A253565 Permutation of natural numbers: a(0) = 1, a(1) = 2; after which, a(2n) = A253550(a(n)), a(2n+1) = A253560(a(n)).

1, 2, 3, 4, 5, 9, 6, 8, 7, 25, 15, 27, 10, 18, 12, 16, 11, 49, 35, 125, 21, 75, 45, 81, 14, 50, 30, 54, 20, 36, 24, 32, 13, 121, 77, 343, 55, 245, 175, 625, 33, 147, 105, 375, 63, 225, 135, 243, 22, 98, 70, 250, 42, 150, 90, 162, 28, 100, 60, 108, 40, 72, 48, 64, 17, 169, 143, 1331, 91, 847, 539, 2401, 65, 605, 385, 1715, 275, 1225, 875, 3125, 39

Offset: 0

Author

Antti Karttunen, Jan 03 2015

Keywords

• nonn
• tabf
• look

Comments

This sequence can be represented as a binary tree. Each child to the left is obtained by applying A253550 to the parent, and each child to the right is obtained by applying A253560 to the parent:

1

|

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

3 4

5......../ \........9 6......../ \........8

/ \ / \ / \ / \

/ \ / \ / \ / \

/ \ / \ / \ / \

7 25 15 27 10 18 12 16

11 49 35 125 21 75 45 81 14 50 30 54 20 36 24 32

etc.

Sequence A253563 is the mirror image of the same tree. Also in binary trees A005940 and A163511 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. Of these four trees, this is the one where the left child is always smaller than the right child.

Note that the indexing of sequence starts from 0, although its range starts from one.

The term a(n) is the Heinz number of the adjusted partial sums of the n-th composition in standard order, where (1) the k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again, (2) the Heinz number of a partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k), and (3) we define the adjusted partial sums of a composition to be obtained by subtracting one from all parts, taking partial sums, and adding one back to all parts. See formula for a simplification. A triangular form is A242628. The inverse is A253566. The non-adjusted version is A358170. - Gus Wiseman, Dec 17 2022

Examples

			From _Gus Wiseman_, Dec 23 2022: (Start)
This represents the following bijection between compositions and partitions. The n-th composition in standard order together with the reversed prime indices of a(n) are:
   0:        () -> ()
   1:       (1) -> (1)
   2:       (2) -> (2)
   3:     (1,1) -> (1,1)
   4:       (3) -> (3)
   5:     (2,1) -> (2,2)
   6:     (1,2) -> (2,1)
   7:   (1,1,1) -> (1,1,1)
   8:       (4) -> (4)
   9:     (3,1) -> (3,3)
  10:     (2,2) -> (3,2)
  11:   (2,1,1) -> (2,2,2)
  12:     (1,3) -> (3,1)
  13:   (1,2,1) -> (2,2,1)
  14:   (1,1,2) -> (2,1,1)
  15: (1,1,1,1) -> (1,1,1,1)
(End)
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..8191
• Index entries for sequences that are permutations of the natural numbers

Crossrefs

Inverse: A253566.

Cf. A252737 (row sums), A252738 (row products).

Cf. A122111, A163511, A253550, A253560, A252464, A253563, A054429.

Applying A001222 gives A000120.

A reverse version is A005940.

These are the Heinz numbers of the rows of A242628.

Sum of prime indices of a(n) is A359043, reverse A161511.

A048793 gives partial sums of reversed standard comps, Heinz number A019565.

A066099 lists standard compositions.

A112798 list prime indices, sum A056239.

A358134 gives partial sums of standard compositions, Heinz number A358170.

Cf. A029837, A059893, A070939, A241916, A358135, A358137, A358195, A359042.

Programs

Mathematica

stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Times@@Prime/@#&/@Table[Accumulate[stc[n]-1]+1,{n,0,60}] (* Gus Wiseman, Dec 17 2022 *)

Formula

a(0) = 1, a(1) = 2; after which, a(2n) = A253550(a(n)), a(2n+1) = A253560(a(n)).

As a composition of related permutations:

a(n) = A122111(A163511(n)).

a(n) = A253563(A054429(n)).

Other identities and observations. For all n >= 0:

a(2n+1) - a(2n) > 0. [See the comment above.]

If n = 2^(x_1)+...+2^(x_k) then a(n) = Product_{i=1..k} prime(x_k-x_{i-1}-k+i) where x_0 = 0. - Gus Wiseman, Dec 23 2022