A006318 -id:A006318 - cp's OEIS frontend

A162543 A Chebyshev transform of the large Schroeder numbers A006318.

1, 2, 5, 18, 73, 312, 1391, 6406, 30235, 145478, 710951, 3519248, 17608681, 88914250, 452512229, 2318774506, 11953427329, 61948592936, 322570037543, 1686777086942, 8854240330363, 46638995523598, 246443050810895

Offset: 0

Paul Barry, Jul 05 2009

Hankel transform is the Somos-4 variant A162546.

Fung Lam, Table of n, a(n) for n = 0..1335

Cf. A162548.

a:=[2,5,18,73,312,1391];; for n in [7..30] do a[n]:=(3*(2*n-1)*a[n-1] - (4*n-5)*a[n-2] +12*(n-2)*a[n-3] -(4*n-11)*a[n-4] +3*(2*n-7)*a[n-5] -(n-5)*a[n-6])/(n+1); od; Concatenation([1], a); # G. C. Greubel, Feb 23 2019

m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!( (1-x+x^2 - Sqrt(1-6*x+3*x^2-6*x^3+x^4))/( 2*x*(1+x^2)) )); // G. C. Greubel, Feb 23 2019

CoefficientList[Series[(1-x+x^2 - Sqrt[1-6*x+3*x^2-6*x^3+x^4])/(2*x*(1+x^2)), {n,0,30}], x] (* G. C. Greubel, Feb 23 2019 *)

my(x='x+O('x^30)); Vec((1-x+x^2 - sqrt(1-6*x+3*x^2-6*x^3+x^4))/( 2*x*(1+x^2))) \\ G. C. Greubel, Feb 23 2019

((1-x+x^2 -sqrt(1-6*x+3*x^2-6*x^3+x^4))/( 2*x*(1+x^2))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Feb 23 2019

G.f.: (1/(1+x^2))*S(x/(1+x^2)), S(x) the g.f. of A006318;
G.f.: (1-x+x^2 - sqrt(1-6*x+3*x^2-6*x^3+x^4))/(2*x*(1+x^2)).
G.f.: 1/(1+x^2-2*x/(1-x/(1+x^2-2*x/(1-x/(1+x^2-2*x/(1-x/(1+x+2*x^2/(1-... (continued fraction);
a(n) = Sum_{k=0..floor(n/2)} (-1)^k*binomial(n-k,k)*A006318(n-2*k).
Recurrence: (n+1)*a(n) = (5-n)*a(n-6) + 3*(2*n-7)*a(n-5) + (11-4*n)*a(n-4)  + 12*(n-2)*a(n-3) + (5-4*n)*a(n-2) + 3*(2*n-1)*a(n-1), n>=6. - Fung Lam, Feb 19 2014

A333481 a(n) = [x^n] S(x)^(2n), where S(x) = (1 - x - sqrt(1 - 6x + x^2))/(2*x) is the o.g.f. of the large Schröder numbers A006318.

Original entry on oeis.org

1, 4, 48, 652, 9344, 138004, 2077968, 31712412, 488793088, 7591462564, 118615816048, 1862444310060, 29361743698304, 464472032918196, 7368841675386960, 117200150284494652, 1868129273410953216, 29834667873867329348, 477283242733227391152, 7647021589988643092428

Offset: 0

Peter Bala, Mar 24 2020

Compare with the sequence A103885(n) = [x^n] S(x)^n, where S(x) is the o.g.f. of the large Schröder numbers A006318. See also A333482.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) hold for prime p and positive integers n and k.
We conjecture that the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) hold for prime p >= 5 and positive integers n and k.
More generally, we conjecture that for any positive integer a and any integer b the sequence u(a,b;n) := [x^(a*n)] S(x)^(b*n) also satisfies the above congruences.

			Examples of congruences:
a(17) - a(1) = 29834667873867329348 - 4 = (2^6)*(17^3)*401*236619262717 == 0 ( mod 17^3 ).
a(2*7) - a(2) = 7368841675386960 - 48 = (2^5)*3*(7^4)*577*3229*17159 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 132585158051380247023537388004 - 138004 = (2^4)*(5^6)* 39461*13439614612035199009 == 0 ( mod 5^6 )

Cf. A006318, A103885, A333482.

[1, seq((2/3)*add(binomial(3*n,k)*binomial(4*n-k-1,3*n-1), k = 0..n), n = 1..25)];
# alternative program
S := x -> (1/2)*(1-x-sqrt(1-6*x+x^2))/x:
G := (x, n) -> series(S(x)^(2*n), x, 76):
seq(coeff(G(x, n), x, n), n = 0..25);

Join[{1}, Table[2*Binomial[4*n-1, 3*n-1] * Hypergeometric2F1[-3*n, -n, 1 - 4*n, -1]/3, {n,1,20}]] (* Vaclav Kotesovec, Mar 28 2020 *)

a(n) = (2/3) * Sum_{k = 0..n} C(3*n,k)*C(4*n-k-1,3*n-1) for n >= 1.
P-recursive: P(n)*a(n + 1) = 4*(7805*n^6 - 7132*n^4 + 1559*n^2 - 72)*a(n) -
P(-n)*a(n - 1), where P(n) =(2*n - 1)*(3*n + 1)*(3*n + 2)*(3*n + 3)*(35*n^2 - 35*n + 6).
a(n) ~ 2^(1/4) * (223 + 70*sqrt(10))^n / (5^(1/4) * sqrt(Pi*n) * 3^(3*n + 1/2)). - Vaclav Kotesovec, Mar 28 2020
a(n) = (1/2) * [x^n] 1/S(-x)^(4*n) for n >= 1. - Peter Bala, Sep 16 2024

A122538 Riordan array (1, x*f(x)) where f(x)is the g.f. of A006318.

Original entry on oeis.org

1, 0, 1, 0, 2, 1, 0, 6, 4, 1, 0, 22, 16, 6, 1, 0, 90, 68, 30, 8, 1, 0, 394, 304, 146, 48, 10, 1, 0, 1806, 1412, 714, 264, 70, 12, 1, 0, 8558, 6752, 3534, 1408, 430, 96, 14, 1, 0, 41586, 33028, 17718, 7432, 2490, 652, 126, 16, 1, 0, 206098, 164512, 89898, 39152, 14002, 4080, 938, 160, 18, 1

Offset: 0

Philippe Deléham, Sep 18 2006

Triangle T(n,k), 0<=k<=n, read by rows, given by [0, 2, 1, 2, 1, 2, 1, ...] DELTA [1, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938 . Inverse is Riordan array (1, x*(1-x)/(1+x)).

T(n, r) gives the number of [0,r]-covering hierarchies with n segments terminating at r (see Kreweras work). - Michel Marcus, Nov 22 2014

			Triangle begins:
  1;
  0,    1:
  0,    2,    1;
  0,    6,    4,    1;
  0,   22,   16,    6,    1;
  0,   90,   68,   30,    8,   1;
  0,  394,  304,  146,   48,  10,  1;
  0, 1806, 1412,  714,  264,  70, 12,  1;
  0, 8558, 6752, 3534, 1408, 430, 96, 14, 1;
Production matrix is:
  0...1
  0...2...1
  0...2...2...1
  0...2...2...2...1
  0...2...2...2...2...1
  0...2...2...2...2...2...1
  0...2...2...2...2...2...2...1
  0...2...2...2...2...2...2...2...1
  0...2...2...2...2...2...2...2...2...1
  ... - _Philippe Deléham_, Feb 09 2014

G. C. Greubel, Rows n = 0..100 of the triangle, flattened
G. Kreweras, Sur les hiérarchies de segments, Cahiers du Bureau Universitaire de Recherche Opérationnelle, Institut de Statistique, Université de Paris, #20 (1973), see page 15.

Another version : A080247, A080245, A033877.
Columns: A000007, A006318, A006319, A006320, A006321.
Diagonals: A000012, A005843, A054000.
Sums include: A001003 (row and alternating sign), A006603 (diagonal).
Cf. A103885.

function T(n,k) // T = A122538
  if k eq 0 then return 0^n;
  elif k eq n then return 1;
  else return T(n-1,k-1) + T(n-1,k) + T(n,k+1);
  end if;
end function;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 27 2024

T[n_, n_]= 1; T[, 0]= 0; T[n, k_]:= T[n, k]= T[n-1, k-1] + T[n-1, k] + T[n, k+1];
Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten (* Jean-François Alcover, Jun 13 2019 *)

def A122538_row(n):
    @cached_function
    def prec(n, k):
        if k==n: return 1
        if k==0: return 0
        return prec(n-1,k-1)-2*sum(prec(n,k+i-1) for i in (2..n-k+1))
    return [(-1)^(n-k)*prec(n, k) for k in (0..n)]
for n in (0..12): print(A122538_row(n)) # Peter Luschny, Mar 16 2016

T(n,k) = T(n-1,k-1) + T(n-1,k) + T(n,k+1) if k > 0, with T(n, 0) = 0^n, and T(n, n) = 1.
Sum_{k=0..n} T(n, k) = A001003(n).
From G. C. Greubel, Oct 27 2024: (Start)
T(2*n, n) = A103885(n).
Sum_{k=0..n} (-1)^k*T(n, k) = -A001003(n-1).
Sum_{k=0..floor(n/2)} T(n-k, k) = [n=0] + 0*[n=1] + A006603(n-2)*[n>1]. (End)

A138463 A bisection of A006318.

Original entry on oeis.org

2, 22, 394, 8558, 206098, 5293446, 142078746, 3937603038, 111818026018, 3236724317174, 95149655201962

Offset: 0

N. J. A. Sloane, May 08 2008

nonn

A138463 := proc(n)
    A006318(2*n+1) ;
end proc:
seq(A138463(n),n=0..50) ; # R. J. Mathar, Mar 25 2024

a(n)=A006318(2n+1).

D-finite with recurrence +(n+1)*(2*n+1)*a(n) +(-70*n^2+3*n+4)*a(n-1) +(70*n^2-207*n+149)*a(n-2) -(n-2)*(2*n-5)*a(n-3)=0. - R. J. Mathar, Mar 25 2024

A144156 Eigentriangle, row sums = A006318.

Original entry on oeis.org

1, 1, 1, 3, 1, 2, 11, 3, 2, 6, 45, 11, 6, 6, 22, 197, 45, 22, 18, 22, 90, 903, 197, 90, 66, 66, 90, 394, 4279, 903, 394, 270, 242, 270, 394, 1806, 20793, 4279, 1806, 1182, 990, 990, 1182, 1806, 8558, 103049, 20793, 8558, 5418, 4334, 4050, 4334, 5418, 8558

Offset: 0

Gary W. Adamson, Sep 12 2008

Row sums = A006318. Left border = A001003.

Sum of n-th row terms = rightmost term of next row.

			First few rows of the triangle =
1;
1, 1;
3, 1, 2;
11, 3, 2, 6;
45, 11, 6, 6, 22;
197, 45, 22, 18, 22, 90;
903, 197, 90, 66, 66, 90, 394;
4279, 903, 394, 270, 242, 270, 394, 1806;
...
Row = (11, 3, 2, 6) = termwise products of (11, 3, 1, 1) and (1, 1, 2, 6) = (11*1, 3*1, 1*2, 1*6).

Cf. A001003, A006318.

Triangle read by rows, T(n,k) = A001003(n-k)*A006318(k-1) with A006318(-1)=1. (corrected by R. J. Mathar, Jul 19 2016)

A174808 A transform of the large Schroeder numbers A006318.

Original entry on oeis.org

1, 2, 8, 34, 162, 820, 4338, 23694, 132612, 756594, 4384022, 25729336, 152627730, 913674362, 5512542128, 33486653154, 204639278346, 1257199799116, 7760098104882, 48102326710998, 299309479778956, 1868853597670754

Offset: 0

Paul Barry, Mar 29 2010

Hankel transform is A174809.

			G.f. = 1 + 2*x + 8*x^2 + 34*x^3 + 162*x^4 + 820*x^5 + 4338*x^6 + ...

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A174809.

m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!((1-x-x^2-Sqrt(1-6*x-5*x^2+2*x^3+x^4))/(2*x*(1+x)))); // G. C. Greubel, Sep 22 2018

A174808 := proc(n)
    add(binomial(k,n-k)*A006318(k),k=0..n) ;
end proc: # R. J. Mathar, Feb 10 2015

CoefficientList[Series[(1-x-x^2 -Sqrt[1-6*x-5*x^2+2*x^3+x^4])/(2*x*(1 + x)), {x, 0, 30}], x] (* G. C. Greubel, Sep 22 2018 *)

x='x+O('x^30); Vec((1-x-x^2-sqrt(1-6*x-5*x^2+2*x^3+x^4))/(2*x*(1+x))) \\ G. C. Greubel, Sep 22 2018

G.f.: (1-x-x^2-sqrt(1-6*x-5*x^2+2*x^3+x^4))/(2*x*(1+x)).
G.f.: 1/(1-2x(1+x)/(1-x(1+x)/(1-2x(1+x)/(1-x(1+x)/(1-...))))) (continued fraction).
a(n) = Sum_{k=0..n} C(k,n-k)*A006318(k).
G.f.: 1 / (1 - (x + x^2)*(1 + 1 / (1 - (x + x^2)*(1 + 1 / ...)))). - Michael Somos, Mar 30 2014
Conjecture: (n+1)*a(n) +(-5*n+4)*a(n-1) +(-11*n+13)*a(n-2) +3*(-n+1)*a(n-3) +3*(n-4)*a(n-4) +(n-5)*a(n-5)=0. - R. J. Mathar, Feb 10 2015

A333091 a(n) is the n-th order Taylor polynomial (centered at 0) of S(x)^(2n ) evaluated at x = 1, where S(x) = (1 - x - sqrt(1 - 6x + x^2))/(2*x) is the o.g.f. of the Schröder numbers A006318.

Original entry on oeis.org

1, 5, 57, 761, 10817, 159005, 2386857, 36348401, 559362561, 8676917429, 135445348057, 2125030235113, 33479772021953, 529326516063181, 8393856020704841, 133449301759137761, 2126391547960594433, 33948765589280671589, 542950968447834265209, 8697032976559212532953

Offset: 0

Peter Bala, Mar 22 2020

The sequence satisfies the Gauss congruences: a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k.
We conjecture that the sequence satisfies the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. Examples of these congruences are given below.
More generally, for each integer m, we conjecture that the sequence a_m(n), defined as the n-th order Taylor polynomial of S(x)^(m*n) evaluated at x = 1 satisfies the same congruences. For cases see A333090 (m = 1) and A333092 (m = 3). For similarly defined sequences see A333093 through A333097.

			n-th order Taylor polynomial of S(x)^(2*n):
  n = 0: S(x)^0 = 1 + O(x)
  n = 1: S(x)^2 = 1 + 4*x + O(x^2)
  n = 2: S(x)^4 = 1 + 8*x + 48*x^2 + O(x^3)
  n = 3: S(x)^6 = 1 + 12*x + 96*x^2 + 652*x^3 + O(x^4)
  n = 4: S(x)^8 = 1 + 16*x + 160*x^2 + 1296*x^3 + 9344*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 4 = 5, a(2) = 1 + 8 + 48 = 57, a(3) = 1 + 12 + 96 + 652 = 761 and a(4) = 1 + 16 + 160 + 1296 + 9344 = 10817.
The triangle of coefficients of the n-th order Taylor polynomial of S(x)^(2*n), n >= 0, in descending powers of x begins
                                          row sums
  n = 0 |    1                                1
  n = 1 |    4    1                           5
  n = 2 |   48    8    1                     57
  n = 3 |  652   96   12   1                761
  n = 4 | 9344 1296  160  16   1          10817
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group. The first column sequence [1, 4, 48, 652, 9344, 138004, ...]  = [x^n] S(x)^(2*n), and may also satisfy the above congruences.
Examples of congruences:
a(13) - a(1) = 529326516063181 - 5 = (2^3)*(13^3)*30116438101 == 0 ( mod 13^3 ).
a(3*7) - a(3) = 2240508640665255893197949 - 761 = (2^2)*3*(7^3)*11* 49485569411283149863 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 150633078429259494145205034005 - 159005 = (2^3)*(3^3)*(5^6)*11*23*61*2663*28097*119633*323083 == 0 ( mod 5^6 ).

A006318, A333090 through A333097.

S:= x -> (1/2)*(1-x-sqrt(1-6*x+x^2))/x:
G := (x,n) -> series(S(x)^(2*n), x, 76):
seq(add(coeff(G(x, n), x, n-k), k = 0..n), n = 0..25);

Table[SeriesCoefficient[((1+x) * (1 - 2*x*(1+x) - Sqrt[1 - 4*x*(1+x)]) / (2*x^2))^n, {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Mar 28 2020 *)

a(n) = [x^n] ( (1 + x)*S^2(x/(1 + x)) )^n
O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + 4*x + 32*x^2 + 324*x^3 + 3696*x^4 + ... = (1/x)*Revert( x/S^2(x) ).
Row sums of the Riordan array ( 1 + x*f'(x)/f(x), x*f(x) ) belonging to the Hitting time subgroup of the Riordan group.
a(n) ~ sqrt(120 + 39*sqrt(10)) * (223 + 70*sqrt(10))^n / (30*sqrt(Pi*n) * 3^(3*n)). - Vaclav Kotesovec, Mar 28 2020

A333092 a(n) is the n-th order Taylor polynomial (centered at 0) of S(x)^(3n) evaluated at x = 1, where S(x) = (1 - x - sqrt(1 - 6x + x^2))/(2*x) is the o.g.f. of the Schröder numbers A006318.

Original entry on oeis.org

1, 7, 109, 1951, 36993, 724007, 14457421, 292732671, 5987886081, 123440423047, 2560421160109, 53373725431583, 1117198199782785, 23465732683090471, 494330214846965389, 10440064992542621951, 220978578227187097601, 4686426367646858888711, 99559270036968523118317

Offset: 0

Peter Bala, Mar 22 2020

The sequence satisfies the Gauss congruences: a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k.
We conjecture that the sequence satisfies the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. Examples of these congruences are given below.
More generally, for each integer m, we conjecture that the sequence a_m(n) defined as the n-th order Taylor polynomial of S(x)^(m*n) evaluated at x = 1 satisfies the same congruences. For cases, see A333090 (m = 1) and A333091 (m = 2). For similarly defined sequences see A333093 through A333097.

			n-th order Taylor polynomial of S(x)^(3*n):
  n = 0: S(x)^0 = 1 + O(x)
  n = 1: S(x)^3 = 1 + 6*x + O(x^2)
  n = 2: S(x)^6 = 1 + 12*x + 96*x^2 + O(x^3)
  n = 3: S(x)^9 = 1 + 18*x + 198*x^2 + 1734*x^3 + O(x^4)
  n = 4: S(x)^12 = 1 + 24*x + 336*x^2 + 3608*x^3 + 33024*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 6 = 7, a(2) = 1 + 12 + 96 = 109, a(3) = 1 + 18 + 198 + 1734 = 1951 and a(4) = 1 + 24 + 336 + 3608 + 33024 = 36993.
The triangle of coefficients of the n-th order Taylor polynomial of S(x)^(2*n), n >= 0, in descending powers of x begins
                                           row sums
  n = 0 |     1                                1
  n = 1 |     6    1                           7
  n = 2 |    96   12    1                    109
  n = 3 |  1734  198   18   1               1951
  n = 4 | 33024 3608  336  24   1          36993
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group. The first column sequence [1, 6, 96, 1734, 33024, 648006, ...]  = [x^n] S(x)^(3*n), and may also satisfy the above congruences.
Examples of congruences:
a(13) - a(1) = 23465732683090471 - 7 = (2^5)*(3^4)*(13^3)*83*911*54497 == 0 ( mod 13^3 ).
a(3*7) - a(3) = 962815680123979633351467303 - 1951 = (2^3)*(7^3)*29*41* 1832861*161008076794727  == 0 ( mod 7^3 ).
a(5^2) - a(5) = 201479167004032422703424646224007 - 724007 = (2^5)*(5^6)* 402958334008064845406849291 == 0 ( mod 5^6 ).

Cf. A006318, A333090 through A333097.

S:= x -> (1/2)*(1-x-sqrt(1-6*x+x^2))/x:
G := (x,n) -> series(S(x)^(3*n), x, 101):
seq(add(coeff(G(x, n), x, n-k), k = 0..n), n = 0..25);

Table[SeriesCoefficient[((1+x)*(1 - 3*x*(1+x) + (x^2 + x - 1)*Sqrt[1 - 4*x*(1+x)]) / (2*x^3))^n, {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Mar 28 2020 *)

a(n) = [x^n] ( (1 + x)*S^3(x/(1 + x)) )^n.
O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + 6*x + 66*x^2 + 902*x^3 + 13794*x^4 + ... = (1/x) * series reversion of ( x/S^3(x) ).
Row sums of the Riordan array ( 1 + x*f'(x)/f(x), x*f(x) ) belonging to the Hitting time subgroup of the Riordan group.
a(n) ~ 3*sqrt(85 + 21*sqrt(17)) * (349 + 85*sqrt(17))^n / (68 * sqrt(Pi*n) * 2^(5*n)). - Vaclav Kotesovec, Mar 28 2020

A039975 An example of a d-perfect sequence: a(n) = A006318(n-1) mod 3.

Original entry on oeis.org

1, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 1

N. J. A. Sloane

nonn

Antti Karttunen, Table of n, a(n) for n = 1..11000
D. Kohel, S. Ling and C. Xing, Explicit Sequence Expansions, in Sequences and their Applications, C. Ding, T. Helleseth, and H. Niederreiter, eds., Proceedings of SETA'98 (Singapore, 1998), 308-317, 1999. DOI: 10.1007/978-1-4471-0551-0_23

Cf. A006318.

Cf. also A039969.

A006318(n) = if( n<1, 1, sum( k=0, n, 2^k * binomial( n, k) * binomial( n, k-1)) / n);
A039975(n) = (A006318(n-1) % 3); \\ Antti Karttunen, Feb 13 2019

a(n) = A006318(n-1) mod 3. - Christian G. Bower, Jun 12 2005

More terms from Christian G. Bower, Jun 12 2005

Bower's formula added to the name by Antti Karttunen, Feb 13 2019

A073150 Triangle of numbers {a(n,k), n >= 0, 0<=k<=n} defined by a(0,0)=1, a(1,0)=2, a(n,0)=A006318(n), a(n,n)=A006319(n), a(n+1,0)=a(n,0)+a(n,n), a(n,m+1)= Sum A006318(k)*a(n-k,0), k=0..m.

Original entry on oeis.org

1, 2, 4, 6, 10, 16, 22, 34, 46, 68, 90, 134, 170, 214, 304, 394, 574, 706, 838, 1018, 1412, 1806, 2594, 3134, 3618, 4158, 4946, 6752, 8558, 12170, 14534, 16514, 18494, 20858, 24470, 33028, 41586

Offset: 0

Paul D. Hanna, Jul 18 2002

Compare to A073151. Related to Royal paths in a lattice (A006318, A006319).

			a(4,2)=1*a(3,0)+2*a(2,0)+6*a(1,0)=1*90+2*22+6*6=170. a(4,0)=1+a(3,3)+a(2,2)+a(1,1)+a(0,0)=1+(68+16+4+1)=90. {1}, {2,4}, {6,10,16}, {22,34,46,68}, {90,134,170,214,304},{394,574,706,838,1018,1412}, {1806,2594,3134,3618,4158,4946,6752}, ...

Cf. A073151, A006318, A006319.

a(n, m+1) = Sum A006318(k)*a(n-k, 0), k=0..m.

cp's OEIS Frontend

A162543 A Chebyshev transform of the large Schroeder numbers A006318.

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A333481 a(n) = [x^n] S(x)^(2*n), where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the o.g.f. of the large Schröder numbers A006318.

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A122538 Riordan array (1, x*f(x)) where f(x)is the g.f. of A006318.

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A138463 A bisection of A006318.

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A144156 Eigentriangle, row sums = A006318.

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A174808 A transform of the large Schroeder numbers A006318.

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A333091 a(n) is the n-th order Taylor polynomial (centered at 0) of S(x)^(2*n ) evaluated at x = 1, where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the o.g.f. of the Schröder numbers A006318.

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A333092 a(n) is the n-th order Taylor polynomial (centered at 0) of S(x)^(3*n) evaluated at x = 1, where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the o.g.f. of the Schröder numbers A006318.

A333481 a(n) = [x^n] S(x)^(2n), where S(x) = (1 - x - sqrt(1 - 6x + x^2))/(2*x) is the o.g.f. of the large Schröder numbers A006318.

A333091 a(n) is the n-th order Taylor polynomial (centered at 0) of S(x)^(2n ) evaluated at x = 1, where S(x) = (1 - x - sqrt(1 - 6x + x^2))/(2*x) is the o.g.f. of the Schröder numbers A006318.

A333092 a(n) is the n-th order Taylor polynomial (centered at 0) of S(x)^(3n) evaluated at x = 1, where S(x) = (1 - x - sqrt(1 - 6x + x^2))/(2*x) is the o.g.f. of the Schröder numbers A006318.