OFFSET
2,5
COMMENTS
This is a generalization of A000030.
The first occurrence of a number k in the sequence is given by A(k+1,k).
LINKS
Robert Israel, Table of n, a(n) for n = 2..10012 (first 142 antidiagonals, flattened)
FORMULA
From Robert Israel, May 31 2016: (Start)
A(n,k) = floor(k/n^floor(log_n(k))).
A(n,k) = k if n > k.
A(n,k) = A(n, floor(k/n)) otherwise.
G.f. of row n, G_n(x), satisfies G_n(x) = (1-x^n)/(1-x)^2 - (1+(n-1)*x^n)/(1-x) + (1-x^n)*G_n(x^n)/(1-x). (End)
EXAMPLE
First few rows of the array are:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1...
1, 2, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2...
1, 2, 3, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 1, 1, 1...
1, 2, 3, 4, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3...
1, 2, 3, 4, 5, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3...
1, 2, 3, 4, 5, 6, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2...
1, 2, 3, 4, 5, 6, 7, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2...
Note that the initial row is row 2.
A(3,3) corresponds to row n=3 and column k=3, and k=3 is written as 10 in base n=3, and the leading digit of 10 is 1, so A(3,3)=1.
A(12,11) corresponds to row n=12 and column k=11, and 11 is written as B in base 12, and the leading and only digit of B is B which is number 11 in decimal, so A(12,11)=11.
MAPLE
A:= (n, k) -> floor(k/n^floor(log[n](k))):
seq(seq(A(n-k, k), k=1..n-2), n=2..20); # Robert Israel, May 31 2016
MATHEMATICA
a[n_, k_] := First[IntegerDigits[k, n]];
PROG
(PARI) T(n, k) = digits(k, n)[1];
tabl(10, 10, n, k, n++; T(n, k)); \\ Michel Marcus, Jun 12 2016
CROSSREFS
KEYWORD
AUTHOR
Andrey Zabolotskiy, May 30 2016
STATUS
approved