A056856 -id:A056856 - cp's OEIS frontend

A108267 Triangle read by rows, T(n, k) = [x^k] (1-x)^(n+1)Sum_{j=0..n} binomial(n + nj + j, nj + j)x^j.

1, 1, 1, 1, 7, 1, 1, 31, 31, 1, 1, 121, 381, 121, 1, 1, 456, 3431, 3431, 456, 1, 1, 1709, 26769, 60691, 26769, 1709, 1, 1, 6427, 193705, 848443, 848443, 193705, 6427, 1, 1, 24301, 1343521, 10350421, 19610233, 10350421, 1343521, 24301, 1

Offset: 0

Paul D. Hanna, May 29 2005 and May 31 2005

G.f. of row n divided by (1-x)^(n+1) equals g.f. of row n of table A060543.
Matrix product of this triangle with Pascal's triangle (A007318) equals A108291.
Seeing each row as a polynomial, all roots seem to be negative reals. - F. Chapoton, Nov 01 2022
From Thomas Anton, Jan 05 2023: (Start)
Consider the set [m] := {1, 2, 3, ..., m} ordered cyclically, and then mapped into itself via f. Let us consider a in [m] as the (a-1)th m-th root of unity e^(2*Pi*i*(a-1)/m). Then f may be extended to a continuous map f':S^1 -> S^1 as follows:
For a immediately before b in the cyclic order, map the interval between a and b to S^1 so that a point in it moving clockwise at constant speed has a value moving clockwise at constant speed, and the map travels the shortest distance possible given this condition.
T(n, k) gives the number of f for m = n-1 such that f(1) = 1 and f' has degree k. This is trivially one n-th of the number of f with degree k when f(1) is arbitrary.
Equivalent to having degree k is that there are k values a immediately before b in the cyclic order such that f(a) > f(b) (in the standard order of N).
If we change things so that a immediately before b satisfies f(a) = f(b) corresponds to a full rotation (this is equivalent to using the condition f(a) >= f(b) in the last paragraph), then T(n, k) is the number of f with degree k+1.
T(n, k) is the (k+1)*(n-1)th (n-1)-nomial coefficient of power n - 1.
(End)

			Triangle begins:
  1;
  1,    1;
  1,    7,      1;
  1,   31,     31,      1;
  1,  121,    381,    121,      1;
  1,  456,   3431,   3431,    456,      1;
  1, 1709,  26769,  60691,  26769,   1709,    1;
  1, 6427, 193705, 848443, 848443, 193705, 6427, 1;
  ...
G.f. of row 3: (1 + 31*x + 31*x^2 + x^3) = (1-x)^4*(1 + 35*x + 165*x^2 + 455*x^3 + ... + C(4*j+3,4*j)*x^j + ...).

M. Bayer, B. Goeckner, S. J. Hong, T. McAllister, M. Olsen, C. Pinckney, J. Vega and M. Yip, Lattice polytopes from Schur and symmetric Grothendieck polynomials, Electronic Journal of Combinatorics, Volume 28, Issue 2 (2021). See Proposition 53 and Table 1.
Tanay Wakhare, Iterated Entropy Derivatives and Binary Entropy Inequalities, arXiv:2312.14743 [cs.IT], 2023.
Tanay Wakhare, Two Studies of Constraints in High Dimensions: Entropy Inequalities and the Randomized Symmetric Binary Perceptron, Master's Thesis, MIT (2024). See p. 22.
Raphael Yuster, Almost k-union closed set systems, arXiv:2302.12276 [math.CO], 2023, p. 8.

Cf. A108267, A000169, A048775, A060543, A108291, A108292, A056856.

p := n -> (1-x)^(n+1)*add(binomial(n + n*j + j, n*j + j)*x^j, j = 0..n):
seq(print(seq(coeff(p(n), x, k), k = 0..n)), n = 0..8); # Peter Luschny, Nov 02 2022

T[n_, k_] := Coefficient[(1 - x)^(n + 1)*
     Sum[Binomial[n + n*j + j, n*j + j]*x^j, {j, 0, n}], x, k];
Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 23 2021 *)

T(n,k)=polcoeff((1-x)^(n+1)*sum(j=0,n,binomial(n+n*j+j,n*j+j)*x^j),k)

T(n, 1) = A048775(n) = binomial(2*n + 1, n + 1) - (n + 1).
Sum_{k=0..n} T(n, k) = A000169(n) = (n + 1)^n.
Sum_{k=0..n} T(n, k)*2^k = A108292(n).
From Thomas Anton, Jan 05 2023: (Start)
T(n, k) = Sum_{i=0..k} (-1)^i*binomial(n + 1, i)*binomial(n+(n+1)*(k-i), n).
T(n, k) = T(n, n-k).
(End)

A220883 Triangle read by rows: row n gives coefficients of expansion of Product_{k = 1..n-1} ((n + 1)*x + k), starting with lowest power.

Original entry on oeis.org

1, 1, 3, 2, 12, 16, 6, 55, 150, 125, 24, 300, 1260, 2160, 1296, 120, 1918, 11025, 29155, 36015, 16807, 720, 14112, 103936, 376320, 716800, 688128, 262144, 5040, 117612, 1063692, 4934601, 12859560, 19013778, 14880348, 4782969, 40320, 1095840, 11812400, 67284000, 224490000, 453600000, 546000000, 360000000, 100000000, 362880, 11292336, 141896700, 963218080, 3943187325, 10190179923, 16741251450, 16953838770, 9646149645, 2357947691

Offset: 1

N. J. A. Sloane, Dec 29 2012

Related to Stirling numbers A008275, A008277.

			Triangle begins:
    1
    1     3
    2    12     16
    6    55    150    125
   24   300   1260   2160   1296
  120  1918  11025  29155  36015  16807
  720 14112 103936 376320 716800 688128 262144
  ...

Paolo Xausa, Table of n, a(n) for n = 1..11325 (rows 1..150 of the triangle, flattened).
Peter Bala, Fractional iteration of a series inversion operator
Jean-Christophe Novelli and Jean-Yves Thibon, Duplicial algebras and Lagrange inversion, arXiv preprint arXiv:1209.5959 [math.CO], 2012.

Cf. A220884, A008275, A008277, A056856, A260687.

seq(seq(coeff(mul((n+1)*t + k, k = 1..n-1), t, i), i = 0..n-1), n = 1 .. 10); # Peter Bala, Nov 16 2015
# Alternative:
T := (n, k) -> (-1)^(n-k)*(n+1)^(k-1)*Stirling1(n, k):
seq(print(seq(T(n, k), k=1..n)), n=1..8);
# Peter Luschny, Mar 20 2024

A220883[n_, k_] := (-1)^(n-k)*(n+1)^(k-1)*StirlingS1[n, k];
Table[A220883[n, k], {n, 10}, {k, n}] (* Paolo Xausa, Mar 19 2024 *)

From Peter Bala, Nov 16 2015: (Start)
E.g.f.: A(x,t) = x + (1 + 3*t)*x^2/2! + (1 + 4*t)*(2 + 4*t)*x^3/3! + ....
The function F(x,t) := 1 + t*A(x,t) has several nice properties:
F(x,t) = 1/x*Revert( x*(1 - x)^t ) = 1 + t*x + t*(1 + 3*t)*x^2/2! + t*(2 + 12*t + 16*t^2)*x^3/3! + ..., where Revert denotes the series reversion operator with respect to x.
F(x,t)*(1 - x*F(x,t))^t = 1.
F(x,t)^m = 1 + m*t*x + m*t*((m + 2)*t + 1)*x^2/2! + m*t*((m + 3)*t + 1)*((m + 3)*t + 2)*x^3/3! + m*t*((m + 4)*t + 1)*((m + 4)*t + 2)*((m + 4)*t + 3)*x^4/4! + ....
Log(F(x,t)) = t*x + t*(1 + 2*t)*x^2/2! + t*(1 + 3*t)*(2 + 3*t)*x^3/3! + t*(1 + 4*t)*(2 + 4*t)*(3 + 4*t)*x^4/4! + ... is the e.g.f for A056856.
F(x,t) = G(x,t)^t, where G(x,t) = 1 + x + (2 + 2*t)*x^2/2! + (2 + 3*t)*(3 + 3*t)*x^3/3! + (2 + 4*t)*(3 + 4*t)*(4 + 4*t)*x^4/4! + ... is the o.g.f. for A260687. (End)
T(n, k) = (-1)^(n-k)*(n+1)^(k-1)*Stirling1(n, k). - Peter Luschny, Mar 01 2021 [Corrected by Paolo Xausa, Mar 19 2024]

A260687 Triangular array with n-th row giving coefficients of polynomial Product_{k = 2..n} (k + n*t) for n >= 1.

Original entry on oeis.org

1, 2, 2, 6, 15, 9, 24, 104, 144, 64, 120, 770, 1775, 1750, 625, 720, 6264, 20880, 33480, 25920, 7776, 5040, 56196, 250096, 571095, 708295, 453789, 117649, 40320, 554112, 3127040, 9433088, 16486400, 16744448, 9175040, 2097152, 362880, 5973264, 41229324, 156498804

Offset: 1

Peter Bala, Nov 16 2015

Related to A220883 and A251592.

			Triangle begins
...1
...2      2
...6     15       9
..24    104     144      64
.120    770    1775    1750     625
.720   6264   20880   33480   25920    7776
5040  56196  250096  571095  708295  453789  117649
...

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison-Wesley, Reading, MA, 2nd ed. 1998

P. Bala, Fractional iteration of a series inversion operator

A000142 (column 0), A000169 (main diagonal), A006675 (column 1). Cf. A001700, A025174, A056856, A163456, A220883, A224274, A251592.

seq(seq(coeff(mul(n*t + k, k = 2 .. n), t, i), i = 0..n-1), n = 1..10);

E.g.f. (with constant term 1 included): A(x,t) = [ 1/x*Revert( x*(1 - x)^t ) ]^(1/t) = Sum_{n >= 0} 1/(n*t + 1)*binomial(n*t + n,n)*x^n = 1 + x + (2 + 2*t)*x^2/2! + (2 + 3*t)*(3 + 3*t)*x^3/3! + (2 + 4*t)*(3 + 4*t)*(4 + 4*t)*x^4/4! + ..., where Revert denotes the series reversion operator with respect to x.
In the notation of the Bala link, A(x,t) = I^t(1/(1 - x)) where I^t is a fractional inversion operator.
A(x,t) = B_(1+t)(x), where B_t(x) is the e.g.f. for A251592 and is the generalized binomial series of Lambert. See Graham et al., Section 5.4 and Section 7.5.
A(x,t)^m = Sum_{n >= 0} m/(n*t + m)*binomial(n*t + n + m - 1,n)*x^n = 1 + m*x + m*(2*t + m + 1)*x^2/2! + m*(3*t + m + 1)*(3*t + m + 2)*x^3/3! + m*(4*t + m + 1)*(4*t + m + 2)*(4*t + m + 3)*x^4/4! + ....
A(x,t)^t = 1 + t*x + t(1 + 3*t)*x^2/2! + t*(1 + 4*t)*(2 + 4*t)*x^3/3! + t*(1 + 5*t)*(2 + 5*t)*(3 + 5*t)*x^4/4! + ... is the e.g.f for A220883 with an extra constant term 1 and an extra factor of t included.
t*log( A(x,t) ) = t*x + t*(1 + 2*t)*x^2/2! + t*(1 + 3*t)*(2 + 3*t)*x^3/3! + t*(1 + 4*t)*(2 + 4*t)*(3 + 4*t)*x^4/4! + ... is the e.g.f for A056856.
For n = 1,2,3,..., the sequence [x^n] A(x,t)^n = [1, (2*t + 3), (3*t + 4)*(3*t + 5)/2!, (4*t + 5)*(4*t + 6)*(4*t + 7)/3!, ...]. This sequence has the following specializations:
t = 0: [1, 3, 10, 35, 126, ...] = A001700 (with different offset).
t = 1: [1, 5, 28, 165, 1001, ...] = A025174.
t = 2: [1, 7, 55, 455, 3876, ...] = A224274.
t = 3: [1, 9, 91, 969, 10626, ...] = A163456.

A203904 Triangular array T; for n>0, row n shows the coefficients of a reduced polynomial having zeros -k/(n+1) for k=1,2,...,n.

Original entry on oeis.org

1, 1, 2, 2, 9, 9, 3, 22, 48, 32, 24, 250, 875, 1250, 625, 10, 137, 675, 1530, 1620, 648, 720, 12348, 79576, 252105, 420175, 352947, 117649, 315, 6534, 52528, 216608, 501760, 659456, 458752, 131072, 4480, 109584, 1063116, 5450004, 16365321

Offset: 1

Clark Kimberling, Jan 08 2012

For n>0, the zeros of the polynomial represented by row n+1 interlace the zeros of the polynomial for row n; see the Example section.
...
T(n,1):  A119619
T(n,n):  A056916.

			First five rows(counting the top row as row 0):
1
1...2.................representing 1+2x
1...9...9.............representing 2+9x+9x^2
3...22..48...32
24...250...875...1250...625
Zeros corresponding to rows 1 to 4:
.................-1/2
............-2/3......-1/3
......-3/4.......-1/2.......-1/4
-4/5........-3/5......-2/5.......-1/5
Interlace property for successive rows illustrated by
  1/5 < 1/4 < 2/5 < 1/2 < 3/5 < 3/4 < 4/5.

Cf. A056856, A119619, A056916, A007305/A007306 (Farey fractions).

p[n_, x_] := Product[(n*x + k)/GCD[n, k], {k, 1, n - 1}]
Table[CoefficientList[p[n, x], x], {n, 1, 10}]
TableForm[%]  (* A203904 triangle *)
Flatten[%%]   (* A203904 sequence *)

A355007 Triangle read by rows. T(n, k) = n^k * |Stirling1(n, k)|.

Original entry on oeis.org

1, 0, 1, 0, 2, 4, 0, 6, 27, 27, 0, 24, 176, 384, 256, 0, 120, 1250, 4375, 6250, 3125, 0, 720, 9864, 48600, 110160, 116640, 46656, 0, 5040, 86436, 557032, 1764735, 2941225, 2470629, 823543, 0, 40320, 836352, 6723584, 27725824, 64225280, 84410368, 58720256, 16777216

Offset: 0

Peter Luschny, Jun 17 2022

			Table T(n, k) begins:
[0] 1;
[1] 0,    1;
[2] 0,    2,     4;
[3] 0,    6,    27,     27;
[4] 0,   24,   176,    384,     256;
[5] 0,  120,  1250,   4375,    6250,    3125;
[6] 0,  720,  9864,  48600,  110160,  116640,   46656;
[7] 0, 5040, 86436, 557032, 1764735, 2941225, 2470629, 823543;

A000142 (column 1), A000407 (row sums), A000312 (main diagonal), A355006.

Cf. A133942.

seq(seq(n^k*abs(Stirling1(n, k)), k = 0..n), n = 0..9);

T[n_, k_] := If[n == k == 0, 1, n^k * Abs[StirlingS1[n, k]]]; Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten (* Amiram Eldar, Jun 17 2022 *)

Sum_{k=0..n} (-1)^k * T(n,k) = A133942(n). - Alois P. Heinz, Mar 30 2023

Conjecture: T(n,k) = A056856(n,k)*n. - R. J. Mathar, Mar 31 2023

cp's OEIS Frontend

A108267 Triangle read by rows, T(n, k) = [x^k] (1-x)^(n+1)*Sum_{j=0..n} binomial(n + n*j + j, n*j + j)*x^j.

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A220883 Triangle read by rows: row n gives coefficients of expansion of Product_{k = 1..n-1} ((n + 1)*x + k), starting with lowest power.

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A260687 Triangular array with n-th row giving coefficients of polynomial Product_{k = 2..n} (k + n*t) for n >= 1.

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A203904 Triangular array T; for n>0, row n shows the coefficients of a reduced polynomial having zeros -k/(n+1) for k=1,2,...,n.

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A355007 Triangle read by rows. T(n, k) = n^k * |Stirling1(n, k)|.

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A108267 Triangle read by rows, T(n, k) = [x^k] (1-x)^(n+1)Sum_{j=0..n} binomial(n + nj + j, nj + j)x^j.