A050407 -id:A050407 - cp's OEIS frontend

A144257 Weight array of A086270.

1, 2, 0, 3, 1, 0, 4, 2, 1, 0, 5, 3, 2, 1, 0, 6, 4, 3, 2, 1, 0, 7, 5, 4, 3, 2, 1, 0, 8, 6, 5, 4, 3, 2, 1, 0, 9, 7, 6, 5, 4, 3, 2, 1, 0, 10, 8, 7, 6, 5, 4, 3, 2, 1, 0, 11, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 12, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 13, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 14, 12, 11, 10, 9, 8, 7, 6, 5

Offset: 1

Clark Kimberling, Sep 16 2008

For the definition of weight array, see A144112.
From Gary W. Adamson, Feb 18 2010: (Start)
Identical to an infinite lower triangular matrix with (1,2,3,...) in every column but the leftmost column shifted one row upwards, giving:
  1;
  2, 0;
  3, 1, 0;
  4, 2, 1, 0;
  5, 3, 2, 1, 0;
  ...
Let the triangle = M. Row sums = A000124; M * [1,2,3,...] = A050407 starting with offset 3: (1, 2, 5, 11, 21, 36, ...); and lim_{n->inf} M^n = the odd-indexed Fibonacci numbers, A001519: (1, 2, 5, 13, ...). (End)

			Northwest corner:
  1 2 3 4 5 6 7 8 9
  0 1 2 3 4 5 6 7 8
  0 1 2 3 4 5 6 7 8
  0 1 2 3 4 5 6 7 8
  0 1 2 3 4 5 6 7 8

Cf. A086270.

Cf. A000124, A050407, A001519. - Gary W. Adamson, Feb 18 2010

Row 1 = A000027. All subsequent rows are 0 followed by A000027.

A081913 a(n) = 2^n*(n^3 - 3n^2 + 2n + 48)/48.

Original entry on oeis.org

1, 2, 4, 9, 24, 72, 224, 688, 2048, 5888, 16384, 44288, 116736, 301056, 761856, 1896448, 4653056, 11272192, 27000832, 64028672, 150470656, 350748672, 811597824, 1865416704, 4261412864, 9680453632, 21877489664, 49207574528

Offset: 0

Paul Barry, Mar 31 2003

Binomial transform of A050407, (starting with 1,1,1,2,5,...). 2nd binomial transform of (1,0,0,1,0,0,0,0,...). Case k=2 where a(n,k) = k^n*(n^3 - 3n^2 + 2n + 6k^3)/(6k^3), with g.f. (1 - 3kx + 3k^2x^2 - (k^3-1)x^3)/(1-kx)^4.

Vincenzo Librandi, Table of n, a(n) for n = 0..225
Index entries for linear recurrences with constant coefficients, signature (8,-24,32,-16).

Cf. A081914.

[2^n*(n^3-3*n^2+2*n+48)/48: n in [0..40]]; // Vincenzo Librandi, Apr 27 2011

a(n) = 2^n*(n^3 - 3n^2 + 2n + 48)/48.

G.f.: (1 - 6x + 12x^2 - 7x^3)/(1-2x)^4.

A144680 Triangle read by rows, lower half of an array formed by A004736 * A144328 (transform).

Original entry on oeis.org

1, 2, 3, 3, 5, 7, 4, 7, 11, 14, 5, 9, 15, 21, 25, 6, 11, 19, 28, 36, 41, 7, 13, 23, 35, 47, 57, 63, 8, 15, 27, 42, 58, 73, 85, 92, 9, 17, 31, 49, 69, 89, 107, 121, 129, 10, 19, 35, 56, 80, 105, 129, 150, 166, 175

Offset: 1

Gary W. Adamson, Sep 19 2008

Triangle read by rows, lower half of an array formed by A004736 * A144328 (transform).

			The array is formed by A004736 * A144328 (transform) where A004736 = the natural number decrescendo triangle and A144328 = a crescendo triangle. First few rows of the array =
  1, 1,  1,  1,  1,  1, ...
  2, 3,  3,  3,  3,  3, ...
  3, 5,  7,  7,  7,  7, ...
  4, 7, 11, 14, 14, 14, ...
  5, 9, 15, 21, 25, 25, ...
  ...
Triangle begins as:
   1;
   2,  3;
   3,  5,  7;
   4,  7, 11, 14;
   5,  9, 15, 21, 25;
   6, 11, 19, 28, 36,  41;
   7, 13, 23, 35, 47,  57,  63;
   8, 15, 27, 42, 58,  73,  85,  92;
   9, 17, 31, 49, 69,  89, 107, 121, 129;
  10, 19, 35, 56, 80, 105, 129, 150, 166, 175;
  ...

G. C. Greubel, Rows n = 1..50 of the triangle, flattened

Cf. A004006, A006008, A027965, A050407, A074742, A144328.

T[n_, k_]:= (3*(k^2-k+2)*n - k*(k-1)*(2*k-1))/6;
Table[T[n, k], {n,12}, {k,n}]//Flatten (* G. C. Greubel, Oct 18 2021 *)

def A144680(n,k): return (3*(k^2-k+2)*n - k*(k-1)*(2*k-1))/6
flatten([[A144680(n,k) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, Oct 18 2021

Sum_{k=1..n} T(n, k) = A006008(n).
From G. C. Greubel, Oct 18 2021: (Start)
T(n, k) = (1/6)*( 3*(k^2 - k + 2)*n - k*(k-1)*(2*k-1) ).
T(n, n) = A004006(n).
T(n, n-1) = A050407(n+2).
T(n, n-2) = A027965(n-1) = A074742(n-2). (End)

A159255 Irregular triangle read by rows: row n gives expansion of (1-x+x^2)*(1+x)^n.

Original entry on oeis.org

1, -1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 2, 1, 1, 2, 1, 1, 3, 3, 2, 3, 3, 1, 1, 4, 6, 5, 5, 6, 4, 1, 1, 5, 10, 11, 10, 11, 10, 5, 1, 1, 6, 15, 21, 21, 21, 21, 15, 6, 1, 1, 7, 21, 36, 42, 42, 42, 36, 21, 7, 1, 1, 8, 28, 57, 78, 84, 84, 78, 57, 28, 8, 1, 1, 9, 36, 85, 135, 162, 168, 162, 135, 85, 36, 9, 1

Offset: 0

Philippe Deléham, Apr 07 2009

			Row n=0 : 1, -1, 1 ;
Row n=1 : 1, 0, 0, 1 ;
Row n=2 : 1, 1, 0, 1, 1 ;
Row n=3 : 1, 2, 1, 1, 2, 1 ;
Row n=4 : 1, 3, 3, 2, 3, 3, 1 ;
Row n=5 : 1, 5, 10, 11, 10, 11, 10, 5, 1;
Row n=6 : 1, 6, 15, 21, 21, 21, 21, 15, 6, 1;
...

Cf. A000079, A014626, A024483, A050407

row(n)=Vecrev(polcoef((1 - y + y^2)/(1 - x*(1+y)) + O(x*x^n), n))
{ for(n=0, 10, print(row(n))) } \\ Andrew Howroyd, Mar 03 2023

G.f.: A(x,y) = (1 - y + y^2)/(1 - x*(1+y)). - Andrew Howroyd, Mar 03 2023

A308682 Number of ways of partitioning the set of the first n positive triangular numbers into two subsets whose sums differ at most by 1.

Original entry on oeis.org

1, 1, 0, 0, 1, 1, 1, 1, 2, 7, 6, 8, 13, 42, 33, 52, 105, 318, 310, 485, 874, 3281, 2974, 5240, 9488, 34233, 30418, 55715, 104730, 378529, 352467, 642418, 1193879, 4466874, 4165910, 7762907, 14493951, 54162165, 50621491, 95133799, 179484713, 674845081

Offset: 0

Alois P. Heinz, Jun 16 2019

nonn

			a(4) = 1: 1,3,6/10.
a(5) = 1: 1,6,10/3,15.
a(6) = 1: 1,6,21/3,10,15.
a(7) = 1: 1,3,10,28/6,15,21.
a(8) = 2: 1,6,10,15,28/3,21,36; 1,10,21,28/3,6,15,36.

Alois P. Heinz, Table of n, a(n) for n = 0..250
Wikipedia, Partition problem

Cf. A000217, A000292, A050407, A307877.

s:= proc(n) s(n):= `if`(n=0, 1, n*(n+1)/2+s(n-1)) end:
b:= proc(n, i) option remember; `if`(i=0, `if`(n<=1, 1, 0),
     `if`(n>s(i), 0, (p->b(n+p, i-1)+b(abs(n-p), i-1))(i*(i+1)/2)))
    end:
a:= n-> ceil(b(0, n)/2):
seq(a(n), n=0..45);

s[n_] := s[n] = If[n == 0, 1, n(n+1)/2 + s[n-1]];
b[n_, i_] := b[n, i] = If[i == 0, If[n <= 1, 1, 0], If[n > s[i], 0, Function[p, b[n + p, i-1] + b[Abs[n-p], i-1]][i(i+1)/2]]];
a[n_] := Ceiling[b[0, n]/2];
a /@ Range[0, 45] (* Jean-François Alcover, May 04 2020, translated from Maple *)

A323228 a(n) = binomial(n + 4, n - 1) + 1.

Original entry on oeis.org

1, 2, 7, 22, 57, 127, 253, 463, 793, 1288, 2003, 3004, 4369, 6189, 8569, 11629, 15505, 20350, 26335, 33650, 42505, 53131, 65781, 80731, 98281, 118756, 142507, 169912, 201377, 237337, 278257, 324633, 376993, 435898, 501943, 575758, 658009, 749399, 850669

Offset: 0

Peter Luschny, Feb 12 2019

A040000 (m=1), A000027 (m=2), A000124 (m=3), A050407 (m=4), A145126 (m=5), this sequence (m=6).

aList := proc(len) local gf, ser:
gf := (x - (x - 1)^5)/(x - 1)^6:
ser := series(gf, x, len+2):
seq(coeff(ser, x, n), n=0..len) end: aList(38);

Table[Binomial[n + 4, n - 1] + 1, {n, 0, 37}]

a(n) = 1 + Pochhammer(n, 5)/5!.

a(n) = [x^n] (x - (x - 1)^5)/(x - 1)^6.

A323231 A(n, k) = [x^k] (1/(1-x) + x/(1-x)^n), square array read by descending antidiagonals for n, k >= 0.

Original entry on oeis.org

1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 3, 2, 1, 1, 2, 4, 4, 2, 1, 1, 2, 5, 7, 5, 2, 1, 1, 2, 6, 11, 11, 6, 2, 1, 1, 2, 7, 16, 21, 16, 7, 2, 1, 1, 2, 8, 22, 36, 36, 22, 8, 2, 1, 1, 2, 9, 29, 57, 71, 57, 29, 9, 2, 1, 1, 2, 10, 37, 85, 127, 127, 85, 37, 10, 2, 1

Offset: 0

Peter Luschny, Feb 10 2019

			Array starts:
[0] 1, 2,  1,  1,   1,   1,    1,    1,    1,     1,     1, ...
[1] 1, 2,  2,  2,   2,   2,    2,    2,    2,     2,     2, ... A040000
[2] 1, 2,  3,  4,   5,   6,    7,    8,    9,    10,    11, ... A000027
[3] 1, 2,  4,  7,  11,  16,   22,   29,   37,    46,    56, ... A000124
[4] 1, 2,  5, 11,  21,  36,   57,   85,  121,   166,   221, ... A050407
[5] 1, 2,  6, 16,  36,  71,  127,  211,  331,   496,   716, ... A145126
[6] 1, 2,  7, 22,  57, 127,  253,  463,  793,  1288,  2003, ... A323228
[7] 1, 2,  8, 29,  85, 211,  463,  925, 1717,  3004,  5006, ...
[8] 1, 2,  9, 37, 121, 331,  793, 1717, 3433,  6436, 11441, ...
[9] 1, 2, 10, 46, 166, 496, 1288, 3004, 6436, 12871, 24311, ...
.
Read as a triangle (by descending antidiagonals):
                                     1
                                  2,   1
                                1,   2,   1
                             1,   2,   2,   1
                           1,   2,   3,   2,   1
                        1,   2,   4,   4,   2,   1
                      1,   2,   5,   7,   5,   2,  1
                    1,  2,   6,  11,  11,   6,   2,  1
                  1,  2,   7,  16,  21,  16,   7,  2,  1
                1,  2,  8,  22,  36,  36,  22,   8,  2,  1
              1,  2,  9,  29,  57,  71,  57,  29,  9,  2,  1
.
A(0, 1) = C(-1, 0) + 1 = 2 because C(-1, 0) = 1. A(1, 0) = C(-1, -1) + 1 = 1 because C(-1, -1) = 0. Warning: Some computer algebra programs (for example Maple and Mathematica) return C(n, n) = 1 for n < 0. This contradicts the definition given by Graham et al. (see reference). On the other hand this definition preserves symmetry.

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 154.

G. C. Greubel, Antidiagonals n = 0..50, flattened

Differs from A323211 only in the second term.
Rows include: A040000, A000027, A000124, A050407, A145126, A323228.
Diagonals A(n, n+d): A323230 (d=0), A260878 (d=1), A323229 (d=2).
Antidiagonal sums are A323227(n) if n!=1.
Cf. A007318 (Pascal's triangle).

using AbstractAlgebra
function Arow(n, len)
    R, x = PowerSeriesRing(ZZ, len+2, "x")
    gf = inv(1-x) + divexact(x, (1-x)^n)
    [coeff(gf, k) for k in 0:len-1] end
for n in 0:9 println(Arow(n, 11)) end

Binomial := (n, k) -> `if`(n < 0 and n = k, 0, binomial(n,k)):
A := (n, k) -> Binomial(n + k - 2, k - 1) + 1:
seq(lprint(seq(A(n, k), k=0..10)), n=0..10);

T[n_, k_]:= If[k==0, 1 + Boole[n==1], If[k==n, 1, Binomial[n-2, k-1] + 1]];
Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Dec 27 2021 *)

def Arow(n):
    R. = PowerSeriesRing(ZZ, 20)
    gf = 1/(1-x) + x/(1-x)^n
    return gf.padded_list(10)
for n in (0..9): print(Arow(n))

A(n, k) = binomial(n + k - 2, k - 1) + 1. Note that binomial(n, n) = 0 if n < 0.
A(n, k) = A(k, n) with the exception A(1,0) != A(0,1).
A(n, n) = binomial(2*n-2, n-1) + 1 = A323230(n).
From G. C. Greubel, Dec 27 2021: (Start)
T(n, k) = binomial(n-2, k-1) + 1 with T(n, 0) = 1 + [n=1], T(n, n) = 1.
T(2*n, n) = A323230(n).
Sum_{k=0..n} T(n,k) = n + 1 + 2^(n-2) - [n=0]/4 + [n=1]/2. (End)

A081917 a(0)=1, a(n)= n^(n-2)(7n^2-3n+2)/6 (n>0).

Original entry on oeis.org

1, 1, 4, 28, 272, 3375, 50976, 907578, 18612224, 432061533, 11200000000, 320680885976, 10051252125696, 342302635261067, 12586048547315712, 496928474121093750, 20968759865037029376, 941737183946729395897

Offset: 0

Paul Barry, Mar 31 2003

Main diagonal of square array defined by T(n,k)=k^n(n^3-3n^2+2n+6k^3)/(6k^3), in which rows have g.f. (1-3kx+3k^2x^2-(k^3-1)x^3)/(1-kx)^4.

Vincenzo Librandi, Table of n, a(n) for n = 0..100

Cf. A050407, A081913, A081914, A081915, A081916.

A329523 a(n) = n * (binomial(n + 1, 3) + 1).

Original entry on oeis.org

0, 1, 4, 15, 44, 105, 216, 399, 680, 1089, 1660, 2431, 3444, 4745, 6384, 8415, 10896, 13889, 17460, 21679, 26620, 32361, 38984, 46575, 55224, 65025, 76076, 88479, 102340, 117769, 134880, 153791, 174624, 197505, 222564, 249935, 279756, 312169, 347320, 385359, 426440

Offset: 0

Ilya Gutkovskiy, Nov 15 2019

The n-th centered n-gonal pyramidal number.

			Square array begins:
  (0), 1,  2,   3,   4,    5,  ... A001477
   0, (1), 3,   7,  14,   25,  ... A004006
   0,  1, (4), 11,  24,   45,  ... A006527
   0,  1,  5, (15), 34,   65,  ... A006003 (partial sums of A005448)
   0,  1,  6,  19, (44),  85,  ... A005900 (partial sums of A001844)
   0,  1,  7,  23,  54, (105), ... A004068 (partial sums of A005891)
...
This sequence is the main diagonal of the array.

E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), 142.

Kelvin Voskuijl, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000292, A000330, A002415, A002417, A006000, A006484, A008911, A050407, A060354, A100119, A188475 (first differences).

[ n*(Binomial(n+1,3)+1):n in [0..40]]; // Marius A. Burtea, Nov 15 2019

R:=PowerSeriesRing(Integers(), 41); [0] cat Coefficients(R!(x*(1-x+5*x^2-x^3)/(1-x)^5)); // Marius A. Burtea, Nov 15 2019

Table[n (Binomial[n + 1, 3] + 1), {n, 0, 40}]
nmax = 40; CoefficientList[Series[x (1 - x + 5 x^2 - x^3)/(1 - x)^5, {x, 0, nmax}], x]
LinearRecurrence[{5, -10, 10, -5, 1}, {0, 1, 4, 15, 44}, 41]

G.f.: x * (1 - x + 5*x^2 - x^3) / (1 - x)^5.
E.g.f.: exp(x) * x * (1 + x + x^2 + x^3 / 6).
a(n) = n * (n + 2) * (n^2 - 2*n + 3) / 6.
a(n) = n * (A000292(n-1) + 1).
a(n) = n + 2 * Sum_{k=1..n} A000330(k-1).
a(n) + a(-n) = 4 * A002415(n).

A367313 Triangle read by rows: T(n,k) is the number of permutations of [n] with weighted inversion index k.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 3, 3, 4, 3, 3, 2, 1, 1, 1, 1, 2, 3, 5, 5, 8, 9, 10, 10, 12, 10, 10, 9, 8, 5, 5, 3, 2, 1, 1, 1, 1, 2, 3, 5, 7, 9, 12, 16, 20, 23, 28, 31, 36, 38, 41, 43, 44, 44, 43, 41, 38, 36, 31, 28, 23, 20, 16, 12, 9, 7, 5, 3, 2, 1, 1

Offset: 0

Todd Simpson, Nov 13 2023

T(n,k) represents two statistics that can be shown to be equal:
(1) Permutations of {1,2,...,n} counted by a "weighted inversion index": for a permutation pi, the weighted inversion index is the sum of i over all pairs i,j with i < j and pi(i) > pi(j).
(2) Partitions lambda with at most n-1 parts counted by weight, where the inequality lambda(i) - lambda(i+1) <= n - i holds for 1 <= i < n (with lambda(n) = 0).
Possible values of this index range from 0 to (n-1)*n*(n+1)/6.  The permutation with the largest weighted inversion index is (n,n-1,...,2,1) and the partition with the largest weight is (n(n-1)/2,(n-1)(n-2)/2,...,3,1).
Let t_n(q) be the sum of T(n,k)q^k, for 0 <= k <= (n-1)*n*(n+1)/6.  Then t_n(q) is the product of (1 - q^(k*(n+1-k)))/(1 - q^k), for 1 <= k <= n-1.

			The permutation pi = (2,5,3,1,4) has these inversions, with the given contributions to weighted inversion index:
   (2,1), 1
   (5,3), 2
   (5,1), 2
   (5,4), 2
   (3,1), 3
The corresponding partition can be created as follows.  For each i <= 5, write the number of j > i with pi(i) > pi(j): (1,3,1,0,0).
For each i, the i-th number in this sequence is at most n-i.
Let lambda(i) be the sum of the values of the sequence starting with the i-th value: lambda = (5,4,1,0,0).
This permutation and partition are counted by T(5,10).  In the product expansion of t_5(q), they correspond to the following choice of terms:
   (1 - q^5)/(1 - q) = 1 + q + q^2 + q^3 + q^4:  choose q,
   (1 - q^8)/(1 - q^2) = 1 + q^2 + q^4 + q^6:  choose q^6,
   (1 - q^9)/(1 - q^3) = 1 + q^3 + q^6:  choose q^3,
   (1 - q^8)/(1 - q^4) = 1 + q^4:  choose 1.
Triangle T(n,k) begins:
  1;
  1;
  1, 1;
  1, 1, 2, 1, 1;
  1, 1, 2, 3, 3, 4, 3, 3,  2,  1,  1;
  1, 1, 2, 3, 5, 5, 8, 9, 10, 10, 12, 10, 10, 9, 8, 5, 5, 3, 2, 1, 1;
  ...

Alois P. Heinz, Rows n = 0..25, flattened

Row sums give A000142.
Row n contains A050407(n+2) terms.
T(n+1,n) gives A000041(n).
Cf. A000165, A001754, A175929.

From Alois P. Heinz, Nov 25 2023: (Start)
Sum_{k=0..A050407(n+2)-1} k * T(n,k) = A001754(n+1).
Sum_{k=0..A050407(2n+3)-1} (-1)^k * T(2n+1,k) = A000165(n). (End)

cp's OEIS Frontend

A144257 Weight array of A086270.

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A081913 a(n) = 2^n*(n^3 - 3n^2 + 2n + 48)/48.

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Magma

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A144680 Triangle read by rows, lower half of an array formed by A004736 * A144328 (transform).

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Mathematica

Sage

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A159255 Irregular triangle read by rows: row n gives expansion of (1-x+x^2)*(1+x)^n.

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PARI

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A308682 Number of ways of partitioning the set of the first n positive triangular numbers into two subsets whose sums differ at most by 1.

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Maple

Mathematica

A323228 a(n) = binomial(n + 4, n - 1) + 1.

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Maple

Mathematica

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A323231 A(n, k) = [x^k] (1/(1-x) + x/(1-x)^n), square array read by descending antidiagonals for n, k >= 0.

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Julia

Maple

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Sage

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A081917 a(0)=1, a(n)= n^(n-2)(7n^2-3n+2)/6 (n>0).

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