A246683 Permutation of natural numbers: a(1) = 1, a(n) = A000079(A055396(n+1)-1) * ((2*a(A246277(n+1))) - 1).

1, 2, 3, 4, 5, 8, 7, 6, 9, 16, 15, 32, 13, 10, 11, 64, 17, 128, 31, 18, 29, 256, 63, 12, 25, 14, 19, 512, 21, 1024, 127, 26, 33, 20, 255, 2048, 61, 58, 35, 4096, 57, 8192, 511, 30, 125, 16384, 23, 24, 49, 50, 27, 32768, 37, 36, 1023, 66, 41, 65536, 2047, 131072, 253, 62, 51, 52, 65, 262144, 39, 122, 509, 524288, 4095, 1048576, 121

Offset: 1

Author

Antti Karttunen, Sep 06 2014

Keywords

• nonn

Comments

See the comments in A246675. This is otherwise similar permutation, except that after having reached an even number 2m when we have shifted the prime factorization of n+1 k steps towards smaller primes, here, in contrary to A246675, we don't shift the binary representation of the odd number 2m-1, but instead of an odd number (2*a(m))-1 the same number (k) of bit-positions leftward, i.e. multiply it with 2^k.

See also the comments at the inverse permutation A246684.

Examples

			Consider 44 = 45-1. To find 45's position in array A246278, we start shifting its prime factorization 45 = 3 * 3 * 5 = p_2 * p_2 * p_3, step by step, until we get an even number, which in this case happens immediately after the first step, as p_1 * p_1 * p_2 = 2*2*3 = 12. 12 is in the 6th column of A246278, thus we take [here a(6) is computed recursively in the same way:] (2*a(6))-1 = (2*8)-1 = 15, "1111" in binary, and shift it one bit left (that is, multiply by 2), to give 2*15 = 30, thus a(44) = 30.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..2049
• Index entries for sequences that are permutations of the natural numbers

Crossrefs

Inverse: A246684.

Variants: A246675, A246677.

Cf. A000051, A000079, A005408, A055396, A246277, A246278,

A249813, A250244.

Differs from A249813 for the first time at n=20, where a(20) = 18, while A249813(20) = 14.

Formula

a(1) = 1, a(n) = A000079(A055396(n+1)-1) * ((2*a(A246277(n+1))) - 1).

As a composition of other permutations:

a(n) = A249813(A250244(n)).

Other identities. For all n >= 1, the following holds:

a(n) = (1+a((2*n)-1))/2. [The odd bisection from a(1) onward with one added and then halved gives the sequence back].

For all n >= 0, the following holds: a(A000051(n)) = A000051(n). [Numbers of the form 2^n + 1 are among the fixed points].