A334860 a(0) = 1, a(1) = 2, after which, a(2n) = A334747(a(n)), a(2n+1) = a(n)^2.

1, 2, 3, 4, 6, 9, 8, 16, 5, 36, 18, 81, 12, 64, 32, 256, 10, 25, 72, 1296, 27, 324, 162, 6561, 24, 144, 128, 4096, 48, 1024, 512, 65536, 15, 100, 50, 625, 108, 5184, 2592, 1679616, 54, 729, 648, 104976, 243, 26244, 13122, 43046721, 20, 576, 288, 20736, 192, 16384, 8192, 16777216, 96, 2304, 2048, 1048576, 768, 262144, 131072, 4294967296, 30

Offset: 0

Author

Antti Karttunen, Jun 08 2020

Keywords

• nonn
• tabf

Comments

This irregular table can be represented as a binary tree. Each child to the left is obtained by applying A334747 to the parent, and each child to the right is obtained by squaring the parent:

1

|

...................2...................

3 4

6......../ \........9 8......../ \........16

/ \ / \ / \ / \

/ \ / \ / \ / \

/ \ / \ / \ / \

5 36 18 81 12 64 32 256

10 25 72 1296 27 324 162 6561 24 144 128 4096 48 1024 512 65536

etc.

This is the mirror image of the tree in A334866.

Fermi-Dirac primes, A050376, occur at rightward growing branches that originate from primes situated at the left edge.

The tree illustrated in A163511 is expanded as x -> 2*x for the left child and x -> A003961(x) for the right child, while this tree is expanded as x -> A225546(2*A225546(x)) for the left child, and x -> A225546(A003961(A225546(x))) for the right child.

Links

• Index entries for sequences that are permutations of the natural numbers

Crossrefs

Cf. A000290, A225546, A334204, A334747, A334859 (inverse), A334866 (mirror image).

Cf. A001146 (right edge of the tree), A019565 (left edge), A334110 (the right children of the left edge).

Composition of permutations A163511 and A225546.

Programs

PARI

A334747(n) = { my(c=core(n), m=n); forprime(p=2, , if(c % p, m*=p; break, m/=p)); m; }; \\ From A334747
A334860(n) = if(n<=1,1+n,if(!(n%2),A334747(A334860(n/2)),A334860((n-1)/2)^2));

Formula

a(0) = 1, a(1) = 2; and for n > 0, a(2n) = A334747(a(n)), a(2n+1) = a(n)^2.

a(n) = A225546(A163511(n)).

For n >= 0, a(2^n) = A019565(1+n), a(2^((2^n)-1)) = A000040(1+n).

A334109(a(n)) = A334204(n).

It seems that for n >= 1, A048675(a(n)) = A135529(n) = A048675(A163511(n)).