A101362 -id:A101362 - cp's OEIS frontend

A019583 a(n) = n*(n-1)^4/2.

0, 0, 1, 24, 162, 640, 1875, 4536, 9604, 18432, 32805, 55000, 87846, 134784, 199927, 288120, 405000, 557056, 751689, 997272, 1303210, 1680000, 2139291, 2693944, 3358092, 4147200, 5078125, 6169176, 7440174, 8912512, 10609215, 12555000, 14776336, 17301504, 20160657

Offset: 0

N. J. A. Sloane

a(n) = n(n-1)^4/2 is half the number of colorings of 5 points on a line with n colors. - R. H. Hardin, Feb 23 2002

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Milan Janjic and Boris Petkovic, A Counting Function, arXiv 1301.4550 [math.CO], 2013.
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A101362.

[n*(n-1)^4/2: n in [0..30]]; // Vincenzo Librandi, Apr 20 2012

CoefficientList[Series[x^2*(1+18*x+33*x^2+8*x^3)/(1-x)^6,{x,0,40}],x] (* Vincenzo Librandi, Apr 20 2012 *)
a[n_] := n*(n - 1)^4/2; Array[a, 30, 0] (* Amiram Eldar, Feb 13 2023 *)

Sum_{j>=2} 1/a(j) = hypergeom([1, 1, 1, 1, 1], [ 2, 2, 2, 3], 1) = -2 + 2*zeta(2) - 2*zeta(3) + 2*zeta(4). - Stephen Crowley, Jun 28 2009
G.f.: x^2*(1 + 18*x + 33*x^2 + 8*x^3)/(1 - x)^6. - Colin Barker, Feb 23 2012
From Amiram Eldar, Feb 13 2023: (Start)
a(n) = A101362(n-1)/2.
Sum_{n>=2} (-1)^n/a(n) = 2 + Pi^2/6 + 7*Pi^4/360 - 4*log(2) - 3*zeta(3)/2. (End)

A158497 Triangle T(n,k) formed by the coordination sequences and the number of leaves for trees.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 3, 6, 12, 1, 4, 12, 36, 108, 1, 5, 20, 80, 320, 1280, 1, 6, 30, 150, 750, 3750, 18750, 1, 7, 42, 252, 1512, 9072, 54432, 326592, 1, 8, 56, 392, 2744, 19208, 134456, 941192, 6588344, 1, 9, 72, 576, 4608, 36864, 294912, 2359296, 18874368, 150994944, 1, 10, 90, 810, 7290, 65610, 590490, 5314410, 47829690, 430467210, 3874204890

Offset: 0

Thomas Wieder, Mar 20 2009

Consider the k-fold Cartesian products CP(n,k) of the vector A(n) = [1, 2, 3, ..., n].
An element of CP(n,k) is a n-tuple T_t of the form T_t = [i_1, i_2, i_3, ..., i_k] with t=1, .., n^k.
We count members T of CP(n,k) which satisfy some condition delta(T_t), so delta(.) is an indicator function which attains values of 1 or 0 depending on whether T_t is to be counted or not; the summation sum_{CP(n,k)} delta(T_t) over all elements T_t of CP produces the count.
For the triangle here we have delta(T_t) = 0 if for any two i_j, i_(j+1) in T_t one has i_j = i_(j+1): T(n,k) = Sum_{CP(n,k)} delta(T_t) = Sum_{CP(n,k)} delta(i_j = i_(j+1)).
The test on i_j > i_(j+1) generates A158498. One gets the Pascal triangle A007318 if the indicator function tests whether for any two i_j, i_(j+1) in T_t one has i_j >= i_(j+1).
Use of other indicator functions can also calculate the Bell numbers A000110, A000045 or A000108.

			Array, A(n, k) = n*(n-1)^(k-1) for n > 1, A(n, k) = 1 otherwise, begins as:
  1,  1,   1,    1,     1,      1,       1,        1,        1, ... A000012;
  1,  1,   1,    1,     1,      1,       1,        1,        1, ... A000012;
  1,  2,   2,    2,     2,      2,       2,        2,        2, ... A040000;
  1,  3,   6,   12,    24,     48,      96,      192,      384, ... A003945;
  1,  4,  12,   36,   108,    324,     972,     2916,     8748, ... A003946;
  1,  5,  20,   80,   320,   1280,    5120,    20480,    81920, ... A003947;
  1,  6,  30,  150,   750,   3750,   18750,    93750,   468750, ... A003948;
  1,  7,  42,  252,  1512,   9072,   54432,   326592,  1959552, ... A003949;
  1,  8,  56,  392,  2744,  19208,  134456,   941192,  6588344, ... A003950;
  1,  9,  72,  576,  4608,  36864,  294912,  2359296, 18874368, ... A003951;
  1, 10,  90,  810,  7290,  65610,  590490,  5314410, 47829690, ... A003952;
  1, 11, 110, 1100, 11000, 110000, 1100000, 11000000, ............. A003953;
  1, 12, 132, 1452, 15972, 175692, 1932612, 21258732, ............. A003954;
  1, 13, 156, 1872, 22464, 269568, 3234816, 38817792, ............. A170732;
  ... ;
The triangle begins as:
  1
  1, 1;
  1, 2,  2;
  1, 3,  6,  12;
  1, 4, 12,  36,  108;
  1, 5, 20,  80,  320,  1280;
  1, 6, 30, 150,  750,  3750,  18750;
  1, 7, 42, 252, 1512,  9072,  54432, 326592;
  1, 8, 56, 392, 2744, 19208, 134456, 941192, 6588344;
  ...;
T(3,3) = 12 counts the triples (1,2,1), (1,2,3), (1,3,1), (1,3,2), (2,1,2), (2,1,3), (2,3,1), (2,3,2), (3,1,2), (3,1,3), (3,2,1), (3,2,3) out of a total of 3^3 = 27 triples in the CP(3,3).

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Array rows n: A170733 (n=14), ..., A170769 (n=50).
Columns k: A000012(n) (k=0), A000027(n) (k=1), A002378(n-1) (k=2), A011379(n-1) (k=3), A179824(n) (k=4), A101362(n-1) (k=5), 2*A168351(n-1) (k=6), 2*A168526(n-1) (k=7), 2*A168635(n-1) (k=8), 2*A168675(n-1) (k=9), 2*A170783(n-1) (k=10), 2*A170793(n-1) (k=11).
Diagonals k: A055897 (k=n), A055541 (k=n-1), A373395 (k=n-2), A379612 (k=n-3).
Sums: (-1)^n*A065440(n) (signed row).
Cf. A000045, A000108, A000110, A007318, A079901, A158498.

A158497:= func< n,k | k le 1 select n^k else n*(n-1)^(k-1) >;
[A158497(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Mar 18 2025

A158497[n_, k_]:= If[n<2 || k==0, 1, n*(n-1)^(k-1)];
Table[A158497[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Mar 18 2025 *)

def A158497(n,k): return n^k if k<2 else n*(n-1)^(k-1)
print(flatten([[A158497(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Mar 18 2025

T(n, k) = (n-1)^(k-1) + (n-1)^k = n*A079901(n-1,k-1), k > 0.

Sum_{k=0..n} T(n,k) = (n*(n-1)^n - 2)/(n-2), n > 2.

Edited by R. J. Mathar, Mar 31 2009

More terms added by G. C. Greubel, Mar 18 2025

cp's OEIS Frontend

A019583 a(n) = n*(n-1)^4/2.

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