A364963 -id:A364963 - cp's OEIS frontend

A163511 a(0)=1. a(n) = p(A000120(n)) * Product_{m=1..A000120(n)} p(m)^A163510(n,m), where p(m) is the m-th prime.

1, 2, 4, 3, 8, 9, 6, 5, 16, 27, 18, 25, 12, 15, 10, 7, 32, 81, 54, 125, 36, 75, 50, 49, 24, 45, 30, 35, 20, 21, 14, 11, 64, 243, 162, 625, 108, 375, 250, 343, 72, 225, 150, 245, 100, 147, 98, 121, 48, 135, 90, 175, 60, 105, 70, 77, 40, 63, 42, 55, 28, 33, 22, 13, 128

Offset: 0

Leroy Quet, Jul 29 2009

This is a permutation of the positive integers.
From Antti Karttunen, Jun 20 2014: (Start)
Note the indexing: the domain starts from 0, while the range excludes zero, thus this is neither a bijection on the set of nonnegative integers nor on the set of positive natural numbers, but a bijection from the former set to the latter.
Apart from that discrepancy, this could be viewed as yet another "entanglement permutation" where the two complementary pairs to be interwoven together are even and odd numbers (A005843/A005408) which are entangled with the complementary pair even numbers (taken straight) and odd numbers in the order they appear in A003961: (A005843/A003961). See also A246375 which has almost the same recurrence.
Note how the even bisection halved gives the same sequence back. (For a(0)=1, take ceiling of 1/2).
(End)
From Antti Karttunen, Dec 30 2017: (Start)
This irregular table can be represented as a binary tree. Each child to the left is obtained by doubling the parent, and each child to the right is obtained by applying A003961 to the parent:
                                     1
                                     |
                  ...................2...................
                 4                                       3
       8......../ \........9                   6......../ \........5
      / \                 / \                 / \                 / \
     /   \               /   \               /   \               /   \
    /     \             /     \             /     \             /     \
  16       27         18       25         12       15         10       7
32  81   54  125    36  75   50  49     24  45   30  35     20  21   14 11
etc.
Sequence A005940 is obtained by scanning the same tree level by level in mirror image fashion. Also in binary trees A253563 and A253565 the terms on level of the tree are some permutation of the terms present on the level n of this tree. A252464(n) gives the distance of n from 1 in all these trees, and A252463 gives the parent of the node containing n.
A252737(n) gives the sum and A252738(n) the product of terms on row n (where 1 is on row 0, 1 on row 1, 3 and 4 on row 2, etc.). A252745(n) gives the number of nodes on level n whose left child is smaller than the right child, and A252744(n) is an indicator function for those nodes.
(End)
Note that the idea behind maps like this (and the mirror image A005940) admits also using alternative orderings of primes, not just standard magnitude-wise ordering (A000040). For example, A332214 is a similar sequence but with primes rearranged as in A332211, and A332817 is obtained when primes are rearranged as in A108546. - Antti Karttunen, Mar 11 2020
From Lorenzo Sauras Altuzarra, Nov 28 2020: (Start)
This sequence is generated from A228351 by applying the following procedure: 1) eliminate the compositions that end in one unless the first one, 2) subtract one unit from every component, 3) replace every tuple [t_1, ..., t_r] by Product_{k=1..r} A000040(k)^(t_k) (see the examples).
Is it true that a(n) = A337909(n+1) if and only if a(n+1) is not a term of A161992?
Does this permutation have any other cycle apart from (1), (2) and (6, 9, 16, 7)? (End)
From Antti Karttunen, Jul 25 2023: (Start)
(In the above question, it is assumed that the starting offset would be 1 instead of 0).
Questions:
Does a(n) = 1+A054429(n) hold only when n is of the form 2^k times 1, 3 or 7, i.e., one of the terms of A029748?
It seems that A007283 gives all fixed points of map n -> a(n), like A335431 seems to give all fixed points of map n -> A332214(n). Is there a general rule for mappings like these that the fixed points (if they exist) must be of the form 2^k times a certain kind of prime, i.e., that any odd composite (times 2^k) can certainly be excluded? See also note in A029747.
(End)
If the conjecture given in A364297 holds, then it implies the above conjecture about A007283. See also A364963. - Antti Karttunen, Sep 06 2023
Conjecture: a(n^k) is never of the form x^k, for any integers n > 0, k > 1, x >= 1. This holds at least for squares, cubes, seventh and eleventh powers (see A365808, A365801, A366287 and A366391). - Antti Karttunen, Sep 24 2023, Oct 10 2023.
See A365805 for why the above holds for any n^k, with k > 1. - Antti Karttunen, Nov 23 2023

			For n=3, whose binary representation is "11", we have A000120(3)=2, with A163510(3,1) = A163510(3,2) = 0, thus a(3) = p(2) * p(1)^0 * p(2)^0 = 3*1*1 = 3.
For n=9, "1001" in binary, we have A000120(9)=2, with A163510(9,1) = 0 and A163510(9,2) = 2, thus a(9) = p(2) * p(1)^0 * p(2)^2 = 3*1*9 = 27.
For n=10, "1010" in binary, we have A000120(10)=2, with A163510(10,1) = 1 and A163510(10,2) = 1, thus a(10) = p(2) * p(1)^1 * p(2)^1 = 3*2*3 = 18.
For n=15, "1111" in binary, we have A000120(15)=4, with A163510(15,1) = A163510(15,2) = A163510(15,3) = A163510(15,4) = 0, thus a(15) = p(4) * p(1)^0 * p(2)^0 * p(3)^0 * p(4)^0 = 7*1*1*1*1 = 7.
[1], [2], [1,1], [3], [1,2], [2,1] ... -> [1], [2], [3], [1,2], ... -> [0], [1], [2], [0,1], ... -> 2^0, 2^1, 2^2, 2^0*3^1, ... = 1, 2, 4, 3, ... - _Lorenzo Sauras Altuzarra_, Nov 28 2020

Inverse: A243071.
Cf. A000040, A000120, A000225, A000788, A003961, A007814, A054429, A055396, A064216, A135523, A161992, A163510, A245605, A245612, A246375, A246378, A246681, A161511, A228351, A243499, A243503, A243504, A269854, A280873, A285727, A290251, A293437, A337909.
Cf. A007283 (known positions where a(n)=n), A029747, A029748, A364255 [= gcd(n,a(n))], A364258 [= a(n)-n], A364287 (where a(n) < n), A364292 (where a(n) <= n), A364494 (where n|a(n)), A364496 (where a(n)|n), A364963, A364297.
Cf. A365808 (positions of squares), A365801 (of cubes), A365802 (of fifth powers), A365805 [= A052409(a(n))], A366287, A366391.
Cf. A005940, A332214, A332817, A366275 (variants).

f[n_] := Reverse@ Map[Ceiling[(Length@ # - 1)/2] &, DeleteCases[Split@ Join[Riffle[IntegerDigits[n, 2], 0], {0}], {k__} /; k == 1]]; {1}~Join~
Table[Function[t, Prime[t] Product[Prime[m]^(f[n][[m]]), {m, t}]][DigitCount[n, 2, 1]], {n, 120}] (* Michael De Vlieger, Jul 25 2016 *)

from sympy import prime
def A163511(n):
    if n:
        k, c, m = n, 0, 1
        while k:
            c += 1
            m *= prime(c)**(s:=(~k&k-1).bit_length())
            k >>= s+1
        return m*prime(c)
    return 1 # Chai Wah Wu, Jul 17 2023

For n >= 1, a(2n) is even, a(2n+1) is odd. a(2^k) = 2^(k+1), for all k >= 0.
From Antti Karttunen, Jun 20 2014: (Start)
a(0) = 1, a(1) = 2, a(2n) = 2*a(n), a(2n+1) = A003961(a(n)).
As a more general observation about the parity, we have:
For n >= 1, A007814(a(n)) = A135523(n) = A007814(n) + A209229(n). [This permutation preserves the 2-adic valuation of n, except when n is a power of two, in which cases that value is incremented by one.]
For n >= 1, A055396(a(n)) = A091090(n) = A007814(n+1) + 1 - A036987(n).
For n >= 1, a(A000225(n)) = A000040(n).
(End)
From Antti Karttunen, Oct 11 2014: (Start)
As a composition of related permutations:
a(n) = A005940(1+A054429(n)).
a(n) = A064216(A245612(n))
a(n) = A246681(A246378(n)).
Also, for all n >= 0, it holds that:
A161511(n) = A243503(a(n)).
A243499(n) = A243504(a(n)).
(End)
More linking identities from Antti Karttunen, Dec 30 2017: (Start)
A046523(a(n)) = A278531(n). [See also A286531.]
A278224(a(n)) = A285713(n). [Another filter-sequence.]
A048675(a(n)) = A135529(n) seems to hold for n >= 1.
A250245(a(n)) = A252755(n).
A252742(a(n)) = A252744(n).
A245611(a(n)) = A253891(n).
A249824(a(n)) = A275716(n).
A292263(a(n)) = A292264(n). [A292944(n) + A292264(n) = n.]
--
A292383(a(n)) = A292274(n).
A292385(a(n)) = A292271(n). [A292271(n) +  A292274(n) = n.]
--
A292941(a(n)) = A292942(n).
A292943(a(n)) = A292944(n).
A292945(a(n)) = A292946(n). [A292942(n) + A292944(n) + A292946(n) = n.]
--
A292253(a(n)) = A292254(n).
A292255(a(n)) = A292256(n). [A292944(n) + A292254(n) + A292256(n) = n.]
--
A279339(a(n)) = A279342(n).
a(A071574(n)) = A269847(n).
a(A279341(n)) = A279338(n).
a(A252756(n)) = A250246(n).
(1+A008836(a(n)))/2 = A059448(n).
(End)
From Antti Karttunen, Jul 26 2023: (Start)
For all n >= 0, a(A007283(n)) = A007283(n).
A001222(a(n)) = A290251(n).
(End)

More terms computed and examples added by Antti Karttunen, Jun 20 2014

A364496 Numbers k such that k is a multiple of A163511(k).

Original entry on oeis.org

0, 3, 6, 12, 24, 48, 96, 192, 384, 768, 1536, 3072, 6144, 12288, 16383, 24576, 32766, 49152, 65532, 98304, 131064, 196608, 262128, 393216, 524256, 786432, 1048512, 1572864, 2097024, 3145728, 4194048, 6291456, 8388096, 12582912, 16776192, 25165824, 33552384, 50331648, 67104768, 100663296, 134209536, 201326592

Offset: 1

Antti Karttunen, Jul 27 2023

nonn

If n is present, then 2*n is also present, and vice versa.
A007283 is included as a subsequence, because it gives the known fixed points of map n -> A163511(n).
Sequence A243071(A364497(.)) sorted into ascending order.

			16383 is present, because A163511(16383) = 43, as 16383 = 2^14 - 1 and A000040(14) = 43, and 43 is a factor of 16383 = 3*43*127.
536870895 is present, because A163511(536870895) = 1177 (11*107), which divides 536870895 (3*5*11*47*107*647). See also example in A364498.

Antti Karttunen, Table of n, a(n) for n = 1..53

Positions of 1's in A364492.
Subsequence of A364292.
Cf. A007283 (subsequence), A163511, A364963 (odd terms).
Cf. also A364295, A364494, A364497, A364498.

A005940(n) = { my(p=2, t=1); n--; until(!n\=2, if((n%2), (t*=p), p=nextprime(p+1))); t };
A054429(n) = ((3<<#binary(n\2))-n-1);
A163511(n) = if(!n,1,A005940(1+A054429(n)))
isA364496(n) = !(n%A163511(n));

A364495 Odd numbers k such that k divides A163511(k).

Original entry on oeis.org

1, 3, 9, 105, 429, 1365, 1617, 3887, 4235, 10829, 14025, 17745, 21125, 22627, 38025, 54587, 70805, 100555, 115159, 147875, 168751, 169065, 175769, 181447, 181545, 291525, 297297, 303875, 338675, 350987, 501787, 513825, 518035, 549081, 560947, 566865, 594473, 624169, 676039, 735875, 745147, 831875, 869193, 957125

Offset: 1

Antti Karttunen, Jul 27 2023

nonn

			For n = 513825 = 3 * 5^2 * 13 * 17 * 31, A163511(n) = 13873275 = 3^4 * 5^2 * 13 * 17 * 31, so A163511(n)/n = 27 (which is an integer), and therefore 513825 is included in this sequence.

Antti Karttunen, Table of n, a(n) for n = 1..801

Odd terms in A364494.
After 1, sequence A243071(A364965(n)), for n>=1, sorted into ascending order.
Cf. A163511.
Cf. also A364296, A364498, A364545, A364963.

A005940(n) = { my(p=2, t=1); n--; until(!n\=2, if((n%2), (t*=p), p=nextprime(p+1))); t };
A054429(n) = ((3<<#binary(n\2))-n-1);
A163511(n) = if(!n,1,A005940(1+A054429(n)))
A163511(n) = if(!n,1,my(p=2, t=1); while(n>1, if(!(n%2), (t*=p), p=nextprime(1+p)); n >>= 1); (t*p));
isA364495(n) = ((n%2)&&!(A163511(n)%n));

A364297 a(n) = A348717(A163511(n)).

Original entry on oeis.org

1, 2, 4, 2, 8, 4, 6, 2, 16, 8, 18, 4, 12, 6, 10, 2, 32, 16, 54, 8, 36, 18, 50, 4, 24, 12, 30, 6, 20, 10, 14, 2, 64, 32, 162, 16, 108, 54, 250, 8, 72, 36, 150, 18, 100, 50, 98, 4, 48, 24, 90, 12, 60, 30, 70, 6, 40, 20, 42, 10, 28, 14, 22, 2, 128, 64, 486, 32, 324, 162, 1250, 16, 216, 108, 750, 54, 500, 250, 686, 8, 144

Offset: 0

Antti Karttunen, Aug 15 2023

nonn

For all i, j: a(i) = a(j) => A278531(i) = A278531(j).
As the underlying sequence A163511 can be represented as a binary tree, so can this be also:
                                       1
                                       |
                    ...................2...................
                   4                                       2
         8......../ \........4                   6......../ \........2
        / \                 / \                 / \                 / \
       /   \               /   \               /   \               /   \
      /     \             /     \             /     \             /     \
    16       8          18       4          12       6          10       2
  32  16   54 8       36  18   50 4       24  12   30 6       20  10   14 2
etc.
Each rightward leaning branch stays constant, because a(2n+1) = a(n).
Conjecture: Mersenne primes (A000668) gives all such odd numbers k for which a(k) = A348717(k). If true, then it immediately implies that map n -> A163511(n) [or equally: map n ->  A243071(n)] has no other fixed points than those given by A007283. But see also A364959. - Edited Sep 03 2023

Cf. A000265, A000668, A007283, A163511, A243071, A348717, A364949, A364956, A365423.

Cf. also A245611, A245612, A278531, A286531, A364959, A364963.

A163511(n) = if(!n,1,my(p=2, t=1); while(n>1, if(!(n%2), (t*=p), p=nextprime(1+p)); n >>= 1); (t*p));
A348717(n) = if(1==n, 1, my(f = factor(n), k = primepi(f[1, 1])-1); for (i=1, #f~, f[i, 1] = prime(primepi(f[i, 1])-k)); factorback(f));
A364297(n) = A348717(A163511(n));

a(0) = 1, a(1) = 2, a(2n) = A163511(2n) = 2*A163511(n), and for n > 0, a(2n+1) = a(n).

A364547 Odd numbers k such that k is a multiple of A005940(k).

Original entry on oeis.org

1, 3, 5, 1035, 524295, 16777217

Offset: 1

Antti Karttunen, Jul 28 2023

Sequence A005941(A364549(.)) sorted into ascending order.
Those terms of A000051 (= 2^k + 1) are included that have A000040(1+k) as one of their prime factors.
a(7) > 402653184.
See also comments in A364963. - Antti Karttunen, Jan 12 2024

			1035 is included because 1034 in binary is "10000001010", which Doudna isomorphism maps to 345 = 3*5*23, which thus divides 1035 (= 3^2 * 5 * 23). Note that there are six 0's in the binary representation between its most significant bit and the trailing "1010", thus we get the prime factors A000040(1+1) = 3, A000040(1+1+1) = 5 and A000040(1+1+1+6) = 23.
524295 is included because 524294 in binary is "10000000000000000110", which Doudna isomorphism maps to 549 = 3^2 * 61, which thus divides 524295 (= 3^2 * 5 * 61 * 191). Note that there are sixteen 0's in the binary representation between its most significant bit and the trailing "110", thus we get the prime factors A000040(2) = 3 and A000040(2+16) = 61.
16777217 = 2^24 + 1 is included because A000040(1+24) = 97, and 16777217 = 97*257*673.

Odd terms in A364546.
Cf. A000040, A000051, A005940, A005941, A364502, A364549.
Cf. also A364545, A364551, A364963.

nn = 2^20 + 2; Array[Set[a[#], #] &, 2]; {1}~Join~Reap[Do[If[EvenQ[n], Set[a[n], 2 a[n/2]], a[n] = k = Times @@ Power @@@ Map[{Prime[PrimePi[#1] + 1], #2} & @@ # &, FactorInteger[a[(n + 1)/2]]]; If[Divisible[n, a[n]], Sow[n]]], {n, 3, nn}] ][[-1, 1]] (* Michael De Vlieger, Jul 28 2023 *)

A005940(n) = { my(p=2, t=1); n--; until(!n\=2, if((n%2), (t*=p), p=nextprime(p+1))); t };
isA364547(n) = ((n%2)&&!(n%A005940(n)));

cp's OEIS Frontend

A163511 a(0)=1. a(n) = p(A000120(n)) * Product_{m=1..A000120(n)} p(m)^A163510(n,m), where p(m) is the m-th prime.

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A364496 Numbers k such that k is a multiple of A163511(k).

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A364495 Odd numbers k such that k divides A163511(k).

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A364297 a(n) = A348717(A163511(n)).

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A364547 Odd numbers k such that k is a multiple of A005940(k).

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