A074909 -id:A074909 - cp's OEIS frontend

A249548 Coefficients of reduced partition polynomials of A134264 for computing Lagrange compositional inversion.

1, 0, 1, 1, 1, 2, 1, 5, 1, 6, 3, 5, 1, 7, 7, 21, 1, 8, 8, 4, 28, 28, 14, 1, 9, 9, 9, 36, 72, 12, 84, 1, 10, 10, 10, 5, 45, 90, 45, 45, 120, 180, 42, 1, 11, 11, 11, 11, 55, 110, 110, 55, 55, 165, 495, 165, 330, 1, 12, 12, 12, 12, 6, 66, 132, 132, 66, 66, 132, 22, 220, 660, 330, 660, 55, 495, 990, 132

Offset: 0

Tom Copeland, Oct 31 2014

Coefficients of reduced partition polynomials of A134264 for computing the complete partition polynomials for the Lagrange compositional inversion of A134264 (see Oct 2014 comment by Copeland there). Umbrally,
e^(x*t) * exp[Prt(.;1,0,h_2,..) * t] = exp[Prt(.;1,x,h_2,..) * t], where Prt(n;1,0,h_2,..,h_n) are the reduced (h_0 = 1 and h_1 = 0) partition polynomials of the complete polynomials Prt(n;h_0,h_1,h_2,..,h_n) of A134264.
Partitions are given in the order of those on page 831 of Abramowitz and Stegun. Formulas for the coefficients of the partitions are given in A134264.
Row sums are the Motzkin sums or Riordan numbers A005043. - Tom Copeland, Nov 09 2014
From Tom Copeland, Jul 03 2018: (Start)
The matrix and operator formalism for Sheffer Appell sequences leads to the following relations with D = d/dh_1.
Exp[Prt(.;1,0,h_2,..) * D] (h_1)^n = [h_1 + Prt(.;1,0,h_2,...)]^n = Prt(n;1,h_1,h_2,...), the partition polynomials of A134264 for g(t)/t with h_0 = 1.
For the umbral compositional inverses described below,
Exp[UPrt(.;1,0,h_2,..) * D] (h_1)^n = [h_1 + UPrt(.;1,0,h_2,...)]^n = UPrt(n;1,h_1,h_2,...).
The respective e.g.f.s are multiplicative inverses; that is, exp[Prt(.;1,0,h_2,..) * t] = 1/exp[UPrt(.;1,0,h_2,..) * t], so the formalism of A133314 applies.
The raising operator R such that R Prt(n;1,h_1,h_2,...) = Prt(n+1;1,h_1,h_2,...) is R = exp[Prt(.;1,0,h_2,...)*D] h_1 exp[UPrt(.;1,0,h_2,..)*D] since R Prt(n+1;1,h_1,h_2,...) = exp[Prt(.;1,0,h_2,...)*D] h_1 (h_1)^n = Prt(n+1;h_1,h_2,...) from the definition of the umbral compositional inverse. This may also be expressed as R = h_1 + d/dD log[exp[Prt(.;1,0,h_2,...) * D]], so, using A127671, R = h_1 + h_2 D + h_3 D^2/2! + (h_4 - h_2^2) D^3/3! + (h_5 - 5 h_2 h_3) D^4/4! + (h_6 - 9 h_2 h_4 + 5 h_2^3 - 7 h_3^2) D^5/5! + (h_7 - 28 h_3 h_4 - 14 h_2 h_5 + 56 h_2^2 h_3) D^6/6! + ... . (End)

			Prt(0) = 1
Prt(1;1,0) = 0
Prt(2;1,0,h_2) = 1 h_2
Prt(3;1,0,h_2,h_3) = 1 h_3
Prt(4;1,0,h_2,..,h_4) = 1 h_4 + 2 (h_2)^2
Prt(5;1,0,h_2,..,h_5) = 1 h_5 + 5 h_2 h_3
Prt(6;1,0,h_2,..,h_6) = 1 h_6 + 6 h_2 h_4 + 3 (h_3)^2 + 5 (h_2)^3
Prt(7;1,0,h_2,..,h_7) = 1 h_7 + 7 h_3 h_4 + 7 h_2 h_5 + 21 h_3 (h_2)^2
...
------------
With h_n denoted by (n'), the first seven partition polynomials of A134264 with h_0=1 are given by the first seven coefficients of the truncated Taylor series expansion of the Euler binomial transform
e^[(1') * t] * {1 + 1 (2') *  t^2/2! + 1 (3') *  t^3/3! + [1 (4') + 2 (2')^2] *  t^4/4! + [1 (5') + 5 (2')(3')] *  t^5/5! + [1 (6') + 6 (2')(4') + 3 (3')^2 + 5 (2')^3] *  t^6/6!}, giving the truncated expansion
1 + 1 (1') * t + [1 (2') + 1 (1')^2] * t^2/2! + ... + [1 (6') + 6 (1')(5') + 6 (2')(4') + 3 (3')^2 + 15 (1')^2(4') + 30 (1')(2')(3') + 5 (2')^3 + 20 (1')^3(3') + 30 (1')^2(2')^2 + 15 (1')^4(2') + 1 (1')^6] * t^6/6!.
Extending the number of reduced partition polynomials of the transform allows for further complete polynomials of A134264 to be computed.

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].

Cf. A134264, A005043, A133314, A049019, A074909, A135278.
Cf. A127671.
Rows lengths are given by A002865 (except for row 1).

rows[n_] := {{1}, {0}}~Join~Module[
    {g = 1 / D[t / (1 + Sum[h[k] t^k, {k, 2, n}] + O[t]^(n+1)), t], p = t, r},
    r = Reap[Do[p = g D[p, t]/k; Sow[Expand[Normal@p /. {t -> 0}]], {k, n+1}]][[2, 1, 2 ;;]];
    Table[Coefficient[r[[k]], Product[h[t], {t, p}]], {k, 2, n}, {p, Sort[Sort /@ IntegerPartitions[k, k, Range[2, k]]]}]];
rows[12] // Flatten (* Andrey Zabolotskiy, Feb 18 2024 *)

From Tom Copeland, Nov 10 2014: (Start)
Terms may be computed symbolically up to order n by using an iterated derivative evaluated at t=0:
with g(t) = 1/{d/dt [t/(1 + 0 t + h_2 t^2 + h_3 t^3 + ... + h_n t^n)]},
evaluate 1/n! * [g(t) d/dt]^n t at t=0, i.e., ask a symbolic math app for the first term in a series expansion of this iterated derivative, to obtain Prt(n-1).
Alternatively, the explicit formula in A134264 for the numerical coefficients of each partition can be used. (End)
From Tom Copeland, Nov 12 2014: (Start)
The first few partitions polynomials formed by taking the reciprocal of the e.g.f. of this entry's e.g.f. (cf. A133314) are
UPrt(0) = 1
UPrt(1;1,0) = 0
UPrt(2;1,0,h_2) = -1 h_2
UPrt(3;1,0,h_2,h_3) = -1 h_3
UPrt(4;1,0,h_2,..,h_4) = -1 h_4 + 4 (h_2)^2
UPrt(5;1,0,h_2,..,h_5) = -1 h_5 + 15 h_2 h_3
UPrt(6;1,0,h_2,..,h_6) = -1 h_6 + 24 h_2 h_4 + 17 (h_3)^2 + -35 (h_2)^3
...
Therefore, umbrally, [Prt(.;1,0,h_2,...) + UPrt(.;1,0,h_2,...)]^n = 0 for n>0 and unity for n=0.
Example of the umbral operation:
(a. +  b.)^2 = a.^2 + 2 a.* b. + b.^2 = a_2 + 2 a_1 * b_1 + b_2.
This implies that the umbral compositional inverses (see below) of the partition polynomials of the Lagrange inversion formula (LIF) of A134264 with h_0=1 are given by UPrt(n;1,h_1,h_2,...,h_n) = [UPrt(.;1,0,h_2,...,h_n) + h_1]^n, so that the sequence of polynomials UPrt(n;1,h_1,h_2,...,h_n) is an Appell sequence in the indeterminate h_1. So, if one calculates UPrt(n;1,h_1,...,h_n), the lower order UPrt(n-1;1,h_1,...,h_(n-1)) can be found by taking the derivative w.r.t. h_1 and dividing by n. Same applies for Prt(n;1,h_1,h_2,...,h_n).
This connects the combinatorics of the permutohedra through A133314 and A049019, or their duals, to the noncrossing partitions, Dyck lattice paths, etc. that are isomorphic with the LIF of A134264.
An Appell sequence P(.,x) with the e.g.f. e^(x*t)/w(t) possesses an umbral inverse sequence UP(.,x) with the e.g.f. w(t)e^(x*t), i.e., polynomials such that P(n,UP(.,x))= x^n = UP(n,P(.,x)) through umbral substitution, as in the binomial example. The Bernoulli polynomials with w(t) = t/(e^t - 1) are a good example with the umbral compositional inverse sequence UP(n,x) = [(x+1)^(n+1)-x^(n+1)] / (n+1) (cf. A074909 and A135278). (End)

Formula for Prt(7,..) and a(12)-a(15) added by Tom Copeland, Jul 22 2016

Rows 8-12 added by Andrey Zabolotskiy, Feb 18 2024

A090971 Sierpiński's triangle, read by rows, starting from 1: T(n,k) = (T(n-1,k) + T(n-1,k-1)) mod 2.

Original entry on oeis.org

1, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1

Offset: 1

Benoit Cloitre, Feb 28 2004

Row sums give A038573.

			Triangle begins with:
  1;
  0, 1;
  1, 1, 1;
  0, 0, 0, 1;
  1, 0, 0, 1, 1;
  0, 1, 0, 1, 0, 1;
  1, 1, 1, 1, 1, 1, 1; ...

G. C. Greubel, Rows n = 1..100 of triangle, flattened

Cf. A007318, A038573, A047999, A062534, A074909.

T[n_, k_]:= T[n, k]= If[k<0 || k>n, 0, If[k==n, 1, Mod[T[n-1,k] + T[n-1, k-1], 2]]]; Table[T[n, k], {n,1,10}, {k,1,n}] (* G. C. Greubel, Feb 03 2019 *)
Table[Boole[BitAnd[n, k] == k], {n, 1, 14}, {k, 1, n}] // Flatten (* Amiram Eldar, Aug 24 2024 *)

T(n,k)=if(k<0 || k>n, 0, if(n==0, 1, (T(n-1,k)+T(n-1,k-1))%2))

From Philippe Deléham, Feb 29 2004: (Start)
Triangle A047999(n, k) for n,k > 0; A047999: Pascal's triangle mod 2.
a(n) = A062534(n-1) mod 2.
T(n-1, k-1) = A074909(n, n-k) mod 2. (End)
T(n, k) = 1 if bitand(n, k) = k, and 0 otherwise. - Amiram Eldar, Aug 24 2024

A132681 Infinitesimal generator matrix for a diagonally-shifted Pascal matrix, binomial(n+m,k+m), for m=1, related to Laguerre(n,x,m).

Original entry on oeis.org

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

Offset: 0

Tom Copeland, Nov 15 2007, Nov 16 2007, Nov 27 2007

Analogous to the infinitesimal Pascal matrix (m=0), A132440.
In general the matrix T begins (here m=1)
0;
m+1,0;
0, m+2, 0;
0, 0, m+3, 0;
0, 0, 0, m+4, 0;
Let LM(t) = exp(t*T) = limit [1 + t*T/n]^n as n tends to infinity.
Laguerre matrix(m) = [bin(n+m,k+m)] = LM(1) = exp(T) = [ revert of A074909 for m=1 ]. Truncating the series gives the n X n submatrices. In fact, the submatrices of T are nilpotent with [Tsub_n]^(n+1) = 0 for n=0,1,2,....
Inverse Lag matrix(m) = LM(-1) = exp(-T)
Umbrally LM[b(.)] = exp(b(.)*T) = [ bin(n+m,k+m) * b(n-k) ]
A(j) = T^j / j! equals the matrix [bin(n+m,k+m) * delta(n-k-j)] where delta(n) = 1 if n=0 and vanishes otherwise (Kronecker delta); i.e. A(j) is a matrix with all the terms 0 except for the j-th lower (or main for j=0) diagonal which equals that of the Laguerre(m) matrix. Hence the A(j)'s form a linearly independent basis for all matrices of the form [bin(n+m,k+m) d(n-k)].
For sequences with b(0) = 1, umbrally,
LM[b(.)] = exp(b(.)*T) = [ bin(n+m,k+m)] * b(n-k) ] .
[LM[b(.)]]^(-1) = exp(c(.)*T) = [ bin(n+m,k+m)] * c(n-k) ] where c = LPT(b) with LPT the list partition transform of A133314. Or,
[LM[b(.)]]^(-1) = exp[LPT(b(.))*T] = LPT[LM(b(.))] = LM[LPT(b(.))] = LM[c(.)] .
The matrix operation b = T*a can be characterized in several ways in terms of the coefficients a(n) and b(n), their o.g.f.'s A(x) and B(x), or e.g.f.'s EA(x) and EB(x).
1) b(0) = 0, b(n) = (n+m) * a(n-1),
2) B(x) = x^(-m) (xDx) x^m A(x)
3) B(x) = x * Lag(1,-:xD:,m) A(x) = x * [(m+1) + xD] A(x)
4) EB(x) = D^(m) * (x) * D^(-m) EA(x) where D is the derivative w.r.t. x, (:xD:)^j = x^j*D^j, Lag(n,x,m) is the associated Laguerre polynomial and D^(-m) x^n / n! = x^(m+n) / (m+n)! are Riemann-Liouville integrals.
So the exponentiated operator can be characterized (with loose notation) as
5) exp(t*T) A(x) = x^(-m) exp(t*xDx) x^m A(x) = [sum(n=0,1,...) (t*x)^n * Lag(n,-:xD:m)] A(x) = [exp{[t*u/(1-t*u)]*:xD:} / (1-t*u)^(m+1) ] A(x) (eval. at u=x) = A[x/(1-t*x)]/(1-t*x)^(m+1), a generalized Euler transformation for an o.g.f.,
6) exp(t*T) EA(x) = D^(m) exp(t*x) D^(-m) EA(x) = [D/(D-1)]^m exp[(t+a(.))*x] = exp(t*x) [(t+D)/D]^m EA(x)
7) exp(t*T) * a = LM(t) * a = [sum(k=0,...,n) bin(n+m,k+m)* t^(n-k) * a(k)], a vector array.
With t=1 and a(k) = (-x)^k / k!, then LM(1) * a = [Laguerre(n,x,m)], a vector array with index n and the o.g.f. A(x) becomes transformed into the e.g.f. for the associated Laguerre polynomials of order m.
The exponential formulas can be umbrally extended as in A132440. And, the formulas can be extended to non-integer m.

T. Copeland, Infinitesimal Generators, the Pascal Pyramid, and the Witt and Virasoro Algebras
G. Hetyei, Meixner polynomials of the second kind and quantum algebras representing su(1,1), arXiv preprint arXiv:0909.4352 [math.QA], 2009.
M. Janjic, Some classes of numbers and derivatives, JIS 12 (2009) 09.8.3

Table[PadLeft[{n, 0}, n], {n, 0, 12}] // Flatten (* Jean-François Alcover, Apr 30 2014 *)

Given a polynomial sequence p_n(x) with p_0(x)=1 and the lowering and raising operators L and R defined by L P_n(x) = n * P_(n-1)(x) and
  R P_n(x) = P_(n+1)(x), the matrix T represents the action of
  R[(m+1)+ RL] in the p_n(x) basis. For p_n(x) = x^n, L = D = d/dx and R = x. For p_n(x)  = x^n/n!, L = DxD and R = D^(-1). - Tom Copeland, Oct 25 2012

Missing 0 added to array by Tom Copeland, Aug 02 2013

A139526 Triangle A061356 read right to left.

Original entry on oeis.org

1, 1, 2, 1, 6, 9, 1, 12, 48, 64, 1, 20, 150, 500, 625, 1, 30, 360, 2160, 6480, 7776, 1, 42, 735, 6860, 36015, 100842, 117649, 1, 56, 1344, 17920, 143360, 688128, 1835008, 2097152, 1, 72, 2268, 40824, 459270, 3306744, 14880348, 38263752, 43046721, 1, 90, 3600, 84000, 1260000, 12600000, 84000000, 360000000, 900000000, 1000000000

Offset: 2

Alford Arnold, Apr 24 2008

Related to the two Appell sequences the Bernoulli polynomials B(n,x) and their umbral compositional inverses (cf. A074909) Up(n,x) = [(x+1)^(n+1)-x^(n+1)] / (n+1). With offset 0, the row polynomials of this entry P(n,x) = (Up(n,0))^(-n) * [x + Up(n,0)]^n = (n+1)^n * [x + 1/(n+1)]^n. Compare to the Abel polynomials of A061356, which are also an Appell sequence. - Tom Copeland, Nov 14 2014

			(1) times (1) = (1)
(1 1) * (1 2) = (1 2)
(1 2 1 ) * (1 3 9) = (1 6 9)
(1 3 3 1) * (1 4 16 64) = (1 12 48 64)
etc.

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA. Second ed. 1994.
Peter D. Schumer (2004), Mathematical Journeys, page 168, Proposition 16.1 (c)

P. Bala, Fractional iteration of a series inversion operator
Wikipedia, Lambert W function

Cf. A000272 (row sums), A061356 (row reverse), A028421, A074909, A000169 (main diagonal), A251592, A260687.

A061356 := proc(n,k) binomial(n-2,k-1)*(n-1)^(n-k-1); end: A139526 := proc(n,k) A061356(n,n-k-1) ; end: for n from 2 to 14 do for k from 0 to n-2 do printf("%d,",A139526(n,k)) ; od: od: # R. J. Mathar, May 22 2008

T[n_, k_] := (n - 1)^k*Binomial[n - 2, n - k - 2];
Table[T[n, k], {n, 2, 11}, {k, 0, n - 2}] // Flatten (* Jean-François Alcover, Jun 13 2023 *)

for(n=2,12,forstep(k=n-1,1,-1,print1(binomial(n-2, k-1)*(n-1)^(n-k-1)","))) \\ Herman Jamke (hermanjamke(AT)fastmail.fm), May 10 2008

E.g.f. (with offset 1) Sum_{n >= 1} (1 + n*t)^(n-1)*x^n/n! = x + (1 + 2*t)*x^2/2! + (1 + 6*t + 9*t^2)*x^3/3! + .... For properties of this function see Graham et al., equations 5.60, 5.61 and 7.71. The e.g.f. is the series reversion with respect to x of the function log(1 + x)/(1 + x)^t, which is the e.g.f. for a signed version of A028421. - Peter Bala, Jul 18 2013
From Peter Bala, Nov 16 2015: (Start)
E.g.f. with offset 0 and constant term 1: A(x,t) = ( Sum_{n >= 0} (n + 1)^(n-1)*t^n*x^n/n! )^(1/t). This is the generalized exponential series E_t(x) in the terminology of Graham et al., Section 5.4.
A(x,t)^m = 1 + Sum_{n >= 1} m*(m + n*t)^(n-1)*x^n/n!.
log(A(x,t)) = Sum_{n >= 1} (n*t)^(n-1)*x^n/n! = 1/t*T(t*x), where T(z) is Euler's tree function. See A000169.
A(x,t) = ( 1/x* Revert( x*exp(-x*t)) )^(1/t), where Revert is the series reversion operator with respect to x.
In the notation of the Bala link the e.g.f. is I^t(e^x), where I^t is a fractional series inversion operator. Cf. A251592, which has o.g.f. I^t(1 + x), and A260687, which has o.g.f. I^t(1/(1 - x)). (End)

More terms from R. J. Mathar and Herman Jamke (hermanjamke(AT)fastmail.fm), May 01 2008

A178252 Triangle T(n,m) read by rows: the numerator of the coefficient [x^m] of the umbral inverse Bernoulli polynomials B^{-1}(n,x), 0 <= m <= n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 2, 2, 1, 1, 1, 5, 10, 5, 1, 1, 1, 3, 5, 5, 3, 1, 1, 1, 7, 7, 35, 7, 7, 1, 1, 1, 4, 28, 14, 14, 28, 4, 1, 1, 1, 9, 12, 21, 126, 21, 12, 9, 1, 1, 1, 5, 15, 30, 42, 42, 30, 15, 5, 1, 1, 1, 11, 55, 165, 66, 77, 66, 165, 55, 11, 1, 1, 1, 6, 22, 55, 99, 132, 132, 99, 55, 22, 6, 1

Offset: 0

Paul Curtz, May 24 2010

The fractions A053382(n,m)/A053383(n,m) give the triangle of the coefficients of the Bernoulli polynomials:
    1;
  -1/2,   1;
   1/6,  -1,    1;
    0,   1/2, -3/2,  1;
  -1/30,  0,    1,  -2,    1;
    0,  -1/6,   0,  5/3, -5/2,  1;
   1/42,  0,  -1/2,  0,   5/2, -3, 1;
The matrix inverse of this triangle defines coefficients of the umbral inverse Bernoulli polynomials B^{-1}(n,x) in row n:
   1;
  1/2,  1;
  1/3,  1,  1;
  1/4,  1, 3/2,   1;
  1/5,  1,  2,    2,   1;
  1/6,  1, 5/2, 10/3, 5/2,   1;
  1/7,  1,  3,    5,   5,    3,   1;
  1/8,  1, 7/2,   7, 35/4,   7,  7/2,  1;
  1/9,  1,  4,  28/3, 14,   14, 28/3,  4,  1;
  1/10, 1, 9/2,  12,  21, 126/5, 21,  12, 9/2, 1;
  1/11, 1,  5,   15,  30,   42,  42,  30, 15,  5, 1;
The current triangle T(n,m) is the numerator of the entry in row n and column m.
In the majority of cases, T(n,m) = A050169(n,m), but since we use the numerators of the reduced fractions, an integer factor may be missing in this equation.
Umbral composition (e.g., B(.,x)^k = B(k,x)) gives B^(-1)(n,B(.,x)) = x^n = B(n,B^(-1)(.,x)). - Tom Copeland, Aug 25 2015

			The triangle T(n,k) begins:
n\k 0 1  2  3   4   5   6    7   8   9  10 11 12 13
0:  1
1:  1 1
2:  1 1  1
3:  1 1  3  1
4:  1 1  2  2   1
5:  1 1  5 10   5   1
6:  1 1  3  5   5   3   1
7:  1 1  7  7  35   7   7    1
8:  1 1  4 28  14  14  28    4   1
9:  1 1  9 12  21 126  21   12   9   1
10: 1 1  5 15  30  42  42   30  15   5   1
11: 1 1 11 55 165  66  77   66 165  55  11  1
12: 1 1  6 22  55  99 132  132  99  55  22  6  1
13: 1 1 13 26 143 143 429 1716 429 143 143 26 13  1
... reformatted. - _Wolfdieter Lang_, Aug 25 2015

Cf. A178340 (denominators).

Cf. A074909, A258820, A003989, A053382, A053383.

nm := 15 : eM := Matrix(nm,nm) :
for n from 0 to nm-1 do for m from 0 to n do eM[n+1,m+1] :=coeff(bernoulli(n,x),x,m) ; end do: for m from n+1 to nm-1 do eM[n+1,m+1] := 0 ; end do: end do:
eM := LinearAlgebra[MatrixInverse](eM) :
for n from 1 to nm do for m from 1 to n do printf("%a,", numer(eM[n,m])) ; end do: end do: # R. J. Mathar, Dec 21 2010

max = 13; coes = Table[ PadRight[ CoefficientList[ BernoulliB[n, x], x], max], {n, 0, max-1}]; inv = Inverse[coes]; Table[ Take[inv[[n]], n], {n, 1, max}] // Flatten // Numerator (* Jean-François Alcover, Aug 09 2012 *)

tabl(nn) = {for (n=0, nn, for (k=0, n, print1(numerator(binomial(n+1,k)/(n+1)), ", ");); print(););} \\ after Tom Copeland comment; Michel Marcus, Jul 25 2015

"Palindromic:" T(n,m+1) = T(n,n-m). T(n,0)=1.
From Tom Copeland, Jun 18 2015: (Start)
The umbral inverse Bernoulli polynomials are Binv(n,x) = [(1+x)^(n+1)-x^(n+1)]/(n+1) with the e.g.f. e^(t*x) * (e^t-1)/t. (See A074909 for more details.) Therefore, T(n,k) is the numerator of the reduced fraction C(n+1,k)/(n+1) for 0 <= k < (n+1).
The reversed rows are presented as the diagonals of A258820.
T(n,k) = A258820(2n-k,n-k) = A003989(n+1,n+1-k) * n! / [ k! (n+1-k)! ], where A003989(j,k) = gcd(j,k). (End)
From Wolfdieter Lang, Aug 26 2015: (Start)
The following refers to the rational triangle TBinv with entries T(n,k)/A178340(n, m), n >= m >= 0.
The inverse of the Bernoulli triangle TB(n, m) with entries A196838(n,m)/A196839(n,m), n >= m >= 0, is the Sheffer triangle (z/(exp(z)-1),z). Therefore, the inverse triangle TBinv is the Sheffer triangle ((exp(z)-1)/z, z). This means that the e.g.f. of the sequence of column m of TBinv ((exp(x)-1)/x)*x^m/m! for m >= 0.
The e.g.f. of the row polynomials of TBinv, called Binv(n, x) = Sum_{m=0..n} TBinv(n,m)*x^m, is gBinv(z,x) = ((exp(z)-1)/z)*exp(x*z) (of the so-called Appell type).
The e.g.f. of the row sums is gBinv(x,1).
The e.g.f. of the alternating row sums is gBinv(x,-1) = (1 - exp(-x))/x.
The e.g.f. of the a-sequence of this Sheffer triangle is 1, and the e.g.f. of the z-sequence is (exp(x) - x -1)/((exp(x) -1)*x). This is the sequence 1/2, -1/12, 0, 1/120, 0, -1/252, 0, 1/240, 0, -1/132, .... For a- and z-sequences of Sheffer triangles and the corresponding recurrences see A006232.
The convolution property of the row polynomials Binv(n, x) is Binv(n, x+y) = Sum_{k=0..n} binomial(n, k)*Binv(n-k, x)*y^n (or with x and y exchanged).
The row polynomials satisfy (d/dx)Binv(n, x) = n*Binv(n-1, x), with Binv(0, x) = 1 (from Meixner's identity).
(End)

Redefined based on reduced fractions by R. J. Mathar, Dec 21 2010

The term umbral was added by Tom Copeland, Aug 25 2015

A208101 Triangle read by rows: T(n,0) = 1; for n > 0: T(n,1) = n, for n>1: T(n,n) = T(n-1,n-2); T(n,k) = T(n-2,k-1) + T(n-1,k) for k: 1 < k < n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 2, 2, 1, 4, 3, 5, 2, 1, 5, 4, 9, 5, 5, 1, 6, 5, 14, 9, 14, 5, 1, 7, 6, 20, 14, 28, 14, 14, 1, 8, 7, 27, 20, 48, 28, 42, 14, 1, 9, 8, 35, 27, 75, 48, 90, 42, 42, 1, 10, 9, 44, 35, 110, 75, 165, 90, 132, 42, 1, 11, 10, 54, 44, 154, 110

Offset: 0

Reinhard Zumkeller, Mar 04 2012

Another variant of Pascal's triangle, cf. A007318.

			The triangle begins:
0:                    1
1:                  1   1
2:                1   2   1
3:              1   3   2   2
4:            1   4   3   5   2
5:          1   5   4   9   5   5
6:        1   6   5  14   9  14   5
7:      1   7   6  20  14  28  14  14
8:    1   8   7  27  20  48  28  42  14
9:  1   9   8  35  27  75  48  90  42  42

Reinhard Zumkeller, Rows n = 0..150 of triangle, flattened
Index entries for triangles and arrays related to Pascal's triangle

Cf. A208976 (row sums), A101461 (row max), A208983 (central), A208355 (right edge), A074909.

a208101 n k = a208101_tabl !! n !! k
a208101_row n = a208101_tabl !! n
a208101_tabl =  iterate
   (\row -> zipWith (+) ([0,1] ++ init row) (row ++ [0])) [1]

T[, 0] = 1; T[n, 1] := n; T[n_, n_] := T[n-1, n-2]; T[n_, k_] /; 1Jean-François Alcover, Feb 03 2018 *)

A198321 Triangle read by rows: T(n, k) = binomial(n, k-1) for 1 <= k <= n, and T(n, 0) = 0^n.

Original entry on oeis.org

1, 0, 1, 0, 1, 2, 0, 1, 3, 3, 0, 1, 4, 6, 4, 0, 1, 5, 10, 10, 5, 0, 1, 6, 15, 20, 15, 6, 0, 1, 7, 21, 35, 35, 21, 7, 0, 1, 8, 28, 56, 70, 56, 28, 8, 0, 1, 9, 36, 84, 126, 126, 84, 36, 9, 0, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 0, 1, 11, 55, 165, 330

Offset: 0

Philippe Deléham, Nov 01 2011

			Triangle begins :
1
0, 1
0, 1, 2
0, 1, 3, 3
0, 1, 4, 6, 4
0, 1, 5, 10, 10, 5
0, 1, 6, 15, 20, 15, 6

Paolo Xausa, Table of n, a(n) for n = 0..11475 (rows 0..150 of triangle, flattened).

Variant of A074909, A135278.

Cf. A007318.

A198321[n_, k_] := If[k == 0, Boole[n == 0], Binomial[n, k - 1]];
Table[A198321[n, k], {n, 0, 10}, {k, 0, n}] (* Paolo Xausa, Oct 23 2024 *)

T(n, k) is given by (0,1,0,0,0,0,0,0,0,0,0,...) DELTA (1,1,-1,1,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938.
Sum_{k=0..n} T(n, k)*x^k = x*((x+1)^n - x^n) for n > 0.
G.f.: (1 - (1+y)*x + y*(1+y)*x^2)/((1 - (1+y)*x)*(1-y*x)).
T(n, k) = T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k-1) - T(n-2,k-2), T(0,0) = 1, T(1,0) = 0, T(1,1) = 1, T(2,0) = 0, T(2,1) = 1, T(2,2) = 2, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Feb 12 2014

New name using a formula of the author by Peter Luschny, Oct 23 2024

A247976 Triangle read by rows: T(n,k) generated by m-gon expansions in the case of odd m with "vertex to vertex" version or even m with "vertex to side" version. (See comment for details.)

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 4, 6, 4, 1, 4, 6, 4, 1, 4, 6, 4, 1, 4, 6, 4, 1, 5, 10, 10, 5, 1, 5, 10, 10, 5, 1, 5, 10, 10, 5, 1, 5, 10, 10, 5, 1, 6, 15, 20, 15, 6, 1, 6, 15, 20, 15, 6, 1, 6, 15, 20, 15, 6, 1, 6, 15, 20, 15, 6, 1, 7, 21, 35, 35, 21, 7

Offset: 1

Kival Ngaokrajang, Sep 28 2014

Refer to triangle expansions in A061777 and A101946 (and their companions for m-gons) which are "vertex to vertex" and "vertex to side" versions respectively. The label values at each iteration can be arranged as triangle. Any m-gon can also be arranged as the same triangle with conditions: (i) m is odd and expansion is "vertex to vertex" version or (ii) m is even and expansion is "vertex to side" version. m*Sum_{i=1..k}T(n,k) gives the total label value in n-th iteration. See illustration.

			Triangle begins:
  1;
  1,  1;
  1,  1,  2;
  1,  2,  1,  2;
  1,  2,  1,  3,  3;
  1,  3,  3,  1,  3,  3;
  1,  3,  3,  1,  4,  6,  4;
  1,  4,  6,  4,  1,  4,  6,  4;
  1,  4,  6,  4,  1,  5, 10, 10,  5;
  1,  5, 10, 10,  5,  1,  5, 10, 10, 5;
  ...

G. C. Greubel, Rows n = 1..50 of the triangle, flattened
Kival Ngaokrajang, Illustration of initial terms

Rows sum: A027383.
Column (start from 1s): c3=A008805, c4=A058187, c5=A000332 repeated, c6=A000389 repeated, c7=A000579 repeated.
Vertex to vertex: A061777, A247618, A247619, A247620.
Vertex to side: A101946, A247903, A247904, A247905.
Cf. A074909.

T[n_, k_]:= T[n, k]= If[k==1, 1, If[k==n, Floor[(n+1)/2], If[OddQ[n], If[k<=(n+ 1)/2, T[n-1, k], T[n-1, k-1] + T[n-1, k]], If[kG. C. Greubel, Feb 18 2022 *)

@CachedFunction
def T(n,k): # A247976
    if (k==1): return 1
    elif (k==n): return (n+1)//2
    elif (n%2==1): return T(n-1,k) if (k <= (n+1)/2) else T(n-1,k-1) + T(n-1,k)
    else: return T(n-1,k-1)+T(n-1,k) if (k < (n+2)/2) else T(n,k-n/2)
flatten([[T(n,k) for k in (1..n)] for n in (1..15)]) # G. C. Greubel, Feb 18 2022

T(n, k) = ( T(n-1, k) if k <= (n+1)/2 otherwise T(n-1, k-1) + T(n-1, k) ) for odd n rows, ( T(n-1, k-1) + T(n-1, k) if k < (n+2)/2 otherwise T(n, k - n/2) ) for even n rows, with T(n, 1) = 1 and T(n, n) = floor((n+1)/2). - G. C. Greubel, Feb 18 2022

cp's OEIS Frontend

A249548 Coefficients of reduced partition polynomials of A134264 for computing Lagrange compositional inversion.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A090971 Sierpiński's triangle, read by rows, starting from 1: T(n,k) = (T(n-1,k) + T(n-1,k-1)) mod 2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A132681 Infinitesimal generator matrix for a diagonally-shifted Pascal matrix, binomial(n+m,k+m), for m=1, related to Laguerre(n,x,m).

Views

Author

Keywords

Comments

Links

Programs

Mathematica

Formula

Extensions

A139526 Triangle A061356 read right to left.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

Extensions

A178252 Triangle T(n,m) read by rows: the numerator of the coefficient [x^m] of the umbral inverse Bernoulli polynomials B^{-1}(n,x), 0 <= m <= n.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

Extensions

A208101 Triangle read by rows: T(n,0) = 1; for n > 0: T(n,1) = n, for n>1: T(n,n) = T(n-1,n-2); T(n,k) = T(n-2,k-1) + T(n-1,k) for k: 1 < k < n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Mathematica

A198321 Triangle read by rows: T(n, k) = binomial(n, k-1) for 1 <= k <= n, and T(n, 0) = 0^n.

Views

Author

Keywords

Examples

Links

A117965 Triangle read by rows: T(i,j) = F(i)F(j)C(i,j) for 1 <= j <= i, where F(n) is the n-th Fibonacci number and C(n,m) is a binomial coefficient.