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.

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

Offset

0,5

Comments

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.

Links

Table of n, a(n) for n=0..86.

Davis Smith, A Sufficient Condition for Normalcy.

Formula

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

Example

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, ...]

Prog

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

Crossrefs

Cf. A055089, A084558, A353962.

Sequence in context: A171099 A337253 A127173 * A035160 A353328 A027414

Adjacent sequences: A362864 A362865 A362866 * A362868 A362869 A362870

Keyword

nonn,tabf

Author

Davis Smith, May 06 2023

Status

approved