OFFSET
1,3
COMMENTS
The reduction is done by scanning the factorial base representation of k [see A007623] from its most significant end [where the most significant digit occurs at the one-based position A084558(k)], by adding the exponent of prime(1+A084558(k)) of n to the current exponent (possibly zero) of prime(1+A084558(k)-A099563(k)) in the prime factorization of n, after which the exponent of prime(1+A084558(k)) is changed to zero. Thus the total number of prime factors of n [A001222(n)] never changes. This single step of reduction is performed with a bivariate function A273673. The reduction then proceeds to the next digit to the right, effectively skipping any zeros until all factorial base digits of k have been scanned through and the prime factorization of n has been changed accordingly. See the examples.
This bivariate function is used to compute A275725.
LINKS
FORMULA
EXAMPLE
The top left 7 X 15 corner of the array:
1, 1, 1, 1, 1, 1, 1
2, 2, 2, 2, 2, 2, 2
3, 2, 3, 2, 3, 2, 3
4, 4, 4, 4, 4, 4, 4
5, 5, 3, 2, 2, 2, 5
6, 4, 6, 4, 6, 4, 6
7, 7, 7, 7, 7, 7, 5
8, 8, 8, 8, 8, 8, 8
9, 4, 9, 4, 9, 4, 9
10, 10, 6, 4, 4, 4, 10
11, 11, 11, 11, 11, 11, 11
12, 8, 12, 8, 12, 8, 12
13, 13, 13, 13, 13, 13, 13
14, 14, 14, 14, 14, 14, 10
15, 10, 9, 4, 6, 4, 15
For row 15 (above), we have 15 = 3*5 = prime(2)*prime(3) and the terms for columns 0 - 6 (in factorial base: 0, 1, 10, 11, 20, 21, 100, see A007623) are computed as:
When k=0, we do nothing and n stays as n (thus column 0 of array is A000027).
When k=1 (with the length 1), we transfer the exponent of prime(2) to prime(1), to get prime(1)*prime(3) = 2*5 = 10.
When k=2, in factorial base "10", with the length 2, we transfer (add) the exponent of prime(3) to prime(2), to get prime(2)*prime(2) = 9.
When k=3, in factorial base "11", we first do as above, to get 9 = prime(2)^2, and for the least significant one, we transfer (add) the exponent of prime(2) to prime(1), to get prime(1)*prime(1) = 4.
When k=4, in factorial base "20", with the length 2, we transfer (add) the exponent of prime(3) to prime(1), to get prime(2)*prime(1) = 6.
When k=5, in factorial base "21", we first do as above, to get 6 = prime(2)*prime(1), and for the remaining "1" in factorial base representation of k, we transfer (add) the exponent of prime(2) to prime(1), to get prime(1)*prime(1) = 4.
When k=6, in factorial base "100", with the length 3, we transfer (add) the exponent of prime(4) to prime(3), but prime(4) = 7 whose exponent is zero in 15, thus the result is also 15.
PROG
CROSSREFS
KEYWORD
AUTHOR
Antti Karttunen, Aug 09 2016
STATUS
approved