A187075 -id:A187075 - cp's OEIS frontend

A156919 Table of coefficients of polynomials related to the Dirichlet eta function.

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

Offset: 0

Johannes W. Meijer, Feb 20 2009, Jun 24 2009

Essentially the same as A185411. Row reverse of A185410. - Peter Bala, Jul 24 2012
The SF(z; n) formulas, see below, were discovered while studying certain properties of the Dirichlet eta function.
From Peter Bala, Apr 03 2011: (Start)
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)

			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;
  ...

D. H. Lehmer, Interesting Series Involving the Central Binomial Coefficient, Am. Math. Monthly 92 (1985) 449-457, Polynomial V in eq (17). [R. J. Mathar, Feb 24 2009]
Shi-Mei Ma, A family of two-variable derivative polynomials for tangent and secant, arXiv: 1204.4963v3 [math.CO], 2012.
Shi-Mei Ma, A family of two-variable derivative polynomials for tangent and secant, El J. Combinat. 20 (1) (2013) P11
S.-M. Ma and T. Mansour, The 1/k-Eulerian polynomials and k-Stirling permutations, arXiv preprint arXiv:1409.6525 [math.CO], 2014.
S.-M. Ma and Y.-N. Yeh, Stirling permutations, cycle structures of permutations and perfect matchings, arXiv preprint arXiv:1503.06601 [math.CO], 2015.
Carla D. Savage and Gopal Viswanathan, The 1/k-Eulerian polynomials, Elec. J. of Comb., Vol. 19, Issue 1, #P9 (2012).
Eric Weisstein's World of Mathematics, Dirichlet Eta Function

A142963 and this sequence can be mapped onto the A156920 triangle.
FP1 sequences A000340, A156922, A156923, A156924.
FP2 sequences A050488, A142965, A142966, A142968.
Appears in A162005, A000182, A162006 and A162007.
Cf. A186695, A187075.
Cf. A185410 (row reverse), A185411.

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

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 *)

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
From Peter Bala, Apr 03 2011 (Start)
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.
Row sums are A001147(n+1). (End)
Sum_{k = 0..n} (-1)^k*T(n,k) = (-1)^binomial(n,2)*A012259(n+1). - Johannes W. Meijer, Sep 27 2011

Minor edits from Johannes W. Meijer, Sep 27 2011

A186695 A Galton triangle: T(n,k) = (2k-1)*(T(n-1,k) + T(n-1,k-1)): a type B analog of the ordered Bell numbers A019538.

Original entry on oeis.org

1, 1, 3, 1, 12, 15, 1, 39, 135, 105, 1, 120, 870, 1680, 945, 1, 363, 4950, 17850, 23625, 10395, 1, 1092, 26565, 159600, 373275, 374220, 135135, 1, 3279, 138285, 1303155, 4795875, 8222445, 6621615, 2027025

Offset: 1

Peter Bala, Mar 26 2011

The row polynomials R(n,x) of A019538 satisfy the recurrence relation R(n+1,x) = x*d/dx((1+x)*R(n,x)), and have the expansion R(n,x) = Sum_{k = 1..n} k!*Stirling2(n,k)*x^k.
Here we consider a sequence of polynomials P(n,x) (n>=1) defined by means of the similar recursion P(n+1,x) = x*d/dx((1+x^2)*P(n,x)), with starting value P(1,x) = x.
The first few polynomials are P(1,x) = x, P(2,x) = x + 3*x^3, P(3,x) = x + 12*x^3 + 15*x^5, and P(4,x) = x + 39*x^3 + 135*x^5 + 105*x^7.
Clearly, the P(n,x) are odd polynomials of the form P(n,x) = Sum_{k = 1..n} T(n,k)*x^(2*k-1).
This triangle lists the coefficients T(n,k). They are related to A039755, the type B Stirling numbers of Suter, by T(n,k) = (2*k-1)!!*A039755(n-1,k-1).

			Triangle begins
  n\k.|..1.....2.....3......4......5......6
  =========================================
  ..1.|..1
  ..2.|..1.....3
  ..3.|..1....12....15
  ..4.|..1....39...135....105
  ..5.|..1...120...870...1680....945
  ..6.|..1...363..4950..17850..23625..10395
  ..
Examples of recurrence relation
  T(4,3) = 5*(T(3,3)+T(3,2)) = 5*(15+12) = 135;
  T(6,4) = 7*(T(5,4)+T(5,3)) = 7*(1680+870) = 17850.

Paweł Hitczenko, A class of polynomial recurrences resulting in (n/log n, n/log^2 n)-asymptotic normality, arXiv:2403.03422 [math.CO], 2024. See p. 8.
Erich Neuwirth, Recursively defined combinatorial functions: Extending Galton's board, Tech Report TR 99-05, 1999.
Erich Neuwirth, Recursively defined combinatorial functions: Extending Galton's board, Discrete Math. 239 No. 1-3, 33-51 (2001).

Cf. A001147, A019538, A039755, A124212, A145901, A156919, A187075.

A186695 := proc(n, k) option remember; if k < 1 or k > n then 0; elif k = 1 then 1; else (2*k-1)*(procname(n-1, k) + procname(n-1, k-1)) ; end if; end proc: seq(seq(A186695(n,k),k = 1..n),n = 1..10);

T[n_, k_] := (2k-1)! Sum[(-1)^(k-j-1) (2j+1)^(n-1) Binomial[k-1, j], {j, 0, k-1}] / (2^(k-1) (k-1)!)^2;
Table[T[n, k], {n, 1, 8}, {k, 1, n}] // Flatten (* Jean-François Alcover, Nov 02 2019 *)

T(n+1,k+1) = ((2*k+1)!/(2^k*k!)^2)*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*(2*j+1)^n.
Recurrence relation: T(n,k) = (2k-1)*(T(n-1,k) + T(n-1,k-1)) with boundary conditions T(n,1) = 1, T(1,k) = 0 for k >= 2.
E.g.f.: F(x,t) = (x/(1+x))*(exp(t)/sqrt((1+x) - x*exp(2*t)) - 1) = Sum_{n>=1} R(n,x)*t^n/n! = x*t + (x + 3*x^2)*t^2/2! + (x + 12*x^2 + 15*x^3)*t^3/3! + ....
Compare with the e.g.f. for A019538, which is (x/(1+x))*(exp(t)/((1+x) - x*exp(t))-1).
The row polynomials R(n,x) are related to the polynomials P(n,x) of the comments section by P(n,x) = 1/x*R(n,x^2).
The generating function F(x,t) satisfies the partial differential equation d/dt(F) = 2*x*(1+x)*d/dx(F) + (x-1)*F + x.
It follows that the polynomials P(n,x) := Sum_{k = 1..n} T(n,k)*x^(2*k-1) satisfy the recurrence P(n+1,x) = x*d/dx((1+x^2)*P(n,x)), with P(1,x) = x. (Cf. the recurrence relation for row polynomials of A185896.)
The recurrence relation for T(n,k) given above now follows.
The row polynomials R(n,x) = Sum_{k = 1..n} T(n,k)*x^k satisfy R(n,-x-1) = (-1)^n*((1+x)/x)*S(n,x), where S(n,x) is the n-th row polynomial of A187075.
In addition, R(n,1/(x-1)) = (1/(x-1)^n)*Q(n-1,x), where Q(n,x) is the n-th row polynomial of A156919.
Row sums are [1,4,28,280,3616...] = 1/2*A124212(n) for n >= 1.
Main diagonal is [1,3,15,105,...] = A001147(k) for k >= 1.
Put S(n) = sum {k = 1..n} (-1)^k*T(n,k)/(k+1). Then for m>=2, S(2*m-1) = S(2*m) = (4^m-1)*Bernoulli(2*m)/m.
From Peter Bala, Aug 30 2016: (Start)
n-th row polynomial R(n,x) = 1/(1 + x)^(3/2) * Sum_{k >= 0} (1/4)^k*(x/(1 + x))^k*binomial(2*k,k)*(2*k + 1)^n.
R(n,x) = (1/(1 + x))*Sum_{k = 0..n} binomial(2*k,k)*A145901(n,k)*(x/4)^k. (End)

A211402 Triangle T(n,k), 0<=k<=n, given by (0,2,0,4,0,6,0,8,0,10,0,...) DELTA (1, 2, 3, 4, 5, 6, 7, 8, 9,...) where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 0, 1, 0, 2, 3, 0, 4, 18, 15, 0, 8, 84, 180, 105, 0, 16, 360, 1500, 2100, 945, 0, 32, 1488, 10800, 27300, 28350, 10395, 0, 64, 6048, 72240, 294000, 529200, 436590, 135135, 0, 128, 24384, 463680, 2857680, 7938000, 11060280, 7567560, 2027025

Offset: 0

Philippe Deléham, Feb 09 2013

A Galton triangle. Essentially the same as A187075.

			Triangle begins :
1
0, 1
0, 2, 3
0, 4, 18, 15
0, 8, 84, 180, 105
0, 16, 360, 1500, 2100, 945
0, 32, 1488, 10800, 27300, 28350, 10395

Cf. A000079, A001147, A048993.

T(n,k) = 2*k*T(n-1,k) + (2*k-1)*T(n-1,k-1), T(0,0) = 1, T(n,k) = 0 if k<0 or if k>n.
T(n,k) = 2^(n-k)*A001147(k)*A048993(n,k).
G.f.: F(x,t) = 1 + x*t + (2*x+3*x^2)*t^2/2! + (4*x+18*x^2+15*x^3)*t^3/3!+ ... = Sum_{n = 0..inf}R(n,x)*t^n/n!.
The row polynomials R(n,x) satisfy the recursion R(n+1,x) = 2*(x+x^2)*R'(n,x) + x*R(n,x) where ' indicates differentiation with respect to x.

A211608 Triangle T(n,k), 0 <= k <= n, given by (0, 1, 0, 2, 0, 3, 0, 4, 0, 5, ...) DELTA (1, 2, 3, 4, 5, 6, 7, 8, 9, ...) where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 0, 1, 0, 1, 3, 0, 1, 9, 15, 0, 1, 21, 90, 105, 0, 1, 45, 375, 1050, 945, 0, 1, 93, 1350, 6825, 14175, 10395, 0, 1, 189, 4515, 36750, 132300, 218295, 135135, 0, 1, 381, 14490, 178605, 992250, 2765070, 3783780, 2027025

Offset: 0

Philippe Deléham, Feb 10 2013

			Triangle begins :
1
0, 1
0, 1,  3
0, 1,  9,   15
0, 1, 21,   90,  105
0, 1, 45,  375, 1050,   945
0, 1, 93, 1350, 6825, 14175, 10395

K. N. Boyadzhiev, Series with central binomial coefficients, Catalan numbers, and harmonic numbers, J. Integer Seq. 15 (2012), Article 12.1.7.

Cf. A000079, A001147, A048993, A187075, A211402, A142963

T(n,k) = A048993(n,k)*A001147(k).
T(n,k) = A211402(n,k)/(2^(n-k)).
T(n,k) = k*T(n-1,k) + (2*k-1)*T(n-1,k-1), T(0,0) = 1, T(n,k) = 0 if k<0 or if k>n.
G.f.: F(x,t) = 1 + x*t + (x+3*x^2)*t^2/2! + (x+9*x^2+15*x^3)*t^3/3! + ... = Sum_{n = 0..inf} R(n,x)* t^n/n!.
The row polynomials R(n,x) satisfy the recursion R(n+1,x) = (x+2*x^2)*R'(n,x) + x*R(n,x) where ' indicates differentiation with respect to x.
R(n,x) = 1/sqrt(1 + 2*x)*Sum_{k >= 0} binomial(2*k,k)/2^k*k^n * x^k/(1 + 2*x)^k (see Boyadzhiev, eqn. 19). - Peter Bala, Jan 18 2018

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A156919 Table of coefficients of polynomials related to the Dirichlet eta function.

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A186695 A Galton triangle: T(n,k) = (2k-1)*(T(n-1,k) + T(n-1,k-1)): a type B analog of the ordered Bell numbers A019538.

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A211402 Triangle T(n,k), 0<=k<=n, given by (0,2,0,4,0,6,0,8,0,10,0,...) DELTA (1, 2, 3, 4, 5, 6, 7, 8, 9,...) where DELTA is the operator defined in A084938.

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A211608 Triangle T(n,k), 0 <= k <= n, given by (0, 1, 0, 2, 0, 3, 0, 4, 0, 5, ...) DELTA (1, 2, 3, 4, 5, 6, 7, 8, 9, ...) where DELTA is the operator defined in A084938.

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