A005581 -id:A005581 - cp's OEIS frontend

A387152 Array read by ascending antidiagonals: A(n, k) = Sum_{j=0..n} binomial(k, j)*|Stirling1(n, j)|.

1, 0, 1, 0, 1, 1, 0, 1, 2, 1, 0, 2, 3, 3, 1, 0, 6, 7, 6, 4, 1, 0, 24, 23, 16, 10, 5, 1, 0, 120, 98, 57, 30, 15, 6, 1, 0, 720, 514, 257, 115, 50, 21, 7, 1, 0, 5040, 3204, 1407, 546, 205, 77, 28, 8, 1, 0, 40320, 23148, 9076, 3109, 1021, 336, 112, 36, 9, 1

Offset: 0

Peter Luschny, Aug 27 2025

			Array begins:
  [0]  1,     1,      1,      1,       1,       1,       1, ...
  [1]  0,     1,      2,      3,       4,       5,       6, ...
  [2]  0,     1,      3,      6,      10,      15,      21, ...
  [3]  0,     2,      7,     16,      30,      50,      77, ...
  [4]  0,     6,     23,     57,     115,     205,     336, ...
  [5]  0,    24,     98,    257,     546,    1021,    1750, ...
  [6]  0,   120,    514,   1407,    3109,    6030,   10696, ...
  [7]  0,   720,   3204,   9076,   20695,   41330,   75356, ...
  [8]  0,  5040,  23148,  67456,  157865,  323005,  602517, ...
  [9]  0, 40320, 190224, 567836, 1358564, 2837549, 5396650, ...

Rows: A000012 [0], A001477 [1], A000217 [2], A005581 [3], A387204 [4].
Columns: A000007 [0], A000142 [shifted, 1], A387205 [2].
Contains A271700 in transpose.
Cf. A211210 (main diagonal), A130534.

A := (n, k) -> add(binomial(k, j)*abs(Stirling1(n, j)), j = 0..n):
seq(seq(A(n-k, k), k = 0..n), n = 0..10);
# Expanding rows or columns:
RowSer := n -> series((1+x)^k*GAMMA(x + n)/GAMMA(x), x, 12):
Trow := n -> k -> coeff(RowSer(n), x, k):
ColSer := n -> series(orthopoly:-L(n, log(1 - x)), x, 12):
Tcol := k -> n -> n! * coeff(ColSer(k), x, n):
seq(lprint(seq(Trow(n)(k), k = 0..7)), n = 0..9);
seq(lprint(seq(Tcol(k)(n), n = 0..7)), k = 0..9);

from functools import cache
@cache
def T(n: int, k: int) -> int:
    if n == 0: return 1
    if k == 0: return 0
    return (n - 1) * T(n - 1, k) + T(n, k - 1) - (n - 2) * T(n - 1, k - 1)
for n in range(7): print([T(n, k) for k in range(7)])

T(n, k) = n! * [x^n] Laguerre(k, log(1 - x)).
From Natalia L. Skirrow, Aug 27 2025: (Start)
D-finite with T(n,k) = (n-1)*T(n-1,k)+T(n,k-1)-(n-2)*T(n-1,k-1).
O.g.f.: hypergeom([1,y/(1-y)],[],x)/(1-y).
Row o.g.f.: (y/(1-y))_n/(1-y), where (x)_n is the Pochhammer symbol/rising factorial.
Row o.g.f. is also 0^n + y/(1-y)^(n+1)*Prod_{j=1..n-2}(j+1-j*y).
E.g.f.: 1/((1-y)*(1-x)^(y/(1-y))).
Column e.g.f.: hypergeom([-k],[1],log(1-y)).
T(n,k) = [x^k] (1+x)^k*(x)_n.
(End)

A141429 Triangle T(n, k) = (k+1)*(n-k+1), read by rows.

Original entry on oeis.org

2, 4, 3, 6, 6, 4, 8, 9, 8, 5, 10, 12, 12, 10, 6, 12, 15, 16, 15, 12, 7, 14, 18, 20, 20, 18, 14, 8, 16, 21, 24, 25, 24, 21, 16, 9, 18, 24, 28, 30, 30, 28, 24, 18, 10, 20, 27, 32, 35, 36, 35, 32, 27, 20, 11, 22, 30, 36, 40, 42, 42, 40, 36, 30, 22, 12, 24, 33, 40, 45, 48, 49, 48, 45, 40, 33, 24, 13

Offset: 1

Roger L. Bagula and Gary W. Adamson, Aug 06 2008

			Triangle begins as:
   2;
   4,  3;
   6,  6,  4;
   8,  9,  8,  5;
  10, 12, 12, 10,  6;
  12, 15, 16, 15, 12,  7;
  14, 18, 20, 20, 18, 14,  8;
  16, 21, 24, 25, 24, 21, 16,  9;
  18, 24, 28, 30, 30, 28, 24, 18, 10;
  20, 27, 32, 35, 36, 35, 32, 27, 20, 11;

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

Cf. A003991, A004247, A005581, A144329, A158823.

[(k+1)*(n-k+1): k in [1..n], n in [1..15]]; // G. C. Greubel, Mar 30 2021

A141429 := proc(n,k)
        (k+1)*(n-k+1) ;
end proc:
seq(seq(A141429(n,m),m=1..n),n=1..14) ; # R. J. Mathar, Nov 10 2011

Table[(k+1)*(n-k+1), {n,15}, {k,n}]//Flatten

flatten([[(k+1)*(n-k+1) for k in (1..n)] for n in (1..15)]) # G. C. Greubel, Mar 30 2021

T(n, k) = (k+1)*(n-k+1).
T(n, k) = A158823(n+2, k+2).
Sum_{k=1..n} T(n, k) = A005581(n+1).

Edited by G. C. Greubel, Mar 30 2021

A180174 Triangle read by rows of the numbers C(n,k) of k-subsets of a quadratically populated n-multiset M.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 3, 5, 7, 9, 10, 10, 10, 10, 10, 9, 7, 5, 3, 1, 1, 4, 9, 16, 25, 35, 45, 55, 65, 75, 84, 91, 96, 99, 100, 100, 100, 99, 96, 91, 84, 75, 65, 55, 45, 35, 25, 16, 9, 4, 1, 1, 5, 14, 30, 55, 90, 135, 190, 255, 330, 414, 505, 601, 700, 800, 900, 1000, 1099

Offset: 0

Thomas Wieder, Aug 15 2010

The multiplicity m(i) of the i-th element with 1 <= i <= n is m(i)=i^2.
Thus M=[1,2,2,2,2,...,i^2 x i,...,n^2 x n].
Row sum is equal to A028361.
Column for k=2 is equal to AA000096.
Column for k=3 is equal to AA005581.
Column for k=4 is equal to AA005582.
The number of coefficients C(n,k) for given n is equal to A056520.

			For n=4 one has M=[1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4].
For k=7 we have 55 subsets from M:
[1, 2, 2, 3, 3, 4, 4], [1, 2, 3, 3, 4, 4, 4], [1, 2, 3, 3, 3, 4, 4],
[1, 2, 2, 3, 4, 4, 4], [1, 2, 2, 3, 3, 3, 4], [1, 2, 2, 2, 3, 4, 4],
[1, 2, 2, 2, 3, 3, 4], [2, 2, 3, 3, 4, 4, 4], [2, 2, 3, 3, 3, 4, 4],
[2, 2, 2, 3, 3, 4, 4], [1, 2, 2, 2, 3, 3, 3], [1, 2, 2, 2, 4, 4, 4],
[1, 3, 3, 3, 4, 4, 4], [2, 3, 3, 3, 4, 4, 4], [2, 2, 2, 3, 4, 4, 4],
[2, 2, 2, 3, 3, 3, 4], [1, 2, 3, 4, 4, 4, 4], [1, 2, 3, 3, 3, 3, 4],
[1, 2, 2, 2, 2, 3, 4], [1, 2, 2, 3, 3, 3, 3], [1, 2, 2, 2, 2, 3, 3],
[1, 2, 2, 4, 4, 4, 4], [1, 2, 2, 2, 2, 4, 4], [1, 3, 3, 4, 4, 4, 4],
[1, 3, 3, 3, 3, 4, 4], [2, 3, 3, 4, 4, 4, 4], [2, 3, 3, 3, 3, 4, 4],
[2, 2, 3, 4, 4, 4, 4], [2, 2, 3, 3, 3, 3, 4], [2, 2, 2, 2, 3, 4, 4],
[2, 2, 2, 2, 3, 3, 4], [2, 2, 2, 3, 3, 3, 3], [2, 2, 2, 2, 3, 3, 3],
[2, 2, 2, 4, 4, 4, 4], [2, 2, 2, 2, 4, 4, 4], [3, 3, 3, 4, 4, 4, 4],
[3, 3, 3, 3, 4, 4, 4], [1, 2, 3, 3, 3, 3, 3], [1, 2, 4, 4, 4, 4, 4],
[1, 3, 4, 4, 4, 4, 4], [1, 3, 3, 3, 3, 3, 4], [2, 3, 4, 4, 4, 4, 4],
[2, 3, 3, 3, 3, 3, 4], [2, 2, 3, 3, 3, 3, 3], [2, 2, 4, 4, 4, 4, 4],
[3, 3, 4, 4, 4, 4, 4], [3, 3, 3, 3, 3, 4, 4], [1, 3, 3, 3, 3, 3, 3],
[1, 4, 4, 4, 4, 4, 4], [2, 3, 3, 3, 3, 3, 3], [2, 4, 4, 4, 4, 4, 4],
[3, 4, 4, 4, 4, 4, 4], [3, 3, 3, 3, 3, 3, 4], [3, 3, 3, 3, 3, 3, 3],
[4, 4, 4, 4, 4, 4, 4].

Cf. A007318, A008302, A028361, A056520, A000096, A005581, A005582.

with(combinat)
kend := 4;
Liste := NULL;
for k from 0 to kend do
Liste := Liste, `$`(k, k^2)
end do;
Liste := [Liste];
for k from 0 to 2^(kend+1)-1 do
Teilergebnis[k] := choose(Liste, k)
end do;
seq(nops(Teilergebnis[k]), k = 0 .. 2^(kend+1)-1)
' Excel VBA
Sub A180174()
Dim n As Long, nend As Long, k As Long, kk As Long, length_row As Long, length_sum As Long
Dim ATable(10, -1000 To 1000) As Double, Summe As Double
Dim offset_row As Integer, offset_column As Integer
Worksheets("Tabelle2").Select
Cells.Select
Selection.ClearContents
Range("A1").Select
offset_row = 1
offset_column = 1
nend = 7
ATable(0, 0) = 1
Cells(0 + offset_row, 0 + offset_column) = 1
For n = 1 To nend
length_row = n * (n + 1) * (2 * n + 1) / 6
length_sum = n ^ 2 + 1
For k = 0 To length_row / 2
Summe = 0
For kk = k - length_sum + 1 To k
Summe = Summe + ATable(n - 1, kk)
Next kk
ATable(n, k) = Summe
Cells(n + offset_row, k + offset_column) = ATable(n, k)
ATable(n, length_row - k) = Summe
Cells(n + offset_row, length_row - k + 0 + offset_column) = ATable(n, k)
Next k
Next n
End Sub

C(0,0) = 0.
C(n,k) = sum_{j=(k-LS+1)}^{k} C(n-1,j).
for n > 0 and k=1,...,LR with LS = n^2+1 and LR = n*(n+1)*(2*n+1)/6.
C(n,k) = C(n,LR-k).

A271700 Triangle read by rows, T(n,k) = Sum_{j=0..n} (-1)^(n-j)C(-j-1,-n-1)S1(k,j), S1 the Stirling cycle numbers A132393, for n>=0 and 0<=k<=n.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 6, 16, 1, 4, 10, 30, 115, 1, 5, 15, 50, 205, 1021, 1, 6, 21, 77, 336, 1750, 10696, 1, 7, 28, 112, 518, 2814, 17766, 128472, 1, 8, 36, 156, 762, 4308, 28050, 207942, 1734447, 1, 9, 45, 210, 1080, 6342, 42528, 322860, 2746815, 25937683

Offset: 0

Peter Luschny, Apr 14 2016

			Triangle starts:
[1]
[1, 1]
[1, 2, 3]
[1, 3, 6,  16]
[1, 4, 10, 30,  115]
[1, 5, 15, 50,  205, 1021]
[1, 6, 21, 77,  336, 1750, 10696]
[1, 7, 28, 112, 518, 2814, 17766, 128472]

A000027 (col. 1), A000217, A161680 (col. 2), A005581 (col. 3), A211210 (diag. n,n), A211211 (diag. n,n-1).

T := (n,k) -> add(abs(Stirling1(k,j))*binomial(-j-1,-n-1)*(-1)^(n-j),j=0..n);
seq(seq(T(n,k), k=0..n), n=0..9);

Flatten[Table[Sum[(-1)^(n-j)Binomial[-j-1,-n-1] Abs[StirlingS1[k,j]],{j,0,n}], {n,0,9},{k,0,n}]]

A346038 Triangle read by rows T(n, k) such that Fib(n, x+1) = Sum_{k=1..n} T(n, k)*Fib(k, x) where Fib(n, x) is the n-th Fibonacci polynomial.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 0, 3, 3, 1, -2, 2, 6, 4, 1, -4, -3, 7, 10, 5, 1, -3, -12, 0, 16, 15, 6, 1, 5, -18, -21, 11, 30, 21, 7, 1, 20, -4, -50, -24, 35, 50, 28, 8, 1, 29, 48, -51, -98, -9, 78, 77, 36, 9, 1, 1, 124, 45, -164, -150, 42, 147, 112, 45, 10, 1, -94, 128, 282, -67, -365, -177, 154, 250, 156, 55, 11, 1

Offset: 1

Michel Marcus, Jul 02 2021

			Triangle begins:
   1;
   1,  1;
   1,  2, 1;
   0,  3, 3,  1;
  -2,  2, 6,  4, 1;
  -4, -3, 7, 10, 5, 1;
  ...
The first 3 Fibonacci polynomials are 1, x, x^2 + 1. So F3(n, x+1) = x^2 + 2*x + 2  = 1*1 + 2*x + 1*(x^2+1) = 1*F(1,x) + 2*F(2, x) + 1*F(3,x), so the 3rd row is [1, 2, 1].

Michel Marcus, Rows n=1..100 of triangle, flattened
Math StackExchange, Variable transformation x->x+1 in Fibonacci-polynomial Fn(x)
Eric Weisstein's World of Mathematics, Fibonacci Polynomial.

Cf. A000012, A000027, A000217, A005581: diagonals.

Cf. A162515 and A168561 (Fibonacci polynomials coefficients).

rowV(n) = my(v= if (n==0, [0], n--; vector(n+1, k, k--; if (k%2==0, binomial(n-k/2, k/2))))); Pol(v); \\ A162515
rowT(n, vfp, vfp1) = {my(vp1 = vfp1[n], vc = vector(n), i=n); forstep (k = poldegree(vp1), 0, -1, vc[i] = polcoef(vp1, k)/polcoef(vfp[k+1], k); vp1 -= vfp[k+1]*vc[i]; i--;); vc;}
tabl(nn) = {my(vfp = vector(nn, k, rowV(k))); my(vfp1 = vector(nn, k, subst(vfp[k], x, x+1))); for(n=1, nn, print((rowT(n, vfp, vfp1))););}

A363378 Third Lie-Betti number of a cycle graph on n vertices.

Original entry on oeis.org

12, 25, 41, 68, 105, 152, 210, 280, 363, 460, 572, 700, 845, 1008, 1190, 1392, 1615, 1860, 2128, 2420, 2737, 3080, 3450, 3848, 4275, 4732, 5220, 5740, 6293, 6880, 7502, 8160, 8855, 9588, 10360, 11172, 12025, 12920, 13858

Offset: 3

Samuel J. Bevins, Jun 01 2023

nonn

Sequence T(n,3) in A360572.

M. Aldi and S. Bevins, L_oo-algebras and hypergraphs, arXiv:2212.13608 [math.CO], 2022. See page 9.
M. Mainkar, Graphs and two step nilpotent Lie algebras, arXiv:1310.3414 [math.DG], 2013. See page 1.
Eric Weisstein's World of Mathematics, Cycle Graph.

Cf. A005581, A054000, A028347, A000027, A360572 (cycle graph triangle)

def A363378(n):
    values = [12,25,41]
    for i in range(6, n+1):
        result = (i*(i+11)*(i-2))/6
        values.append(result)
    return values

a(3) = 12, a(4) = 25, a(5) = 41, a(n) = n*(n+11)*(n-2)/6 for n >= 6.
a(n) = A005581(n-4) + A054000(n-1) + A028347(n-2) + A000027(n) for n >= 6.
a(n) = A106058(n+1) - 2 for n >= 6. - Hugo Pfoertner, Jun 02 2023

A191532 Triangle T(n,k) read by rows: T(n,n) = 2n+1, T(n,k)=k for k

Original entry on oeis.org

1, 0, 3, 0, 1, 5, 0, 1, 2, 7, 0, 1, 2, 3, 9, 0, 1, 2, 3, 4, 11, 0, 1, 2, 3, 4, 5, 13, 0, 1, 2, 3, 4, 5, 6, 15, 0, 1, 2, 3, 4, 5, 6, 7, 17, 0, 1, 2, 3, 4, 5, 6, 7, 8, 19, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 21, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 23, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 25, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 27

Offset: 0

Paul Curtz, Jun 05 2011

We can build products of linear polynomials with these T(n,k) defining the absolute terms:
1+n = A000027(1+n)                               =2,  3,  4,   5,   6,   7,
n*(3+n)/2 = A000096(1+n)                         =2,  5,  9,  14,  20,  27,
n*(1+n)*(5+n)/6 = A005581(2+n)                   =2,  7, 16,  30,  50,  77,
n*(1+n)*(2+n)*(7+n)/24 = A005582(1+n)            =2,  9, 25,  55, 105, 182,
n*(1+n)*(2+n)*(3+n)*(9+n)/120 = A005583(n)       =2, 11, 36,  91, 196, 378,
n*(1+n)*(2+n)*(3+n)*(4+n)*(11+n)/720 = A005584(n)=2, 13, 49, 140, 336, 714,

			1;
0,3;
0,1,5;
0,1,2,7;
0,1,2,3,9;
0,1,2,3,4,11;

Cf. A191302.

T(n,k) = A002262(n-1,k).

sum_{k=0..n} T(n,k) = A000217(1+n).

A238156 Triangle T(n,k), 0<=k<=n, read by rows, given by (0, 1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (2, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 0, 2, 0, 2, 3, 0, 2, 7, 4, 0, 2, 11, 16, 5, 0, 2, 15, 36, 30, 6, 0, 2, 19, 64, 91, 50, 7, 0, 2, 23, 100, 204, 196, 77, 8, 0, 2, 27, 144, 385, 540, 378, 112, 9, 0, 2, 31, 196, 650, 1210, 1254, 672, 156, 10, 0, 2, 35, 256, 1015, 2366, 3289, 2640, 1122, 210, 11

Offset: 0

Philippe Deléham, Feb 18 2014

Row sums are A001519(n+1) = A122367(n).

Diagonal sums are A052969(n).

			Triangle begins:
1;
0, 2;
0, 2, 3;
0, 2, 7, 4;
0, 2, 11, 16, 5;
0, 2, 15, 36, 30, 6;
0, 2, 19, 64, 91, 50, 7;
0, 2, 23, 100, 204, 196, 77, 8;
0, 2, 27, 144, 385, 540, 278, 112, 9;
0, 2, 31, 196, 650, 1210, 1254, 672, 156, 10;
0, 2, 35, 256, 1015, 2366, 3289, 2640, 1122, 210, 11;
...

Cf. A016742, A004767, A005581, A005583

G.f.: (1-x)/(1-x-2*x*y+x^2*y^2).
Sum_{k=0..n} T(n,k)*2^k = 4^n = A000302(n).
T(n,k) = T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k-2), T(0,0) = 1, T(1,0) = 0, T(1,1) = 2, T(n,k) = 0 if k<0 or if k>n.

A320657 a(n) is the number of non-unimodal sequences with n nonzero terms that arise as a convolution of sequences of binomial coefficients preceded by a finite number of ones.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 5, 7, 12, 16, 24, 30, 41, 50, 65, 77, 96, 112, 136, 156, 185, 210, 245, 275, 316, 352, 400, 442, 497, 546, 609, 665, 736, 800, 880, 952, 1041, 1122, 1221, 1311, 1420, 1520, 1640, 1750, 1881, 2002, 2145, 2277, 2432, 2576, 2744, 2900, 3081, 3250, 3445, 3627, 3836, 4032

Offset: 1

Tricia Muldoon Brown, Oct 17 2018

nonn

For integers x,y,p,q >= 0, set (s_i){i>=1} to be the sequence of p ones followed by the binomial coefficients C(x,j) for 0 <= j <= x followed by an infinite string of zeros, and set (t_i){i>=1} to be the sequence of q ones followed by the binomial coefficients C(y,j) for 0 <= j <= y followed by an infinite string of zeros. Then a(n) is the number of non-unimodal sequences (r_i){i>=1} where r_i = Sum{j=1..i} s_j*t_{i-j} for some(s_i) and (t_i) such that x + y + p + q + 1 = n.

Let T be a rooted tree created by identifying the root vertices of two broom graphs. a(n) is the number of trees T on n vertices whose poset of connected, vertex-induced subgraphs is not rank unimodal.

T. M. Brown, On the unimodality of convolutions of sequences of binomial coefficients, arXiv:1810.08235 [math.CO] (2018). See table 1 on page 17.
M. Jacobson, A. E. Kézdy, and S. Seif, The poset on connected induced subgraphs of a graph need not be Sperner, Order, 12 (1995) 315-318.

Cf. A005993, A024206. Equals A005581 for n even.

Table[If[EvenQ[n], 2*(Sum[Floor[i(i+4)/4], {i,0,(n/2)}]) - Floor[n^2/16], 2*(Sum[Floor[i(i+4)/4], {i,0,(n-1)/2}]) - Floor[(n-1)^2/16] + Floor[(n+1)(n+9)/16]], {n,0,40}]

a(n+10) = 2*(Sum_{i=1..n/2} floor(i*(i+4)/4)) - floor(n^2/16) for n even.

a(n+10) = 2*(Sum_{i=1..(n-1)/2} floor(i(i+4)/4)) - floor((n-1)^2/16) + floor((n+1)*(n+9)/16) for n odd.

cp's OEIS Frontend

A387152 Array read by ascending antidiagonals: A(n, k) = Sum_{j=0..n} binomial(k, j)*|Stirling1(n, j)|.

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A141429 Triangle T(n, k) = (k+1)*(n-k+1), read by rows.

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Magma

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A180174 Triangle read by rows of the numbers C(n,k) of k-subsets of a quadratically populated n-multiset M.

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Maple

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A271700 Triangle read by rows, T(n,k) = Sum_{j=0..n} (-1)^(n-j)*C(-j-1,-n-1)*S1(k,j), S1 the Stirling cycle numbers A132393, for n>=0 and 0<=k<=n.

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A346038 Triangle read by rows T(n, k) such that Fib(n, x+1) = Sum_{k=1..n} T(n, k)*Fib(k, x) where Fib(n, x) is the n-th Fibonacci polynomial.

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PARI

A363378 Third Lie-Betti number of a cycle graph on n vertices.

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A191532 Triangle T(n,k) read by rows: T(n,n) = 2n+1, T(n,k)=k for k

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A238156 Triangle T(n,k), 0<=k<=n, read by rows, given by (0, 1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (2, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.

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A320657 a(n) is the number of non-unimodal sequences with n nonzero terms that arise as a convolution of sequences of binomial coefficients preceded by a finite number of ones.

A271700 Triangle read by rows, T(n,k) = Sum_{j=0..n} (-1)^(n-j)C(-j-1,-n-1)S1(k,j), S1 the Stirling cycle numbers A132393, for n>=0 and 0<=k<=n.