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A285315
Numbers n for which A019565(n) < n.
8
8, 16, 32, 33, 64, 65, 66, 128, 129, 130, 131, 132, 136, 256, 257, 258, 259, 260, 261, 264, 272, 512, 513, 514, 515, 516, 517, 518, 520, 521, 528, 544, 576, 640, 768, 1024, 1025, 1026, 1027, 1028, 1029, 1030, 1031, 1032, 1033, 1034, 1040, 1041, 1042, 1056, 1057, 1088, 1089, 1152, 1280, 1536, 2048, 2049, 2050, 2051
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
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)
(define A285315 (MATCHING-POS 1 0 (lambda (n) (< (A019565 n) n))))
(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
Complement: A285316.
Cf. A285317, A285319 (subsequences).
Sequence in context: A018922 A290287 A303981 * A020948 A219547 A259751
KEYWORD
nonn
AUTHOR
Antti Karttunen, Apr 18 2017
STATUS
approved