A147309 -id:A147309 - cp's OEIS frontend

A147315 L-matrix for Euler numbers A000111(n+1).

1, 1, 1, 2, 3, 1, 5, 11, 6, 1, 16, 45, 35, 10, 1, 61, 211, 210, 85, 15, 1, 272, 1113, 1351, 700, 175, 21, 1, 1385, 6551, 9366, 5901, 1890, 322, 28, 1, 7936, 42585, 70055, 51870, 20181, 4410, 546, 36, 1, 50521, 303271, 563970, 479345, 218925, 58107, 9240, 870, 45, 1

Offset: 0

Paul Barry, Nov 05 2008

This is the inverse of the coefficient array for the orthogonal polynomials p(n,x) defined by: p(n,x)=if(n=-1,0,if(n=0,1,(x-n)p(n-1,x)-C(n,2)p(n-2,x))).
The Hankel array H for A000111(n+1) satisfies H=L*D*U with U the transpose of L.
Row sums are A000772(n+1) with e.g.f. dif(exp(-1)exp(sec(x)+tan(x)),x).
From Peter Bala, Jan 31 2011: (Start)
The following comments refer to the table with an offset of 1: i.e., both the row and column indexing starts at 1.
An increasing tree is a labeled rooted tree with the property that the sequence of labels along any path starting from the root is increasing. A000111(n) for n>=1 enumerates the number of increasing unordered trees on the vertex set {1,2,...,n}, rooted at 1, in which all outdegrees are <=2 (plane unary-binary trees in the notation of [Bergeron et al.])
The entry T(n,k) of the present table gives the number of forests of k increasing unordered trees on the vertex set {1,2,...,n} in which all outdegrees are <=2. See below for some examples.
For ordered forests of such trees see A185421. For forests of increasing ordered trees on the vertex set {1,2,...,n}, rooted at 1, in which all outdegrees are <=2, see A185422.
The Stirling number of the second kind Stirling2(n,k) is the number of partitions of the set [n] into k blocks. Arranging the elements in each block in ascending numerical order provides an alternative combinatorial interpretation for Stirling2(n,k) as counting forests of k increasing unary trees on n nodes. Thus we may view the present array, which counts increasing unary-binary trees, as generalized Stirling numbers of the second kind associated with A000111 or with the zigzag polynomials Z(n,x) of A147309 - see especially formulas (2) and (3) below.
See A145876 for generalized Eulerian numbers associated with A000111. (End)
The Bell transform of A000111(n+1). For the definition of the Bell transform see A264428. - Peter Luschny, Jan 18 2016

			Triangle begins
    1;
    1,    1;
    2,    3,    1;
    5,   11,    6,   1;
   16,   45,   35,  10,   1;
   61,  211,  210,  85,  15,  1;
  272, 1113, 1351, 700, 175, 21, 1;
  ...
The production array for L is the tridiagonal array
  1,  1;
  1,  2,  1;
  0,  3,  3,  1;
  0,  0,  6,  4,  1;
  0,  0,  0, 10,  5,  1;
  0,  0,  0,  0, 15,  6,  1;
  0,  0,  0,  0,  0, 21,  7,  1;
  0,  0,  0,  0,  0,  0, 28,  8,  1,;
  0,  0,  0,  0,  0,  0,  0, 36,  9,  1;
From _Peter Bala_, Jan 31 2011: (Start)
Examples of forests:
The diagrams below are drawn so that the leftmost child of a binary node has the maximum label.
T(4,1) = 5. The 5 forests consisting of a single non-plane increasing unary-binary tree on 4 nodes are
...4... ........ .......... ........... ...........
...|... ........ .......... ........... ...........
...3... .4...3.. .4........ ........4.. ........3..
...|... ..\./... ..\....... ......./... ......./...
...2... ...2.... ...3...2.. ..3...2.... ..4...2....
...|... ...|.... ....\./... ...\./..... ...\./.....
...1... ...1.... .....1.... ....1...... ....1......
T(4,2) = 11. The 11 forests consisting of two non-plane increasing unary-binary trees on 4 nodes are
......... ...3.....
.3...2... ...|.....
..\./.... ...2.....
...1...4. ...|.....
......... ...1...4.
.
......... ...4.....
.4...2... ...|.....
..\./.... ...2.....
...1...3. ...|.....
......... ...1...3.
.
......... ...4.....
.4...3... ...|.....
..\./.... ...3.....
...1...2. ...|.....
......... ...1...2.
.
......... ...4.....
.4...3... ...|.....
..\./.... ...3.....
...2...1. ...|.....
......... ...2...1.
.
......... ......... ..........
..2..4... ..3..4... ..4...3...
..|..|... ..|..|... ..|...|...
..1..3... ..1..2... ..1...2...
......... ......... .......... (End)

Alois P. Heinz, Rows n = 0..140, flattened
Paul Barry, A Note on Three Families of Orthogonal Polynomials defined by Circular Functions, and Their Moment Sequences, Journal of Integer Sequences, Vol. 15 (2012), #12.7.2.
F. Bergeron, Ph. Flajolet and B. Salvy, Varieties of increasing trees, Lecture Notes in Computer Science vol. 581, ed. J.-C. Raoult, Springer 1992, pp. 24-48.
Tom Copeland, Mathemagical Forests
Vladimir Kruchinin, Composition of ordinary generating functions, arXiv:1009.2565 [math.CO], 2010.

Cf. A145876, A147309, A185421, A185422, A185424.

A147315 := proc(n,k) n!*exp(x*(sec(t)+tan(t)-1)) - 1: coeftayl(%,t=0,n) ; coeftayl(%,x=0,k) ; end proc:
seq(seq(A147315(n,k),k=1..n),n=0..12) ; # R. J. Mathar, Mar 04 2011
# second Maple program:
b:= proc(u, o) option remember;
      `if`(u+o=0, 1, add(b(o-1+j, u-j), j=1..u))
    end:
g:= proc(n) option remember; expand(`if`(n=0, 1, add(
      g(n-j)*x*binomial(n-1, j-1)*b(j, 0), j=1..n)))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=1..n+1))(g(n+1)):
seq(T(n), n=0..10);  # Alois P. Heinz, May 19 2021

t[n_, k_] := t[n, k] = t[n-1, k-1] + (k+1)*t[n-1, k] + 1/2*(k+1)*(k+2)*t[n-1, k+1]; t[n_, k_] /; (n < 0 || k < 0 || k > n) = 0; t[0, 0] = t[1, 0] = 1; Flatten[Table[t[n, k], {n, 0, 9}, {k, 0, n}]][[1 ;; 47]] (* Jean-François Alcover, Jun 21 2011, after PARI prog. *)

Co(n,k):=sum(binomial(k,j)*(if oddp(n-k+j) then 0 else if (n-k+j)/2A147315(n,m):=1/m!*sum((if oddp(n-k) then 0 else 2^(1-k)*sum((-1)^(floor((n+k)/2)-i)*binomial(k,i)*(2*i-k)^n,i,0,floor(k/2)))*(sum(Co(i,m)*binomial(k-i+m-1,m-1),i,1,k)),k,m,n); /* Vladimir Kruchinin, Feb 17 2011 */

T(n,m):=(sum(binomial(k+m,m)*((-1)^(n-k-m)+1)*sum(binomial(j+k+m,k+m)*(j+k+m+1)!*2^(-j-k-1)*(-1)^((n+k+m)/2+j+k+m)*stirling2(n+1,j+k+m+1), j,0,n-k-m), k,0,n-m))/(m+1)!; /* Vladimir Kruchinin, May 17 2011 */

{T(n,k)=if(k<0||k>n,0,if(n==0,1,T(n-1,k-1)+(k+1)*T(n-1,k)+(k+1)*(k+2)/2*T(n-1,k+1)))} /* offset=0 */

{T(n,k)=local(X=x+x*O(x^(n+2)));(n+1)!*polcoeff(polcoeff(exp(y*((1+sin(X))/cos(X)-1))-1,n+1,x),k+1,y)} /* offset=0 */

/* Generate from the production matrix P: */
{T(n,k)=local(P=matrix(n,n,r,c,if(r==c-1,1,if(r==c,c,if(r==c+1,c*(c+1)/2)))));if(k<0||k>n,0,if(n==k,1,(P^n)[1,k+1]))}

# uses[bell_matrix from A264428, A000111]
# Adds a column 1,0,0,0, ... at the left side of the triangle.
bell_matrix(lambda n: A000111(n+1), 10) # Peter Luschny, Jan 18 2016

From Peter Bala, Jan 31 2011: (Start)
The following formulas refer to the table with an offset of 1: i.e., both the row n and column k indexing start at 1.
GENERATING FUNCTION
E.g.f.:
(1)... exp(x*(sec(t)+tan(t)-1)) - 1 = Sum_{n>=1} R(n,x)*t^n/n!
= x*t + (x+x^2)*t^2/2! + (2*x+3*x^2+x^3)*t^3/3! + ....
TABLE ENTRIES
(2)... T(n,k) = (1/k!)*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*Z(n,j),
where Z(n,x) denotes the zigzag polynomials as described in A147309.
Compare (2) with the formula for the Stirling numbers of the second kind
(3)... Stirling2(n,k) = (1/k!)*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*j^n.
Recurrence relation
(4)... T(n+1,k) = T(n,k-1) + k*T(n,k) + (1/2)*k(k+1)*T(n,k+1).
ROW POLYNOMIALS
The row polynomials R(n,x) begin
R(1,x) = x
R(2,x) = x+x^2
R(3,x) = 2*x+3*x^2+x^3
They satisfy the recurrence
(5)... R(n+1,x) = x*{R(n,x)+R'(n,x) + (1/2)*R''(n,x)},
where ' indicates differentiation with respect to x. This should be compared with the recurrence satisfied by the Bell polynomials Bell(n,x)
(6)... Bell(n+1,x) = x*(Bell(n,x) + Bell'(n,x)). (End)
From Vladimir Kruchinin, Feb 17 2011: (Start)
Sum_{m=1..n} T(n,m) = A000772(n).
Sum_{m=1..2n-1} T(2n-1,m)* Stirling1(m,1) = A000364(n).
Let Co(n,k) = Sum_{j=1..k} binomial(k,j)*(if (n-k+j) is odd then 0 else if (n-k+j)/2T(n,m) = m!* Sum_{k=m..n} (if n-k is odd then 0 else 2^(1-k)) * Sum_{i=0..floor(k/2)} (-1)^(floor((n+k)/2)-i) * binomial(k,i) * (2*i-k)^n)))) * Sum_{i=1..k} Co(i,m) * binomial(k-i+m-1,m-1), n>0.
(End)
T(n,m) = Sum_{k = 0..n-m} binomial(k+m,m)*((-1)^(n-k-m)+1)*Sum_{j=0..n-k-m} binomial(j+k+m,k+m)*(j+k+m+1)!*2^(-j-k-1)*(-1)^((n+k+m)/2+j+k+m)* Stirling2(n+1,j+k+m+1)/(m+1)!. - Vladimir Kruchinin, May 17 2011
The row polynomials R(n,x) are given by D^n(exp(x*t)) evaluated at t = 0, where D is the operator (1+t+t^2/2!)*d/dt. Cf. A008277 and A094198. See also A185422. - Peter Bala, Nov 25 2011

More terms from Michel Marcus, Mar 01 2014

A185415 Table of coefficients of a polynomial sequence of binomial type related to A080635.

Original entry on oeis.org

1, 0, 1, 2, 0, 1, 0, 8, 0, 1, 18, 0, 20, 0, 1, 0, 148, 0, 40, 0, 1, 378, 0, 658, 0, 70, 0, 1, 0, 5040, 0, 2128, 0, 112, 0, 1, 14562, 0, 33992, 0, 5628, 0, 168, 0, 1, 0, 277164, 0, 158480, 0, 12936, 0, 240, 0, 1

Offset: 1

Peter Bala, Jan 27 2011

Define a sequence of polynomials P(n,x) by means of the recurrence relation
(1)... P(n+1,x) = x*{P(n,x-1)-P(n,x)+P(n,x+1)}
with starting value P(0,x) = 1. The first few polynomials are
  P(1,x) = x
  P(2,x) = x^2
  P(3,x) = x*(x^2+2),
  P(4,x) = x^2*(x^2+8),
  P(5,x) = x*(x^4+20*x^2+18).
This triangle lists the coefficients of these polynomials in ascending powers of x. The triangle has links with A080635, which gives the number of ordered increasing 0-1-2 trees on n nodes (plane unary-binary trees in the notation of [BERGERON et al.]). The number of forests of k such trees on n nodes is given by the formula
... 1/k!*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*P(n,j).
See A185422.
We also have A080635(n) = P(n,1), which can be used to calculate the terms of A080635 - see A185416.
For similarly defined polynomial sequences for other families of trees see A147309 and A185419. See also A185417.
Exponential Riordan array [(3/2)*(1-sqrt(3)*tan((Pi+3*sqrt(3)*x)/6))/(-1+2*sin((Pi-6*sqrt(3)*x)/6)), log((1/2)*(1+sqrt(3)*tan(sqrt(3)*x/2+Pi/6)))]. Production matrix is the exponential Riordan array [2*cosh(x)-1,x] beheaded. A185422*A008277^{-1}.

			Triangle begins:
  n\k|....1......2......3......4......5......6......7......8
  ==========================================================
  ..1|....1
  ..2|....0......1
  ..3|....2......0......1
  ..4|....0......8......0......1
  ..5|...18......0.....20......0......1
  ..6|....0....148......0.....40......0......1..
  ..7|..378......0....658......0.....70......0......1
  ..8|....0...5040......0...2128......0....112......0......1

F. Bergeron, Ph. Flajolet and B. Salvy, Varieties of Increasing Trees, in Lecture Notes in Computer Science vol. 581, ed. J.-C. Raoult, Springer 1922, pp. 24-48.

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
F. Bergeron, Ph. Flajolet and B. Salvy, Varieties of increasing trees

Cf. A080635, A147309, A185417, A185419, A185422, A185423.

#A185415
P := proc(n,x)
description 'polynomial sequence P(n,x)'
if n = 0
return 1
else return
x*(P(n-1,x-1)-P(n-1,x)+P(n-1,x+1))
end proc:
with(PolynomialTools):
for n from 1 to 10
CoefficientList(P(n,x),x);
end do;

p[0][x_] = 1; p[n_][x_] := p[n][x] = x*(p[n-1][x-1] - p[n-1][x] + p[n-1][x+1]) // Expand; row[n_] := CoefficientList[ p[n][x], x]; Table[row[n] // Rest, {n, 1, 10}] // Flatten (* Jean-François Alcover, Sep 11 2012 *)

GENERATING FUNCTION
The e.g.f. is
(1)... F(x, t) = E(t)^x = Sum_{n >= 0} P(n, x) * t^n/n!,
where
E(t) = 1/2+sqrt(3)/2*tan[sqrt(3)/2*t+Pi/6] = 1 + t + t^2/2! + 3*t^3/3! + 9*t^4/4! + ... is the e.g.f. for A080635.
ROW POLYNOMIALS
One easily checks that
... d/dt(F(x,t)) = x*(F(x-1,t)-F(x,t)+F(x+1,t))
and hence the row generating polynomials P(n,x) satisfy the recurrence relation
(2)... P(n+1,x) = x*{P(n,x-1)-P(n,x)+P(n,x+1)}.
RELATIONS WITH OTHER SEQUENCES
A080635(n) = P(n,1).
A185422(n,k) = 1/k!*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*P(n,j).
A185423(n,k) = Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*P(n,j).

A185414 Square array, read by antidiagonals, used to recursively calculate the zigzag numbers A000111.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 5, 5, 3, 1, 16, 16, 10, 4, 1, 61, 61, 39, 17, 5, 1, 272, 272, 176, 80, 26, 6, 1, 1385, 1385, 903, 421, 145, 37, 7, 1, 7936, 7936, 5200, 2464, 880, 240, 50, 8, 1, 50521, 50521, 33219, 15917, 5825, 1661, 371, 65, 9, 1

Offset: 1

Peter Bala, Jan 26 2011

The table entries T(n,k), for n,k>=1, are defined by means of the recurrence relation (1)... T(n+1,k) = 1/2*{(k-1)*T(n,k-1)+(k+1)*T(n,k+1)}, with boundary condition T(1,k) = 1.
The first column of the table produces the sequence of zigzag numbers A000111. Cf. A185416, A185418 and A185420.
Diagonal T(n,n+1) = A290579(n) for n>=1. - Paul D. Hanna, Aug 07 2017

			The array begins:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...;
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ...;
2, 5, 10, 17, 26, 37, 50, 65, 82, ...;
5, 16, 39, 80, 145, 240, 371, 544, 765, ...;
16, 61, 176, 421, 880, 1661, 2896, 4741, 7376, ...;
61, 272, 903, 2464, 5825, 12336, 23947, 43328, 73989, ...;
272, 1385, 5200, 15917, 41936, 98377, 210320, 416765, ...;
1385, 7936, 33219, 112640, 326965, 840960, 1962191, ...; ...
Examples of the recurrence:
T(4,4) = 80 = (3*T(3,3) + 5*T(3,5))/2 = (3*10 + 5*26)/2;
T(5,3) = 176 = (2*T(4,2) + 4*T(4,4))/2 = (2*16 + 4*80)/2;
T(6,2) = 272 = (1*T(5,1) + 3*T(5,3))/2 = (1*16 + 3*176)/2.

Cf. A000111, A147309, A185416, A185418, A185420, A290579.

#A185414 Z := proc(n,x)
description 'zigzag polynomials A147309'
if n = 0 return 1 else return 1/2*x*(Z(n-1,x-1)+Z(n-1,x+1))
end proc:
# values of Z(n,x)/x
for n from 1 to 10 do seq(Z(n,k)/k, k = 1..10);
end do;

{T(n,k)=if(n==1,1,((k-1)*T(n-1,k-1)+(k+1)*T(n-1,k+1))/2)}
for(n=1,10, for(k=1,10, print1(T(n,k),", ")); print(""))

(1)... T(n,k) = Z(n,k)/k with Z(n,x) the zigzag polynomials described in A147309.

A185417 Table of coefficients of a polynomial sequence related to the Springer numbers.

Original entry on oeis.org

1, 1, 2, 3, 4, 4, 11, 26, 12, 8, 57, 120, 136, 32, 16, 361, 970, 760, 560, 80, 32, 2763, 7052, 8860, 3680, 2000, 192, 64, 24611, 72530, 72884, 58520, 15120, 6496, 448, 128, 250737, 716528, 976464, 538048, 314720, 55552, 19712, 1024, 256

Offset: 1

Peter Bala, Jan 28 2011

Define a polynomial sequence S(n,x) recursively by
(1)... S(n+1,x) = x*S(n,x-1)+(x+1)*S(n,x+1) with S(0,x) = 1.
This table lists the coefficients of these polynomials (for n>=1) in ascending powers of x.
The first few polynomials are
S(0,x) = 1
S(1,x) = 2*x+1
S(2,x) = 4*x^2+4*x+3
S(3,x) = 8*x^3+12*x^2+26*x+11.
The sequence [1,1,3,11,57,...] of constant terms of the polynomials is the sequence of Springer numbers A001586. The zeros of the polynomials S(n,-x) lie on the vertical line Re x = 1/2 in the complex plane.
Compare the recurrence (1) with the recurrence relation satisfied by the coefficients T(n,k) of the polynomials of A104035, namely
(2)... T(n+1,k) = k*T(n,k-1)+(k+1)*T(n,k+1).

			Table begin
n\k|.....0.....1.....2.....3.....4.....5......6
===============================================
0..|.....1
1..|.....1.....2
2..|.....3.....4.....4
3..|....11....26....12.....8
4..|....57...120...136....32...16
5..|...361...970...760...560...80.....32
6..|..2763..7052..8860..3680..2000...192....64
...

Peter Bala, The zeros of the row polynomials of A185417

Cf A001586 (1st column and row sums), A104035, A126156, A147309, A185415, A185418, A185419

#A185417
S := proc(n,x) option remember;
description 'polynomials S(n,x)'
if n = 0 return 1 else return x*S(n-1,x-1)+(x+1)*S(n-1,x+1)
end proc:
with(PolynomialTools):
for n from 1 to 10 CoefficientList(S(n,x),x); end do;

S[0, ] = 1; S[n, x_] := S[n, x] = x*S[n-1, x-1] + (x+1)*S[n-1, x+1]; Table[ CoefficientList[S[n, x], x], {n, 0, 8}] // Flatten (* Jean-François Alcover, Apr 15 2015 *)

E.g.f: F(x,t) = 1/(cos(t)-sin(t))*(tan(2*t)+sec(2*t))^x
= (cos(t)+sin(t))^x/(cos(t)-sin(t))^(x+1)
= 1 + (2*x+1)*t + (4*x^2+4*x+3)*t^2/2! + ....
Note that (tan(t)+sec(t))^x is the e.g.f for table A147309.
ROW POLYNOMIALS
The easily checked identity d/dt F(x,t) = x*F(x-1,t)+(x+1)*F(x+1,t) shows that the row generating polynomials of this table are the polynomials S(n,x) described in the Comments section above.
The polynomials S(n,-x) satisfy a Riemann hypothesis: that is, the zeros of S(n,-x) lie on the vertical line Re(x) = 1/2 in the complex plane - see the link.
RELATION WITH OTHER SEQUENCES
1st column [1,1,3,11,57,...] is A001586.
Row sums sequence [1,3,11,57,...] is also A001586.
For n>=1, the values 1/2^n*P(2*n,-1/2) = [1,7,139,5473,...] appear to be A126156.

A185419 Table of coefficients of a polynomial sequence of binomial type related to the enumeration of minimax trees A080795.

Original entry on oeis.org

1, 3, 1, 10, 9, 1, 42, 67, 18, 1, 248, 510, 235, 30, 1, 1992, 4378, 2835, 605, 45, 1, 19600, 44268, 34888, 10605, 1295, 63, 1, 222288, 524748, 461748, 178913, 31080, 2450, 84, 1, 2851712, 7103088, 6728428, 3069612, 690753, 77112, 4242, 108, 1

Offset: 1

Peter Bala, Feb 07 2011

DEFINITION
Define a sequence of polynomials M(n,x) by means of the recurrence relation
(1)... M(n+1,x) = x*{2*M(n,x+1)-M(n,x-1)},
with starting value M(0,x) = 1. We call these the minimax polynomials.
The first few polynomials are
M(1,x) = x
M(2,x) = x*(x   +  3)
M(3,x) = x*(x^2 +  9*x   +  10)
M(4,x) = x*(x^3 + 18*x^2 +  67*x   +  42)
M(5,x) = x*(x^4 + 30*x^3 + 235*x^2 + 510*x + 248).
This triangle lists the coefficients of these polynomials (apart from M(0,x)) in ascending powers of x.
RELATION TO MINIMAX TREES
The value M(n,1) equals the number of minimax trees on n nodes - A080795(n). This result can be used to recursively calculate the entries of A080795 - see A185420.
In addition, the minimax polynomials M(n,x) occur in the formula for the number T(n,k) of forests of k minimax trees on n nodes. ... T(n,k) = 1/k!*sum {j = 0..k} (-1)^(k-j)*binomial(k,j)*M(n,j).
ANALOGIES WITH THE MONOMIALS
{M(n,x)}n>=0 is a polynomial sequence of binomial type and so is analogous to the sequence of monomials x^n. Denoting M(n,x) by x^[n] to emphasize this analogy, we have, for example, the following analog of Bernoulli's formula for the sum of integer powers:
(2)... 1^[p]+...+(n-1)^[p] = -2*n^[p]+ 1/(p+1)*Sum_{k = 0..floor(p/2)} 8^k*binomial(p+1,2k)*B_(2k)*n^[p+1-2k], where {B_k}k>=0 = [1, -1/2, 1/6, 0, -1/30, ...] is the sequence of Bernoulli numbers.
For other polynomial sequences defined by recurrences similar to (1), and related to the zigzag numbers A000111 and the Springer numbers A001586, see A147309 and A185417, respectively. See also A185415.
The Bell transform of A143523(n). For the definition of the Bell transform see A264428. - Peter Luschny, Jan 18 2016

			Triangle begins
n\k|.....1......2......3......4......5......6......7
====================================================
..1|.....1
..2|.....3......1
..3|....10......9......1
..4|....42.....67.....18......1
..5|...248....510....235.....30......1
..6|..1992...4378...2835....605.....45......1
..7|.19600..44268..34888..10605...1295.....63......1
..
Example of the generalized Bernoulli summation formula:
The second row of the triangle gives x^[2] = 3*x+x^2.
Then 1^[2]+2^[2]+...+(n-1)^[2] = (n^3+3*n^2-4*n)/3 = 1/3*(MB(3,n)-MB(3,0)).
From _R. J. Mathar_, Mar 15 2013: (Start)
The matrix inverse starts
       1;
      -3,       1;
      17,      -9,        1;
    -147,      95,      -18,      1;
    1697,   -1245,      305,    -30,      1;
  -24483,   19687,    -5670,    745,    -45,    1;
  423857, -365757,   118237, -18690,   1540,  -63,   1;
-8560947, 7819287, -2761122, 498197, -50190, 2842, -84, 1; (End)

Cf. A080253 (coeffs. of delta operator), A080795 (row sums), A143523 (column 1), A147309, A185415, A185417, A185420.

#A185419
M := proc(n,x) option remember;
if n = 0 then
return 1
else return
x*(2*M(n-1,x+1)-M(n-1,x-1))
end if;
end proc:
with(PolynomialTools):
for n from 1 to 10 do
CoefficientList(M(n,x),x);
end do;

M[0, ] = 1; M[n, x_] := M[n, x] = x (2 M[n-1, x+1] - M[n-1, x-1]);
Table[CoefficientList[M[n, x], x] // Rest, {n, 1, 10}] (* Jean-François Alcover, Jun 26 2019 *)

# uses[bell_matrix from A264428]
# Adds a column 1,0,0,0, ... at the left side of the triangle.
bell_matrix(lambda n: A143523(n), 10) # Peter Luschny, Jan 18 2016

GENERATING FUNCTION
Let a = 3-2*sqrt(2). Let f(t) = (1/2)*sqrt(2)*((1+a*exp(2*sqrt(2)*t))/ (1-a*exp(2*sqrt(2)*t))) = 1 + t + 4*t^2/2! + 20*t^3/3! + ... be the e.g.f. for A080795. Then the e.g.f. for the current table, including a constant 1, is
(1)... F(x,t) = f(t)^x = Sum_{n>=0} M(n,x)*t^n/n! = 1 + x*t + (3*x+x^2)*t^2/2! + (10*x+9*x^2+x^3)*t^3/3! + ....
ROW POLYNOMIALS
One easily checks that d/dt(F(x,t)) = x*(2*F(x+1,t)-F(x-1,t)) and hence the row generating polynomials M(n,x) satisfy the recurrence relation
(2)... M(n+1,x) = x*{2*M(n,x+1)-M(n,x-1)}.
The form of the e.g.f shows that the row polynomials are a polynomial sequence of binomial type. The associated delta operator D* is given by
(3)... D* = sqrt(2)/4*log((3+2*sqrt(2))*(sqrt(2)*exp(D)-1)/(sqrt(2)*exp(D)+1)),
where D is the derivative operator d/dx. This expands to
(4)... D* = D - 3*D^2/2! + 17*D^3/3! - 147*D^4/4! + ....
The sequence of coefficients [1,3,17,147,...] is A080253.
The delta operator D* acts as a lowering operator for the minimax polynomials
(5)...(D*) M(n,x) = n*M(n-1,x).
In what follows it will be convenient to denote M(n,x) by x^[n].
ANALOG OF THE LITTLE FERMAT THEOREM
For integer x and odd prime p
(6)... x^[p] = (-1)^((p^2-1)/8)*x (mod p).
More generally, for k = 1,2,...
(7)... x^[p+k-1] = (-1)^((p^2-1)/8)*x^[k] (mod p).
GENERALIZED BERNOULLI POLYNOMIALS ASSOCIATED WITH THE MINIMAX POLYNOMIALS
The generalized Bernoulli polynomial MB(k,x) associated with the minimax polynomial x^[k] (= M(k,x)) may be defined as the result of applying the differential operator D*/(exp(D)-1) to the polynomial x^[k]:
(8)... MB(k,x) := {D*/(exp(D)-1)} x^[k].
The first few generalized Bernoulli polynomials are
MB(0,x) = 1,
MB(1,x) = x   -  2,
MB(2,x) = x^2 -    x   + 4/3,
MB(3,x) = x^3 +  3*x^2 - 4*x,
MB(4,x) = x^4 + 10*x^3 + 3*x^2 - 14*x - 32/15.
Since exp(D)-1 is the forward difference operator it follows from (5) and (8) that
(9)... MB(k,x+1) - MB(k,x) = k*x^[k-1].
Summing (9) from x = 1 to x = n-1 and telescoping we find a closed form expression for the finite sums
(10)... 1^[p]+2^[p]+...+(n-1)^[p] = 1/(p+1)*{MB(p+1,n)-MB(p+1,1)}.
The generalized Bernoulli polynomials can be expanded in terms of the minimax polynomials x^[k]. Use (3) to express exp(D)-1 in terms of D*.
Substitute the resulting expression in (8) and expand as a power series in D* to arrive at the expansion:
(11)... MB(k,x) = -2*k*x^[k-1] + Sum_{j=0..floor(k/2)} 2^(3*j) * binomial(k,2j)*B_(2j)*x^[k-2j], where {B_j}j>=0 = [1,-1/2,1/6,0,-1/30,...] denotes the Bernoulli number sequence.
RELATION WITH OTHER SEQUENCES
Column 1 [1, 3, 10, 42, 248, ...] = A143523 with an offset of 1.
Row sums [1, 1, 4, 20, 128, 1024, ...] = A080795.

A185421 Ordered forests of k increasing unordered trees on the vertex set {1,2,...,n} in which all outdegrees are <= 2.

Original entry on oeis.org

1, 1, 2, 2, 6, 6, 5, 22, 36, 24, 16, 90, 210, 240, 120, 61, 422, 1260, 2040, 1800, 720, 272, 2226, 8106, 16800, 21000, 15120, 5040, 1385, 13102, 56196, 141624, 226800, 231840, 141120, 40320, 7936, 85170, 420330, 1244880, 2421720, 3175200, 2751840, 1451520, 362880

Offset: 1

Peter Bala, Jan 31 2011

An increasing tree is a labeled rooted tree with the property that the sequence of labels along any path starting from the root is increasing. A000111(n) for n >= 0 enumerates increasing unordered trees on the vertex set {1,2,...,n}, rooted at 1, in which all outdegrees are <= 2 (plane unary binary trees in the notation of [Bergeron et al.]).

The entry T(n,k) of the present table counts ordered forests of k such trees having n nodes in total. See below for an example. For unordered forests see A147315. For a table of ordered forests of increasing ordered trees in which all outdegrees are <=2 see A185423.

			Triangle begins
n\k|....1......2......3......4......5......6......7
===================================================
..1|....1
..2|....1......2
..3|....2......6......6
..4|....5.....22.....36.....24
..5|...16.....90....210....240....120
..6|...61....422...1260...2040...1800....720
..7|..272...2226...8106..16800..21000..15120...5040
..
Examples of recurrence relation for table entries:
T(5,2) = 2*{T(4,1)+T(4,2)+1/2*T(4,3)} = 2*(5+22+18) = 90;
T(6,1) = 1*{T(5,0)+T(5,1)+1/2*T(5,2)} = 16 + 1/2*90 = 61.
Examples of forests:
T(4,2) = 22. The 11 unordered forests consisting of 2 trees on 4 nodes are shown in the example section of A147315. Putting an order on the trees in a forest produces 2!*11 = 22 ordered forests.

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
F. Bergeron, Ph. Flajolet and B. Salvy, Varieties of increasing trees, Lecture Notes in Computer Science vol. 581, ed. J.-C. Raoult, Springer 1992, pp. 24-48.

Cf. A147315, A147309, A185423.

#A185421 E := t -> sec(t)+tan(t)-1:
F := (x,t) -> 1/(1-x*E(t)) - 1:
Fser := series(F(x,t),t=0,12):
for n from 1 to 7 do
seq(coeff(n!*coeff(Fser,t,n),x,i),i=1..n) od;

nmax = 9; t[n_ /; n > 0, k_ /; k > 0] := t[n, k] = k*(t[n-1, k-1] + t[n-1, k] + 1/2*t[n-1, k+1]);
t[1, 1] = 1; t[0, ] = 0; t[, 0] = 0; Flatten[Table[t[n, k], {n, 1, nmax}, {k, 1, n}]]
(* Jean-François Alcover, Jun 22 2011, after recurrence *)

{T(n,k)=if(n<1||k<1||k>n,0,if(n==1,1,k*(T(n-1,k-1)+T(n-1,k)+T(n-1,k+1)/2)))}

{T(n,k)=local(X=x+x*O(x^n));n!*polcoeff(polcoeff(1/(1-y*((1+sin(X))/cos(X)-1))-1,n,x),k,y)}

TABLE ENTRIES
(1)... T(n,k) = k!*A147315(n-1,k-1).
(2)... T(n,k) = Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*Z(n,j), where Z(n,x) denotes the zigzag polynomials as described in A147309.
Recurrence relation
(3)... T(n+1,k) = k*{T(n,k-1)+T(n,k)+1/2*T(n,k+1)}.
GENERATING FUNCTION
Let E(t) = sec(t)+tan(t)-1. E(t) is the egf for the enumeration of increasing unordered trees on the vertex set {1,2,...,n}, rooted at 1, in which all outdegrees are <=2 (plane unary binary trees in the notation of [Bergeron et al.]).
The egf of the present array is
(4)... 1/(1-x*E(t)) - 1 = Sum_{n >= 1} R(n,x)*t^n/n! = x*t + x*(1+2*x)*t^2/2! + x*(2+6*x+6*x^2)*t^3/3! + ...
ROW POLYNOMIALS
The row generating polynomials R(n,x) begin.
... R(1,x) = x
... R(2,x) = x*(1+2*x)
... R(3,x) = x*(2+6*x+6*x^2)
... R(4,x) = x*(5+22*x+36*x^2+24*x^3).
The ordered Bell polynomials OB(n,x) are the row polynomials of A019538 given by the formula
(5)... OB(n,x) = Sum_{k = 1..n} k!*Stirling2(n,k)*x^k.
By comparing the e.g.f.s for A019538 and the present table we obtain the surprising identity
(6)... (-i)^(n-1)*OB(n,x)/x = R(n,y)/y, where i = sqrt(-1) and x = i*y + (-1/2+i/2). It follows that the zeros of the polynomial R(n,y)/y lie on the vertical line Re(y) = -1/2 in the complex plane.
RELATIONS WITH OTHER SEQUENCES
(7)... T(n,1) = A000111(n).
Setting y = 0 in (6) yields
(8)... A000111(n) = i^(n+1)*Sum_{k=1..n} (-1)^k*k!*Stirling2(n,k) *((1+i)/2)^(k-1).

Maple program corrected by Peter Luschny, Aug 02 2011

A185424 Numerators of generalized Bernoulli numbers associated with the zigzag numbers A000111.

Original entry on oeis.org

1, -1, 1, -1, 19, -5, 253, -61, 3319, -1385, 222557, -50521, 422152729, -2702765, 59833795, -199360981, 439264083023, -19391512145, 76632373664299, -2404879675441, 4432283799315809, -370371188237525

Offset: 0

Peter Bala, Feb 18 2011

DEFINITION
Let E(t) = sec(t)+tan(t) denote the generating function for the zigzag numbers A000111. The zigzag Bernoulli numbers, denoted ZB(n), are defined by means of the generating function
(1)... log E(t)/(E(t)-1) = sum {n = 0..inf} ZB(n)*t^n/n!.
Notice that if we were to take E(t) equal to exp(t) then (1) would be the defining function for the classical Bernoulli numbers B_n. The first few even-indexed values of ZB(n) are
....n..|..0...2.....4.......6........8..........10...........12....
===================================================================
.ZB(n).|..1..1/6..19/30..253/42..3319/30..222557/66..422152729/2730
while the odd-indexed values begin
....n..|. ..1......3......5.......7........9.........11..
=========================================================
.ZB(n).|. -1/2...-1/2...-5/2...-61/2...-1385/2...-50521/2
The present sequence gives the numerators of the zigzag Bernoulli numbers. It is not difficult to show that the odd-indexed value ZB(2*n+1) equals -1/2*A000364(n). The numerators of the even-indexed values ZB(2*n) are shown separately in A185425.
VON STAUDT-CLAUSEN THEOREM
The following analog of the von Staudt-Clausen theorem holds:
(2)... ZB(2*n) + 1/2 + S(1) + (-1)^(n+1)*S(3) equals an integer, where
... S(1) = sum {prime p, p = 1 (mod 4), p-1|2*n} 1/p,
... S(3) = sum {prime p, p = 3 (mod 4), p-1|2*n} 1/p.
For example,
(3)... ZB(12) + 1/2 + (1/5+1/13) - (1/3+1/7) = 154635.
Further examples are given below.

			Examples of von Staudt and Clausen's theorem for ZB(2*n):
ZB(2) = 1/6 = 1 - 1/2 - 1/3;
ZB(4) = 19/30 = 1 - 1/2 + 1/3 - 1/5;
ZB(6) = 253/42 = 7 - 1/2 - 1/3 - 1/7;
ZB(8) = 3319/30 = 111 - 1/2 + 1/3 - 1/5;
ZB(10) = 222557/66 = 3373 - 1/2 - 1/3 - 1/11.

Cf. A000111, A027641, A027642, A147309, A185425.

Sequence of denominators is A141056.

#A185424
a:= n-> numer((-1)^(n*(n-1)/2)*add(binomial(n,k)/(k+1)* bernoulli(n-k) *euler(k), k = 0..n)):
seq(a(n), n = 0..20);

Numerator[ Range[0, 30]! CoefficientList[ Series[Log(Sec[x]+Tan[x])/(Sec[x] +Tan[x] - 1), {x, 0, 30}], x]]

SEQUENCE ENTRIES
a(n) = numerator of the rational number ZB(n) where
(1)... ZB(n) = (-1)^(n*(n-1)/2)*sum {k = 0..n} binomial(n,k)/(k+1)* Bernoulli(n- k)*Euler(k).
For odd indices this simplifies to
(2)... ZB(2*n+1) = (-1)^n*Euler(2*n)/2, where Euler(2*n) = A028296(n).
For even indices we have
(3)... ZB(2*n) = (-1)^n*sum {k = 0..n} binomial(2*n,2*k)/(2*k+1)* Bernoulli(2*n- 2*k)*Euler(2*k).
GENERATING FUNCTION
E.g.f:
(4)... log(sec(t)+tan(t))/(sec(t)+tan(t)-1) =
   1 -1/2*t +1/6*t^2/2! -1/2*t^3/3! + ....
RELATION WITH ZIGZAG POLYNOMIALS OF A147309
The classical Bernoulli numbers B_n are given by the double sum
(5)... B_n = sum {k=0..n} sum {j=0..k} (-1)^j*binomial(k,j)*j^n/(k+1).
The corresponding formula for the zigzag Bernoulli numbers is
(6)... ZB(n) = sum {k=0..n} sum {j=0..k}(-1)^j*binomial(k,j)*Z(n,j)/(k+1), where Z(n,x) is a zigzag polynomial as defined in A147309. Umbrally, we can express this as
(7)... ZB(n) = Z(n,B), where on the lhs the understanding is that in the expansion of the zigzag polynomial Z(n,x) a term such as c_k*x^k is to be replaced with c_k*B_k. For example, Z(6,x) = 40*x^2+20*x^4+x^6 and so ZB(6) = 40*B_2+20*B_4+B_6 = 40*(1/6)+20*(-1/30)+(1/42) = 253/42.

A012259 Expansion of e.g.f. exp(arctanh(tan(x))).

Original entry on oeis.org

1, 1, 1, 5, 17, 121, 721, 6845, 58337, 698161, 7734241, 111973685, 1526099057, 25947503401, 419784870961, 8200346492525, 153563504618177, 3389281372287841, 72104198836466881, 1774459993676715365, 42270463533824671697, 1147649139272698443481

Offset: 0

Patrick Demichel (patrick.demichel(AT)hp.com)

nonn

			 exp(arctanh(tan(x))) = 1 + x + x^2/2! + 5*x^3/3! + 17*x^4/4! + 121*x^5/5! + ...

G. C. Greubel, Table of n, a(n) for n = 0..425
Shi-Mei Ma, Jun Ma, and Yeong-Nan Yeh, The alternating run polynomials of permutations, arXiv:1904.11437 [math.CO], 2019.
Shi-Mei Ma, Jun Ma, and Yeong-Nan Yeh, David-Barton type identities and alternating run polynomials, Academia Sinica (Taipei, 2019).
Mathematics Stack Exchange, Mystery regarding power series of 1/sqrt(1+x^x), Question 6939.

Cf. A012077, A012085, A185411, A202038 (signed version).

m:=30; R:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!( Sqrt((1+Tan(x))/(1-Tan(x))) )); [Factorial(n-1)*b[n]: n in [1..m]]; // G. C. Greubel, Jun 06 2019

With[{nn=30}, CoefficientList[Series[Sqrt[(1+Tan[x])/(1-Tan[x])], {x, 0, nn}], x]*Range[0,nn]!] (* Vaclav Kotesovec, Oct 23 2013 *)

{a(n)=local(A=1); for(i=0, n, A = exp( intformal( (A^2 + subst(A^2, x, -x))/2 +x*O(x^n)) )); n!*polcoeff(A, n)}
for(n=0, 25, print1(a(n), ", ")) \\ Paul D. Hanna, Feb 04 2017

my(x='x+O('x^30)); Vec(serlaplace( sqrt((1+tan(x))/(1-tan(x))) )) \\ G. C. Greubel, Jun 06 2019

m = 30; T = taylor(sqrt((1+tan(x))/(1-tan(x))), x, 0, m); [factorial(n)*T.coefficient(x, n) for n in (0..m)] # G. C. Greubel, Jun 06 2019

Alternative form of e.g.f: sqrt(sec(2*x) + tan(2*x)) = 1 + x + x^2/2! + 5*x^3/3! + 17*x^4/4! + ... (where sec(x)=1/cos(x)). - Peter Bala, Jan 11 2011
a(n) = 2^n*Z(n,1/2), where Z(n,x) is the n-th zigzag polynomial as defined in A147309.
Put y = x*log(x)/4. The connection between the expansion sqrt(2/(1+x^x)) = 1 - y - y^2/2! + 5*y^3/3! + 17*y^4/4! - 121*y^5/5! ... and the present sequence is explained in the answer to Mathematics Stack Exchange Question 6939. - Peter Bala, Jul 10 2011
exp(arctanh(tan(x))) = sqrt( (1 + tan(x))/(1 - tan(x) ) ) = sqrt( tan(x+pi/4) ). - David Callan, Dec 13 2011
a(n) ~ 2^(2*n+3/2) * n^n / (Pi^(n+1/2) * exp(n)). - Vaclav Kotesovec, Oct 23 2013
E.g.f. A(x) satisfies: A(x) = exp( Integral (A(x)^2 + A(-x)^2)/2 dx ). - Paul D. Hanna, Feb 04 2017
E.g.f. A(x) satisfies: A'(x) = A(x) * (A(x)^2 + A(-x)^2)/2. - Paul D. Hanna, Feb 04 2017

Showing 1-10 of 13 results. Next

cp's OEIS Frontend

A109449 Triangle read by rows, T(n,k) = binomial(n,k)*A000111(n-k), 0 <= k <= n.

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A145876 Triangle read by rows: T(n,k) is the number of permutations of [n] having k-1 alternating descents (1<=k<=n). The index i is an alternating descent of a permutation p if either i is odd and p(i)>p(i+1), or i is even and p(i)

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A147315 L-matrix for Euler numbers A000111(n+1).

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A185415 Table of coefficients of a polynomial sequence of binomial type related to A080635.

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A185414 Square array, read by antidiagonals, used to recursively calculate the zigzag numbers A000111.

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A185417 Table of coefficients of a polynomial sequence related to the Springer numbers.

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