A085390 -id:A085390 - cp's OEIS frontend

A228273 T(n,k) is the number of s in {1,...,n}^n having longest ending contiguous subsequence with the same value of length k; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

1, 0, 1, 0, 2, 2, 0, 18, 6, 3, 0, 192, 48, 12, 4, 0, 2500, 500, 100, 20, 5, 0, 38880, 6480, 1080, 180, 30, 6, 0, 705894, 100842, 14406, 2058, 294, 42, 7, 0, 14680064, 1835008, 229376, 28672, 3584, 448, 56, 8, 0, 344373768, 38263752, 4251528, 472392, 52488, 5832, 648, 72, 9

Offset: 0

Alois P. Heinz, Aug 19 2013

			T(0,0) = 1: [].
T(1,1) = 1: [1].
T(2,1) = 2: [1,2], [2,1].
T(2,2) = 2: [1,1], [2,2].
T(3,1) = 18: [1,1,2], [1,1,3], [1,2,1], [1,2,3], [1,3,1], [1,3,2], [2,1,2], [2,1,3], [2,2,1], [2,2,3], [2,3,1], [2,3,2], [3,1,2], [3,1,3], [3,2,1], [3,2,3], [3,3,1], [3,3,2].
T(3,2) = 6: [1,2,2], [1,3,3], [2,1,1], [2,3,3], [3,1,1], [3,2,2].
T(3,3) = 3: [1,1,1], [2,2,2], [3,3,3].
Triangle T(n,k) begins:
  1;
  0,        1;
  0,        2,       2;
  0,       18,       6,      3;
  0,      192,      48,     12,     4;
  0,     2500,     500,    100,    20,    5;
  0,    38880,    6480,   1080,   180,   30,   6;
  0,   705894,  100842,  14406,  2058,  294,  42,  7;
  0, 14680064, 1835008, 229376, 28672, 3584, 448, 56,  8;

Alois P. Heinz, Rows n = 0..140, flattened

Row sums give: A000312.
Columns k=0-4 give: A000007, A066274(n) = 2*A081131(n) for n>1, A053506(n) for n>2, A055865(n-1) = A085389(n-1) for n>3, A085390(n-1) for n>4.
Main diagonal gives: A028310.
Lower diagonals include (offsets may differ): A002378, A045991, A085537, A085538, A085539.
Cf. A228154, A228617.

T:= (n, k)-> `if`(n=0 and k=0, 1, `if`(k<1 or k>n, 0,
             `if`(k=n, n, (n-1)*n^(n-k)))):
seq(seq(T(n,k), k=0..n), n=0..12);

f[0,0]=1;
f[n_,k_]:=Which[1<=k<=n-1,n^(n-k)*(n-1),k<1,0,k==n,n,k>n,0];
Table[Table[f[n,k],{k,0,n}],{n,0,10}]//Grid (* Geoffrey Critzer, May 19 2014 *)

T(0,0) = 1, else T(n,k) = 0 for k<1 or k>n, else T(n,n) = n, else T(n,k) = (n-1)*n^(n-k).
Sum_{k=0..n}   T(n,k) = A000312(n).
Sum_{k=0..n} k*T(n,k) = A031972(n).

A085389 a(1) = 1; for n >= 2, a(n) = (n*(n+1)^(n-1))/(n+1).

Original entry on oeis.org

1, 2, 12, 100, 1080, 14406, 229376, 4251528, 90000000, 2143588810, 56757583872, 1654301902188, 52644347205632, 1816448730468750, 67553994410557440, 2694045224950414864, 114692890480116793344, 5191945444217181018258, 249036800000000000000000, 12617615847934310595791220

Offset: 1

Paul Barry, Jun 30 2003

Paolo Xausa, Table of n, a(n) for n = 1..350
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets

Cf. A053506, A085390.

A085389[n_] := If[n == 1, 1, n*(n+1)^(n-2)];
Array[A085389, 25] (* Paolo Xausa, Aug 07 2025 *)

Main subdiagonal of A085388.
a(n) = A055865(n), n>1. - R. J. Mathar, Sep 12 2008
a(n) = [x^n] x*(1 - x)/(1 - x - n*x). - Ilya Gutkovskiy, Oct 02 2017

Name edited by Paolo Xausa, Aug 07 2025

A085388 First differences of n^k.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 2, 0, 1, 3, 6, 4, 0, 1, 4, 12, 18, 8, 0, 1, 5, 20, 48, 54, 16, 0, 1, 6, 30, 100, 192, 162, 32, 0, 1, 7, 42, 180, 500, 768, 486, 64, 0, 1, 8, 56, 294, 1080, 2500, 3072, 1458, 128, 0, 1, 9, 72, 448, 2058, 6480, 12500, 12288, 4374, 256, 0, 1, 10, 90, 648

Offset: 1

Paul Barry, Jun 30 2003

T(n,k) is the number of k-digit numbers in base n; n,k >= 2. - Mohammed Yaseen, Nov 11 2022

			Rows begin
  1,   0,   0,   0,   0, ...
  1,   1,   2,   4,   8, ...
  1,   2,   6,  18,  54, ...
  1,   3,  12,  48, 192, ...
  1,   4,  20, 100, 500, ...

Rows include A011782, A025192, A002001, A005054, A052934, A055272, A055274, A055275, A052268, A055276.
Diagonals include A053506, A085389, A085390.
Row-wise binomial transform is A083064.

T(n,k) = (n-1)*n^(k-1) + 0^k/n. - Corrected by Mohammed Yaseen, Nov 11 2022

T(n,0) = 1; T(n,k) = n^k - n^(k-1) for k >= 1. - Mohammed Yaseen, Nov 11 2022

Offset corrected by Mohammed Yaseen, Nov 11 2022

cp's OEIS Frontend

A228273 T(n,k) is the number of s in {1,...,n}^n having longest ending contiguous subsequence with the same value of length k; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

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A085389 a(1) = 1; for n >= 2, a(n) = (n*(n+1)^(n-1))/(n+1).

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A085388 First differences of n^k.

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