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

0,2

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.

Links

Antti Karttunen, Table of n, a(n) for n = 0..40320

Index entries for sequences related to factorial base representation

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)

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 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.

Prog

(Scheme, with memoization-macro definec)
(definec (A275735 n) (if (zero? n) 1 (* (A000079 (A257511 n)) (A003961 (A275735 (A257684 n))))))
(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))])
print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 19 2017
(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

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.

Sequence in context: A248746 A227154 A324655 * A328835 A076435 A257010

Adjacent sequences: A275732 A275733 A275734 * A275736 A275737 A275738

Keyword

nonn,base,look

Author

Antti Karttunen, Aug 09 2016

Status

approved