A350383 -id:A350383 - cp's OEIS frontend

A213684 Logarithmic derivative of A001002.

1, 5, 22, 105, 511, 2534, 12720, 64449, 328900, 1688115, 8705060, 45064110, 234054198, 1219053680, 6364813192, 33302104593, 174570695175, 916628799380, 4820160541350, 25381091113455, 133808636072595, 706211862466500, 3730964595817680, 19729042153581150

Offset: 1

Paul D. Hanna, Jun 22 2012

A001002(n) is the number of dissections of a convex (n+2)-gon into triangles and quadrilaterals by nonintersecting diagonals.
The g.f. of A001002 satisfies: G(x) = 1 + x*G(x)^2 + x^2*G(x)^3.
Central terms in A155161: a(n) = A155161(2*n,n). - Reinhard Zumkeller, Apr 17 2013
a(n) is the 2n-th term of the n-fold self-convolution of the Fibonacci numbers. - Alois P. Heinz, Feb 07 2021

			L.g.f.: L(x) = x + 5*x^2/2 + 22*x^3/3 + 105*x^4/4 + 511*x^5/5 +...
such that
L(x) = x*(1+x) + d/dx x^3*(1+x)^2/2! + d^2/dx^2 x^5*(1+x)^3/3! + d^3/dx^3 x^7*(1+x)^4/4! +...
The g.f. of A001002 begins:
exp(L(x)) = 1 + x + 3*x^2 + 10*x^3 + 38*x^4 + 154*x^5 + 654*x^6 +...

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
Dmitry V. Kruchinin and Vladimir V. Kruchinin, A Generating Function for the Diagonal T2n,n in Triangles, Journal of Integer Sequence, Vol. 18 (2015), article 15.4.6.

Cf. A000045, A001002, A037027, A155161, A298610, A350383.

a213684 n = a155161 (2*n) n  -- Reinhard Zumkeller, Apr 17 2013

with(orthopoly): seq(add(i, i in [seq((-1)^iquo(n-k,2)*coeff(G(n,n,x/2), x, k), k=0..n)]), n=1..24); # Peter Luschny, Jan 26 2018

Table[n*Sum[Binomial[k+n-1,n]*Binomial[k,n-k]/k,{k,1,n}],{n,1,20}] (* Vaclav Kotesovec, Oct 20 2012 *)

{a(n)=n*sum(r=1,n,binomial(r+n-1,n)*binomial(r,n-r)/r)}
for(n=1, 30, print1(a(n), ", "))

{Dx(n, F)=local(D=F); for(i=1, n, D=deriv(D)); D}
{a(n)=local(A=1); A=(sum(m=1, n+1, Dx(m-1, x^(2*m-1)*(1+x)^m/m!)+x*O(x^n))); n*polcoeff(A, n)}
for(n=1, 30, print1(a(n), ", "))

a(n) = n * Sum_{r=1..n} binomial(r+n-1,n) * binomial(r,n-r) / r.
L.g.f.: Sum_{n>=1} d^(n-1)/dx^(n-1) x^(2*n-1)*(1+x)^n/n! = Sum_{n>=1} a(n)*x^n/n.
Recurrence: 75*(n-1)*n*a(n) = 5*(n-1)*(59*n-12)*a(n-1) + (559*n^2-1503*n+1100)* a(n-2) + 21*(3*n-8)*(3*n-7)*a(n-3). - Vaclav Kotesovec, Oct 20 2012
a(n) ~ 3^(3*n)/(2*5^(n-1/2)*sqrt(6*Pi*n)). - Vaclav Kotesovec, Oct 20 2012
a(n) = A037027(2*n-1,n-1). - Vladimir Kruchinin, Feb 28 2013
a(n) = Sum_{k=0..n} (-1)^floor((n-k)/2) [x^k] G(n,n,x/2), where G(n,a,x) denotes the n-th Gegenbauer polynomial; row sums of A298610. - Peter Luschny, Jan 26 2018
a(n) = [x^n] (1/(1-x-x^2))^n. - Alois P. Heinz, Feb 07 2021
From Peter Bala, Mar 11 2025: (Start)
a(n) = Sum_{k = 0..n} n/(2*n-k) * binomial(2*n-k, k)*binomial(2*n-2*k, n).
a(n) = (1/2)*binomial(2*n, n)*hypergeom([-n/2, (-n+1)/2], [-2*n+1], -4). Cf. A350383.
Second-order recurrence: 5*n*(n-1)*(8*n-13)*a(n) = 2*(n-1)*(88*n^2-187*n+75)*a(n-1) + 3*(8*n-5)*(3*n-4)*(3*n-5)*a(n-2) with a(1) = 1 and a(2) = 5. (End)

A103779 Series reversion of y + y^2 + y^3.

Original entry on oeis.org

0, 1, -1, 1, 0, -4, 14, -30, 33, 55, -429, 1365, -2652, 1428, 12920, -64600, 178296, -277932, -152950, 2770350, -10785390, 25312650, -26053020, -84847620, 576753450, -1856900682, 3566658438, -843350102, -24973594296, 117328602840, -317641049880, 455822225496

Offset: 0

Paul Barry, Feb 15 2005

Second column of A103778 (inverse of trinomial triangle A071675).

			G.f.: A(x) = x - x^2 + x^3 - 4*x^5 + 14*x^6 - 30*x^7 + 33*x^8 + 55*x^9 - 429*x^10 + 1365*x^11 - 2652*x^12 + 1428*x^13 + 12920*x^14 + ... where A(x + x^2 + x^3) = x.

R. J. Mathar, Table of n, a(n) for n = 0..103
Elżbieta Liszewska and Wojciech Młotkowski, Some relatives of the Catalan sequence, arXiv:1907.10725 [math.CO], 2019.

Cf. A350383.

CoefficientList[ InverseSeries[ Series[y + y^2 + y^3, {y, 0, 28}], x], x] (* Robert G. Wilson v *)

a(n):=if n=1 then 1 else -sum(sum(binomial(j,n-k-j) *binomial(k,j),j,0,k)*a(k),k,1,n-1); /* Vladimir Kruchinin, Apr 08 2011 */

a(n):=if n=1 then 1 else 1/n*sum(binomial(k,n-1-k)*(-1)^k *binomial(n+k-1,n-1),k,1,n-1); /* Vladimir Kruchinin, May 12 2012 */

Vec(serreverse(x*(1+x+x^2)+O(x^66))) /* Joerg Arndt, Aug 19 2012 */

/* G.f. A(x) satisfies: A(x)^2 = A( x^2 - 2*x*A(x)^2 ) */
{a(n) = my(A=x+x^2, X=x+x*O(x^n)); for(i=1, n, A = subst(A, x, x^2 - 2*X*A^2)^(1/2) ); polcoeff(A, n)}
for(n=1, 40, print1(a(n), ", ")) \\ Paul D. Hanna, Apr 17 2016

G.f.: -2^(2/3) * ((3*sqrt(3)*sqrt(27*x^2+14*x+3)-27*x-7)^(1/3) -(3*sqrt(3) * sqrt(27*x^2+14*x+3)+27*x+7)^(1/3) +2^(1/3))/6.
a(0) = 0, a(1) = 1, and, for n >= 2, a(n) = -Sum_{k = 1..n-1} a(k) * Sum_{j = 0..k} C(j,n-k-j) * C(k,j). - Vladimir Kruchinin, Apr 08 2011
a(n) = (1/n)*Sum_{k = 1..n-1} (-1)^k * C(k,n-1-k) * C(n+k-1,n-1), with a(1) = 1. - Vladimir Kruchinin, May 12 2012
D-finite with recurrence 3*n*(n-1)*a(n) +7*(n-1)*(2*n-3)*a(n-1) +3*(3*n-5)*(3*n-7)*a(n-2)=0. - R. J. Mathar, Oct 06 2012
G.f. A(x) satisfies: A(x)^2 = A( x^2 - 2*x*A(x)^2 ). - Paul D. Hanna, Apr 17 2016
From Paul D. Hanna, Sep 06 2022: (Start)
G.f. A(x) satisfies:
A(x)^5 = A( x^5 - 5*x*(1+x)^2*A(x)^5 ), and
A(x)^5 = ( x^5 - 5*x*(1+x)^2*A(x)^5 ) * (1 - A(x)^5) / (1 - A(x)^15). (End)
A(x) = x*exp(Sum_{n >= 1} A350383(n)*x^n/n). - Peter Bala, Mar 11 2025

A228960 a(n) = [x^n] (1 + x + x^3 + x^4)^n.

Original entry on oeis.org

1, 1, 4, 17, 51, 136, 393, 1233, 3865, 11851, 36301, 112520, 351352, 1098189, 3433704, 10758609, 33794505, 106344793, 335061790, 1056924667, 3338026857, 10554163533, 33402840615, 105809430024, 335444908176, 1064268538776, 3379009937161, 10735253448349, 34127137228747

Offset: 1

Paul D. Hanna, Sep 10 2013

Equals the Logarithmic derivative of A198951, where the g.f. of A198951 satisfies: G(x) = (1 + x*G(x))*(1 + x^3*G(x)^3).

			L.g.f.: L(x) = x + x^2/2 + 4*x^3/3 + 17*x^4/4 + 51*x^5/5 + 136*x^6/6 +...
Given G(x) = exp(L(x)), which is the g.f. of A198951:
G(x) = 1 + x + x^2 + 2*x^3 + 6*x^4 + 16*x^5 + 39*x^6 + 99*x^7 + 271*x^8 + 763*x^9 + 2146*x^10 +...+ A198951(n)*x^n +...
then the l.g.f. L(x) satisfies the series:
L(x) = (1 + x^2*G(x)^2)*x
+ (1 + 2^2*x^2*G(x)^2 + x^4*G(x)^4)*x^2/2
+ (1 + 3^2*x^2*G(x)^2 + 3^2*x^4*G(x)^4 + x^6*G(x)^6)*x^3/3
+ (1 + 4^2*x^2*G(x)^2 + 6^2*x^4*G(x)^4 + 4^2*x^6*G(x)^6 + x^8*G(x)^8)*x^4/4
+ (1 + 5^2*x^2*G(x)^2 + 10^2*x^4*G(x)^4 + 10^2*x^6*G(x)^6 + 5^2*x^8*G(x)^8 + x^10*G(x)^10)*x^5/5 +...
The table of coefficients in (1 + x + x^3 + x^4)^n begins:
n=1: [1,(1), 0,  1,   1,   0,   0,    0,    0,    0,    0, ...];
n=2: [1, 2, (1), 2,   4,   2,   1,    2,    1,    0,    0, ...];
n=3: [1, 3,  3, (4),  9,   9,   6,    9,    9,    4,    3, ...];
n=4: [1, 4,  6,  8, (17), 24,  22,   28,   36,   28,   22, ...];
n=5: [1, 5, 10, 15,  30, (51), 60,   75,  105,  110,  100, ...];
n=6: [1, 6, 15, 26,  51,  96,(136), 180,  261,  326,  345, ...];
n=7: [1, 7, 21, 42,  84, 168, 273, (393), 588,  819,  987, ...];
n=8: [1, 8, 28, 64, 134, 280, 504,  792,(1233),1848, 2472, ...];
n=9: [1, 9, 36, 93, 207, 450, 876, 1494, 2439,(3865),5616, ...]; ...
the terms in parenthesis forms the initial terms of this sequence.

G. C. Greubel, Table of n, a(n) for n = 1..1000

Cf. A198951, A350383, A350406, A350407.

A228960 := proc(n)
    (1+x+x^3+x^4)^n ;
    coeftayl(%,x=0,n) ;
end proc: # R. J. Mathar, Sep 15 2013

Table[Coefficient[(1 + x + x^3 + x^4)^n,x,n],{n,1,30}] (* Vaclav Kotesovec, Dec 27 2013 *)
Table[HypergeometricPFQ[{1/3 - n/3, 2/3 - n/3, -n, -n/3}, {1/3, 2/3, 1}, 1], {n, 20}] (* Vladimir Reshetnikov, Oct 04 2016 *)

{a(n)=polcoeff((1+x+x^3+x^4+x*O(x^n))^n, n)}
for(n=1,30,print1(a(n),", "))

{a(n)=local(A=1/x*serreverse(x/(1+x+x^3+x^4+x*O(x^n)))); n*polcoeff(log(A), n)}
for(n=1,30,print1(a(n),", "))

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

{a(n)=local(A=1+x); for(i=1, n, A=sum(m=1, n, (1-x^2*exp(2*A))^(2*m+1)*sum(j=0, n\2, binomial(m+j, j)^2*x^(2*j)*exp(2*j*A+x*O(x^n)))*x^m/m)); n*polcoeff(A, n, x)}
for(n=1,30,print1(a(n),", "))

L.g.f. L(x) satisfies:
(1) L(x) = log( (1/x)*Series_Reversion(x/((1+x)*(1+x^3))) ).
(2) L(x) = Sum_{n>=1} x^n/n * Sum_{k=0..n} C(n,k)^2 * x^(2*k) * exp(2*k*L(x)).
(3) L(x) = Sum_{n>=1} x^n/n * (1 - x^2*exp(2*L(x)))^(2*n+1) * Sum_{k>=0} C(n+k,k)^2 * x^(2*k) * exp(2*k*L(x)).
Conjecture: 3*n*(1661*n-1820) *(3*n-1) *(3*n-2) *a(n) +(1927258*n^4 -17091925*n^3 +50171975*n^2 -59448794*n +24442440) *a(n-1) +12*(-705605*n^4 +6363374*n^3 -20228575*n^2 +27219817*n -13185456) *a(n-2) +54*(n-2) *(250934*n^3 -1892927*n^2 +4367836*n -3055963) *a(n-3) -486*(n-2)*(n-3) *(26090*n-34343) *(2*n-7) *a(n-4)=0. - R. J. Mathar, Sep 15 2013.
Recurrence (of order 3): 3*n*(3*n-2)*(3*n-1)*(238*n^3 - 1302*n^2 + 2285*n - 1293)*a(n) = 2*(13328*n^6 - 92904*n^5 + 249452*n^4 - 329211*n^3 + 224408*n^2 - 74649*n + 9360)*a(n-1) - 18*(n-1)*(2380*n^5 - 15400*n^4 + 36710*n^3 - 39398*n^2 + 18345*n - 2880)*a(n-2) + 162*(n-2)*(n-1)*(2*n-5)*(238*n^3 - 588*n^2 + 395*n - 72)*a(n-3). - Vaclav Kotesovec, Dec 27 2013
a(n) ~ c*d^n/sqrt(n), where d = 1/81*((2144134 + 520506*sqrt(17))^(2/3) - 2036 + 112*(2144134 + 520506*sqrt(17))^(1/3))*(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 = 1/102*sqrt(17)*sqrt((34102 + 8262*sqrt(17))^(1/3)*((34102+8262*sqrt(17))^(2/3) + 136 + 136*(34102 + 8262*sqrt(17))^(1/3)))/((34102 + 8262*sqrt(17))^(1/3)*sqrt(Pi)) = 0.3061270429417747... - Vaclav Kotesovec, Dec 27 2013
From Peter Bala, Jun 15 2015: (Start)
a(n) = [x^(3*n)](1 + x + x^3 + x^4)^n.
a(n) = Sum_{k = 0..floor(n/3)} binomial(n,k)*binomial(n,3*k). Applying Maple's sumrecursion command to this formula gives the above recurrence of Kotesovec. (End)
a(n) = hypergeom([1/3-n/3, 2/3-n/3, -n, -n/3], [1/3, 2/3, 1], 1). - Vladimir Reshetnikov, Oct 04 2016
From Peter Bala, Apr 15 2023: (Start)
Conjecture 1: 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. A350383.
Conjecture 2: let k be a positive integer, m an integer and let f(x) = g(x)/h(x), where both g(x) and h(x) are finite products of cyclotomic polynomials. Then 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)

A350406 a(n) = [x^n] 1/(1 + x + x^2 + x^3)^n.

Original entry on oeis.org

1, -1, 1, -1, 5, -26, 91, -246, 597, -1540, 4576, -14521, 44915, -132328, 380290, -1102076, 3268437, -9838428, 29616364, -88538500, 263489380, -785026110, 2348923875, -7053379710, 21204016275, -63716916276, 191394838116, -575200476046, 1730575897202

Offset: 0

Seiichi Manyama, Dec 29 2021

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

Cf. A005725, A348410, A350383, A350407.

a := proc (n) option remember; if n = 0 then 1 elif n = 1 then -1 elif n = 2 then 1 else -(4*n*(n-1)*(5*n-9)*(550*n^3-1045*n^2+164*n+223)*a(n-1) + (n-1)*(5*n+1)*(5075*n^4-22330*n^3+33771*n^2-20232*n+4032)*a(n-2) + (16*n-24)*(4*n-5)*(4*n-7)*(5*n+1)*(5*n-3)*(5*n-4)*a(n-3))/(16*n*(n+1)*(n-1)*(5*n-4)*(5*n-8)*(5*n-9)) end if; end:
seq(a(n), n = 0..30); # Peter Bala, Mar 18 2023
a := n -> (-1)^n*hypergeom([1/4 - n/4, 1/2 - n/4, 3/4 - n/4, -n/4, n], [1/4, 1/2, 3/4, 1], 1): seq(simplify(a(n)), n = 0..28); # Peter Luschny, Mar 19 2023

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

a(n) = (-1)^n*sum(k=0, n, binomial(n-1+k, k)*binomial(n, 4*k));

a(n) = (-1)^n * Sum_{k=0..n} binomial(n-1+k,k) * binomial(n,4*k).
From Peter Bala, Mar 17 2023: (Start)
a(n) = Sum_{k = 0..floor(n/2)} (-1)^(n-k)*binomial(n+k-1,k)*binomial(2*n-2*k-1,n-2*k).
16*n*(n+1)*(n-1)*(5*n-4)*(5*n-8)*(5*n-9)*a(n) = - ( 4*n*(n-1)*(5*n-9)* (550*n^3-1045*n^2+164*n+223)*a(n-1) + (n-1)*(5*n+1)*(5075*n^4-22330*n^3 +33771*n^2-20232*n+4032)*a(n-2) + 8*(2*n-3)*(4*n-5)*(4*n-7)*(5*n+1)*(5*n-3)*(5*n-4)*a(n-3) ).
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. (End)
a(n) = (-1)^n*hypergeom([1/4 - n/4, 1/2 - n/4, 3/4 - n/4, -n/4, n], [1/4, 1/2, 3/4, 1], 1). - Peter Luschny, Mar 19 2023

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)

cp's OEIS Frontend

A213684 Logarithmic derivative of A001002.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

PARI

PARI

Formula

A103779 Series reversion of y + y^2 + y^3.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Maxima

Maxima

PARI

PARI

Formula

A228960 a(n) = [x^n] (1 + x + x^3 + x^4)^n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

PARI

PARI

PARI

Formula

A350406 a(n) = [x^n] 1/(1 + x + x^2 + x^3)^n.

Views

Author

Keywords

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

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

Views

Author

Keywords

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula