A285319 -id:A285319 - cp's OEIS frontend

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.

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

Offset: 0

Marc LeBrun

A permutation of the squarefree numbers A005117. The missing positive numbers are in A013929. - Alois P. Heinz, Sep 06 2014
From Antti Karttunen, Apr 18 & 19 2017: (Start)
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 A019565-cycles) 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 two-way 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 A019565-trajectory 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 1-bit 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

			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.
From _Philippe Deléham_, Jun 03 2015: (Start)
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)
From _Peter Munn_, Jun 14 2020: (Start)
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)

Reinhard Zumkeller, Table of n, a(n) for n = 0..8191

Row 1 of A285321.
Equivalent sequences for k-th-power-free numbers: A101278 (k=3), A101942 (k=4), A101943 (k=5), A054842 (k=10).
Cf. A007088, A030308, A000040, A013929, A005117, A103785, A103786, A110765, A064273, A246353, A283475, A283477, A285319, A285331, A285332, A288569, A293442.
Cf. A109162 (iterates).
Cf. also A048675 (a left inverse), A087207, A097248, A260443, A054841.
Cf. A285315 (numbers for which a(n) < n), A285316 (for which a(n) > n).
Cf. A276076, A276086 (analogous sequences for factorial and primorial bases), A334110 (terms squared).
For partial sums see A288570.
A003961, A003987, A004198, A059897, A089913, A331590, A334747 are used to express relationships between sequence terms.
Column 1 of A329332.
Even bisection (which contains the odd terms): A332382.
A160102 composed with A052330, and subsequence of the latter.
Related to A000079 via A225546, to A057335 via A122111, to A008578 via A336322.
Least prime index of a(n) is A001511.
Greatest prime index of a(n) is A029837 or A070939.
Taking prime indices gives A048793, reverse A272020, row sums A029931.
A112798 lists prime indices, length A001222, sum A056239.
Cf. A000120, A000720, A005940, A066099, A358137, A358170.

a019565 n = product $ zipWith (^) a000040_list (a030308_row n)
-- Reinhard Zumkeller, Apr 27 2013

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:
seq(a(n), n=0..60);  # Alois P. Heinz, Sep 06 2014

Do[m=1;o=1;k1=k;While[ k1>0, k2=Mod[k1, 2];If[k2\[Equal]1, m=m*Prime[o]];k1=(k1-k2)/ 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] }];
  a = b[6] (* Fred Daniel Kline, Jun 26 2017 *)

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

from operator import mul
from functools import reduce
from sympy import prime
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
# Chai Wah Wu, Dec 25 2014

(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

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
For all n >= 0: A048675(a(n)) = n; A013928(a(n)) = A064273(n). - Antti Karttunen, Jul 29 2015
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
From Antti Karttunen, Apr 18 2017 and Jun 18 2017: (Start)
a(n) = A097248(A260443(n)), a(A005187(n)) = A283475(n), A108951(a(n)) = A283477(n).
A055396(a(n)) = A001511(n), a(A087207(n)) = A007947(n). (End)
a(2^n - 1) = A002110(n). - Michael De Vlieger, Jul 05 2017
a(n) = A225546(A000079(n)). - Peter Munn, Oct 31 2019
From Peter Munn, Mar 04 2022: (Start)
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).
a(n+1) = A334747(a(n)).
a(x+y) = A331590(a(x), a(y)).
a(n) = A336322(A008578(n+1)).
(End)

Definition corrected by Klaus-R. Löffler, Aug 20 2014

New name from Peter Munn, Jun 14 2020

A285332 a(0) = 1, a(1) = 2, a(2n) = A019565(a(n)), a(2n+1) = A065642(a(n)).

Original entry on oeis.org

1, 2, 3, 4, 6, 9, 5, 8, 15, 12, 14, 27, 10, 25, 7, 16, 210, 45, 35, 18, 105, 28, 462, 81, 21, 20, 154, 125, 30, 49, 11, 32, 10659, 420, 910, 75, 78, 175, 33, 24, 3094, 315, 385, 56, 780045, 924, 374, 243, 110, 63, 55, 40, 4389, 308, 170170, 625, 1155, 60, 286, 343, 42, 121, 13, 64, 54230826, 31977, 28405, 630, 1330665, 1820, 714

Offset: 0

Antti Karttunen, Apr 17 2017

Note the indexing: the domain starts from 0, while the range excludes zero.
This sequence can be represented as a binary tree. Each left hand child is produced as A019565(n), and each right hand child as A065642(n), when the parent node contains n >= 2:
                                     1
                                     |
                  ...................2...................
                 3                                       4
       6......../ \........9                   5......../ \........8
      / \                 / \                 / \                 / \
     /   \               /   \               /   \               /   \
    /     \             /     \             /     \             /     \
  15       12         14       27         10       25          7       16
210 45   35  18    105  28  462  81     21  20  154  125     30 49   11  32
etc.
Where will 38 appear in this tree? It is a reasonable assumption that by iterating A087207 starting from 38, as A087207(38) = 129, A087207(129) = 8194, A087207(8194) = 1501199875790187, ..., we will eventually hit a prime A000040(k), most likely with a largish index k. This prime occurs at the penultimate edge at right, as a(A000918(k)) = a((2^k)-2), and thus 38 occurs somewhere below it as a(m) = 38, m > k. All the numbers that share prime factors with 38, namely 76, 152, 304, 608, 722, ..., occur similarly late in this tree, as they form the rightward branch starting from 38. Alternatively, by iterating A285330 (each iteration moves one step towards the root) starting from 38, we might instead first hit some power of 3, or say, one of the terms of A033845 (the rightward branch starting from 6), in which case the first prime encountered would be a(2)=3 and 38 would appear on the left-hand side instead of the right-hand side subtree.
As long as it remains conjecture that A019565 has no cycles, it is certainly also an open question whether this is a permutation of the natural numbers: If A019565 has any cycles, then neither any of the terms in those cycles nor any A065642-trajectories starting from those terms (that is, numbers sharing same prime factors) may occur in this tree.
Sequence exhibits some outrageous swings, for example, a(703) = 224, but a(704) is 1427 decimal digits (4739 binary digits) long, thus it no longer fits into a b-file.
However, the scatter plot of A286543 gives some flavor of the behavior of this sequence even after that point. - Antti Karttunen, Dec 25 2017

Antti Karttunen, Table of n, a(n) for n = 0..703
Michael De Vlieger, Diagram of the binary tree of a(n) showing 1 <= n <= 2^8.
Antti Karttunen and David J. Seal, Discussion on SeqFan mailing list
Index entries for sequences that are permutations of the natural numbers

Inverse: A285331.
Cf. A007947, A019565, A033845, A048675, A065642, A087207, A109162, A285319, A285320, A285328, A285330, A285333, A286542, A286543.
Compare also to permutation A285112 and array A285321.

Block[{a = {1, 2}}, Do[AppendTo[a, If[EvenQ[i], Times @@ Prime@ Flatten@ Position[#, 1] &@ Reverse@ IntegerDigits[a[[i/2 + 1]], 2], If[# == 1, 1, Function[{n, c}, SelectFirst[Range[n + 1, n^2], Times @@ FactorInteger[#][[All, 1]] == c &]] @@ {#, Times @@ FactorInteger[#][[All, 1]]}] &[a[[(i - 1)/2 + 1]] ] ]], {i, 2, 70}]; a] (* Michael De Vlieger, Mar 12 2021 *)

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
A007947(n) = factorback(factorint(n)[, 1]); \\ From Andrew Lelechenko, May 09 2014
A065642(n) = { my(r=A007947(n)); if(1==n,n,n = n+r; while(A007947(n) <> r, n = n+r); n); };
A285332(n) = { if(n<=1,n+1,if(!(n%2),A019565(A285332(n/2)),A065642(A285332((n-1)/2)))); };
for(n=0, 4095, write("b285332.txt", n, " ", A285332(n)));

from operator import mul
from sympy import prime, primefactors
def a007947(n): return 1 if n<2 else reduce(mul, primefactors(n))
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 # This function from Chai Wah Wu
def a065642(n):
    if n==1: return 1
    r=a007947(n)
    n = n + r
    while a007947(n)!=r:
        n+=r
    return n
def a(n):
    if n<2: return n + 1
    if n%2==0: return a019565(a(n//2))
    else: return a065642(a((n - 1)//2))
print([a(n) for n in range(51)]) # Indranil Ghosh, Apr 18 2017

;; With memoization-macro definec.
(definec (A285332 n) (cond ((<= n 1) (+ n 1)) ((even? n) (A019565 (A285332 (/ n 2)))) (else (A065642 (A285332 (/ (- n 1) 2))))))

a(0) = 1, a(1) = 2, a(2n) = A019565(a(n)), a(2n+1) = A065642(a(n)).
For n >= 0, a(2^n) = A109162(2+n). [The left edge of the tree.]
For n >= 0, a(A000225(n)) = A000079(n). [Powers of 2 occur at the right edge of the tree.]
For n >= 2, a(A000918(n)) = A000040(n). [And the next vertices inwards contain primes.]
For n >= 2, a(A036563(1+n)) = A001248(n). [Whose right children are their squares.]
For n >= 0, a(A055010(n)) = A000244(n). [Powers of 3 are at the rightmost edge of the left subtree.]
For n >= 2, a(A129868(n-1)) = A062457(n).
A048675(a(n)) = A285333(n).
A046523(a(n)) = A286542(n).

A285315 Numbers n for which A019565(n) < n.

Original entry on oeis.org

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

Antti Karttunen, Apr 18 2017

nonn

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 .. ∞.

Antti Karttunen, Table of n, a(n) for n = 1..10000

Complement: A285316.
Cf. A019565, A048675.
Cf. A285317, A285319 (subsequences).

a019565[n_]:=Times @@ Prime@ Flatten@ Position[#, 1] &@ Reverse@ IntegerDigits[n, 2] ; Select[Range[3000], a019565[#]<# &] (* Indranil Ghosh, Apr 18 2017, after Michael De Vlieger *)

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));

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)Indranil Ghosh, Apr 18 2017, after Chai Wah Wu

;; With Antti Karttunen's IntSeq-library.
(define A285315 (MATCHING-POS 1 0 (lambda (n) (< (A019565 n) n))))

A285320 If n == 0 or A008683(n) == 0, then a(n) = 0, otherwise a(n) = 1+a(A048675(n)); number of iterations of A048675 needed before the result is either zero or nonsquarefree number (A013929).

Original entry on oeis.org

0, 1, 2, 3, 0, 1, 4, 1, 0, 0, 2, 1, 0, 1, 1, 5, 0, 1, 0, 1, 0, 3, 2, 1, 0, 0, 2, 0, 0, 1, 2, 1, 0, 1, 2, 1, 0, 1, 3, 3, 0, 1, 2, 1, 0, 0, 2, 1, 0, 0, 0, 3, 0, 1, 0, 1, 0, 3, 1, 1, 0, 1, 1, 0, 0, 1, 2, 1, 0, 3, 2, 1, 0, 1

Offset: 0

Antti Karttunen, Apr 18 2017

Conjecture: all terms are well-defined (finite). This implies also the conjecture I have made in A019565.

			a(38) = 3 because 38 = 2*19 (thus squarefree), A048675(38) = 129 (= 3*43), A048675(129) = 8194 (= 2*17*241) and A048675(8194) = 4503599627370561 (= 3^2 * 37 * 71 * 190483425427), so three steps were needed before nonsquarefree number was reached.
a(74) >= 3 as A048675(74) = 2049 (squarefree), A048675(2049) =  10633823966279326983230456482242756610 (squarefree), A048675(10633823966279326983230456482242756610) = ???

A left inverse of A109162.
Cf. A008683, A005117, A013929, A048675.
Cf. also A285319, A285331, A285332.

A048675(n) = my(f = factor(n)); sum(k=1, #f~, f[k, 2]*2^primepi(f[k, 1]))/2; \\ Michel Marcus, Oct 10 2016
A285320(n) = if(!n || !moebius(n),0,1+A285320(A048675(n)));

(define (A285320 n) (if (or (zero? n) (zero? (A008683 n))) 0 (+ 1 (A285320 (A048675 n)))))

If n == 0 or A008683(n) == 0, then a(n) = 0, otherwise a(n) = 1+a(A048675(n)).

a(A109162(n)) = n.

A285331 Inverse for A285332: a(1) = 0, a(2) = 1, a(A019565(n)) = 2a(n), a(A065642(n)) = 1 + 2a(n).

Original entry on oeis.org

0, 1, 2, 3, 6, 4, 14, 7, 5, 12, 30, 9, 62, 10, 8, 15, 126, 19, 254, 25, 24, 252, 510, 39, 13, 76, 11, 21, 1022, 28, 2046, 31, 38, 316, 18, 79, 4094

Offset: 1

Antti Karttunen, Apr 17 2017, comments edited Apr 19 2017

Note the indexing: the domain starts from 1, while the range includes also zero.
For the question whether this sequence and A285332 are permutations of natural numbers, see comments in A285332 and the conjecture stated in A019565.
As a practical problem, it seems next-to-impossible to compute even the value of a(38). Even though we know that 38 certainly is not in a finite cycle of A019565, because A048675(38) = 129, A048675(129) = 8194 and A048675(8194) = 4503599627370561 which factorizes as 3^2 * 37 * 71 * 190483425427 (thus is not squarefree and A285320(38) = 3), the value of a(38) is most likely so huge that it will not fit into the data section or even into a b-file. The same problem applies to all numbers that share prime factors with 38, namely 76, 152, 304, 608, 722, ...
Terms a(39) .. a(61) are [632, 51, 8190, 60, 16382, 505, 17, 72057594037927932, 32766, 159, 29, 103, 1016, 153, 65534, 319, 50, 43, 16376, 131014, 131070, 57, 262142].
The name is slightly misleading. The given definition of a(n) is not always very helpful to compute the terms (cf. example of n = 38), it is actually not clear whether the sequence is well defined. - M. F. Hasler, Mar 01 2018

			a(1) = 0 and a(2) = 1 by definition.
a(3) = a(prime(2)) = a(A019565(2^1)) = 2*a(2) = 2.
a(4) = a(2^2) = a(A065642(2)) = 1 + 2*a(2) = 3.
a(5) = a(prime(3)) = a(A019565(2^2)) = 2*a(4) = 6.
a(9) = a(3^2) = a(A065642(3)) = 1 + 2*a(3) = 5.
a(10) = a(2*5) = a(prime(1)*prime(3)) = a(A019565(2^0+2^2)) = 2*a(1+4) = 12.
To compute a(38), write 38 = prime(1)*prime(8) = A019565(2^7+2^0), so a(38) = 2*a(129). To compute this, use 129 = prime(2)*prime(14) = A019565(2^13+2^1), so a(129) = 2*a(8194). But 8194 = prime(1)*prime(7)*prime(53) = A019565(2^0+2^6+2^52), so a(8194) = 2*a(4503599627370561)...

Inverse: A285332.
Cf. A005117, A008683, A019565, A048675, A065642, A087207, A285319, A285320, A285328.
Compare also to permutation A285111.

A285331(n)={ if(n<=2,n-1,if(moebius(n)<>0, 2*A285331(A048675(n)), 1+2*A285331(A285328(n))))} \\ See A048675 & A285328 for respective PARI code. (We avoid content duplication leading to obsolete code.)

;; With memoization-macro definec.
(definec (A285331 n) (cond ((<= n 2) (- n 1)) ((not (zero? (A008683 n))) (* 2 (A285331 (A048675 n)))) (else (+ 1 (* 2 (A285331 (A285328 n)))))))

a(1) = 0, a(2) = 1, and for n > 2, if A008683(n) <> 0 [when n is squarefree], a(n) = 2*a(A048675(n)), otherwise a(n) = 1 + 2*a(A285328(n)).

For all n >= 0, a(A285332(n)) = n.

A285317 Squarefree numbers n for which A019565(n) < n.

Original entry on oeis.org

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

Offset: 1

Antti Karttunen, Apr 18 2017

nonn

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 .. oo.

Antti Karttunen, Table of n, a(n) for n = 1..10000

Intersection of A005117 and A285315.

Cf. A005117, A008683, A019565, A048675, A285318, A285319 (subsequence), A285332.

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 *)

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));

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
print([n for n in range(1, 5201) if core(n) == n and a019565(n) < n]) # Indranil Ghosh, Apr 18 2017, after Chai Wah Wu

;; With Antti Karttunen's IntSeq-library.
(define A285317 (MATCHING-POS 1 0 (lambda (n) (and (< (A019565 n) n) (not (zero? (A008683 n)))))))

a(n) = A019565(A285318(n)).

cp's OEIS Frontend

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.

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A285332 a(0) = 1, a(1) = 2, a(2n) = A019565(a(n)), a(2n+1) = A065642(a(n)).

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A285315 Numbers n for which A019565(n) < n.

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A285320 If n == 0 or A008683(n) == 0, then a(n) = 0, otherwise a(n) = 1+a(A048675(n)); number of iterations of A048675 needed before the result is either zero or nonsquarefree number (A013929).

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A285331 Inverse for A285332: a(1) = 0, a(2) = 1, a(A019565(n)) = 2*a(n), a(A065642(n)) = 1 + 2*a(n).

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A285317 Squarefree numbers n for which A019565(n) < n.

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A286612 Numbers n for which A019565(n) <= A087207(n) < n.

A285331 Inverse for A285332: a(1) = 0, a(2) = 1, a(A019565(n)) = 2a(n), a(A065642(n)) = 1 + 2a(n).