A080923 -id:A080923 - cp's OEIS frontend

A118264 Coefficient of q^n in (1-q)^3/(1-3q); dimensions of the enveloping algebra of the derived free Lie algebra on 3 letters.

1, 0, 3, 8, 24, 72, 216, 648, 1944, 5832, 17496, 52488, 157464, 472392, 1417176, 4251528, 12754584, 38263752, 114791256, 344373768, 1033121304, 3099363912, 9298091736, 27894275208, 83682825624, 251048476872, 753145430616

Offset: 0

Mike Zabrocki, Apr 20 2006

a(n) is the number of generalized compositions of n when there are i^2-1 different types of i, (i=1,2,...). - Milan Janjic, Sep 24 2010

			The enveloping algebra of the derived free Lie algebra is characterized as the intersection of the kernels of all partial derivative operators in the space of non-commutative polynomials, a(0) = 1 since all constants are killed by derivatives, a(1) = 0 since no polys of degree 1 are killed, a(2) = 3 since all Lie brackets [x1,x2], [x1,x3], [x2, x3] are killed by all derivative operators.

C. Reutenauer, Free Lie algebras. London Mathematical Society Monographs. New Series, 7. Oxford Science Publications. The Clarendon Press, Oxford University Press, New York, 1993. xviii+269 pp.

Michael De Vlieger, Table of n, a(n) for n = 0..2097
Nantel Bergeron, Christophe Reutenauer, Mercedes Rosas, and Mike Zabrocki, Invariants and Coinvariants of the Symmetric Group in Noncommuting Variables, arXiv:math.CO/0502082, 2005. See also Canad. J. Math. 60 (2008), no. 2, 266-296.
Joscha Diehl, Rosa Preiß, and Jeremy Reizenstein, Conjugation, loop and closure invariants of the iterated-integrals signature, arXiv:2412.19670 [math.RA], 2024. See pp. 17, 20.
Index entries for linear recurrences with constant coefficients, signature (3).

Cf. A080923, A027376, A118265, A118266.

f:=n->coeftayl((1-q)^3/(1-3*q),q=0,n):seq(f(i),i=0..15);

CoefficientList[Series[(1-q)^3/(1-3q),{q,0,30}],q] (* or *) Join[{1,0,3}, NestList[3#&,8,30]] (* Harvey P. Dale, Jun 28 2011 *)
Join[{1, 0, 3}, LinearRecurrence[{3}, {8}, 24]] (* Jean-François Alcover, Sep 23 2017 *)

G.f.: (1-x)^3/(1-3x).
a(n) = 3^{n-1}-3^{n-3} for n>=3.
a(n) = A080923(n-1), n>1.
If p[i]=i^2-1 and if A is Hessenberg matrix of order n defined by: A[i,j]=p[j-i+1], (i<=j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det A. - Milan Janjic, May 02 2010
For a(n)>=8, a(n+1)=3*a(n). - Harvey P. Dale, Jun 28 2011

Formula corrected Mike Zabrocki, Jul 22 2010

A110168 Riordan array ((1-x^2)/(1+3x+x^2),x/(1+3x+x^2)).

Original entry on oeis.org

1, -3, 1, 7, -6, 1, -18, 24, -9, 1, 47, -84, 50, -12, 1, -123, 275, -225, 85, -15, 1, 322, -864, 900, -468, 129, -18, 1, -843, 2639, -3339, 2219, -840, 182, -21, 1, 2207, -7896, 11756, -9528, 4610, -1368, 244, -24, 1, -5778, 23256, -39825, 38121, -22518, 8532, -2079, 315, -27, 1, 15127, -67650, 130975

Offset: 0

Paul Barry, Jul 14 2005

Inverse of A110165. Row sums are 1,-2,2,-2,... with g.f. (1-x)/(1+x). Diagonal sums are (-1)^n*A080923. Product of A110162 and inverse binomial transform (1/(1+x),x/(1+x)).

			Rows begin
1;
-3,1;
7,-6,1;
-18,24,-9,1;
47,-84,50,-12,1;
-123,275,-225,85,-15,1;

T(n,k) = T(n-1,k-1) - 3*T(n-1,k) - T(n-2,k), T(0,0) = T(1,1) = T(2,2) = 1, T(1,0) = -3, T(2,0) = 7, T(2,1) = -6, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Jan 22 2014

A220673 Coefficients of formal series in powers of (tan(x))^2 for tan(5*x)/tan(x).

Original entry on oeis.org

5, 40, 376, 3560, 33720, 319400, 3025400, 28657000, 271443000, 2571145000, 24354235000, 230686625000, 2185095075000, 20697517625000, 196049700875000, 1857009420625000, 17589845701875000, 166613409915625000, 1578184870646875000

Offset: 0

Wolfdieter Lang, Jan 16 2013

Formally Sum_{n>=0} a(n)*(tan(x))^(2*n) = tan(5*x)/ tan(x).
Convergence holds for x from the two open intervals (1-sqrt(6/5), 1-2/sqrt(5)) and (1+2/sqrt(5), 1+sqrt(6/5)), namely (-0.095445115, 0.105572809) and (1.894427191, 2.095445115) (10 digits).
These intervals follow from the denominator of the o.g.f. G(5,x) = (5 - 10*x + x^2)/(1 - 10*x + 5*x^2).
If one replaces x by (tan(x))^2 in this o.g.f. one obtains the formula for tan(5*x)/tan(x) in terms of (tan(x))^2. This formula is the special (n=5) solution of a general recurrence derivable from the addition theorem for tan(n*x) = tan(x + (n-1)*x), namely, with Q(n,x) := tan(n,x)/tan(x), Q(n,x) = (1 + Q(n-1,x))/(1 - v*Q(n-1,x)), where v = v(x) =(tan(x))^2, and the input is Q(1,x) = 1. Read as function of v the solution for Q(5,x) is just G(5,v) with replaced v=v(x).
See the irregular triangles A034867 and A034839 whose row polynomials N(n,x) and D(n,x), respectively, give for n >= 1 the solution to the recurrences N(n,x) = D(n-1,x) + N(n-1,x), D(n,x) = D(n-1,x) + x*N(n-1,x), with inputs N(1,x) = 1 and D(1,x) = 1. The proof by the Pascal triangle A007318 recurrence is trivial. Therefore, Q(n,x) from the preceding comment is given by Q(n,x) = N(n,-v)/D(n,-v) with v=v(x) = (tan(x))^2.
One also has, with Chebyshev's S polynomials (see A049310) Q(n,x) = tan(n*x)/tan(x) = (S(n,y) + S(n-2,y))/(S(n,y) - S(n-2,y)) = y*S(n-1,y)/(S(n,y) - S(n-2,y)) = 1/(1 - (2/y)*S(n-2,y)/S(n-1,y)), where y = y(x) = 2/sqrt(1 + (tan(x))^2). This derives from sin(n*x)/cos(n*x) in terms of Chebyshev S polynomials with argument 2*cos(x) = y(x). Note that S(n-2,y)/S(n-1,y) has the continued fraction representation 1/(y-1/(y- ... -1/(y )..(n-1)brackets..), i.e. (n-1) y's.
These calculations have been motivated by e-mails from Thomas Olsen.

			Q(5,x=0.1) = tan(0.5)/tan(0.1) = 5.444802663 (Maple 10 digits);
G(5,tan(0.1)^2) = 5.444802664;
Sum_{n>=0} a(n)*(tan(0.1))^(2*n) = 5.444802664.

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. 2*A000012 (case n=2), A080923(n+1), n>=0 (case n=3), A077445(n+1), n>=0 (case n=4), A034867, A034839.

m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((5-10*x+x^2)/(1-10*x+5*x^2))); // G. C. Greubel, Mar 06 2018

CoefficientList[Series[(5-10*x+x^2)/(1-10*x+5*x^2), {x,0,50}], x] (* G. C. Greubel, Mar 06 2018 *)

my(x='x+O('x^30)); Vec((5-10*x+x^2)/(1-10*x+5*x^2)) \\ G. C. Greubel, Mar 06 2018

O.g.f.: G(5,x) = (5 - 10*x + x^2)/(1 - 10*x + 5*x^2).
a(n) = delta(n,0)/5 - 8*b(n) + 24*b(n+1)/5, n>=0, with Kronecker's delta and b(n):= A190987(n).
E.g.f.: (1 + 8*exp(5*x)*(3*cosh(2*sqrt(5)*x) + sqrt(5)*sinh(2*sqrt(5)*x)))/5. - Stefano Spezia, May 23 2025

A155118 Array T(n,k) read by antidiagonals: the k-th term of the n-th iterated differences of A140429.

Original entry on oeis.org

0, 1, 1, 1, 2, 3, 3, 4, 6, 9, 5, 8, 12, 18, 27, 11, 16, 24, 36, 54, 81, 21, 32, 48, 72, 108, 162, 243, 43, 64, 96, 144, 216, 324, 486, 729, 85, 128, 192, 288, 432, 648, 972, 1458, 2187, 171, 256, 384, 576, 864, 1296, 1944, 2916, 4374, 6561, 341, 512, 768, 1152, 1728, 2592, 3888, 5832, 8748, 13122, 19683

Offset: 0

Paul Curtz, Jan 20 2009

Deleting column k=0 and reading by antidiagonals yields A036561.

Deleting column k=0 and reading the antidiagonals downwards yields A175840.

			The array starts in row n=0 with columns k>=0 as:
   0   1    3    9    27    81    243    729    2187  ... A140429;
   1   2    6   18    54   162    486   1458    4374  ... A025192;
   1   4   12   36   108   324    972   2916    8748  ... A003946;
   3   8   24   72   216   648   1944   5832   17496  ... A080923;
   5  16   48  144   432  1296   3888  11664   34992  ... A257970;
  11  32   96  288   864  2592   7776  23328   69984  ...
  21  64  192  576  1728  5184  15552  46656  139968  ...
Antidiagonal triangle begins as:
   0;
   1,   1;
   1,   2,   3;
   3,   4,   6,   9;
   5,   8,  12,  18,  27;
  11,  16,  24,  36,  54,  81;
  21,  32,  48,  72, 108, 162, 243;
  43,  64,  96, 144, 216, 324, 486, 729;
  85, 128, 192, 288, 432, 648, 972, 1458, 2187; - _G. C. Greubel_, Mar 25 2021

Nathaniel Johnston, Table of n, a(n) for n = 0..10000

Cf. A001045, A004054, A046936.

t:= func< n,k | k eq 0 select (2^(n-k) -(-1)^(n-k))/3 else 2^(n-k)*3^(k-1) >;
[t(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Mar 25 2021

T:=proc(n,k)if(k>0)then return 2^n*3^(k-1):else return (2^n - (-1)^n)/3:fi:end:
for d from 0 to 8 do for m from 0 to d do print(T(d-m,m)):od:od: # Nathaniel Johnston, Apr 13 2011

t[n_, k_]:= If[k==0, (2^(n-k) -(-1)^(n-k))/3, 2^(n-k)*3^(k-1)];
Table[t[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Mar 25 2021 *)

def A155118(n,k): return (2^(n-k) -(-1)^(n-k))/3 if k==0 else 2^(n-k)*3^(k-1)
flatten([[A155118(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Mar 25 2021

For the square array:
T(n,k) = 2^n*3^(k-1), k>0.
T(n,k) = T(n-1,k+1) - T(n-1,k), n>0.
Rows:
T(0,k) = A140429(k) = A000244(k-1).
T(1,k) = A025192(k).
T(2,k) = A003946(k).
T(3,k) = A080923(k+1).
T(4,k) = A257970(k+3).
Columns:
T(n,0) = A001045(n) (Jacobsthal numbers J_{n}).
T(n,1) = A000079(n).
T(n,2) = A007283(n).
T(n,3) = A005010(n).
T(n,4) = A175806(n).
T(0,k) - T(k+1,0) = 4*A094705(k-2).
From G. C. Greubel, Mar 25 2021: (Start)
For the antidiagonal triangle:
t(n, k) = T(n-k, k).
t(n, k) = (2^(n-k) - (-1)^(n-k))/3 (J_{n-k}) if k = 0 else 2^(n-k)*3^(k-1).
Sum_{k=0..n} t(n, k) = 3^n - J_{n+1}, where J_{n} = A001045(n).
Sum_{k=0..n} t(n, k) = A004054(n-1) for n >= 1. (End)

a(22) - a(57) from Nathaniel Johnston, Apr 13 2011

cp's OEIS Frontend

A118264 Coefficient of q^n in (1-q)^3/(1-3q); dimensions of the enveloping algebra of the derived free Lie algebra on 3 letters.

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A110168 Riordan array ((1-x^2)/(1+3x+x^2),x/(1+3x+x^2)).

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A220673 Coefficients of formal series in powers of (tan(x))^2 for tan(5*x)/tan(x).

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A155118 Array T(n,k) read by antidiagonals: the k-th term of the n-th iterated differences of A140429.

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