A086246 -id:A086246 - cp's OEIS frontend

A348957 G.f. A(x) satisfies A(x) = (1 + x * A(-x)) / (1 - x * A(x)).

1, 2, 2, 10, 18, 98, 210, 1194, 2786, 16258, 39906, 236938, 601458, 3615330, 9399858, 57024426, 150947010, 922283522, 2475603138, 15212318730, 41290579410, 254909413218, 698230131858, 4327273358250, 11943274468770, 74260741616514, 206279837823650, 1286199407132554

Offset: 0

Ilya Gutkovskiy, Nov 05 2021

nonn

Cf. A006318, A086246.
Cf. A349310, A361638, A364407, A366266, A366364.
Cf. A112478, A364394, A364395, A366363.

nmax = 27; A[] = 0; Do[A[x] = (1 + x A[-x])/(1 - x A[x]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
a[0] = 1; a[n_] := a[n] = -(-1)^n a[n - 1] + Sum[a[k] a[n - k - 1], {k, 0, n - 1}]; Table[a[n], {n, 0, 27}]
CoefficientList[y/.AsymptoticSolve[y-y^2+x(1+y^3)==0,y->1,{x,0,27}][[1]],x] (* Alexander Burstein, Nov 26 2021 *)

a(0) = 1; a(n) = -(-1)^n * a(n-1) + Sum_{k=0..n-1} a(k) * a(n-k-1).
a(n) ~ c * 3^(3*n/4) * (1 + sqrt(3))^n / (sqrt(2*Pi) * n^(3/2) * 2^(n/2)), where c = 3^(1/4) if n is even and c = (1 + sqrt(3))/sqrt(2) if n is odd. - Vaclav Kotesovec, Nov 14 2021
From Alexander Burstein, Nov 26 2021: (Start)
G.f.: A(-x) = 1/A(x).
G.f.: A(x) = 1 + x*(1+A(x)^3)/A(x). (End)
a(n) = (-1)^(n-1) * (1/n) * Sum_{k=0..n} binomial(n,k) * binomial(2*n-3*k-2,n-1) for n > 0. - Seiichi Manyama, Apr 11 2024

A247364 Riordan array (f(x), (f(x)-1)/f(x)) where f(x) = (1 + x - sqrt(1 - 2x - 3x^2))/(2*x).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 4, 4, 3, 1, 1, 9, 9, 6, 4, 1, 1, 21, 21, 15, 8, 5, 1, 1, 51, 51, 36, 22, 10, 6, 1, 1, 127, 127, 91, 54, 30, 12, 7, 1, 1, 323, 323, 232, 142, 75, 39, 14, 8, 1, 1, 835, 835, 603, 370, 205, 99, 49, 16, 9, 1, 1, 2188, 2188, 1585, 983

Offset: 0

Philippe Deléham, Sep 14 2014

This is essentially the reversal of the triangle in A034928, and A204849 with a duplicated first column.

Row sums are A005554(n+1).

			Triangle begins:
1
1, 1
1, 1, 1
2, 2, 1, 1
4, 4, 3, 1, 1
9, 9, 6, 4, 1, 1
21, 21, 15, 8, 5, 1, 1
51, 51, 36, 22, 10, 6, 1, 1
Production matrix begins:
1, 1
0, 0, 1
1, 1, 0, 1
1, 1, 1, 0, 1
1, 1, 1, 1, 0, 1
1, 1, 1, 1, 1, 0, 1
1, 1, 1, 1, 1, 1, 0, 1

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened

Cf. A001006, A005043, A005554, A034928, A086246.

a247364 n k = a247364_tabl !! n !! k
a247364_row n = a247364_tabl !! n
a247364_tabl = [1] : (map reverse a034928_tabf)
-- Reinhard Zumkeller, Sep 20 2014

T(n,0) = A086246(n+1), T(n+1,1) = A001006(n), T(n+2,2) = A005043(n+2).

A185010 a(n) = A000108(n)A015518(n+1), where A000108 are the Catalan numbers and A015518(n) = 2A015518(n-1) + 3*A015518(n-2).

Original entry on oeis.org

1, 2, 14, 100, 854, 7644, 72204, 703560, 7037030, 71772844, 743844452, 7810307960, 82909630972, 888316731800, 9593823377880, 104332819539600, 1141523825614470, 12556761952114380, 138785264158902900, 1540516430396559000, 17165754516697206420, 191944345934966132040

Offset: 0

Paul D. Hanna, Dec 26 2012

nonn

More generally, given {S} such that: S(n) = b*S(n-1) + c*S(n-2), |b|>0, |c|>0, S(0)=1, then

Sum_{n>=0} S(n)*Catalan(n)*x^n = sqrt( (1-2*b*x - sqrt(1-4*b*x-16*c*x^2))/(2*b^2+8*c) )/x.

			G.f.: A(x) = 1 + 1*2*x + 2*7*x^2 + 5*20*x^3 + 14*61*x^4 + 42*182*x^5 + 132*547*x^6 +...+ A000108(n)*A015518(n+1)*x^n +...

G. C. Greubel, Table of n, a(n) for n = 0..925
S. B. Ekhad and M. Yang, Proofs of Linear Recurrences of Coefficients of Certain Algebraic Formal Power Series Conjectured in the On-Line Encyclopedia Of Integer Sequences, (2017).

Cf. A000108, A001006, A015518.

CoefficientList[Series[Sqrt[(1 - 4*x - Sqrt[1 - 8*x - 48*x^2])/32]/x, {x, 0, 50}], x] (* G. C. Greubel, Jun 09 2017 *)

{A000108(n)=binomial(2*n,n)/(n+1)}
{A015518(n)=polcoeff(x/(1-2*x-3*x^2 +x*O(x^n)),n)}
{a(n)=A000108(n)*A015518(n+1)}
for(n=0,30,print1(a(n),", "))

G.f.: sqrt( (1-4*x - sqrt(1-8*x-48*x^2))/32 )/x.
G.f.: sqrt( M(4*x) ), where M(x) is g.f. of A001006. - Werner Schulte, Aug 10 2015
Conjecture: n*(n+1)*a(n) -4*n*(2*n-1)*a(n-1) -12*(2*n-1)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Oct 08 2016
G.f: B(m(4z)/4), where B(x) is the g.f. of A000984 and m(x) is the g.f. of A086246. - Alexander Burstein, May 20 2021

A333472 a(n) = [x^n] ( c (x/(1 + x)) )^n, where c(x) = (1 - sqrt(1 - 4x))/(2x) is the o.g.f. of the Catalan numbers A000108.

Original entry on oeis.org

1, 1, 3, 13, 59, 276, 1317, 6371, 31131, 153292, 759428, 3780888, 18900389, 94805959, 476945913, 2405454213, 12158471195, 61574325840, 312365992620, 1587052145492, 8074474510884, 41131551386120, 209760563456920, 1070822078321520, 5471643738383781, 27982867986637151

Offset: 0

Peter Bala, Mar 23 2020

Let F(x) = 1 + f(1)*x + f(2)*x^2 + ... be a power series with integer coefficients. The associated sequence u(n) := [x^n] F(x)^n is known to satisfy the Gauss congruences: u(n*p^k) == u(n*p^(k-1)) ( mod p^k ) for any prime p and positive integers n and k. For certain power series F(x), stronger congruences may hold. Examples include F(x) = (1 + x)^2, F(x) = 1/(1 - x) and F(x) = c(x), where c(x) is the o.g.f. of the Catalan numbers A000108. The associated sequences (with some differences of offset) are A000984, A001700 and A025174, respectively.
Here we take F(x) = c(x/(1 + x)) = 1 + x + x^2 + 2*x^3 + 4*x^4 + 9*x^5 + 21*x^6 + ... (cf. A001006 and A086246) and conjecture that the associated sequence a(n) = [x^n] ( c(x/(1 + x)) )^n satisfies the supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(2*k) ) for prime p >= 5 and positive integers n and k. Cf. A333473.
More generally, we conjecture that for any positive integer r and any integer s the sequence a(r,s;n) := [x^(r*n)] ( c(x/(1 + x)) )^(s*n) also satisfies the above congruences.
Note that the sequence b(n) := [x^n] c(x)^n = A025174(n) satisfies the stronger congruences b(n*p^k) == b(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. The sequence d(n) := [x^n] ( (1 + x)*c(x/(1 + x)) )^n = A333093(n) appears to satisfy the same congruences.

			Examples of congruences:
a(11) - a(1) = 3780888 - 1 = (11^2)*31247 == 0 ( mod 11^2 ).
a(3*7) - a(3) = 41131551386120 - 13 = (7^2)*13*23671*2727841 == 0 ( mod 7^2 ).
a(5^2) - a(5) = 27982867986637151 - 276 = (5^4)*13*74687*46113049 == 0 ( mod 5^4 ).

Cf. A000108, A001006, A086246, A106228, A219537, A333093, A333473.

Cat := x -> (1/2)*(1-sqrt(1-4*x))/x:
G := x -> Cat(x/(1+x)):
H := (x,n) -> series(G(x)^n, x, 51):
seq(coeff(H(x, n), x, n), n = 0..25);

Table[SeriesCoefficient[((1 + x - Sqrt[1 - 2*x - 3*x^2]) / (2*x))^n, {x, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Mar 29 2020 *)

a(n) = [x^n] ( (1 + x - sqrt(1 - 2*x - 3*x^2)) / (2*x) )^n.

a(n) ~ sqrt(((9386 + 1026*sqrt(57))^(1/3) + (9386 - 1026*sqrt(57))^(1/3) - 19)/228) * (((1261 + 57*sqrt(57))^(1/3) + (1261 - 57*sqrt(57))^(1/3) + 10)/6)^n / sqrt(Pi*n). - Vaclav Kotesovec, Mar 29 2020

A337991 Triangle read by rows: T(n,m) = Sum_{i=1..n} C(n,i-m)C(n+m-i,i-1)C(n+m-i,m)/n, with T(0,0)=1.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 5, 4, 1, 4, 13, 15, 7, 1, 9, 35, 52, 36, 11, 1, 21, 96, 175, 160, 75, 16, 1, 51, 267, 576, 655, 415, 141, 22, 1, 127, 750, 1869, 2541, 2030, 952, 245, 29, 1, 323, 2123, 6000, 9492, 9156, 5488, 1988, 400, 37, 1, 835, 6046, 19107, 34476, 38976, 28476, 13356, 3852, 621, 46, 1

Offset: 0

Vladimir Kruchinin, Oct 06 2020

			Triangle begins as:
   1;
   1,   1;
   1,   2,   1;
   2,   5,   4,   1;
   4,  13,  15,   7,   1;
   9,  35,  52,  36,  11,   1;
  21,  96, 175, 160,  75,  16,  1;
  51, 267, 576, 655, 415, 141, 22,  1;
  ...

G. C. Greubel, Rows n = 0..100 of the triangle, flattened

Cf. A001006, A005773, A086246, A132894.
Diagonals include: A000124, A006008.
Sums include: A000007 (signed row), A019590 (signed diagonal), A025227 (row), A102407 (diagonal).

B:=Binomial;
A337991:= func< n,k | n eq 0 select 1 else (1/n)*(&+[B(n, j-k)*B(n+k-j, j-1)*B(n+k-j, k): j in [1..n]]) >;
[A337991(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 31 2024

T[0, 0] = 1; T[n_, m_] := Sum[Binomial[n, i - m] * Binomial[n + m - i, i - 1] * Binomial[n + m - i, m]/n, {i, 1, n}]; Table[T[n, m], {n, 0, 10}, {m, 0, n}] // Flatten (* Amiram Eldar, Oct 06 2020 *)

T(n,m):=if m=n then 1 else if n=0 then 0 else sum(binomial(n,i-m)*binomial(n+m-i,i-1)*binomial(n+m-i,m),i,1,n)/n;

def A337991(n,k):
    b=binomial
    if n==0: return 1
    else: return (1/n)*sum(b(n, j-k)*b(n+k-j, j-1)*b(n+k-j, k) for j in range(1,n+1))
# SageMath
flatten([[A337991(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Oct 31 2024

G.f.: ( 1 - x*(y-1)- sqrt(x^2*(y^2-2*y-3) - 2*x*(y+1) + 1) )/(2*x).
From G. C. Greubel, Oct 31 2024: (Start)
T(n, k) = binomial(n, 1-k)*binomial(n+k-1, k)*Hypergeometric3F2([1-n, (1 -n -k)/2, (2-n-k)/2], [2-k, 1-n-k], 4), with T(0, 0) = 1.
T(n, 0) = A086246(n+1).
T(n, n-1) = A000124(n-1), n >= 1.
T(n, n-2) = A006008(n-1), n >= 2.
T(n, n-3) = (1/72)*(n^4 -6*n^3 +47*n^2 -114*n +144)*binomial(n-1,2), n >= 3.
T(n, n-4) = (1/480)*(n-2)*(n^4 -8*n^3 +99*n^2 -332*n +960)*binomial(n-1,3), n >= 4.
Sum_{k=0..n} T(n, k) = A025227(n+1).
Sum_{k=0..n} (-1)^k*T(n, k) = A000007(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = A102407(n).
Sum_{k=0..floor(n/2)} (-1)^k*T(n-k, k) = A019590(n+1). (End)

A348189 Pseudo-involutory Riordan companion of 1 + 2xM(x), where M(x) is the g.f. of A001006.

Original entry on oeis.org

1, 0, 0, 2, 0, 6, 8, 24, 60, 148, 396, 1026, 2744, 7350, 19872, 54102, 148104, 407682, 1127328, 3130542, 8726256, 24407634, 68482776, 192698124, 543642476, 1537443024, 4357677516, 12376868254, 35221087656, 100409367690, 286730523104, 820078634232, 2348966799132

Offset: 1

Alexander Burstein, Oct 06 2021

nonn

Alexander Burstein and Louis W. Shapiro, Pseudo-involutions in the Riordan group, arXiv:2112.11595 [math.CO], 2021.
Alexander Burstein and Louis W. Shapiro, Pseudo-involutions in the Riordan group and Chebyshev polynomials, arXiv:2502.13673 [math.CO], 2025.

Cf. A001006, A007971, A086246.

a[n_] := SeriesCoefficient[(1 - Sqrt[1-2*x-3*x^2])/(x * (2 + x - Sqrt[1-2*x-3*x^2])), {x, 0, n}]; Array[a, 33, 0] (* Amiram Eldar, Oct 06 2021 *)

my(x='x+O('x^35)); Vec((1-sqrt(1-2*x-3*x^2))/(x*(2+x-sqrt(1-2*x-3*x^2)))) \\ Michel Marcus, Oct 06 2021

G.f.: A(x) = (1 - sqrt(1 - 2*x - 3*x^2))/(x*(2 + x - sqrt(1 - 2*x - 3*x^2))).
If M(x) is the g.f. of A001006, then A(x) = (1 + 2*x*M(x))/(1 + 2*x + 2*x^2*M(x)).
Let M(x) be the g.f. of A001006 and F(x) = 1 + 2*x*M(x) (equivalently, x*F(x) = g.f. of A007971). Then F(-x*A(x)) = 1/F(x).
A(-x*A(x)) = 1/A(x).
G.f.: Let F(x) be the g.f. of A107264, then A(x) = 1 + 2*x^3*A(x)^2*F(x^2*A(x)). - Alexander Burstein, Feb 14 2022

cp's OEIS Frontend

A348957 G.f. A(x) satisfies A(x) = (1 + x * A(-x)) / (1 - x * A(x)).

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A247364 Riordan array (f(x), (f(x)-1)/f(x)) where f(x) = (1 + x - sqrt(1 - 2x - 3x^2))/(2*x).

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A185010 a(n) = A000108(n)*A015518(n+1), where A000108 are the Catalan numbers and A015518(n) = 2*A015518(n-1) + 3*A015518(n-2).

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A333472 a(n) = [x^n] ( c (x/(1 + x)) )^n, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. of the Catalan numbers A000108.

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A337991 Triangle read by rows: T(n,m) = Sum_{i=1..n} C(n,i-m)*C(n+m-i,i-1)*C(n+m-i,m)/n, with T(0,0)=1.

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A348189 Pseudo-involutory Riordan companion of 1 + 2*x*M(x), where M(x) is the g.f. of A001006.

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A185010 a(n) = A000108(n)A015518(n+1), where A000108 are the Catalan numbers and A015518(n) = 2A015518(n-1) + 3*A015518(n-2).

A333472 a(n) = [x^n] ( c (x/(1 + x)) )^n, where c(x) = (1 - sqrt(1 - 4x))/(2x) is the o.g.f. of the Catalan numbers A000108.

A337991 Triangle read by rows: T(n,m) = Sum_{i=1..n} C(n,i-m)C(n+m-i,i-1)C(n+m-i,m)/n, with T(0,0)=1.

A348189 Pseudo-involutory Riordan companion of 1 + 2xM(x), where M(x) is the g.f. of A001006.