A057335 - OEIS

A057335

a(0) = 1, and for n > 0, a(n) = A000040(A000120(n)) * a(floor(n/2)); essentially sequence A055932 generated using A000120, hence sorted by number of factors.

1, 2, 4, 6, 8, 12, 18, 30, 16, 24, 36, 60, 54, 90, 150, 210, 32, 48, 72, 120, 108, 180, 300, 420, 162, 270, 450, 630, 750, 1050, 1470, 2310, 64, 96, 144, 240, 216, 360, 600, 840, 324, 540, 900, 1260, 1500, 2100, 2940, 4620, 486, 810, 1350, 1890, 2250, 3150, 4410

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OFFSET

0,2

COMMENTS

Note that for n>0 the prime divisors of a(n) are consecutive primes starting with 2. All of the least prime signatures (A025487) are included; with the other values forming A056808.

Using the formula, terms of b(n)= a(n)/A057334(n) are: 1, 1, 2, 2, 4, 4, 6, 6, 8, ..., indeed a(n) repeated. - Michel Marcus, Feb 09 2014

a(n) is the unique normal number whose unsorted prime signature is the k-th composition in standard order (graded reverse-lexicographic). This composition (row k of 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. A number is normal if its prime indices cover an initial interval of positive integers. Unsorted prime signature is the sequence of exponents in a number's prime factorization. - Gus Wiseman, Apr 19 2020

LINKS

Michael De Vlieger, Table of n, a(n) for n = 0..10000

FORMULA

a(n) = A057334(n) * a (repeated).

A334032(a(n)) = n; a(A334032(n)) = A071364(n). - Gus Wiseman, Apr 19 2020

a(n) = A122111(A019565(n)); A019565(n) = A122111(a(n)). - Peter Munn, Jul 18 2020

a(n) = A336321(2^n). - Peter Munn, Mar 04 2022

EXAMPLE

From Gus Wiseman, Apr 19 2020: (Start)

The sequence of terms together with their prime indices begins:

1: {}

2: {1}

4: {1,1}

6: {1,2}

8: {1,1,1}

12: {1,1,2}

18: {1,2,2}

30: {1,2,3}

16: {1,1,1,1}

24: {1,1,1,2}

36: {1,1,2,2}

60: {1,1,2,3}

54: {1,2,2,2}

90: {1,2,2,3}

150: {1,2,3,3}

210: {1,2,3,4}

32: {1,1,1,1,1}

48: {1,1,1,1,2}

For example, the 27th composition in standard order is (1,2,1,1), and the normal number with prime signature (1,2,1,1) is 630 = 2*3*3*5*7, so a(27) = 630.

(End)

MATHEMATICA

Table[Times @@ Map[If[# == 0, 1, Prime@ #] &, Accumulate@ IntegerDigits[n, 2]], {n, 0, 54}] (* Michael De Vlieger, May 23 2017 *)

PROG

(PARI) mg(n) = if (n==0, 1, prime(hammingweight(n))); \\ A057334

lista(nn) = {my(v = vector(nn)); v[1] = 1; for (i=2, nn, v[i] = mg(i-1)*v[(i+1)\2]; ); v; } \\ Michel Marcus, Feb 09 2014

(PARI) A057335(n) = if(0==n, 1, prime(hammingweight(n))*A057335(n\2)); \\ Antti Karttunen, Jul 20 2020

CROSSREFS

Cf. A000120, A057334, A055932 and A056808.

Cf. A324939.

Unsorted prime signature is A124010.

Numbers whose prime signature is aperiodic are A329139.