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A362867
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Irregular triangle read by rows; the n-th row is the n-th permutation of 0 to infinity, in reversed colexicographic ordering, terminating when the rest of the row equals k.
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0
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0, 1, 0, 0, 2, 1, 2, 0, 1, 1, 2, 0, 2, 1, 0, 0, 1, 3, 2, 1, 0, 3, 2, 0, 3, 1, 2, 3, 0, 1, 2, 1, 3, 0, 2, 3, 1, 0, 2, 0, 2, 3, 1, 2, 0, 3, 1, 0, 3, 2, 1, 3, 0, 2, 1, 2, 3, 0, 1, 3, 2, 0, 1, 1, 2, 3, 0, 2, 1, 3, 0, 1, 3, 2, 0, 3, 1, 2, 0, 2, 3, 1, 0, 3, 2, 1, 0
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OFFSET
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0,5
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COMMENTS
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Every row of this triangle represents a number which is normal in base b >= 2. Since there are infinitely many bases, every row represents a set of infinitely many numbers, each of which is normal in at least one base.
Treating A(n,k) as equal to k when k is greater than A084558(n) (A084558(n) is the length of the n-th row minus one), for any base b >= 2, the concatenation of the base-b expansion of the n-th row is normal in base b. In other words, for any n >= 0 and b >= 2, C(n,b) is normal in base b.
C(n,b) = Sum_{k >= 0} A(n,k)/(b^(Sum_{m=0..k} ceiling(log_b(A(n,m)+1))))
(The equation for Champernowne's constants using the n-th row of this triangle rather than 0,1,2,...)
C(0,b) is Champernowne's constant for base b (C_b).
Even though for any b and m >= 2, b != m, C(n,b) != C(n,m), it is possible for C(n,b) = C(m,p) where n != m and b != p. In such a case, C(n,b) is normal in two different bases. m will likely be significantly larger than n and p will likely be a power of b.
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LINKS
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FORMULA
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EXAMPLE
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The irregular triangle begins (the numbers enclosed in square brackets are the continuation of the row):
n/k 0, 1, 2, 3, 4, 5, ...
0 | 0 [1, 2, 3, 4, 5, ...]
1 | 1, 0 [2, 3, 4, 5, ...]
2 | 0, 2, 1 [3, 4, 5, ...]
3 | 2, 0, 1 [3, 4, 5, ...]
4 | 1, 2, 0 [3, 4, 5, ...]
5 | 2, 1, 0 [3, 4, 5, ...]
6 | 0, 1, 3, 2 [4, 5, ...]
7 | 1, 0, 3, 2 [4, 5, ...]
8 | 0, 3, 1, 2 [4, 5, ...]
9 | 3, 0, 1, 2 [4, 5, ...]
10 | 1, 3, 0, 2 [4, 5, ...]
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PROG
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(PARI) A362867_row(n)=my(N=n, m=0); while(N\=m++, ); Vecrev(vecextract(abs([-(m-1)..0]), numtoperm(m, n)))
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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