return prod(prime(i+1) for i, v in enumerate(bin(n)[:1:-1]) if v == '1') if n > 0 else 1
[n for n in range(1, 3001) if a019565(n)<n] # Indranil Ghosh, Apr 18 2017, after Chai Wah Wu
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, Apr 18 2017
STATUS
approved