A228960 -id:A228960 - cp's OEIS frontend

A198951 G.f. satisfies: A(x) = (1 + xA(x))(1 + x^3*A(x)^3).

1, 1, 1, 2, 6, 16, 39, 99, 271, 763, 2146, 6062, 17359, 50337, 147057, 431874, 1275273, 3786649, 11298031, 33846202, 101762937, 306997821, 929038518, 2819426688, 8578433304, 26163061776, 79970186791, 244938841096, 751646959402, 2310683396056, 7115199919151

Offset: 0

Paul D. Hanna, Oct 31 2011

a(n) is also the number of rooted labeled trees on n nodes such that each node has 0, 1, 3, or 4 children. - Patrick Devlin, Mar 04 2012

			G.f.: A(x) = 1 + x + x^2 + 2*x^3 + 6*x^4 + 16*x^5 + 39*x^6 + 99*x^7 + ...
Related expansions:
A(x)^3 = 1 + 3*x + 6*x^2 + 13*x^3 + 36*x^4 + 105*x^5 + 292*x^6 + ...
A(x)^4 = 1 + 4*x + 10*x^2 + 24*x^3 + 67*x^4 + 200*x^5 + 582*x^6 + ...
The logarithm of the g.f. equals the series:
log(A(x)) = (1 + x^2*A(x)^2)*x + (1 + 2^2*x^2*A(x)^2 + x^4*A(x)^4)*x^2/2 +
(1 + 3^2*x^2*A(x)^2 + 3^2*x^4*A(x)^4 + x^6*A(x)^6)*x^3/3 +
(1 + 4^2*x^2*A(x)^2 + 6^2*x^4*A(x)^4 + 4^2*x^6*A(x)^6 + x^8*A(x)^8)*x^4/4 +
(1 + 5^2*x^2*A(x)^2 + 10^2*x^4*A(x)^4 + 10^2*x^6*A(x)^6 + 5^2*x^8*A(x)^8 + x^10*A(x)^10)*x^5/5 + ...
more explicitly,
log(A(x)) = x + x^2/2 + 4*x^3/3 + 17*x^4/4 + 51*x^5/5 + 136*x^6/6 + 393*x^7/7 + 1233*x^8/8 + ...

Alois P. Heinz, Table of n, a(n) for n = 0..600

Cf. A198953, A007863, A036765, A186241, A200731, A228960.

a:= n-> coeff(series(RootOf(A=(1+x*A)*(1+x^3*A^3), A), x, n+1), x, n):
seq(a(n), n=0..30);  # Alois P. Heinz, May 16 2012

InverseSeries[ Series[ x/((1 + x)*(1 + x^3)), {x, 0, 31}], x] // CoefficientList[#, x]& // Rest (* Jean-François Alcover, Sep 10 2013 *)

{a(n)=local(A=1/x*serreverse(x/(1+x+x^3+x^4+x*O(x^n)))); polcoeff(A, n)}

{a(n)=polcoeff((1+x+x^3+x^4+x*O(x^n))^(n+1)/(n+1), n)}

{a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n,sum(j=0, m, binomial(m, j)^2*x^(2*j)*(A+x*O(x^n))^(2*j))*x^m/m))); polcoeff(A, n)}

{a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, (1-x^2*A^2)^(2*m+1)*sum(j=0, n\2, binomial(m+j, j)^2*x^(2*j)*(A^2+x*O(x^n))^j)*x^m/m))); polcoeff(A, n, x)}

G.f. satisfies:
(1) A(x) = exp( Sum_{n>=1} x^n/n * Sum_{k=0..n} C(n,k)^2 * x^(2*k) * A(x)^(2*k) ).
(2) A(x) = (1/x)*Series_Reversion(x/((1+x)*(1+x^3))).
(3) a(n) = [x^n] (1 + x + x^3 + x^4)^(n+1) / (n+1).
(4) A(x) = exp( Sum_{n>=1} x^n/n * (1-x^2*A(x)^2)^(2*n+1)*Sum_{k>=0} C(n+k,k)^2 * x^(2*k) * A(x)^(2*k) ).
D-finite with recurrence: 3*(n+1)*(3*n+2)*(3*n+4)*(119*n^3 - 210*n^2 + 73*n - 6)*a(n) = 2*(6664*n^6 - 1764*n^5 - 11585*n^4 + 426*n^3 + 4129*n^2 - 102*n - 288)*a(n-1) - 18*(n-1)*(1190*n^5 - 910*n^4 - 1937*n^3 + 895*n^2 + 606*n - 216)*a(n-2) + 162*(n-2)*(n-1)*(2*n-3)*(119*n^3 + 147*n^2 + 10*n - 24)*a(n-3). - Vaclav Kotesovec, Sep 09 2013
a(n) ~ c*d^n/n^(3/2), where d = 1/81*((2144134+520506*sqrt(17))^(2/3)+112*(2144134+520506*sqrt(17))^(1/3)-2036)/(2144134+520506*sqrt(17))^(1/3) = 3.23407602060970245... is the root of the equation -324 + 180*d - 112*d^2 + 27*d^3 = 0 and c = 0.6286981954423757284622435... - Vaclav Kotesovec, Sep 09 2013
A(1/d) = 370/243 + (3*sqrt(17)/509 - 3070/123687)*(2144134+520506*sqrt(17))^(1/3) + (141*sqrt(17)/2072648 - 129529/503653464)*(2144134+520506*sqrt(17))^(2/3) = 2.053716618436594614948796... - Vaclav Kotesovec, Sep 10 2013
From Peter Bala, Jun 21 2015: (Start)
a(n) = 1/(n + 1)*Sum_{k = 0..floor(n/3)} binomial(n + 1,k)* binomial(n + 1,n - 3*k). Applying Maple's sumrecursion command to this formula gives the above recurrence of Kotesovec.
More generally, the coefficient of x^n in A(x)^r equals r/(n + r)*Sum_{k = 0..floor(n/3)} binomial(n + r,k)*binomial(n + r,n - 3*k) by the Lagrange-Bürmann formula.
O.g.f. A(x) = exp(Sum_{n >= 1} A228960(n)*x^n/n), where A228960(n) = Sum_{k = 0..floor(n/3)} binomial(n,k)*binomial(n,3*k). Cf. A036765, A186241 and A200731. (End)

A350383 a(n) = [x^n] 1/(1 + x + x^2)^n.

Original entry on oeis.org

1, -1, 1, 2, -15, 49, -98, 48, 561, -2860, 8151, -12948, -9282, 149226, -594320, 1428952, -1448655, -5538975, 37450900, -122995950, 239589735, -37528755, -1886983020, 8939152560, -24579514050, 35197176924, 51580335366, -541312482256, 2033695030128, -4624358661240

Offset: 0

Seiichi Manyama, Dec 29 2021

Seiichi Manyama, Table of n, a(n) for n = 0..1000

Cf. A002426, A027907, A103779, A110556, A165817, A213684, A228960, A350406, A350407, A362087.

a := n -> (-1)^n*hypergeom([-n/3, 1/3 - n/3, 2/3 - n/3, n], [1/3, 2/3, 1], 1): seq(simplify(a(n)), n = 0..30); # Peter Bala, Apr 17 2023

a[n_] := Coefficient[Series[1/(1 + x + x^2)^n, {x, 0, n}], x, n]; Array[a, 30, 0] (* Amiram Eldar, Dec 29 2021 *)

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

a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n-1+k,k) * binomial(n,3*k).
Recurrence: 3*(n-1)*n*(4*n - 7)*a(n) = -2*(n-1)*(28*n^2 - 63*n + 27)*a(n-1) - 3*(3*n - 5)*(3*n - 4)*(4*n - 3)*a(n-2). - Vaclav Kotesovec, Mar 18 2023
From Peter Bala, Apr 15 2023: (Start)
a(n) = (-1)^n*hypergeom([-n/3, 1/3 - n/3, 2/3 - n/3, n], [1/3, 2/3, 1], 1).
Conjecture: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(2*r)) holds for positive integers n and r and all primes p >= 5. Cf. A228960.
More generally, let k be a positive integer, m an integer and let f(x) = g(x)/h(x), where g(x) and h(x) are both finite products of cyclotomic polynomials. Then we conjecture that the same supercongruences hold, except for a finite number of primes p depending on f(x), for the sequence {a_(k,m,f)(n): n >= 0} defined by a_(k,m,f)(n) = [x^(k*n)] f(x)^(m*n). (End)
From Peter Bala, Mar 11 2025: (Start)
G.f.: A(x) = 1 + x*d/dx(log(G(x)/x)), where G(x) = x - x^2 + x^3 - 4*x^5 + 14*x^6 - 30*x^7 + ... is the g.f. of A103779.
The following formulas hold for n >= 1:
a(n) = [x^n] T(2*n, (1 - x)/2), where T(n, x) denotes the n-th Chebyshev polynomial of the first kind.
a(n) = Sum_{k = 0..n} (-1)^(n+k) * n/(2*n-k) * binomial(2*n-k, k)*binomial(2*n-2*k, n).
a(n) = (1/2)*(-1)^n*binomial(2*n, n)*hypergeom([-n/2, (-n+1)/2], [-2*n+1], 4). Cf. A213684. (End)

A350407 a(n) = [x^n] 1/(1 + x + x^2 + x^3 + x^4)^n.

Original entry on oeis.org

1, -1, 1, -1, 1, 4, -35, 146, -447, 1133, -2464, 4355, -4355, -9296, 77078, -314636, 1006145, -2738565, 6398155, -12175809, 14621376, 14828330, -179236815, 786257460, -2625667395, 7412386254, -17983703880, 36057065030, -49553122730, -14585596720

Offset: 0

Seiichi Manyama, Dec 29 2021

Seiichi Manyama, Table of n, a(n) for n = 0..1000

Cf. A187925, A228960, A350383, A350406.

a := n -> (-1)^n*hypergeom([-n/5, 1/5 - n/5, 2/5 - n/5, 3/5 - n/5, 4/5 - n/5, n], [1/5, 2/5, 3/5, 4/5, 1], 1): seq(simplify(a(n)), n = 0..30); # Peter Bala, Apr 16 2023

a[n_] := Coefficient[Series[1/(1 + x + x^2 + x^3 + x^4)^n, {x, 0, n}], x, n]; Array[a, 30, 0] (* Amiram Eldar, Dec 29 2021 *)

a(n) = sum(k=0, n, (-1)^(n-k)*binomial(n-1+k, k)*binomial(n, 5*k));

a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n-1+k,k) * binomial(n,5*k).
Recurrence: 125*(n-3)*(n-2)*(n-1)*n*(2*n - 7)*(3*n - 11)*(3*n - 8)*(6*n - 23)*(6*n - 17)*(6*n - 11)*a(n) = -50*(n-3)*(n-2)*(n-1)*(3*n - 11)*(6*n - 23)*(6*n - 17)*(576*n^4 - 4896*n^3 + 14402*n^2 - 16875*n + 6250)*a(n-1) - 30*(n-3)*(n-2)*(6*n - 23)*(6*n - 5)*(7884*n^6 - 123516*n^5 + 791601*n^4 - 2652565*n^3 + 4894096*n^2 - 4707500*n + 1842500)*a(n-2) - 2*(n-3)*(3*n - 5)*(6*n - 11)*(6*n - 5)*(60192*n^6 - 1113552*n^5 + 8528546*n^4 - 34608379*n^3 + 78470893*n^2 - 94255700*n + 46855500)*a(n-3) - 5*(2*n - 5)*(3*n - 8)*(3*n - 5)*(5*n - 19)*(5*n - 18)*(5*n - 17)*(5*n - 16)*(6*n - 17)*(6*n - 11)*(6*n - 5)*a(n-4). - Vaclav Kotesovec, Mar 18 2023
From Peter Bala, Apr 16 2023: (Start)
a(n) = (-1)^n*hypergeom([-n/5, 1/5 - n/5, 2/5 - n/5, 3/5 - n/5, 4/5 - n/5, n], [1/5, 2/5, 3/5, 4/5, 1], 1).
Conjecture: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(2*r)) holds for positive integers n and r and all primes p >= 3. (End)

A362408 a(n) = [x^n] (F(x)/F(-x))^n where F(x) = (1 + x)*(1 + x^3).

Original entry on oeis.org

1, 2, 8, 44, 256, 1502, 8912, 53510, 324352, 1980332, 12160008, 75015162, 464566144, 2886488906, 17985045464, 112333392044, 703119387648, 4409231140086, 27696141476336, 174229516043630, 1097501783152256, 6921721148337452, 43701895245221848

Offset: 0

Peter Bala, Apr 18 2023

Compare with A228960(n) = [x^n] F(x)^n.
Let k and m be positive integers and let f(x) be a finite product of cyclotomic polynomials. Define b(n) = [x^(k*n)] (f(x)/f(-x))^(m*n). Then we conjecture that the supercongruences a(p) == a(1) (mod p^3) and, for n >= 2, a(n*p) == a(n) (mod p^2) hold for all primes p, with a finite number of exceptions depending on f(x).
The present sequence is the case k = m = 1 and f(x) = (1 + x)*(1 + x^3) = C(2,x)^2 * C(6,x), where C(n,x) denotes the n-th cyclotomic polynomial. See A002003 for the case k = m = 1 and f(x) = (1 + x).

Wikipedia, Cyclotomic polynomial

Cf. A002003, A228960, A333579.

F(x) := (1 + x)*(1 + x^3): G(x) := taylor(F(x)/F(-x),x = 0, 50); seq(coeftayl(G(x)^n, x = 0, n), n = 0..50);

Conjectures: 1) the supercongruence a(p) == 2 (mod p^3) holds for all primes p >= 5 (checked up to p = 47).

2) for n >= 2, the supercongruence a(n*p) == a(n) (mod p^2) holds for all primes p >= 5.

cp's OEIS Frontend

A198951 G.f. satisfies: A(x) = (1 + x*A(x))*(1 + x^3*A(x)^3).

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A350383 a(n) = [x^n] 1/(1 + x + x^2)^n.

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A350407 a(n) = [x^n] 1/(1 + x + x^2 + x^3 + x^4)^n.

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A362408 a(n) = [x^n] (F(x)/F(-x))^n where F(x) = (1 + x)*(1 + x^3).

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A378407 a(n) = Sum_{k=0..floor(n/3)} binomial(n,k) * binomial(n+2*k,n-3*k).

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A198951 G.f. satisfies: A(x) = (1 + xA(x))(1 + x^3*A(x)^3).

A378407 a(n) = Sum_{k=0..floor(n/3)} binomial(n,k) * binomial(n+2k,n-3k).