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1
a(0)=1, a(n) = 3*a(n-1) + n + 1.
(Formerly M3882 N1592)
+10
29
1, 5, 18, 58, 179, 543, 1636, 4916, 14757, 44281, 132854, 398574, 1195735, 3587219, 10761672, 32285032, 96855113, 290565357, 871696090, 2615088290, 7845264891, 23535794695, 70607384108, 211822152348, 635466457069
COMMENTS
Second right hand column (n-m=1) of the A156920 triangle. The generating function of this sequence enabled the analysis of the polynomials A156921 and A156925. (End)
REFERENCES
F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 260.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
Dillan Agrawal, Selena Ge, Jate Greene, Tanya Khovanova, Dohun Kim, Rajarshi Mandal, Tanish Parida, Anirudh Pulugurtha, Gordon Redwine, Soham Samanta, and Albert Xu, Chip-Firing on Infinite k-ary Trees, arXiv:2501.06675 [math.CO], 2025. See pp. 9, 18.
FORMULA
G.f.: 1/((1-3*x)*(1-x)^2).
a(n) = (3^(n+2) - 2*n - 5)/4.
a(n) = Sum_{k=0..n+1} (n-k+1)*3^k = Sum_{k=0..n+1} k*3^(n-k+1). - Paul Barry, Jul 30 2004 a(n) = Sum_{k=0..n} binomial(n+2, k+2)*2^k. - Paul Barry, Jul 30 2004 a(-1)=0, a(0)=1, a(n) = 4*a(n-1) - 3*a(n-2) + 1. - Miklos Kristof, Mar 09 2005 E.g.f.: exp(x)*(9*exp(2*x) - 2*x - 5)/4. - Stefano Spezia, Nov 09 2024
EXAMPLE
G.f. = 1 + 5*x + 18*x^2 + 58*x^3 + 179*x^4 + 543*x^5 + 1636*x^6 + ...
MAPLE
a[ -1]:=0:a[0]:=1:for n from 1 to 50 do a[n]:=4*a[n-1]-3*a[n-2]+1 od: seq(a[n], n=0..50); # Miklos Kristof, Mar 09 2005
MATHEMATICA
a[ n_] := MatrixPower[ {{1, 0, 0}, {1, 1, 0}, {1, 1, 3}}, n + 1][[3, 1]]; (* Michael Somos, May 28 2014 *) RecurrenceTable[{a[0]==1, a[n]==3a[n-1]+n+1}, a, {n, 30}] (* or *) LinearRecurrence[{5, -7, 3}, {1, 5, 18}, 30] (* Harvey P. Dale, Jan 31 2017 *)
CROSSREFS
Equals A156920 second right hand column. Equals A142963 second right hand column divided by 2^n. Equals A156919 second right hand column divided by 2. (End)
Table of coefficients of polynomials related to the Dirichlet eta function.
+10
22
1, 2, 1, 4, 10, 1, 8, 60, 36, 1, 16, 296, 516, 116, 1, 32, 1328, 5168, 3508, 358, 1, 64, 5664, 42960, 64240, 21120, 1086, 1, 128, 23488, 320064, 900560, 660880, 118632, 3272, 1, 256, 95872, 2225728
COMMENTS
The SF(z; n) formulas, see below, were discovered while studying certain properties of the Dirichlet eta function.
Let D be the differential operator 2*x*d/dx. The row polynomials of this table come from repeated application of the operator D to the function g(x) = 1/sqrt(1 - x). For example,
D(g) = x*g^3
D^2(g) = x*(2 + x)*g^5
D^3(g) = x*(4 + 10*x + x^2)*g^7
D^4(g) = x*(8 + 60*x + 36*x^2 + x^3)*g^9.
Thus this triangle is analogous to the triangle of Eulerian numbers A008292, whose row polynomials come from the repeated application of the operator x*d/dx to the function 1/(1 - x). (End)
FORMULA
SF(z; n) = Sum_{m >= 1} m^(n-1)*4^(-m)*z^(m-1)*Gamma(2*m+1)/(Gamma(m)^2) = P(z;n) / (2^(n+1)*(1-z)^((2*n+3)/2)) for n >= 0. The polynomials P(z;n) = Sum_{k = 0..n} a(k)*z^k generate the a(n) sequence.
If we write the sequence as a triangle the following relation holds: T(n,m) = (2*m+2)*T(n-1,m) + (2*n-2*m+1)*T(n-1,m-1) with T(n,m=0) = 2^n and T(n,n) = 1, n >= 0 and 0 <= m <= n.
G.f.: 1/(1-xy-2x/(1-3xy/(1-4x/(1-5xy/(1-6x/(1-7xy/(1-8x/(1-... (continued fraction). - Paul Barry, Jan 26 2011 E.g.f.: exp(z*(x + 2)) * (1 - x)/(exp(2*x*z) - x*exp(2*z))^(3/2) = Sum_{n >= 0} P(x,n)*z^n/n! = 1 + (2 + x)*z + (4 + 10*x + x^2)*z^2/2! + (8 + 60*x + 36*x^2 + x^3)*z^3/3! + ... .
Explicit formula for the row polynomials:
P(x,n-1) = Sum_{k = 1..n} 2^(n-2*k)*binomial(2k,k)*k!*Stirling2(n,k)*x^(k-1)*(1 - x)^(n-k).
The polynomials x*(1 + x)^n * P(x/(x + 1),n) are the row polynomials of A187075. The polynomials x^(n+1) * P((x + 1)/x,n) are the row polynomials of A186695.
EXAMPLE
The first few rows of the triangle are:
[1]
[2, 1]
[4, 10, 1]
[8, 60, 36, 1]
[16, 296, 516, 116, 1]
The first few P(z;n) are:
P(z; n=0) = 1
P(z; n=1) = 2 + z
P(z; n=2) = 4 + 10*z + z^2
P(z; n=3) = 8 + 60*z + 36*z^2 + z^3
The first few SF(z;n) are:
SF(z; n=0) = (1/2)*(1)/(1-z)^(3/2);
SF(z; n=1) = (1/4)*(2+z)/(1-z)^(5/2);
SF(z; n=2) = (1/8)*(4+10*z+z^2)/(1-z)^(7/2);
SF(z; n=3) = (1/16)*(8+60*z+36*z^2+z^3)/(1-z)^(9/2);
In the Savage-Viswanathan paper, the coefficients appear as
1;
1, 2;
1, 10, 4;
1, 36, 60, 8;
1, 116, 516, 296, 16;
1, 358, 3508, 5168, 1328, 32;
1, 1086, 21120, 64240, 42960, 5664, 64;
...
MAPLE
A156919 := proc(n, m) if n=m then 1; elif m=0 then 2^n ; elif m<0 or m>n then 0; else 2*(m+1)*procname(n-1, m)+(2*n-2*m+1)*procname(n-1, m-1) ; end if; end proc: seq(seq( A156919(n, m), m=0..n), n=0..7); # R. J. Mathar, Feb 03 2011
MATHEMATICA
g[0] = 1/Sqrt[1-x]; g[n_] := g[n] = 2x*D[g[n-1], x]; p[n_] := g[n] / g[0]^(2n+1) // Cancel; row[n_] := CoefficientList[p[n], x] // Rest; Table[row[n], {n, 0, 9}] // Flatten (* Jean-François Alcover, Aug 09 2012, after Peter Bala *) Flatten[Table[Rest[CoefficientList[Nest[2 x D[#, x] &, (1 - x)^(-1/2), k] (1 - x)^(k + 1/2), x]], {k, 10}]] (* Jan Mangaldan, Mar 15 2013 *)
1, 1, 1, 1, 5, 1, 1, 15, 18, 1, 1, 37, 129, 58, 1, 1, 83, 646, 877, 179, 1, 1, 177, 2685, 8030, 5280, 543, 1, 1, 367, 10002, 56285, 82610, 29658, 1636, 1, 1, 749, 34777, 335162, 919615, 756218, 159742, 4916, 1
COMMENTS
The Flower Triangle seems to be an appropriate name for the triangular array of this sequence. The zero patterns of the Flower Polynomials of the first, see A156921, the second, see A156925, the third, see A156927, and the fourth kind, see A156933, look like flowers. The first Maple program generates the Flower Triangle sequence.
The second program generates the Right Hand Columns sequences and the third one generates the Left Hand Column sequences. For an explanation of these two algorithms see A142963.
FORMULA
T(n,m) = (m+1)*T(n-1,m) + (2*n-2*m+1)*T(n-1,m-1) with T(n,m=0) = 1 and T(n,n) = 1, n>=0 and 0 <= m <= n.
E.g.f.: 1 + t*x + (t+t^2)*x^2/2! + (t+5*t^2+t^3)*x^3/3! + ... = sqrt(E(x,2*t)), where E(x,t) = (1-t)*exp(x*t)/(exp(x*t)-t*exp(x)) = 1 + t*x + (t+t^2)*x^2/2! + (t+4*t^2+t^3)*x^3/3! + ... is the e.g.f. for the Eulerian numbers A008292. The row polynomials R(n,x) satisfy 1/sqrt(1-2*x)*(x*d/dx)^n(1/sqrt(1-2*x)) = R(n,x)/(1-2*x)^(n+1).
(End)
EXAMPLE
The first few rows of the triangle are:
[1]
[1, 1]
[1, 5 , 1 ]
[1, 15, 18, 1]
[1, 37, 129, 58, 1]
[1, 83, 646, 877, 179, 1]
MAPLE
A156920 := proc(n, m): if n=m then 1; elif m=0 then 1 ; elif m<0 or m>n then 0; else (m+1)*procname(n-1, m)+(2*n-2*m+1)*procname(n-1, m-1) ; end if; end proc: seq(seq( A156920(n, m), m=0..n), n=0..8);
RHCnr:=5; RHCmax:=10; RHCend:=RHCnr+RHCmax: for k from RHCnr to RHCend do for n from 0 to k do S2[k, n]:=sum((-1)^(n+i)*binomial(n, i)*i^k/n!, i=0..n) end do: G(k, x):= sum(S2[k, p]*((2*p)!/p!) *x^p/(1-4*x)^(p+1), p=0..k)/(((-1)^(k+1)*2*x)/(-1+4*x)^(k+1)): fx:=simplify(G(k, x)): nmax:=degree(fx); RHC[k-RHCnr+1]:= coeff(fx, x, k-RHCnr)/2^(k-RHCnr) end do: a:=n-> RHC[n]: seq(a(n), n=1..RHCend-RHCnr);
LHCnr:=5; LHCmax:=10: LHCend:=LHCnr+LHCmax: for k from LHCnr to LHCend do for n from 0 to k do S2[k, n]:=sum((-1)^(n+i)*binomial(n, i)*i^k/n!, i=0..n) end do: G(k, x):= sum(S2[k, p]*((2*p)!/p!)*x^p/(1-4*x)^(p+1), p=0..k)/ (((-1)^(k+1)*2*x)/(-1+4*x)^(k+1)): fx:=simplify(G(k, x)): nmax:=degree(fx); for n from 0 to nmax do d[n]:= coeff(fx, x, n)/2^n end do: LHC[n]:=d[LHCnr-1] end do: a:=n-> LHC[n]: seq(a(n), n=LHCnr..LHCend-1);
MATHEMATICA
T[_, 0] = 1; T[n_, n_] = 1; T[n_, m_] := T[n, m] = (m + 1)*T[n - 1, m] + (2*n - 2*m + 1)*T[n - 1, m - 1];
EXTENSIONS
Minor edits by Johannes W. Meijer, Sep 28 2011
Triangle read by rows, coefficients of the polynomials P(k, x) = (1/2) Sum_{p=0..k-1} Stirling2(k, p+1)*x^p*(1-4*x)^(k-1-p)*(2*p+2)!/(p+1)!.
+10
15
1, 1, 2, 1, 10, 4, 1, 30, 72, 8, 1, 74, 516, 464, 16, 1, 166, 2584, 7016, 2864, 32, 1, 354, 10740, 64240, 84480, 17376, 64, 1, 734, 40008, 450280, 1321760, 949056, 104704, 128, 1, 1498, 139108, 2681296, 14713840, 24198976, 10223488, 629248, 256, 1, 3030, 462264, 14341992
COMMENTS
Previous name: Table of coefficients of row polynomials of certain o.g.f.s.
The o.g.f.s G(k, x) for the k-family of sequences S(k, n):= Sum_{p=0..n} p^k*binomial(2*p, p)*binomial(2*(n-p), n-p), k=0,1,... (convolution of two sequences involving the central binomial coefficients) are 1/(1-4*x) for k=0 and 2*x*P(k, x)/(1-4*x)^(k+1) for k=1,2,..., with the row polynomials P(k, x) = Sum_{m=0..k-1} a(n,m)*x^m).
The author was led to compute the sums S(k, n) by a question asked by M. Greiter, Jun 27 2008.
In order to keep the index k>=1 of Sigma(k, n) also for the polynomials P(k, x), their degree is then k-1.
FORMULA
G(k, x) = Sum_{p=0..k} S2(k, p)*((2*p)!/p!)*x^p/(1-4*x)^(p+1), k >= 0 (here k >= 1), with the Stirling2 triangle S2(k, p):= A048993(k, p). (Proof from the product of the o.g.f.s of the two convoluted sequences and the normal ordering (x^d_x)^k = Sum_{p=0..k} S2(k, p)*x^p*d_x^p, with the derivative operator d_x.) a(k,m) = [x^m]P(k, x) = [x^m] ((1-4*x)^(k+1))*G(k,x)/(2*x), k>=1, m=0,1,...,k-1.
For the triangle coefficients the following relation holds: T(n,m) = (m+1)*T(n-1,m) + (4*n-4*m-2)*T(n-1,m-1) with T(n,m=0) = 1 and T(n,m=n-1) = 2^(n-1), n >= 1 and 0 <= m <= n-1. - Johannes W. Meijer, Feb 20 2009 (x*d/dx)^n (1/(sqrt(1 - 4*x)) = 2*x*P(n,x)/sqrt(1 - 4*x)^(n+1/2) for n >= 1.
x*P(n,x)/(1 - 4*x)^(n+1/2) = (1/2)*Sum_{k >= 1} binomial(2*k,k)* k^n*x^k for n >= 1.
P(n+1,x) = ((4*n - 2)*x + 1)*P(n,x) - x*(4*x - 1)*d/dx(P(n,x)).
Hence the polynomial P(n,x) has all real zeros by Liu et al., Theorem 1.1, Corollary 1.2. (End)
EXAMPLE
Triangle starts:
[1]
[1, 2]
[1, 10, 4]
[1, 30, 72, 8]
[1, 74, 516, 464, 16]
[1, 166, 2584, 7016, 2864, 32]
[1, 354, 10740, 64240, 84480, 17376, 64]
[1, 734, 40008, 450280, 1321760, 949056, 104704, 128]
...
P(3,x) = 1+10*x+4*x^2.
G(3,x) = 2*x*(1+10*x+4*x^2)/(1-4*x)^4.
MAPLE
A142963 := proc(n, m): if n=m+1 then 2^(n-1); elif m=0 then 1 ; elif m<0 or m>n-1 then 0; else (m+1)*procname(n-1, m)+(4*n-4*m-2)*procname(n-1, m-1); end if; end proc: seq(seq( A142963(n, m), m=0..n-1), n=1..9); # Johannes W. Meijer, Sep 28 2011
# Alternatively (assumes offset 0):
p := (n, x) -> (1/2)*add(Stirling2(n+1, k+1)*x^k*(1-4*x)^(n-k)*(2*k+2)!/(k+1)!, k=0..n): for n from 0 to 7 do [n], PolynomialTools:-CoefficientList(p(n, x), x) od;
MATHEMATICA
t[_, 0] = 1; t[n_, m_] /; m == n-1 := 2^m; t[n_, m_] := (m+1)*t[n-1, m] + (4*n-4*m-2)*t[n-1, m-1]; Table[t[n, m], {n, 1, 10}, {m, 0, n-1}] // Flatten (* Jean-François Alcover, Jun 21 2013, after Johannes W. Meijer *)
EXTENSIONS
Name reformulated with offset corrected, edited by Wolfdieter Lang, Aug 23 2019
FP1 polynomials related to the generating functions of the right hand columns of the A156920 triangle.
+10
14
1, 1, 1, 1, -6, 1, 7, -79, 119, 126, -270, 1, 28, -515, 1654, 8689, -65864, 142371, -82242, -99090, 113400, 1, 86, -2255, 5784, 300930, -3904584, 20663714, -41517272, -80232259, 657717054
COMMENTS
The FP1 polynomials appear in the numerators of the GF1 o.g.f.s. of the right hand columns of A156920. The FP1 can be calculated with the formula for the RHC sequence, see A156920, and the formula for the general structure of the generating function GF1, see below. An appropriate name for the FP1 polynomials seems to be the flower polynomials of the first kind because the zero patterns of these polynomials look like flowers. The zero patterns of the FP2, see A156925, and the FP1 resemble each other closely. A Maple program that generates for a right hand column with a certain RHCnr its GF1 and FP1 can be found below. RHCnr stands for right hand column number and starts from 1.
FORMULA
G.f.: GF1(z;RHCnr) := FP1(z;RHCnr)/product((1-(2*m-1)*z)^(RHCnr+1-m),m=1..RHCnr)
Row sums (n) = (-1)^(1+(n+1)*(n+2)/2)* A098695(n).
EXAMPLE
The first few rows of the "triangle" of the coefficients of the FP1 polynomials.
In the columns the coefficients of the powers of z^m, m=0,1,2,... , appear.
[1]
[1]
[1, 1, -6]
[1, 7, -79, 119, 126, -270]
[1, 28, -515, 1654, 8689, -65864, 142371, -82242, -99090, 113400]
Matrix of the coefficients of the FP1 polynomials. The coefficients in the columns of this matrix are the powers of z^m, m=0,1,2,.. .
[1, 0 ,0, 0, 0, 0, 0, 0, 0, 0]
[1, 0 ,0, 0, 0, 0, 0, 0, 0, 0]
[1, 1, -6, 0 ,0, 0, 0, 0, 0, 0]
[1, 7, -79, 119, 126, -270, 0, 0, 0, 0]
[1, 28, -515, 1654, 8689, -65864, 142371, -82242, -99090, 113400]
The first few FP1 polynomials are:
FP1(z; RHCnr=1) = 1
FP1(z; RHCnr=2) = 1
FP1(z; RHCnr =3) = 1+z-6*z^2
Some GF1(z;RHCnr) are:
GF1(z;RHCnr= 3) = (1+z-6*z^2)/((1-5*z)*(1-3*z)^2*(1-z)^3)
GF1(z;RHCnr= 4) = (1+7*z-79*z^2+119*z^3+126*z^4-270*z^5)/((1-7*z)*(1-5*z)^2*(1-3*z)^3*(1-z)^4)
MAPLE
RHCnr:=4: if RHCnr=1 then RHCmax :=1; else RHCmax:=(RHCnr-1)*(RHCnr)/2 end if: RHCend:=RHCnr+RHCmax: for k from RHCnr to RHCend do for n from 0 to k do S2[k, n]:=sum((-1)^(n+i)*binomial(n, i)*i^k/n!, i=0..n) end do: G(k, x):= sum(S2[k, p]*((2*p)!/p!) *x^p/(1-4*x)^(p+1), p=0..k)/(((-1)^(k+1)*2*x)/(-1+4*x)^(k+1)): fx:=simplify(G(k, x)): nmax:=degree(fx); RHC[k-RHCnr+1]:= coeff(fx, x, k-RHCnr)/2^(k-RHCnr) end do: a:=n-> RHC[n]: seq(a(n), n=1..RHCend-RHCnr+1); for nx from 0 to RHCmax do num:=sort(sum(A[t]*z^t, t=0..RHCmax)); nom:=Product((1-(2*u-1)*z)^(RHCnr-u+1), u=1..RHCnr): RHCa:= series(num/nom, z, nx+1); y:=coeff(RHCa, z, nx)-A[nx]; x:=RHC[nx+1]; A[nx]:=x-y; end do: FP1[RHCnr]:=sort(num, z, ascending); GenFun[RHCnr] :=FP1[RHCnr]/product((1-(2*m-1)*z)^(RHCnr-m+1), m=1..RHCnr);
CROSSREFS
The number of FP1 terms follow the triangular numbers A000217, with quite surprisingly one exception here a(0)=1. For the polynomials in the denominators of the GF1(z;RHCnr) see A157702.
Fourth right hand column (n-m=3) of the A156920 triangle
+10
9
1, 37, 646, 8030, 82610, 756218, 6411720, 51586344, 400011435, 3020658295, 22373863774, 163379472214, 1180488191108, 8462445970580, 60305767988960, 427848087263712, 3025286818472661
FORMULA
a(n) = 30*a(n-1)-385*a(n-2)+2776*a(n-3)-12418*a(n-4)+35908*a(n-5)-67818*a(n-6)+82552*a(n-7)-62109*a(n-8)+26190*a(n-9)-4725*a(n-10)
a(n) = (-8*n^3+972*n^2*3^n-84*n^2+7776*n*3^n-11250*n*5^n-286*n+15309*3^n-50625*5^n+36015*7^n-315)/384
G.f.: GF1(z;RHCnr=4) = (1+7*z-79*z^2+119*z^3+126*z^4-270*z^5)/((1-7*z)*(1-5*z)^2*(1-3*z)^3*(1-z)^4)
CROSSREFS
Equals A156920 fourth right hand column. Equals A156919 fourth right hand column divided by 8. Equals A142963 fourth right hand column divided by 2^n
Fifth right hand column (n-m=4) of the A156920 triangle
+10
9
1, 83, 2685, 56285, 919615, 12813843, 160206627, 1854550395, 20291056470, 212826091180, 2161547322134, 21414479565774, 208076662576370, 1991164206775450, 18825064380813450
FORMULA
a(n)=55*a(n-1)-1365*a(n-2)+20251*a(n-3)-200557*a(n-4)+1402203*a(n-5)-7137473*a(n-6)+26886431*a(n-7)-75433971*a(n-8)+157376597*a(n-9)-241846607*a(n-10)+268663713*a(n-11)-208880991*a(n-12)+107416665*a(n-13)-32730075*a(n-14)+4465125*a(n-15)
a(n)= (16*n^4-7776*n^3*3^n+256*n^3-104976*n^2*3^n+225000*n^2*5^n+ 1496*n^2- 464616*n*3^n+ 2250000*n*5^n-2016840*n*7^n+3776*n-673596*3^n+5568750*5^n-11092620*7^n+6200145*9^n+3465)/6144
G.f.: GF1(z;RHCnr=5) = (1+28*z-515*z^2+1654*z^3+8689*z^4-65864*z^5+142371*z^6-82242*z^7-99090*z^8+113400*z^9)/((1-9*z)*(1-7*z)^2*(1-5*z)^3*(1-3*z)^4*(1-z)^5)
CROSSREFS
Equals A156920 fifth right hand column. Equals A156919 fifth right hand column divided by 16. Equals A142963 fifth right hand column divided by 2^n
Triangle T(n,k), 0 <= k <= n, read by rows: given by [ 1, 0, 3, 0, 5, 0, 7, 0, 9, 0, ...] DELTA [ 0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 0, ...] where DELTA is the operator defined in A084938.
+10
3
1, 1, 0, 1, 1, 0, 1, 5, 1, 0, 1, 18, 15, 1, 0, 1, 58, 129, 37, 1, 0, 1, 179, 877, 646, 83, 1, 0, 1, 543, 5280, 8030, 2685, 177, 1, 0, 1, 1636, 29658, 82610, 56285, 10002, 367, 1, 0, 1, 4916, 159742, 756218, 919615, 335162, 34777, 749, 1, 0
COMMENTS
Generalized Eulerian numbers A008292.
FORMULA
T(n, k) = (n-k)*T(n-1, k-1) + (2*k+1)*T(n-1, k) with T(0, 0) = 1, T(0, k) = 0 if k > 0, T(n, k) = 0 if k < 0.
Sum_{k>=0} T(n, k)*2^k = A001147(n).
EXAMPLE
Triangle begins:
1;
1, 0;
1, 1, 0;
1, 5, 1, 0;
1, 18, 15, 1, 0;
1, 58, 129, 37, 1, 0; ...
MATHEMATICA
T[0, 0] := 1; T[n_, -1] := 0; T[n_, n_] := 0; T[n_, k_] := T[n, k] = (n - k)*T[n - 1, k - 1] + (2*k + 1)*T[n - 1, k]; Join[{1}, Table[If[k < 0, 0, If[k >= n, 0, T[n, k]]], {n, 1, 5}, {k, 0, n}] // Flatten] (* G. C. Greubel, Jun 30 2017 *)
Triangle T(n,k), 0 <= k <= n, given by (0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 0, ...) DELTA (1, 0, 3, 0, 5, 0, 7, 0, 9, 0, ...) where DELTA is the operator defined in A084938.
+10
3
1, 0, 1, 0, 1, 1, 0, 1, 5, 1, 0, 1, 15, 18, 1, 0, 1, 37, 129, 58, 1, 0, 1, 83, 646, 877, 179, 1, 0, 1, 177, 2685, 8030, 5280, 543, 1, 0, 1, 367, 10002, 56285, 82610, 29658, 1636, 1, 0, 1, 749, 34777, 335162
FORMULA
T(n,k) = k*T(n-1,k) + (2n-2k+1)*T(n-1,k-1) , T(n,n) = 1, T(n,k) = 0 if k<0 or if k>n.
G.f.: 1/(1-xy/(1-x/(1-3xy/(1-2x/(1-5xy/(1-3x/(1-7xy/(1- ...(continued fraction).
EXAMPLE
Triangle begins :
1
0, 1
0, 1, 1
0, 1, 5, 1
0, 1, 15, 18, 1
0, 1, 37, 129, 58, 1
0, 1, 83, 646, 877, 179, 1
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