A275735 Prime-factorization representations of "factorial base level polynomials": a(0) = 1; for n >= 1, a(n) = 2^A257511(n) * A003961(a(A257684(n))).

1, 2, 2, 4, 3, 6, 2, 4, 4, 8, 6, 12, 3, 6, 6, 12, 9, 18, 5, 10, 10, 20, 15, 30, 2, 4, 4, 8, 6, 12, 4, 8, 8, 16, 12, 24, 6, 12, 12, 24, 18, 36, 10, 20, 20, 40, 30, 60, 3, 6, 6, 12, 9, 18, 6, 12, 12, 24, 18, 36, 9, 18, 18, 36, 27, 54, 15, 30, 30, 60, 45, 90, 5, 10, 10, 20, 15, 30, 10, 20, 20, 40, 30, 60, 15, 30, 30, 60, 45, 90, 25, 50, 50, 100, 75

Offset: 0

Author

Antti Karttunen, Aug 09 2016

Keywords

• nonn
• base
• look

Comments

These are prime-factorization representations of single-variable polynomials where the coefficient of term x^(k-1) (encoded as the exponent of prime(k) in the factorization of n) is equal to the number of times a nonzero digit k occurs in the factorial base representation of n. See the examples.

Examples

			For n = 0 whose factorial base representation (A007623) is also 0, there are no nonzero digits at all, thus there cannot be any prime present in the encoding, and a(0) = 1.
For n = 1 there is just one 1, thus a(1) = prime(1) = 2.
For n = 2 ("10"), there is just one 1-digit, thus a(2) = prime(1) = 2.
For n = 3 ("11") there are two 1-digits, thus a(3) = prime(1)^2 = 4.
For n = 18 ("300") there is just one 3, thus a(18) = prime(3) = 5.
For n = 19 ("301") there is one 1 and one 3, thus a(19) = prime(1)*prime(3) = 2*5 = 10.
For n = 141 ("10311") there are three 1's and one 3, thus a(141) = prime(1)^3 * prime(3) = 2^3 * 5^1 = 40.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..40320
• Index entries for sequences related to factorial base representation

Crossrefs

Cf. A000079, A001221, A001222, A003961, A007623, A008683, A181819, A225901, A257511, A257684, A265349, A265350, A264990, A275729, A275806, A351954.

Cf. also A275725, A275733, A275734 for other such prime factorization encodings of A060117/A060118-related polynomials, and also A276076.

Differs from A227154 for the first time at n=18, where a(18) = 5, while A227154(18) = 4.

Programs

PARI

A276076(n) = { my(i=0,m=1,f=1,nextf); while((n>0),i=i+1; nextf = (i+1)*f; if((n%nextf),m*=(prime(i)^((n%nextf)/f));n-=(n%nextf));f=nextf); m; };
A181819(n) = factorback(apply(e->prime(e),(factor(n)[,2])));
A275735(n) = A181819(A276076(n)); \\ Antti Karttunen, Apr 03 2022

Python

from sympy import prime
from operator import mul
import collections
def a007623(n, p=2): return n if n
Indranil Ghosh, Jun 19 2017

Formula

a(0) = 1; for n >= 1, a(n) = 2^A257511(n) * A003961(a(A257684(n))).

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

a(n) = A275734(A225901(n)).

A001221(a(n)) = A275806(n).

A001222(a(n)) = A060130(n).

A048675(a(n)) = A275729(n).

A051903(a(n)) = A264990(n).

A008683(a(A265349(n))) = -1 or +1 for all n >= 0.

A008683(a(A265350(n))) = 0 for all n >= 1.

From Antti Karttunen, Apr 03 2022: (Start)

A342001(a(n)) = A351954(n).

a(n) = A181819(A276076(n)). (End)