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A275723
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Square array A(n,k) = prime factorization of n (= 1..) completely reduced by factorial base representation of k (= 0..), read by descending antidiagonals as A(1,0), A(1,1), A(2,0), A(1,2), A(2,1), A(3,0), etc. See the Comments section for the meaning of reduction in this context.
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4
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1, 1, 2, 1, 2, 3, 1, 2, 2, 4, 1, 2, 3, 4, 5, 1, 2, 2, 4, 5, 6, 1, 2, 3, 4, 3, 4, 7, 1, 2, 2, 4, 2, 6, 7, 8, 1, 2, 3, 4, 2, 4, 7, 8, 9, 1, 2, 2, 4, 2, 6, 7, 8, 4, 10, 1, 2, 3, 4, 5, 4, 7, 8, 9, 10, 11, 1, 2, 2, 4, 5, 6, 7, 8, 4, 6, 11, 12, 1, 2, 3, 4, 3, 4, 5, 8, 9, 4, 11, 8, 13, 1, 2, 2, 4, 2, 6, 5, 8, 4, 4, 11, 12, 13, 14, 1, 2, 3, 4, 2, 4, 3, 8, 9, 4, 11, 8, 13, 14, 15
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OFFSET
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1,3
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COMMENTS
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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.
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LINKS
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FORMULA
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Other identities. For all n >= 1 and k >= 0:
A001222(A(n,k)) = A001222(n). [This reduction doesn't change the total number of prime factors of n.]
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EXAMPLE
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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.
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PROG
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(Scheme)
(define (A275723bi n fex) (let loop ((n n) (fex fex)) (cond ((zero? fex) n) (else (loop (A273673bi n fex) (A257687 fex))))))
;; Code for A273673bi given in A273673.
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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