A362867 - OEIS

%I #26 Jul 15 2023 06:09:32

%S 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,

%T 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,

%U 1,3,0,1,3,2,0,3,1,2,0,2,3,1,0,3,2,1,0

%N 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.

%C 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.

%C 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 C(n,b) = Sum_{k >= 0} A(n,k)/(b^(Sum_{m=0..k} ceiling(log_b(A(n,m)+1))))

%C (The equation for Champernowne's constants using the n-th row of this triangle rather than 0,1,2,...)

%C C(0,b) is Champernowne's constant for base b (C_b).

%C 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.

%H Davis Smith, <a href="/A353962/a353962_3.txt">A Sufficient Condition for Normalcy</a>.

%F A(n,k) = A055089(n,k) - 1.

%e The irregular triangle begins (the numbers enclosed in square brackets are the continuation of the row):

%e n/k  0, 1, 2, 3, 4, 5, ...

%e 0  | 0 [1, 2, 3, 4, 5, ...]

%e 1  | 1, 0 [2, 3, 4, 5, ...]

%e 2  | 0, 2, 1 [3, 4, 5, ...]

%e 3  | 2, 0, 1 [3, 4, 5, ...]

%e 4  | 1, 2, 0 [3, 4, 5, ...]

%e 5  | 2, 1, 0 [3, 4, 5, ...]

%e 6  | 0, 1, 3, 2 [4, 5, ...]

%e 7  | 1, 0, 3, 2 [4, 5, ...]

%e 8  | 0, 3, 1, 2 [4, 5, ...]

%e 9  | 3, 0, 1, 2 [4, 5, ...]

%e 10 | 1, 3, 0, 2 [4, 5, ...]

%o (PARI) A362867_row(n)=my(N=n,m=0);while(N\=m++, );Vecrev(vecextract(abs([-(m-1)..0]),numtoperm(m,n)))

%Y Cf. A055089, A084558, A353962.

%K nonn,tabf

%O 0,5

%A _Davis Smith_, May 06 2023