A134432 -id:A134432 - cp's OEIS frontend

A271705 Triangle read by rows, T(n,k) = Sum_{j=0..n} C(n,j)*L(j,k), L the unsigned Lah numbers A271703, for n>=0 and 0<=k<=n.

1, 1, 1, 1, 4, 1, 1, 15, 9, 1, 1, 64, 66, 16, 1, 1, 325, 490, 190, 25, 1, 1, 1956, 3915, 2120, 435, 36, 1, 1, 13699, 34251, 23975, 6755, 861, 49, 1, 1, 109600, 328804, 283136, 101990, 17696, 1540, 64, 1, 1, 986409, 3452436, 3534636, 1554966, 342846, 40404, 2556, 81, 1

Offset: 0

Peter Luschny, Apr 14 2016

This is the Sheffer (aka exponential Riordan) matrix T = P*L = A007318*A271703 = (exp(x), x/(1-x)). Note that P = A007318 is Sheffer (exp(t), t) (of the Appell type). The Sheffer a-sequence is [1,1,repeat(0)] and the z-sequence has e.g.f. (x/(1+x))*(1 - exp(-x/(1+x)) given in A288869 / A000027. Because the column k=0 has only entries 1, the z-sequence gives fractional representations of 1. See A288869. - Wolfdieter Lang, Jun 20 2017

			Triangle starts:
  1;
  1,    1;
  1,    4,    1;
  1,   15,    9,    1;
  1,   64,   66,   16,   1;
  1,  325,  490,  190,  25,  1;
  1, 1956, 3915, 2120, 435, 36, 1;
  ...
Recurrence: T(3, 2) = (3/2)*4 + 3*1 = 9. - _Wolfdieter Lang_, Jun 20 2017

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
Marin Knežević, Vedran Krčadinac, and Lucija Relić, Matrix products of binomial coefficients and unsigned Stirling numbers, arXiv:2012.15307 [math.CO], 2020.

Cf. A000290 (diag n, n-1), A062392 (diag n, n-2).
Cf. A007526 (col. 1), A134432 (col. 2).
Cf. A052844 (row sums), A059110 (matrix inverse).
Cf. A007318, A271703, A288869.

B:=Binomial;
A271705:= func< n,k | k eq 0 select 1 else (&+[B(n, j+k)*B(j+k, k)*B(j+k-1, k-1)*Factorial(j): j in [0..n-k]]) >;
[A271705(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 09 2022

L := (n,k) -> `if`(k<0 or k>n,0,(n-k)!*binomial(n,n-k)*binomial(n-1,n-k)):
T := (n,k) -> add(L(j,k)*binomial(-j-1,-n-1)*(-1)^(n-j), j=0..n):
seq(seq(T(n,k), k=0..n), n=0..9);

T[n_, k_]:= If[k==0, 1, Sum[((k*j!)/(j+k))*Binomial[n, j+k]*Binomial[j+k, k]^2, {j,0,n-k}]];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jan 09 2022 *)

b=binomial
def A271705(n,k): return 1 if (k==0) else sum(factorial(j-k)*b(n, j)*b(j, k)*b(j-1, k-1) for j in (k..n))
flatten([[A271705(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jan 09 2022

From Wolfdieter Lang, Jun 20 2017: (Start)
T(n, k) = Sum_{m=k..n} A007318(n, m)*A271703(m, k), n >= k >= 0, and 0 for k < m. See also the name.
E.g.f. of column k: exp(x)*(x/(1-x))^k/k! (Sheffer property), k >= 0.
E.g.f. of triangle (or row polynomials in x): exp(z)*exp(x*z/(1-z)).
Recurrence for T(n, k), k >= 1, with T(n, 0) = 1, T(n, k) = 0 if n < k: T(n, k) = (n/k)*T(n-1, k-1) + n*T(n-1, k), n >= 1, k = 1..n. (From the a-sequence with column k=0 as input.) (End)
T(n, k) = Sum_{j=0..n-k} j!*binomial(n, j+k)*binomial(j+k, k)*binomial(j+k-1, k-1) with T(n, 0) = 1. - G. C. Greubel, Jan 09 2022
From Natalia L. Skirrow, Jun 11 2025: (Start)
T(n, k) = C(n, k)*hypergeom([k-n, k], [], -1), which equals C(n, k)*A143409(n-k, k-1) for k>0.
By the saddle point method upon the e.g.f., n-th row polynomial converges with n (for all y) to n^n*exp(2*sqrt(n*y) - n - y/2 + 1)/sqrt(2*sqrt(n/y)); as such, the n-th row's expectation is ~ sqrt(n)-1/4 and the n-th row's variance is ~ (sqrt(n)-1)/2. (End)

A326659 T(n,k) = [0=0]; triangle T(n,k), n >= 0, 0 <= k <= n, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 4, 2, 1, 15, 18, 6, 1, 64, 132, 96, 24, 1, 325, 980, 1140, 600, 120, 1, 1956, 7830, 12720, 10440, 4320, 720, 1, 13699, 68502, 143850, 162120, 103320, 35280, 5040, 1, 109600, 657608, 1698816, 2447760, 2123520, 1108800, 322560, 40320

Offset: 0

Alois P. Heinz, Sep 12 2019

[] is an Iverson bracket.

			Triangle T(n,k) begins:
  1;
  1,     1;
  1,     4,     2;
  1,    15,    18,      6;
  1,    64,   132,     96,     24;
  1,   325,   980,   1140,    600,    120;
  1,  1956,  7830,  12720,  10440,   4320,   720;
  1, 13699, 68502, 143850, 162120, 103320, 35280, 5040;
  ...

Alois P. Heinz, Rows n = 0..140, flattened
Wikipedia, Iverson bracket

Columns k=0-2 give: A000012, A007526, 2*A134432(n-1).
Main diagonal gives A000142.
Row sums give A308876.
Cf. A000522, A073474, A093964, A143409, A196347, A271705, A327606.

T:= proc(n, k) option remember;
      `if`(0=0, 1, 0)
    end:
seq(seq(T(n, k), k=0..n), n=0..10);

T[n_ /; n >= 0, k_ /; k >= 0] := T[n, k] = Boole[0 < k <= n]*n*(T[n-1, k-1] + T[n-1, k]) + Boole[k == 0 && n >= 0];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 09 2021 *)

E.g.f. of column k: exp(x)*(x/(1-x))^k.
T(n,k) = k! * A271705(n,k).
T(n,k) = n * A073474(n-1,k-1) for n,k >= 1.
T(n,1) = n * A000522(n-1) for n >= 1.
T(n,2) = n * A093964(n-1) for n >= 1.
Sum_{k=1..n} k * T(n,k) = A327606(n).

A134431 Triangle read by rows: T(n,k) is the number of arrangements of the set {1,2,...,n} in which the sum of the entries is equal to k (n >= 0, k >= 0; to n=0 there corresponds the empty set).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 2, 2, 6, 1, 1, 1, 3, 3, 4, 8, 8, 6, 6, 24, 1, 1, 1, 3, 3, 5, 10, 10, 14, 14, 36, 30, 30, 24, 24, 120, 1, 1, 1, 3, 3, 5, 11, 12, 16, 22, 44, 44, 66, 60, 78, 174, 168, 144, 144, 120, 120, 720, 1, 1, 1, 3, 3, 5, 11, 13, 18, 24, 52, 52, 80, 98, 120, 234

Offset: 0

Emeric Deutsch, Nov 16 2007

Row n has 1 + n(n+1)/2 terms (n >= 0). Row sums yield the arrangement numbers (A000522). T(n, n(n+1)/2) = n!. Sum_{k=0..n(n+1)/2} k*T(n,k) = A134432(n).

			T(4,7)=8 because we have 34,43 and the six permutations of {1,2,4}.
Triangle starts:
  1;
  1, 1;
  1, 1, 1, 2;
  1, 1, 1, 3, 2, 2, 6;
  1, 1, 1, 3, 3, 4, 8, 8, 6, 6, 24;

Alois P. Heinz, Rows n = 0..48, flattened

Cf. A000522, A134432.

Q[0]:=1: for n to 7 do Q[n]:=sort(simplify(Q[n-1]+t^n*x*(diff(x*Q[n-1],x))), t) end do: for n from 0 to 7 do P[n]:=sort(subs(x=1,Q[n])) end do: for n from 0 to 7 do seq(coeff(P[n],t,j),j=0..(1/2)*n*(n+1)) end do; # yields sequence in triangular form
# second Maple program:
b:= proc(n, s, t) option remember;
      `if`(n=0, t!*x^s, b(n-1, s, t)+b(n-1, s+n, t+1))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n, 0$2)):
seq(T(n), n=0..8);  # Alois P. Heinz, Dec 22 2017

b[n_, s_, t_] := b[n, s, t] = If[n == 0, t!*x^s, b[n - 1, s, t] + b[n - 1, s + n, t + 1]];
T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}]] @ b[n, 0, 0];
T /@ Range[0, 8] // Flatten (* Jean-François Alcover, Feb 19 2020, after Alois P. Heinz *)

The row generating polynomials P[n](t) are equal to Q[n](t,1), where the polynomials Q[n](t,x) are defined by Q[0]=1 and Q[n]=Q[n-1] + xt^n (d/dx)xQ[n-1]. [Q[n](t,x) is the bivariate generating polynomial of the arrangements of {1,2,...,n}, where t (x) marks the sum (number) of the entries; for example, Q[2](t,x)=1+tx + t^2*x + 2t^3*x^2, corresponding to: empty, 1, 2, 12 and 21, respectively.]

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A271705 Triangle read by rows, T(n,k) = Sum_{j=0..n} C(n,j)*L(j,k), L the unsigned Lah numbers A271703, for n>=0 and 0<=k<=n.

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A326659 T(n,k) = [0=0]; triangle T(n,k), n >= 0, 0 <= k <= n, read by rows.

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A134431 Triangle read by rows: T(n,k) is the number of arrangements of the set {1,2,...,n} in which the sum of the entries is equal to k (n >= 0, k >= 0; to n=0 there corresponds the empty set).

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Maple

Mathematica

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