A057335 -id:A057335 - cp's OEIS frontend

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.

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

A358171 The a(n)-th composition in standard order (A066099) is the first differences plus one of the prime indices of n (A112798).

Original entry on oeis.org

0, 0, 0, 1, 0, 2, 0, 3, 1, 4, 0, 6, 0, 8, 2, 7, 0, 5, 0, 12, 4, 16, 0, 14, 1, 32, 3, 24, 0, 10, 0, 15, 8, 64, 2, 13, 0, 128, 16, 28, 0, 20, 0, 48, 6, 256, 0, 30, 1, 9, 32, 96, 0, 11, 4, 56, 64, 512, 0, 26, 0, 1024, 12, 31, 8, 40, 0, 192, 128, 18, 0, 29, 0

Offset: 1

Gus Wiseman, Dec 21 2022

nonn

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. This gives a bijective correspondence between nonnegative integers and integer compositions.

A prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798.

			The prime indices of 36 are {1,1,2,2}, with first differences plus one (1,2,1), which is the 13th composition in standard order, so a(36) = 13.

Gus Wiseman, Statistics, classes, and transformations of standard compositions

See link for sequences related to standard compositions.
Prepend 1 to indices: A253566 (cf. A358169), inverse A253565 (cf. A242628).
Taking Heinz number instead of standard composition number gives A325352.
These compositions minus one are listed by A355536, sums A243055.
A001222 counts prime indices, distinct A001221.
A066099 lists standard compositions, lengths A000120, sums A070939.
A112798 lists prime indices, sum A056239.
A355534 = augmented diffs. of rev. prime indices, Heinz numbers A325351.
Cf. A000720, A005940, A055932, A057335, A066312, A286470, A287352, A325390.

primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
stcinv[q_]:=Total[2^(Accumulate[Reverse[q]])]/2;
Table[stcinv[Differences[primeMS[n]]+1],{n,100}]

cp's OEIS Frontend

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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