A341915 -id:A341915 - cp's OEIS frontend

A341916 Inverse permutation to A341915.

0, 1, 3, 2, 7, 4, 6, 5, 15, 8, 12, 11, 14, 9, 13, 10, 31, 16, 24, 23, 28, 19, 27, 20, 30, 17, 25, 22, 29, 18, 26, 21, 63, 32, 48, 47, 56, 39, 55, 40, 60, 35, 51, 44, 59, 36, 52, 43, 62, 33, 49, 46, 57, 38, 54, 41, 61, 34, 50, 45, 58, 37, 53, 42, 127, 64, 96

Offset: 0

Rémy Sigrist, Feb 24 2021

			A341915(5) = 7, so a(7) = 5.

Rémy Sigrist, Table of n, a(n) for n = 0..8191
Index entries for sequences related to binary expansion of n
Index entries for sequences that are permutations of the natural numbers
Rémy Sigrist, PARI program for A341916

Cf. A006068, A059893, A341915 (inverse), A341943 (fixed points).

PARI
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See Links section.
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a(n) = A006068(A059893(n)).

a(n) < 2^k for any n < 2^k.

A341943 Fixed points of A341915.

Original entry on oeis.org

0, 1, 6, 11, 54, 91, 438, 731, 3510, 5851, 28086, 46811, 224694, 374491, 1797558, 2995931, 14380470, 23967451, 115043766, 191739611, 920350134, 1533916891

Offset: 1

Rémy Sigrist, Feb 24 2021

			A341915(6) = 6, so 6 belongs to this sequence.
A341915(7) = 4, so 7 does not belong to this sequence.

Cf. A341915, A341916.

Apparently:
- a(1) = 1,
- a(2) = 6,
- a(2*k+1) = 8*a(2*k-1) + 3 for any k > 0,
- a(2*k+2) = 8*a(2*k) + 6 for any k > 0.

A163755 a(0)=1. For n>=1, write n in binary. Let b(n,m) be the length of the m-th run of 0's or 1's, reading right to left. Then a(n) = product{m=1 to M} p(m)^b(n,m), where p(m) is the m-th prime, and M is the number of runs of 0's and 1's in binary n.

Original entry on oeis.org

1, 2, 6, 4, 12, 30, 18, 8, 24, 90, 210, 60, 36, 150, 54, 16, 48, 270, 1050, 180, 420, 2310, 630, 120, 72, 450, 1470, 300, 108, 750, 162, 32, 96, 810, 5250, 540, 2100, 16170, 3150, 360, 840, 6930, 30030, 4620, 1260, 11550, 1890, 240, 144, 1350, 7350, 900, 2940, 25410

Offset: 0

Leroy Quet, Aug 03 2009

This sequence is a permutation of the terms of sequence A055932.

Clarification: By "run" of 0's or 1's in binary n, it is meant a group either entirely of 0's, and bounded by 1's or the edge of the binary number interpreted as a string, or entirely of 1's, and bounded by 0's or the edge of the string. In other words, the runs of 0's alternate with the runs of 1's.

			13 in binary is 1101. So reading right to left, there is a run of one 1, followed by a run of one 0, followed by a run of two 1's. So the lengths of the runs are 1,1,2. Therefore a(13) = p(1)^1 * p(2)^1 * p(3)^2 = 2^1 * 3^1 * 5^2 = 150.

Andrew Weimholt, Table of n, a(n) for n = 0..1000
Christian Krause, Simon Strandgaard, et al, A mined LODA assembly source for this sequence

Cf. A055932, A057335, A341915.

Cf. also A005940, A163511.

a(n) = A057335(A341915(n)). [Found by LODA miner, should be easy to prove] - Antti Karttunen, Apr 22 2022

cp's OEIS Frontend

A341916 Inverse permutation to A341915.

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A341943 Fixed points of A341915.

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A163755 a(0)=1. For n>=1, write n in binary. Let b(n,m) be the length of the m-th run of 0's or 1's, reading right to left. Then a(n) = product{m=1 to M} p(m)^b(n,m), where p(m) is the m-th prime, and M is the number of runs of 0's and 1's in binary n.

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