A006367 -id:A006367 - cp's OEIS frontend

A210553 Triangle of coefficients of polynomials v(n,x) jointly generated with A210552; see the Formula section.

1, 2, 1, 3, 2, 2, 4, 3, 5, 3, 5, 4, 9, 8, 5, 6, 5, 14, 15, 15, 8, 7, 6, 20, 24, 31, 26, 13, 8, 7, 27, 35, 54, 57, 46, 21, 9, 8, 35, 48, 85, 104, 108, 80, 34, 10, 9, 44, 63, 125, 170, 209, 199, 139, 55, 11, 10, 54, 80, 175, 258, 360, 404, 366, 240, 89, 12, 11, 65, 99

Offset: 1

Clark Kimberling, Mar 22 2012

Let T(n,k) denote the term in row n, column k.
T(n,n):  A000045 (Fibonacci numbers)
T(n,n-1): A006367
T(n,n-2): A105423
T(n,1): 1,2,3,4,5,6,7,8,9,...
T(n,2): 1,2,3,4,5,6,7,8,9,...
T(n,3): A000096
T(n,4): A005563
T(n,5): A055831
T(n,6): A111694
Row sums: A000225
Alternating row sums: A052551
For a discussion and guide to related arrays, see A208510.

			First five rows:
1
2...1
3...2...2
4...3...5...3
5...4...9...8...5
First three polynomials v(n,x): 1, 2 + x , 3 + 2x + 2x^2.

Cf. A210552, A208510.

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + x*v[n - 1, x] + 1;
v[n_, x_] := x*u[n - 1, x] + v[n - 1, x] + 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]   (* A210552 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]   (* A210553 *)
Table[u[n, x] /. x -> 1, {n, 1, z}]  (* A000225 *)
Table[v[n, x] /. x -> 1, {n, 1, z}]  (* A000225 *)
Table[u[n, x] /. x -> -1, {n, 1, z}] (* A094024 *)
Table[v[n, x] /. x -> -1, {n, 1, z}] (* A052551 *)

u(n,x)=x*u(n-1,x)+x*v(n-1,x)+1,
v(n,x)=x*u(n-1,x)+v(n-1,x)+1,
where u(1,x)=1, v(1,x)=1.

A238159 Number of compositions of n with exactly one part equal to 1 or exactly one part equal to 2.

Original entry on oeis.org

0, 1, 1, 2, 5, 11, 15, 35, 70, 124, 234, 447, 827, 1529, 2834, 5222, 9587, 17573, 32137, 58641, 106821, 194280, 352824, 639913, 1159238, 2097759, 3792375, 6849778, 12361822, 22292405, 40172089, 72344671, 130203409, 234200988, 421037335, 756538955, 1358728300

Offset: 0

Geoffrey Critzer, Feb 18 2014

nonn

			a(4) = 5 because we have: 1+3, 3+1, 1+1+2, 1+2+1, 2+1+1.

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (9,-34,73,-109,137,-147,120,-79,40,3,-20,21,-22,8,-3,1,3,0,1).

Cf. A006367 exactly one part equal to 1, A079662 exactly one part equal to 2 (with appropriate offset).

nn=30;a=1/(1-(x/(1-x)-x));b=1/(1-(x/(1-x)-x^2));c=1/(1-(x/(1-x)-x-x^2));CoefficientList[Series[a^2x +b^2x^2-2 c^3x^3,{x,0,nn}],x]
(* or *)
Table[Length[Select[Level[Table[Select[Compositions[n,k],Count[#,0]==0&],{k,1,n}],{2}],Count[#,1]==1||Count[#,2]==1&]],{n,0,10}]

G.f.: x*A(x)^2 + x^2*B(x)^2 - 2*x^3*C(x)^3 where A(x)=1/(1 - (x/(1-x)-x)), B(x)=1/(1 - (x/(1-x)-x^2)), C(x)=1/(1 - (x/(1-x)-x-x^2)).

a(n) ~ c * n / (2^(n-1) * d^n), where c = 0.02749202171174083217... is the root of the equation -1 + 18*c + 552*c^2 + 4232*c^3 = 0 and d = 0.2849201454990266329... is the root of the equation -1 + 4*d - 4*d^2 + 8*d^3 = 0. - Vaclav Kotesovec, May 01 2014

A239366 Triangular array read by rows: T(n,k) is the number of palindromic compositions of n having exactly k 1's, n>=0, 0<=k<=n.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 2, 0, 1, 0, 1, 1, 1, 1, 0, 0, 1, 3, 0, 3, 0, 1, 0, 1, 2, 1, 1, 2, 1, 0, 0, 1, 5, 0, 5, 0, 4, 0, 1, 0, 1, 3, 2, 3, 2, 1, 3, 1, 0, 0, 1, 8, 0, 10, 0, 7, 0, 5, 0, 1, 0, 1, 5, 3, 5, 5, 4, 3, 1, 4, 1, 0, 0, 1, 13, 0, 18, 0, 16, 0, 9, 0, 6, 0, 1, 0, 1, 8, 5, 10, 8, 7, 9, 5, 4, 1, 5, 1, 0, 0, 1

Offset: 0

Geoffrey Critzer, Mar 20 2014

Row sums = 2^floor(n/2).

T(n,0) = A053602(n-1) for n>0, T(n,1) = A079977(n-5) for n>4, T(2n+1,3) = A006367(n-1) for n>0, both bisections of column k=2 contain A010049. - Alois P. Heinz, Mar 21 2014

			1,
0, 1,
1, 0, 1,
1, 0, 0, 1,
2, 0, 1, 0, 1,
1, 1, 1, 0, 0, 1,
3, 0, 3, 0, 1, 0, 1,
2, 1, 1, 2, 1, 0, 0, 1,
5, 0, 5, 0, 4, 0, 1, 0, 1,
3, 2, 3, 2, 1, 3, 1, 0, 0, 1
There are eight palindromic compositions of 6: T(6,0)=3 because we have: 6, 3+3, 2+2+2.  T(6,2)=3 because we have: 1+4+1, 2+1+1+2, 1+2+2+1.  T(6,4)=1 because we have: 1+1+2+1+1. T(6,6)=1 because we have: 1+1+1+1+1+1.

Alois P. Heinz, Rows n = 0..140, flattened

b:= proc(n) option remember;  `if`(n=0, 1, expand(
      add(b(n-j)*`if`(j=1, x^2, 1), j=1..n)))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n))
    (add(b(i)*`if`(n-2*i=1, x, 1), i=0..n/2)):
seq(T(n), n=0..30);  # Alois P. Heinz, Mar 21 2014

nn=15;Table[Take[CoefficientList[Series[((1+x)*(1-x+x^2+x*y-x^2*y))/(1-x^2-x^4-x^2*y^2+x^4*y^2),{x,0,nn}],{x,y}][[n]],n],{n,1,nn}]//Grid

G.f.: G(x,y) = ((1 + x)*(1 - x + x^2 + x*y - x^2*y))/(1 - x^2 - x^4 - x^2*y^2 + x^4*y^2). Satisfies G(x,y) = 1/(1 - x) - x + y*x + (x^2/(1 - x^2) - x^2 +y^2*x^2)*G(x,y).

cp's OEIS Frontend

A210553 Triangle of coefficients of polynomials v(n,x) jointly generated with A210552; see the Formula section.

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A238159 Number of compositions of n with exactly one part equal to 1 or exactly one part equal to 2.

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A239366 Triangular array read by rows: T(n,k) is the number of palindromic compositions of n having exactly k 1's, n>=0, 0<=k<=n.

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