A179865 -id:A179865 - cp's OEIS frontend

A221857 Number A(n,k) of shapes of balanced k-ary trees with n nodes, where a tree is balanced if the total number of nodes in subtrees corresponding to the branches of any node differ by at most one; square array A(n,k), n>=0, k>=0, read by antidiagonals.

1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 2, 1, 0, 1, 1, 3, 1, 1, 0, 1, 1, 4, 3, 4, 1, 0, 1, 1, 5, 6, 1, 4, 1, 0, 1, 1, 6, 10, 4, 9, 4, 1, 0, 1, 1, 7, 15, 10, 1, 27, 1, 1, 0, 1, 1, 8, 21, 20, 5, 16, 27, 8, 1, 0, 1, 1, 9, 28, 35, 15, 1, 96, 81, 16, 1, 0, 1, 1, 10, 36, 56, 35, 6, 25, 256, 81, 32, 1, 0

Offset: 0

Alois P. Heinz, Apr 10 2013

			: A(2,2) = 2  : A(2,3) = 3      : A(3,3) = 3          :
:   o     o   :   o    o    o   :   o      o      o   :
:  / \   / \  :  /|\  /|\  /|\  :  /|\    /|\    /|\  :
: o         o : o      o      o : o o    o   o    o o :
:.............:.................:.....................:
: A(3,4) = 6                                          :
:    o        o        o        o       o        o    :
:  /( )\    /( )\    /( )\    /( )\   /( )\    /( )\  :
: o o      o   o    o     o    o o     o   o      o o :
Square array A(n,k) begins:
  1, 1, 1,  1,   1,   1,  1,  1,  1,   1,   1, ...
  1, 1, 1,  1,   1,   1,  1,  1,  1,   1,   1, ...
  0, 1, 2,  3,   4,   5,  6,  7,  8,   9,  10, ...
  0, 1, 1,  3,   6,  10, 15, 21, 28,  36,  45, ...
  0, 1, 4,  1,   4,  10, 20, 35, 56,  84, 120, ...
  0, 1, 4,  9,   1,   5, 15, 35, 70, 126, 210, ...
  0, 1, 4, 27,  16,   1,  6, 21, 56, 126, 252, ...
  0, 1, 1, 27,  96,  25,  1,  7, 28,  84, 210, ...
  0, 1, 8, 81, 256, 250, 36,  1,  8,  36, 120, ...

Alois P. Heinz, Antidiagonals n = 0..140, flattened
Jeffrey Barnett, Counting Balanced Tree Shapes, 2007
Samuele Giraudo, Intervals of balanced binary trees in the Tamari lattice, arXiv:1107.3472 [math.CO], Apr 2012

Columns k=1-10 give: A000012, A110316, A131889, A131890, A131891, A131892, A131893, A229393, A229394, A229395.
Rows n=0+1, 2-3, give: A000012, A001477, A179865.
Diagonal and upper diagonals give: A028310, A000217, A000292, A000332, A000389, A000579, A000580, A000581, A000582, A001287, A001288.
Lower diagonals give: A000012, A000290, A092364(n) for n>1.

A:= proc(n, k) option remember; local m, r; if n<2 or k=1 then 1
      elif k=0 then 0 else r:= iquo(n-1, k, 'm');
      binomial(k, m)*A(r+1, k)^m*A(r, k)^(k-m) fi
    end:
seq(seq(A(n, d-n), n=0..d), d=0..12);

a[n_, k_] := a[n, k] = Module[{m, r}, If[n < 2 || k == 1, 1, If[k == 0, 0, {r, m} = QuotientRemainder[n-1, k]; Binomial[k, m]*a[r+1, k]^m*a[r, k]^(k-m)]]]; Table[a[n, d-n], {d, 0, 12}, {n, 0, d}] // Flatten (* Jean-François Alcover, Apr 17 2013, translated from Maple *)

A253145 Triangular numbers (A000217) omitting the term 1.

Original entry on oeis.org

0, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666, 703, 741, 780, 820, 861, 903, 946, 990, 1035, 1081, 1128, 1176, 1225, 1275

Offset: 0

Paul Curtz, Mar 23 2015

The full triangle of the inverse Akiyama-Tanigawa transform applied to (-1)^n*A062510(n)=3*(-1)^n*A001045(n) yielding a(n) is
0,      3,    6,  10,   15,  21,  28, 36, ...
-3,    -6,  -12, -20,  -30, -42, -56, ...    essentially -A002378
3,     12,   24,  40,   60,  84, ...         essentially  A046092
-9,   -24,  -48, -80, -120, ...              essentially -A033996
15,    48,   96, 160, ...
-33,  -96, -192, ...
63,   192, ...
-129, ...
etc.
First column: (-1)^n*A062510(n).
The following columns are multiples of A122803(n)=(-2)^n. See A007283(n), A091629(n), A020714(n+1), A110286, A175805(n), 4*A005010(n).
An autosequence of the first kind is a sequence whose main diagonal is A000004 = 0's.
b(n) = 0, 0 followed by a(n) is an autosequence of the first kind.
The successive differences of b(n) are
0,   0,  0, 3, 6, 10, 15, 21, ...
0,   0,  3, 3, 4,  5,  6,  7, ...  see A194880(n)
0,   3,  0, 1, 1,  1,  1,  1, ...
3,  -3,  1, 0, 0,  0,  0,  0, ...
-6,  4, -1, 0, 0,  0,  0,  0, ...
10, -5,  1, 0, 0,  0,  0,  0, ...
-15, 6, -1, 0, 0,  0,  0,  0, ...
21, -7,  1, 0, 0,  0,  0,  0, ...
The inverse binomial transform (first column) is the signed sequence. This is general.
Also generalized hexagonal numbers without 1. - Omar E. Pol, Mar 23 2015

Muniru A Asiru, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A179865, A001045, A062510, A002378, A046092, A033996, A122803, A007283, A091629, A020714, A110286, A161680, A175805, A005010, A194880, A001840, A022003, A255935.

Concatenation([0],List([1..50],n->(n+1)*(n+2)/2)); # Muniru A Asiru, Oct 31 2018

Prepend[Table[(n + 1) (n + 2)/2, {n, 49}], 0] (* Michael De Vlieger, Mar 23 2015 *)

a(n)=if(n,(n+1)*(n+2)/2,0) \\ Charles R Greathouse IV, Mar 23 2015

Inverse Akiyama-Tanigawa transform of (-1)^n*A062510(n).
a(n) = (n+1)*(n+2)/2 for n > 0. - Charles R Greathouse IV, Mar 23 2015
a(n+1) = 3*A001840(n+1) + A022003(n).
a(n) = A161680(n+2) for n >= 1. - Georg Fischer, Oct 30 2018
From Stefano Spezia, May 28 2025: (Start)
G.f.: x*(3 - 3*x + x^2)/(1 - x)^3.
E.g.f.: exp(x)*(2 + 4*x + x^2)/2 - 1. (End)

A372973 Triangle read by rows: the exponential almost-Riordan array ( 1/(1-x) | 1/(1-x), log(1/(1-x)) ).

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 6, 2, 3, 1, 24, 6, 11, 6, 1, 120, 24, 50, 35, 10, 1, 720, 120, 274, 225, 85, 15, 1, 5040, 720, 1764, 1624, 735, 175, 21, 1, 40320, 5040, 13068, 13132, 6769, 1960, 322, 28, 1, 362880, 40320, 109584, 118124, 67284, 22449, 4536, 546, 36, 1

Offset: 0

Stefano Spezia, May 26 2024

			The triangle begins:
    1;
    1,   1;
    2,   1,   1;
    6,   2,   3,   1;
   24,   6,  11,   6,  1;
  120,  24,  50,  35, 10,  1;
  720, 120, 274, 225, 85, 15, 1;
  ...

Y. Alp and E. G. Kocer, Exponential Almost-Riordan Arrays, Results Math 79, 173 (2024). See page 6.

Cf. A000012 (right diagonal), A000254, A000399 (k=3), A000454 (k=4), A000482 (k=5), A001233 (k=6), A001234 (k=7), A098558 (row sums), A179865 (subdiagonal), A243569 (k=8), A243570 (k=9).

Triangle A130534 with 1st column A000142.

T[n_,0]:=n!; T[n_,k_]:=(n-1)!/(k-1)!SeriesCoefficient[1/(1-x)Log[1/(1-x)]^(k-1),{x,0,n-1}]; Table[T[n,k],{n,0,9},{k,0,n}]//Flatten

T(n,0) = n!; T(n,k) = (n-1)!/(k-1)! * [x^(n-1)] log(1/(1-x))^(k-1)/(1-x).
T(n,1) = (n-1)! for n > 0.
T(n,2) = A000254(n-1) for n > 1.

A386319 Triangle read by rows where row n is the start, corner and end vertex numbers of a triangular spiral with n sides on a triangular grid, starting from 1 and working inwards (0 <= k <= n).

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 5, 6, 1, 4, 7, 9, 10, 1, 5, 9, 12, 14, 15, 1, 6, 11, 15, 18, 20, 21, 1, 7, 13, 18, 22, 25, 27, 28, 1, 8, 15, 21, 26, 30, 33, 35, 36, 1, 9, 17, 24, 30, 35, 39, 42, 44, 45, 1, 10, 19, 27, 34, 40, 45, 49, 52, 54, 55, 1, 11, 21, 30, 38, 45, 51, 56, 60, 63, 65, 66, 1, 12, 23, 33, 42, 50, 57, 63, 68, 72, 75, 77, 78

Offset: 0

Binay Krishna Maity, Jul 18 2025

The first 2 sides are length n-1 so that T(n,1) = 1 + (n-1) and T(n,2) = 1 + 2*(n-1) and then the side lengths decrease by 1 each time as it spirals in (ending at triangular number A000217(n) when n>=1).
These sides mesh to fill the triangle as they go inwards, and can also be thought of going outwards tracing out the sides of the triangle.
The resulting vertex numbers are 1 together with row n of A141419.
Row n=1 is taken as a side of length 0 so the start and end numbers are both 1 (which is not really a spiral but is consistent with the formula and two points 1,2 would be even less like a triangle filled by a spiral).

			Triangle begins:
--------------------------------------
   n\k  0   1   2   3   4   5   6   7
--------------------------------------
   0|   1;
   1|   1,  1;
   2|   1,  2,  3;
   3|   1,  3,  5,  6;
   4|   1,  4,  7,  9, 10;
   5|   1,  5,  9, 12, 14, 15;
   6|   1,  6, 11, 15, 18, 20, 21;
   7|   1,  7, 13, 18, 22, 25, 27, 28;
  ...
For n = 2 the spiral is 2 sides of length 1 so row [1, 2, 3],
   1 --- 2
       /
     3
For n = 4 the spiral is:
   1  2  3  4
    9  10  5
      8  6
        7
The start, corner and end vertices are [1, 4, 7, 9, 10].

Columns: A000012 (k=0), A000027 (k=1), A144396 (k=3).
Cf. A179865(n+1) (main diagonal), A056520 (row sums).
Cf. A000217, A141419.

T[n_,k_]:=If[k==0,1,k(2n-k+1)/2];Table[T[n,k],{n,0,12},{k,0,n}]//Flatten (* James C. McMahon, Jul 31 2025 *)

T(n,0) = 1.

T(n,k) = k*(2*n - k + 1)/2 for k >= 1.

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A221857 Number A(n,k) of shapes of balanced k-ary trees with n nodes, where a tree is balanced if the total number of nodes in subtrees corresponding to the branches of any node differ by at most one; square array A(n,k), n>=0, k>=0, read by antidiagonals.

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A253145 Triangular numbers (A000217) omitting the term 1.

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A372973 Triangle read by rows: the exponential almost-Riordan array ( 1/(1-x) | 1/(1-x), log(1/(1-x)) ).

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A386319 Triangle read by rows where row n is the start, corner and end vertex numbers of a triangular spiral with n sides on a triangular grid, starting from 1 and working inwards (0 <= k <= n).

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