cp's OEIS Frontend

A362867 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.

%I A362867 #26 Jul 15 2023 06:09:32
%S A362867 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 A362867 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 A362867 1,3,0,1,3,2,0,3,1,2,0,2,3,1,0,3,2,1,0
%N A362867 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 A362867 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 A362867 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 A362867 C(n,b) = Sum_{k >= 0} A(n,k)/(b^(Sum_{m=0..k} ceiling(log_b(A(n,m)+1))))
%C A362867 (The equation for Champernowne's constants using the n-th row of this triangle rather than 0,1,2,...)
%C A362867 C(0,b) is Champernowne's constant for base b (C_b).
%C A362867 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 A362867 Davis Smith, <a href="/A353962/a353962_3.txt">A Sufficient Condition for Normalcy</a>.
%F A362867 A(n,k) = A055089(n,k) - 1.
%e A362867 The irregular triangle begins (the numbers enclosed in square brackets are the continuation of the row):
%e A362867 n/k  0, 1, 2, 3, 4, 5, ...
%e A362867 0  | 0 [1, 2, 3, 4, 5, ...]
%e A362867 1  | 1, 0 [2, 3, 4, 5, ...]
%e A362867 2  | 0, 2, 1 [3, 4, 5, ...]
%e A362867 3  | 2, 0, 1 [3, 4, 5, ...]
%e A362867 4  | 1, 2, 0 [3, 4, 5, ...]
%e A362867 5  | 2, 1, 0 [3, 4, 5, ...]
%e A362867 6  | 0, 1, 3, 2 [4, 5, ...]
%e A362867 7  | 1, 0, 3, 2 [4, 5, ...]
%e A362867 8  | 0, 3, 1, 2 [4, 5, ...]
%e A362867 9  | 3, 0, 1, 2 [4, 5, ...]
%e A362867 10 | 1, 3, 0, 2 [4, 5, ...]
%o A362867 (PARI) A362867_row(n)=my(N=n,m=0);while(N\=m++, );Vecrev(vecextract(abs([-(m-1)..0]),numtoperm(m,n)))
%Y A362867 Cf. A055089, A084558, A353962.
%K A362867 nonn,tabf
%O A362867 0,5
%A A362867 _Davis Smith_, May 06 2023