A349363 -id:A349363 - cp's OEIS frontend

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

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)

A349334 G.f. A(x) satisfies A(x) = 1 + x * A(x)^7 / (1 - x).

Original entry on oeis.org

1, 1, 8, 85, 1051, 14197, 203064, 3022909, 46347534, 726894786, 11606936525, 188060979332, 3084087347910, 51094209834068, 853859480938095, 14376597494941454, 243649099741045190, 4153091242153905838, 71152973167920086796, 1224593757045581062444

Offset: 0

Ilya Gutkovskiy, Nov 15 2021

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..791

Cf. A002212, A307678, A349331, A349332, A349333, A349335.

Cf. A002296, A346649 (partial sums), A349363.

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 = 19; 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[Binomial[n - 1, k - 1] Binomial[7 k, k]/(6 k + 1), {k, 0, n}], {n, 0, 19}]

{a(n) = my(A=[1]); for(m=1, n, A=concat(A, 0);
A[#A] = 1 + sum(k=1, m-1, (polcoeff(Ser(A)^7, k)) )); A[n+1]}
for(n=0, 30, print1(a(n), ", ")) \\ Vaclav Kotesovec, Nov 23 2024, after Paul D. Hanna

a(n) = Sum_{k=0..n} binomial(n-1,k-1) * binomial(7*k,k) / (6*k+1).

a(n) ~ 870199^(n + 1/2) / (343 * sqrt(Pi) * n^(3/2) * 2^(6*n + 2) * 3^(6*n + 3/2)). - Vaclav Kotesovec, Nov 15 2021

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

cp's OEIS Frontend

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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A349334 G.f. A(x) satisfies A(x) = 1 + x * A(x)^7 / (1 - x).

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