A127897 -id:A127897 - cp's OEIS frontend

A307678 G.f. A(x) satisfies: A(x) = 1 + x*A(x)^3/(1 - x).

1, 1, 4, 19, 101, 578, 3479, 21714, 139269, 912354, 6078832, 41066002, 280636657, 1936569717, 13475408847, 94446518559, 666149216744, 4724705621702, 33676421377532, 241100485812034, 1732999323835918, 12501487280292424, 90478497094713958, 656788523782034248, 4780725762185300389

Offset: 0

Ilya Gutkovskiy, Apr 21 2019

nonn

Convolution square root of A270386.

			G.f.: A(x) = 1 + x + 4*x^2 + 19*x^3 + 101*x^4 + 578*x^5 + 3479*x^6 + 21714*x^7 + 139269*x^8 + 912354*x^9 + 6078832*x^10 + ...

Michael De Vlieger, Table of n, a(n) for n = 0..1000
Guillermo Esteban, Clemens Huemer, and Rodrigo I. Silveira, New production matrices for geometric graphs, arXiv:2003.00524 [math.CO], 2020.

Cf. A001764, A002212, A006013, A127897, A188687 (partial sums), A270386.

terms = 24; A[] = 1; Do[A[x] = 1 + x A[x]^3/(1 - x) + O[x]^(terms + 1) // Normal, terms + 1]; CoefficientList[A[x], x]
a[0] = 1; a[n_] := a[n] = Sum[Sum[Sum[a[k] a[i - k] a[j - i], {k, 0, i}], {i, 0, j}], {j, 0, n - 1}]; Table[a[n], {n, 0, 24}]
terms = 24; CoefficientList[Series[2 Sqrt[(1 - x) Sin[1/3 ArcSin[3/2 Sqrt[3] Sqrt[x/(1 - x)]]]^2/x]/Sqrt[3], {x, 0, terms}], x]

a(n):=sum(binomial(n-1,n-k)*(binomial(3*k,k))/(2*k+1),k,0,n); /* Vladimir Kruchinin, Feb 05 2022*/

{a(n) = my(A=[1]); for(m=1, n, A=concat(A, 0);
A[#A] = 1 + sum(k=1, m-1, (polcoeff(Ser(A)^3, k)) )); A[n+1]}
for(n=0, 30, print1(a(n), ", ")) \\ Vaclav Kotesovec, Nov 23 2024, after Paul D. Hanna

a(0) = 1; a(n) = Sum_{j=0..n-1} Sum_{i=0..j} Sum_{k=0..i} a(k)*a(i-k)*a(j-i).
a(n) ~ 31^(n + 1/2) / (3*sqrt(Pi) * n^(3/2) * 2^(2*n+2)). - Vaclav Kotesovec, May 06 2019
G.f.: (2/sqrt(3*x/(1-x)))*sin((1/3)*asin(sqrt((27*x/(1-x))/4))). - Vladimir Kruchinin, Feb 05 2022
a(n) = Sum_{k=0..n} C(n-1,n-k)*C(3*k,k)/(2*k+1). - Vladimir Kruchinin, Feb 05 2022

A317133 G.f.: Sum_{n>=0} binomial(4(n+1), n)/(n+1) x^n / (1+x)^(n+1).

Original entry on oeis.org

1, 3, 15, 85, 526, 3438, 23358, 163306, 1167235, 8490513, 62648451, 467769217, 3527692298, 26832220834, 205601792340, 1585604105312, 12297768490441, 95861469636203, 750611119223931, 5901214027721577, 46564408929573723, 368644188180241449, 2927350250765841801, 23310167641788680947, 186089697960587977233, 1489085453187335910243

Offset: 0

Paul D. Hanna, Jul 21 2018

nonn

Note that: binomial(4*(n+1), n)/(n+1) = A002293(n+1) for n >= 0, where F(x) = Sum_{n>=0} A002293(n)*x^n satisfies F(x) = 1 + x*F(x)^4.
Compare the g.f. to:
(C1) M(x) = Sum_{n>=0} binomial(2*(n+1), n)/(n+1) * x^n / (1+x)^(n+1) where M(x) = 1 + M(x) + M(x)^2 is the g.f. of Motzkin numbers (A001006).
(C2) 1 = Sum_{n>=0} binomial(m*(n+1), n)/(n+1) * x^n / (1+x)^(m*(n+1)) holds for fixed m.
(C3) If S(x,p,q) = Sum_{n>=0} binomial(p*(n+1),n)/(n+1) * x^n/(1+x)^(q*(n+1)), then Series_Reversion ( x*S(x,p,q) ) = x*S(x,q,p) holds for fixed p and q.

			G.f.: A(x) = 1 + 3*x + 15*x^2 + 85*x^3 + 526*x^4 + 3438*x^5 + 23358*x^6 + 163306*x^7 + 1167235*x^8 + 8490513*x^9 + 62648451*x^10 + ...
such that
A(x) = 1/(1+x) + 4*x/(1+x)^2 + 22*x^2/(1+x)^3 + 140*x^3/(1+x)^4 + 969*x^4/(1+x)^5 + 7084*x^5/(1+x)^6 + ... + A002293(n+1)*x^n/(1+x)^(n+1) + ...
RELATED SERIES.
Series_Reversion( x*A(x) )  =  x/((1+x)^4 - x)  =  x - 3*x^2 + 3*x^3 + 5*x^4 - 22*x^5 + 27*x^6 + 28*x^7 - 163*x^8 + 235*x^9 + 134*x^10 + ...
which equals the sum:
Sum_{n>=0} binomial(n+1, n)/(n+1) * x^(n+1)/(1+x)^(4*(n+1)).

Cf. A316371, A127897, A317134, A349361, A349362, A349363, A349364.

Rest[CoefficientList[InverseSeries[Series[x/((1 + x)^4 - x), {x, 0, 20}], x], x]] (* Vaclav Kotesovec, Jul 22 2018 *)

{a(n) = my(A = sum(m=0, n, binomial(4*(m+1), m)/(m+1) * x^m / (1+x +x*O(x^n))^(1*(m+1)))); polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))

{a(n) = my(A = (1/x) * serreverse( x/((1+x)^4 - x +x*O(x^n)) ) ); polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))

G.f. A(x) satisfies:
(1) A(x) = (1 + x*A(x))^4 / (1+x).
(2) A(x) = (1/x) * Series_Reversion( x/((1+x)^4 - x) ).
(3) A(x) = Sum_{n>=0} binomial(4*(n+1), n)/(n+1) * x^n / (1+x)^(n+1).
a(n) ~ 229^(n + 3/2) / (sqrt(Pi) * 2^(7/2) * n^(3/2) * 3^(3*n + 9/2)). - Vaclav Kotesovec, Jul 22 2018
a(n) = (1/(n+1)) * Sum_{k=0..n} (-1)^k * binomial(n+1,k) * binomial(4*n-4*k+4,n-k). - Seiichi Manyama, Mar 23 2024

A349362 G.f. A(x) satisfies: A(x) = 1 + x * A(x)^6 / (1 + x).

Original entry on oeis.org

1, 1, 5, 40, 370, 3740, 40006, 445231, 5102165, 59799505, 713496815, 8637432580, 105826926716, 1309793896431, 16351672606365, 205665994855320, 2603696877136060, 33151784577226295, 424258396639960591, 5454120586840761631, 70402732493668027775

Offset: 0

Ilya Gutkovskiy, Nov 15 2021

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..500

Cf. A001006, A002295, A127897, A317133, A346065 (binomial transform), A346666, A349333, A349361, A349363, A349364.

a:= n-> coeff(series(RootOf(1+x*A^6/(1+x)-A, A), x, n+1), x, n):
seq(a(n), n=0..20);  # Alois P. Heinz, Nov 15 2021

nmax = 20; A[] = 0; Do[A[x] = 1 + x A[x]^6/(1 + x) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
Table[Sum[(-1)^(n - k) Binomial[n - 1, k - 1] Binomial[6 k, k]/(5 k + 1), {k, 0, n}], {n, 0, 20}]

a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n-1,k-1) * binomial(6*k,k) / (5*k+1).
a(n) = (-1)^(n+1)* F([7/6, 4/3, 3/2, 5/3, 11/6, 1-n], [7/5, 8/5, 9/5, 2, 11/5], 6^6/5^5), where F is the generalized hypergeometric function. - Stefano Spezia, Nov 15 2021
a(n) ~ 43531^(n + 1/2) / (72 * sqrt(3*Pi) * n^(3/2) * 5^(5*n + 3/2)). - Vaclav Kotesovec, Nov 17 2021

A349361 G.f. A(x) satisfies: A(x) = 1 + x * A(x)^5 / (1 + x).

Original entry on oeis.org

1, 1, 4, 26, 194, 1581, 13625, 122120, 1126780, 10631460, 102104845, 994855179, 9809872626, 97710157154, 981636609906, 9935473707279, 101214412755647, 1036991125300748, 10678412226507032, 110459290208905008, 1147261657267290037

Offset: 0

Ilya Gutkovskiy, Nov 15 2021

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..500

Cf. A001006, A002294, A127897, A317133, A345368 (binomial transform), A346665, A349332, A349362, A349363, A349364.

a:= n-> coeff(series(RootOf(1+x*A^5/(1+x)-A, A), x, n+1), x, n):
seq(a(n), n=0..20);  # Alois P. Heinz, Nov 15 2021

nmax = 20; A[] = 0; Do[A[x] = 1 + x A[x]^5/(1 + x) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
Table[Sum[(-1)^(n - k) Binomial[n - 1, k - 1] Binomial[5 k, k]/(4 k + 1), {k, 0, n}], {n, 0, 20}]

a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n-1,k-1) * binomial(5*k,k) / (4*k+1).
a(n) = (-1)^(n+1)*F([6/5, 7/5, 8/5, 9/5, 1-n], [3/2, 7/4, 2, 9/4], 5^5/2^8), where F is the generalized hypergeometric function. - Stefano Spezia, Nov 15 2021
From Vaclav Kotesovec, Nov 17 2021: (Start)
a(n) ~ 2869^(n + 1/2) / (25 * sqrt(Pi) * n^(3/2) * 2^(8*n + 7/2)).
Recurrence: 8*n*(2*n - 1)*(4*n - 1)*(4*n + 1)*a(n) = 3*(615*n^4 - 718*n^3 - 275*n^2 + 618*n - 200)*a(n-1) + 4*(n-2)*(2485*n^3 - 6879*n^2 + 6524*n - 2040)*a(n-2) + 2*(n-3)*(n-2)*(8095*n^2 - 23517*n + 18092)*a(n-3) + 12*(n-4)*(n-3)*(n-2)*(935*n - 1838)*a(n-4) + 2869*(n-5)*(n-4)*(n-3)*(n-2)*a(n-5). (End)

A127896 Expansion of 1/(1 + 2x + 3x^2 + x^3).

Original entry on oeis.org

1, -2, 1, 3, -7, 4, 10, -25, 16, 33, -89, 63, 108, -316, 245, 350, -1119, 943, 1121, -3952, 3598, 3539, -13920, 13625, 10971, -48897, 51256, 33208, -171287, 191694, 97265, -598325, 713161, 271388, -2083934

Offset: 0

Paul Barry, Feb 04 2007

Row sums of A127895. Series reversion is A127897.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Paul Barry, Centered polygon numbers, heptagons and nonagons, and the Robbins numbers, arXiv:2104.01644 [math.CO], 2021.
Index entries for linear recurrences with constant coefficients, signature (-2,-3,-1).

Cf. A127895, A127897.

I:=[1, -2, 1]; [n le 3 select I[n] else -2*Self(n-1) -3*Self(n-2) -Self(n-3): n in [1..50]]; // G. C. Greubel, Apr 29 2018

CoefficientList[Series[1/(1+2x+3x^2+x^3),{x,0,40}],x]  (* Harvey P. Dale, Apr 19 2011 *)
LinearRecurrence[{-2, -3, -1}, {1, -2, 1}, 30] (* G. C. Greubel, Apr 29 2018 *)

x='x+O('x^50); Vec(1/(1+2*x+3*x^2+x^3)) \\ G. C. Greubel, Apr 29 2018

a(n) = Sum_{k=0..n} (-1)^(n-k)*C(n+2k+2,n-k).

a(n) = -2*a(n-1) -3*a(n-2) -a(n-3), n>=3. - Vincenzo Librandi, Mar 22 2011

A349364 G.f. A(x) satisfies: A(x) = 1 + x * A(x)^8 / (1 + x).

Original entry on oeis.org

1, 1, 7, 77, 987, 13839, 205513, 3176747, 50578445, 823779286, 13660621282, 229865812134, 3915003083306, 67361559577578, 1169138502393414, 20444573270374050, 359858503314494318, 6370677542063831319, 113359050598950194801, 2026309136822686950087

Offset: 0

Ilya Gutkovskiy, Nov 15 2021

nonn

In general, for m > 1, Sum_{k=0..n} (-1)^(n-k) * binomial(n-1,k-1) * binomial(m*k,k) / ((m-1)*k+1) ~ (m-1)^(m/2 - 2) * (m^m/(m-1)^(m-1) - 1)^(n + 1/2) / (sqrt(2*Pi) * m^((m-1)/2) * n^(3/2)). - Vaclav Kotesovec, Nov 17 2021

Seiichi Manyama, Table of n, a(n) for n = 0..500

Cf. A001006, A007556, A127897, A317133, A346668, A346672 (binomial transform), A349335, A349361, A349362, A349363.

a:= n-> coeff(series(RootOf(1+x*A^8/(1+x)-A, A), x, n+1), x, n):
seq(a(n), n=0..19);  # Alois P. Heinz, Nov 15 2021

nmax = 19; A[] = 0; Do[A[x] = 1 + x A[x]^8/(1 + x) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
Table[Sum[(-1)^(n - k) Binomial[n - 1, k - 1] Binomial[8 k, k]/(7 k + 1), {k, 0, n}], {n, 0, 19}]

a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n-1,k-1) * binomial(8*k,k) / (7*k+1).
a(n) = (-1)^(n+1)*F([9/8, 5/4, 11/8, 3/2, 13/8, 7/4, 15/8, 1-n], [9/7, 10/7, 11/7, 12/7, 13/7, 2, 15/7], 8^8/7^7), where F is the generalized hypergeometric function. - Stefano Spezia, Nov 15 2021
a(n) ~ 15953673^(n + 1/2) / (2048 * sqrt(Pi) * n^(3/2) * 7^(7*n + 3/2)). - Vaclav Kotesovec, Nov 17 2021

A349363 G.f. A(x) satisfies: A(x) = 1 + x * A(x)^7 / (1 + x).

Original entry on oeis.org

1, 1, 6, 57, 629, 7589, 96942, 1288729, 17643920, 247089010, 3522891561, 50964747400, 746241617226, 11038241689188, 164696773030055, 2475832560808858, 37462189433509758, 570112127356828846, 8720472842436039280, 133997057207982607092, 2067402314984991892461

Offset: 0

Ilya Gutkovskiy, Nov 15 2021

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..500

Cf. A001006, A002296, A127897, A317133, A346667, A346671 (binomial transform), A349334, A349361, A349362, A349364.

a:= n-> coeff(series(RootOf(1+x*A^7/(1+x)-A, A), x, n+1), x, n):
seq(a(n), n=0..20);  # Alois P. Heinz, Nov 15 2021

nmax = 20; A[] = 0; Do[A[x] = 1 + x A[x]^7/(1 + x) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
Table[Sum[(-1)^(n - k) Binomial[n - 1, k - 1] Binomial[7 k, k]/(6 k + 1), {k, 0, n}], {n, 0, 20}]

a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n-1,k-1) * binomial(7*k,k) / (6*k+1).
a(n) = (-1)^(n+1)* F([8/7, 9/7, 10/7, 11/7, 12/7, 13/7, 1-n], [4/3, 3/2, 5/3, 11/6, 2, 13/6], 7^7/6^6), where F is the generalized hypergeometric function. - Stefano Spezia, Nov 15 2021
a(n) ~ 776887^(n + 1/2) / (343 * sqrt(Pi) * n^(3/2) * 2^(6*n + 2) * 3^(6*n + 3/2)). - Vaclav Kotesovec, Nov 17 2021

A371542 G.f. A(x) satisfies A(x) = (1 + x*A(x) / (1+x))^3.

Original entry on oeis.org

1, 3, 9, 34, 141, 621, 2849, 13467, 65127, 320686, 1602294, 8103087, 41397186, 213331026, 1107604764, 5788249329, 30422897664, 160717169622, 852894534042, 4544635033164, 24305345593290, 130423538829518, 701994030831654, 3788979493701069, 20503322609731348

Offset: 0

Seiichi Manyama, Mar 26 2024

nonn

Cf. A127897, A370720, A371494.
Cf. A371495, A371538.
Cf. A371516.

a(n) = sum(k=0, n, (-1)^(n-k)*binomial(n-1, n-k)*binomial(3*k+3, k)/(k+1));

a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n-1,n-k) * binomial(3*k+3,k)/(k+1).
From Seiichi Manyama, Dec 10 2024: (Start)
G.f. A(x) satisfies A(x) = 1/( 1 - x*A(x)^(2/3)/(1 + x) )^3.
G.f. A(x) satisfies A(x) = 1 + x * (1 + A(x)^(4/3) + A(x)^(5/3)).
G.f.: A(x) = (1 + B(x))^3 where B(x) is the g.f. of A127897. (End)

A317134 G.f.: Sum_{n>=0} binomial(4(n+1), n)/(n+1) x^n / (1+x)^(2*(n+1)).

Original entry on oeis.org

1, 2, 9, 44, 236, 1336, 7862, 47608, 294720, 1856748, 11865684, 76731572, 501176237, 3301501694, 21909634763, 146337236580, 982962605577, 6635968279354, 45001173711683, 306406562117884, 2093909763907401, 14356806252396614, 98735015302171955, 680906548260420320, 4707709357806093085, 32625093782844333722, 226588405850230665429, 1576882804780751603092

Offset: 0

Paul D. Hanna, Jul 22 2018

nonn

Note that: binomial(4*(n+1), n)/(n+1) = A002293(n+1) for n >= 0, where F(x) = Sum_{n>=0} A002293(n)*x^n satisfies F(x) = 1 + x*F(x)^4.
Compare the g.f. to:
(C1) 1 = Sum_{n>=0} binomial(m*(n+1), n)/(n+1) * x^n / (1+x)^(m*(n+1)) holds for fixed m.
(C2) If S(x,p,q) = Sum_{n>=0} binomial(p*(n+1),n)/(n+1) * x^n/(1+x)^(q*(n+1)), then Series_Reversion ( x*S(x,p,q) ) = x*S(x,q,p) holds for fixed p and q.

			G.f.: A(x) = 1 + 2*x + 9*x^2 + 44*x^3 + 236*x^4 + 1336*x^5 + 7862*x^6 + 47608*x^7 + 294720*x^8 + 1856748*x^9 + 11865684*x^10 + ...
such that
A(x) = 1/(1+x)^2 + 4*x/(1+x)^4 + 22*x^2/(1+x)^6 + 140*x^3/(1+x)^8 + 969*x^4/(1+x)^10 + 7084*x^5/(1+x)^12 + ... + A002293(n+1)*x^n/(1+x)^(2*(n+1)) + ...
RELATED SERIES.
Series_Reversion( x*A(x) )  =  4*x/((1+x)^2 + sqrt( (1+x)^4 - 4*x ))^2  =  x - 2*x^2 - x^3 + 6*x^4 + 3*x^5 - 20*x^6 - 18*x^7 + 74*x^8 + 111*x^9 - 278*x^10 - 657*x^11 + 980*x^12 + 3739*x^13 + ...
which equals the sum:
Sum_{n>=0} binomial(2*(n+1), n)/(n+1) * x^(n+1)/(1+x)^(4*(n+1)).
The square-root of the g.f. is an integer series:
sqrt(A(x)) = 1 + x + 4*x^2 + 18*x^3 + 92*x^4 + 504*x^5 + 2897*x^6 + 17235*x^7 + 105233*x^8 + 655687*x^9 + 4152461*x^10 + ... + A317135(n)*x^n + ...
which equals the sum:
Sum_{n>=0} binomial(4*n+2, n)/(2*n+1) * x^(n+1)/(1+x)^(2*n+1).

Cf. A316371, A127897, A316987, A317133.

Rest[CoefficientList[InverseSeries[Series[4*x/((1 + x)^2 + Sqrt[(1 + x)^4 - 4*x])^2, {x, 0, 30}], x], x]](* Vaclav Kotesovec, Jul 22 2018 *)

{a(n) = my(A = sum(m=0, n, binomial(4*(m+1), m)/(m+1) * x^m / (1+x +x*O(x^n))^(2*(m+1)))); polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))

{a(n) = my(A = (1/x) * serreverse( 4*x/((1+x)^2 + sqrt( (1+x)^4 - 4*x  + x*O(x^n)))^2 )); polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))

G.f. A(x) satisfies:
(1) A(x) = ( (1 + x*A(x))^2 + sqrt( (1 + x*A(x))^4 - 4*x*A(x) ) )^2 / 4.
(2) A(x) = (1/x) * Series_Reversion( 4*x/((1+x)^2 + sqrt( (1+x)^4 - 4*x ))^2 ).
(3) A(x) = Sum_{n>=0} binomial(4*(n+1), n)/(n+1) * x^n / (1+x)^(2*(n+1)).
a(n) ~ 37^(1/4) * (101 + 16*sqrt(37))^(n+1) / (2*sqrt(Pi) * n^(3/2) * 3^(3*n + 9/2)). - Vaclav Kotesovec, Jul 22 2018

A383118 a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n,k) * binomial(3*k,k).

Original entry on oeis.org

1, 2, 10, 47, 238, 1232, 6499, 34715, 187198, 1016840, 5555560, 30497150, 168073195, 929348396, 5153362231, 28646281502, 159579236014, 890644144580, 4979200476088, 27878225498030, 156298588113088, 877350590047496, 4930273302851830, 27733610884176338

Offset: 0

Ilya Gutkovskiy, Apr 17 2025

nonn

Inverse binomial transform of A005809.

Cf. A002426, A005809, A188686, A346628, A383119.

Cf. A127897.

Table[Sum[(-1)^(n - k) Binomial[n, k] Binomial[3 k, k], {k, 0, n}], {n, 0, 23}]
Table[(-1)^n HypergeometricPFQ[{1/3, 2/3, -n}, {1/2, 1}, 27/4], {n, 0, 23}]
nmax = 23; CoefficientList[Series[(1/x) Sum[Binomial[3 k, k] (x/(1 + x))^(k + 1), {k, 0, nmax}], {x, 0, nmax}], x]

a(n) = sum(k=0, n, (-1)^(n-k)*binomial(n, k)*binomial(3*k, k)); \\ Seiichi Manyama, Apr 17 2025

G.f.: (1/x) * Sum_{k>=0} binomial(3*k,k) * (x/(1 + x))^(k+1).
a(n) = [x^n] (1 + 2*x + 3*x^2 + x^3)^n.
The g.f. x * exp( Sum_{k>=1} a(k) * x^k/k ) has integer coefficients and equals Series_Reversion( x/((1+x)^3 - x) ). See A127897. - Seiichi Manyama, Apr 17 2025
a(n) ~ 23^(n + 1/2) / (3 * sqrt(Pi*n) * 2^(2*n+1)). - Vaclav Kotesovec, Apr 17 2025

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A307678 G.f. A(x) satisfies: A(x) = 1 + x*A(x)^3/(1 - x).

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A317133 G.f.: Sum_{n>=0} binomial(4*(n+1), n)/(n+1) * x^n / (1+x)^(n+1).

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A349362 G.f. A(x) satisfies: A(x) = 1 + x * A(x)^6 / (1 + x).

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A349361 G.f. A(x) satisfies: A(x) = 1 + x * A(x)^5 / (1 + x).

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A127896 Expansion of 1/(1 + 2*x + 3*x^2 + x^3).

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A349364 G.f. A(x) satisfies: A(x) = 1 + x * A(x)^8 / (1 + x).

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A349363 G.f. A(x) satisfies: A(x) = 1 + x * A(x)^7 / (1 + x).

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A371542 G.f. A(x) satisfies A(x) = (1 + x*A(x) / (1+x))^3.

A317133 G.f.: Sum_{n>=0} binomial(4(n+1), n)/(n+1) x^n / (1+x)^(n+1).

A127896 Expansion of 1/(1 + 2x + 3x^2 + x^3).

A317134 G.f.: Sum_{n>=0} binomial(4(n+1), n)/(n+1) x^n / (1+x)^(2*(n+1)).