

A019565


The squarefree numbers ordered lexicographically by their prime factorization (with factors written in decreasing order). a(n) = Product_{k in I} prime(k+1), where I is the set of indices of nonzero binary digits in n = Sum_{k in I} 2^k.


270



1, 2, 3, 6, 5, 10, 15, 30, 7, 14, 21, 42, 35, 70, 105, 210, 11, 22, 33, 66, 55, 110, 165, 330, 77, 154, 231, 462, 385, 770, 1155, 2310, 13, 26, 39, 78, 65, 130, 195, 390, 91, 182, 273, 546, 455, 910, 1365, 2730, 143, 286, 429, 858, 715, 1430, 2145, 4290
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OFFSET

0,2


COMMENTS

Because a(n) toggles the parity of n there are neither fixed points nor any cycles of odd length.
Conjecture: there are no finite cycles of any length. My grounds for this conjecture: any finite cycle in this sequence, if such cycles exist at all, must have at least one member that occurs somewhere in A285319, the terms that seem already to be quite rare. Moreover, any such a number n should satisfy in addition to A019565(n) < n also that A048675^{k}(n) is squarefree, not just for k=0, 1 but for all k >= 0. As there is on average a probability of only 6/(Pi^2) = 0.6079... that any further term encountered on the trajectory of A048675 is squarefree, the total chance that all of them would be squarefree (which is required from the elements of A019565cycles) is soon minuscule, especially as A048675 is not very tightly bounded (many trajectories seem to skyrocket, at least initially). I am also assuming that usually there is no significant correlation between the binary expansions of n and A048675(n) (apart from their least significant bits), or, for that matter, between their prime factorizations.
See also the slightly stronger conjecture in A285320, which implies that there would neither be any twoway infinite cycles.
If either of the conjectures is false (there are cycles), then certainly neither sequence A285332 nor its inverse A285331 can be a permutation of natural numbers. (End)
The conjecture made in A087207 (see also A288569) implies the two conjectures mentioned above. A further constraint for cycles is that in any A019565trajectory which starts from a squarefree number (A005117), every other term is of the form 4k+2, while every other term is of the form 6k+3.  Antti Karttunen, Jun 18 2017
The sequence satisfies the exponential function identity, a(x + y) = a(x) * a(y), whenever x and y do not have a 1bit in the same position, i.e., when A004198(x,y) = 0. See also A283475.  Antti Karttunen, Oct 31 2019
The above identity becomes unconditional if binary exclusive OR, A003987(.,.), is substituted for addition, and A059897(.,.), a multiplicative equivalent of A003987, is substituted for multiplication. This gives us a(A003987(x,y)) = A059897(a(x), a(y)).  Peter Munn, Nov 18 2019
Also the Heinz number of the binary indices of n, where the Heinz number of a sequence (y_1,...,y_k) is prime(y_1)*...*prime(y_k), and a number's binary indices (A048793) are the positions of 1's in its reversed binary expansion.  Gus Wiseman, Dec 28 2022


LINKS



FORMULA

G.f.: Product_{k>=0} (1 + prime(k+1)*x^2^k), where prime(k)=A000040(k).  Ralf Stephan, Jun 20 2003
a(n) = f(n, 1, 1) with f(x, y, z) = if x > 0 then f(floor(x/2), y*prime(z)^(x mod 2), z+1) else y.  Reinhard Zumkeller, Mar 13 2010
a(n) = a(2^x)*a(2^y)*a(2^z)*... = prime(x+1)*prime(y+1)*prime(z+1)*..., where n = 2^x + 2^y + 2^z + ...  Benedict W. J. Irwin, Jul 24 2016
a(2n) = A003961(a(n)); a(2n+1) = 2*a(2n).
a(x XOR y) = A059897(a(x), a(y)) = A089913(a(x), a(y)), where XOR denotes bitwise exclusive OR (A003987).
(End)


EXAMPLE

5 = 2^2+2^0, e_1 = 2, e_2 = 0, prime(2+1) = prime(3) = 5, prime(0+1) = prime(1) = 2, so a(5) = 5*2 = 10.
This sequence regarded as a triangle withs rows of lengths 1, 1, 2, 4, 8, 16, ...:
1;
2;
3, 6;
5, 10, 15, 30;
7, 14, 21, 42, 35, 70, 105, 210;
11, 22, 33, 66, 55, 110, 165, 330, 77, 154, 231, 462, 385, 770, 1155, 2310;
...
(End)
The initial terms are shown below, equated with the product of their prime factors to exhibit the lexicographic order. We start with 1, since 1 is factored as the empty product and the empty list is first in lexicographic order.
n a(n)
0 1 = .
1 2 = 2.
2 3 = 3.
3 6 = 3*2.
4 5 = 5.
5 10 = 5*2.
6 15 = 5*3.
7 30 = 5*3*2.
8 7 = 7.
9 14 = 7*2.
10 21 = 7*3.
11 42 = 7*3*2.
12 35 = 7*5.
(End)


MAPLE

a:= proc(n) local i, m, r; m:=n; r:=1;
for i while m>0 do if irem(m, 2, 'm')=1
then r:=r*ithprime(i) fi od; r
end:


MATHEMATICA

Do[m=1; o=1; k1=k; While[ k1>0, k2=Mod[k1, 2]; If[k2\[Equal]1, m=m*Prime[o]]; k1=(k1k2)/ 2; o=o+1]; Print[m], {k, 0, 55}] (* Lei Zhou, Feb 15 2005 *)
Table[Times @@ Prime@ Flatten@ Position[#, 1] &@ Reverse@ IntegerDigits[n, 2], {n, 0, 55}] (* Michael De Vlieger, Aug 27 2016 *)
b[0] := {1}; b[n_] := Flatten[{ b[n  1], b[n  1] * Prime[n] }];


PROG

(PARI) a(n)=factorback(vecextract(primes(logint(n+!n, 2)+1), n)) \\ M. F. Hasler, Mar 26 2011, updated Aug 22 2014, updated Mar 01 2018
(Haskell)
a019565 n = product $ zipWith (^) a000040_list (a030308_row n)
(Python)
from operator import mul
from functools import reduce
from sympy import prime
return reduce(mul, (prime(i+1) for i, v in enumerate(bin(n)[:1:1]) if v == '1')) if n > 0 else 1
(Scheme) (define (A019565 n) (let loop ((n n) (i 1) (p 1)) (cond ((zero? n) p) ((odd? n) (loop (/ ( n 1) 2) (+ 1 i) (* p (A000040 i)))) (else (loop (/ n 2) (+ 1 i) p))))) ;; (Requires only the implementation of A000040 for prime numbers.)  Antti Karttunen, Apr 20 2017


CROSSREFS

Cf. A007088, A030308, A000040, A013929, A005117, A103785, A103786, A110765, A064273, A246353, A283475, A283477, A285319, A285331, A285332, A288569, A293442.
Cf. A285315 (numbers for which a(n) < n), A285316 (for which a(n) > n).
Even bisection (which contains the odd terms): A332382.
Least prime index of a(n) is A001511.


KEYWORD



AUTHOR



EXTENSIONS

Definition corrected by KlausR. Löffler, Aug 20 2014


STATUS

approved



