A051845 Triangle T(n,k) read by rows, in which row n gives all permutations of digits 1..n interpreted in base n+1.

1, 5, 7, 27, 30, 39, 45, 54, 57, 194, 198, 214, 222, 238, 242, 294, 298, 334, 346, 358, 366, 414, 422, 434, 446, 482, 486, 538, 542, 558, 566, 582, 586, 1865, 1870, 1895, 1905, 1930, 1935, 2045, 2050, 2105, 2120, 2140, 2150, 2255, 2265, 2285, 2300, 2355, 2360

Offset: 1

Author

Antti Karttunen, Dec 13 1999

Keywords

• easy
• nonn
• tabf
• base

Comments

All terms in any odd row 2m+1 are divisible by 2m+1

The n-th row has n! elements.

Variant of permutational numbers with shifted digits 0->1->2->...->p+1 in p+1 positional system -- see A134750. - Artur Jasinski, Nov 08 2007

From Alexander R. Povolotsky, Oct 22 2022: (Start)

All terms in any even-indexed row n=2m are divisible by m, where m>0.

Row n starts with T(n,1) = ((n+1)^(n+1)-n^2-n-1)/n^2 = A023811(n+1).

Row n ends with T(n,n!) = ((n+1)^(n+1)*(n-1)+1)/n^2 = A051846(n).

(End)

Examples

			Triangle begins:
         k=1   k=2   k=3  ...
  n=1:     1;
  n=2      5,    7;
  n=3:    27,   30,  ...,   57;
  n=4:   194,  198,  214,  ..., 586;
  n=5:  1865, 1870, 1905, 1930, ..., 7465;
E.g., the permutations of digits 1, 2 and 3 in lexicographic order are 123, 132, 213, 231, 312, 321, which interpreted in base 4 give the third row of the table: 27, 30, 39, 45, 54, 57.
		

Links

• Samuel Harkness, Table of n, a(n) for n = 1..10000

Crossrefs

Left edge = A023811, right edge = A051846.

Cf. A134640, A134750.

Programs

Maple

with(combinat,permute); compute_u_rows := proc(u) local a,n; a := []; for n from 1 to u do a := [op(a),op(map(list_in_base_b,permute(n),(n+1)))]; od; RETURN(a); end; list_in_base_b := proc(l,b) local k; add(l[nops(l)-k]*(b^k), k=0..(nops(l)-1)); end;

Mathematica

a = {}; b = {}; Do[AppendTo[b, n]; w = Permutations[b]; Do[j = FromDigits[1 + w[[m]], n + 2]; AppendTo[a, j], {m, 1, Length[w]}], {n, 0, 5}]; a (* Artur Jasinski, Nov 08 2007 *)

Python

from itertools import permutations
def fd(d, b): return sum(di*b**i for i, di in enumerate(d[::-1]))
def row(n): return [fd(d, n+1) for d in permutations(range(1, n+1))]
print([an for r in range(1, 6) for an in row(r)]) # Michael S. Branicky, Oct 21 2022