A243505 Permutation of natural numbers, take the odd bisection of A122111 and divide the largest prime factor out: a(n) = A052126(A122111(2n-1)).

1, 2, 4, 8, 3, 16, 32, 6, 64, 128, 12, 256, 9, 5, 512, 1024, 24, 18, 2048, 48, 4096, 8192, 10, 16384, 27, 96, 32768, 36, 192, 65536, 131072, 20, 72, 262144, 384, 524288, 1048576, 15, 54, 2097152, 7, 4194304, 144, 768, 8388608, 108, 1536, 288, 16777216, 40, 33554432, 67108864, 30

Offset: 1

Author

Antti Karttunen, Jun 25 2014

Keywords

• nonn

Links

• Antti Karttunen, Table of n, a(n) for n = 1..1024
• Indranil Ghosh, Python program for computing the sequence
• Index entries for sequences that are permutations of the natural numbers

Crossrefs

Inverse: A243506.

Cf. A000040, A000079, A006254, A007051, A052126, A105560.

Related or similar permutations: A122111, A064216, A241916, A243065-A243066, A244981-A244982, A244983-A244984, A244153-A244154.

Programs

Mathematica

A052126[n_] := n/FactorInteger[n][[-1, 1]];
A122111[n_] := Product[Prime[Sum[If[j < i, 0, 1], {j, #}]], {i, Max[#]}]&[ Flatten[Table[Table[PrimePi[f[[1]]], {f[[2]]}], {f, FactorInteger[n]}]]];
a[n_] := A052126[A122111[2n-1]];
Array[a, 60] (* Jean-François Alcover, Sep 23 2020 *)

Scheme

(define (A243505 n) (A122111 (A064216 n)))

Formula

a(n) = A052126(A122111((2*n)-1)).

a(n) = A122111((2*n)-1) / A105560((2*n)-1).

As a composition of related permutations:

a(n) = A122111(A064216(n)).

a(n) = A241916(A243065(n)).

Other identities:

For all n >= 2, a(n) = A070003(A244984(n)-1) / A105560((2*n)-1).

For all n >= 1, a(A006254(n)) = A000079(n) and a(A007051(n)) = A000040(n).

For all n >= 1, A105560(2n-1) divides a(n).