

A275735


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


21



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
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OFFSET

0,2


COMMENTS

These are primefactorization representations of singlevariable polynomials where the coefficient of term x^(k1) (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.


LINKS



FORMULA

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


EXAMPLE

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 1digit, thus a(2) = prime(1) = 2.
For n = 3 ("11") there are two 1digits, 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.


PROG

(Scheme, with memoizationmacro definec)
(Python)
from sympy import prime
from operator import mul
import collections
def a007623(n, p=2): return n if n<p else a007623(n//p, p+1)*10 + n%p
def a(n):
y=collections.Counter(map(int, list(str(a007623(n)).replace("0", "")))).most_common()
return 1 if n==0 else reduce(mul, [prime(y[i][0])**y[i][1] for i in range(len(y))])
(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])));


CROSSREFS

Cf. A000079, A001221, A001222, A003961, A007623, A008683, A181819, A225901, A257511, A257684, A265349, A265350, A264990, A275729, A275806, A351954.
Differs from A227154 for the first time at n=18, where a(18) = 5, while A227154(18) = 4.


KEYWORD



AUTHOR



STATUS

approved



