A275734 Prime-factorization representations of "factorial base slope polynomials": a(0) = 1; for n >= 1, a(n) = A275732(n) * a(A257684(n)).

1, 2, 3, 6, 2, 4, 5, 10, 15, 30, 10, 20, 3, 6, 9, 18, 6, 12, 2, 4, 6, 12, 4, 8, 7, 14, 21, 42, 14, 28, 35, 70, 105, 210, 70, 140, 21, 42, 63, 126, 42, 84, 14, 28, 42, 84, 28, 56, 5, 10, 15, 30, 10, 20, 25, 50, 75, 150, 50, 100, 15, 30, 45, 90, 30, 60, 10, 20, 30, 60, 20, 40, 3, 6, 9, 18, 6, 12, 15, 30, 45, 90, 30, 60, 9, 18, 27

Offset: 0

Author

Antti Karttunen, Aug 08 2016

Keywords

• nonn
• base

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 nonzero digits that occur on the slope (k-1) levels below the "maximal slope" in the factorial base representation of n. See A275811 for the definition of the "digit slopes" in this context.

Examples

			For n=23 ("321" in factorial base representation, A007623), all three nonzero digits are maximal for their positions (they all occur on "maximal slope"), thus a(23) = prime(1)^3 = 2^3 = 8.
For n=29 ("1021"), there are three nonzero digits, where both 2 and the rightmost 1 are on the "maximal slope", while the most significant 1 is on the "sub-sub-sub-maximal", thus a(29) = prime(1)^2 * prime(4)^1 = 2*7 = 28.
For n=37 ("1201"), there are three nonzero digits, where the rightmost 1 is on the maximal slope, 2 is on the sub-maximal, and the most significant 1 is on the "sub-sub-sub-maximal", thus a(37) = prime(1) * prime(2) * prime(4) = 2*3*7 = 42.
For n=55 ("2101"), the least significant 1 is on the maximal slope, and the digits "21" at the beginning are together on the sub-sub-maximal slope (as they are both two less than the maximal digit values 4 and 3 allowed in those positions), thus a(55) = prime(1)^1 * prime(3)^2 = 2*25 = 50.
		

Links

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

Crossrefs

Cf. A001221, A001222, A002110, A007489, A007814, A048675, A051903, A056169, A056170, A060130, A060502, A225901.

Cf. A257684, A275732.

Cf. A275811.

Cf. A260736, A275728, A275808, A275812, A275946, A275947, A275962.

Cf. A275804 (indices of squarefree terms), A275805 (of terms not squarefree).

Cf. also A275725, A275733, A275735, A276076 for other such prime factorization encodings of A060117/A060118-related polynomials.

Programs

Python

from operator import mul
from sympy import prime, factorial as f
def a007623(n, p=2): return n if n
0 else '0' for i in x)[::-1]
    return 0 if n==1 else sum(int(y[i])*f(i + 1) for i in range(len(y)))
def a(n): return 1 if n==0 else a275732(n)*a(a257684(n))
print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 19 2017

Formula

a(0) = 1; for n >= 1, a(n) = A275732(n) * a(A257684(n)).

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

a(n) = A275735(A225901(n)).

a(A007489(n)) = A002110(n).

A001221(a(n)) = A060502(n).

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

A007814(a(n)) = A260736(n).

A051903(a(n)) = A275811(n).

A048675(a(n)) = A275728(n).

A248663(a(n)) = A275808(n).

A056169(a(n)) = A275946(n).

A056170(a(n)) = A275947(n).

A275812(a(n)) = A275962(n).