OFFSET
1,1
COMMENTS
The n-th row of this triangle contains n uniformly located n-digit numbers, i.e., n terms of an arithmetic progression with 10^(n-1)-1 as the term preceding the first term and (n+1)-th term is the largest possible n-digit term.
Starting with n=2, the n-th row of this triangle can be obtained by deleting the least significant digit, 9, from terms ending in 9 in the (n+1)-th row, and ignoring the main diagonal terms, of the triangle in A093846.
Floor(A093846(4,1)/10) = T(3,1) = 324, but floor(A093846(2,1)/10) = 5 and T(1,1) = 4, floor(A093846(7,1)/10) = 228571 and T(6,1) = 228570, etc. - Michael De Vlieger, Jul 18 2016
LINKS
G. C. Greubel, Rows n = 1..100 of triangle, flattened
EXAMPLE
Triangle begins with:
4;
39, 69;
324, 549, 774;
2799, 4599, 6399, 8199;
24999, 39999, 54999, 69999, 84999;
....
MAPLE
A093850 := proc(n, r)
10^(n-1)-1+r*floor(9*10^(n-1)/(n+1)) ;
end proc:
seq(seq(A093850(n, r), r=1..n), n=1..14) ; # R. J. Mathar, Sep 28 2011
MATHEMATICA
Table[# -1 +r*Floor[9*#/(n+1)] &[10^(n-1)], {n, 8}, {r, n}]//Flatten (* Michael De Vlieger, Jul 18 2016 *)
PROG
(PARI) {T(n, k) = 10^(n-1) -1 +k*floor(9*10^(n-1)/(n+1))}; \\ G. C. Greubel, Mar 21 2019
(Magma) [[10^(n-1) -1 +k*Floor(9*10^(n-1)/(n+1)): k in [1..n]]: n in [1..8]]; // G. C. Greubel, Mar 21 2019
(Sage) [[10^(n-1) -1 +k*floor(9*10^(n-1)/(n+1)) for k in (1..n)] for n in (1..8)] # G. C. Greubel, Mar 21 2019
CROSSREFS
KEYWORD
AUTHOR
Amarnath Murthy, Apr 18 2004
EXTENSIONS
Second comment clarified by Michael De Vlieger, Jul 18 2016
Edited by G. C. Greubel, Mar 21 2019
STATUS
approved