A006131 -id:A006131 - cp's OEIS frontend

A109447 Binomial coefficients C(n,k) with n-k odd, read by rows.

1, 2, 1, 3, 4, 4, 1, 10, 5, 6, 20, 6, 1, 21, 35, 7, 8, 56, 56, 8, 1, 36, 126, 84, 9, 10, 120, 252, 120, 10, 1, 55, 330, 462, 165, 11, 12, 220, 792, 792, 220, 12, 1, 78, 715, 1716, 1287, 286, 13, 14, 364, 2002, 3432, 2002, 364, 14, 1, 105, 1365, 5005, 6435, 3003, 455, 15

Offset: 1

Philippe Deléham, Aug 27 2005

The same as A119900 without 0's. A reflected version of A034867 or A202064. - Alois P. Heinz, Feb 07 2014
From Vladimir Shevelev, Feb 07 2014: (Start)
Also table of coefficients of polynomials  P_1(x)=1, P_2(x)=2, for n>=2, P_(n+1)(x) = 2*P_n(x)+(x-1)* P_(n-1)(x).  The polynomials P_n(x)/2^(n-1) are connected with sequences A000045 (x=5), A001045 (x=9), A006130 (x=13), A006131 (x=17), A015440 (x=21), A015441 (x=25), A015442 (x=29), A015443 (x=33), A015445 (x=37), A015446 (x=41), A015447 (x=45), A053404 (x=49); also the polynomials P_n(x) are connected with sequences A000129, A002605, A015518, A063727, A085449, A002532, A083099, A015519, A003683, A002534, A083102, A015520. (End)

			Starred terms in Pascal's triangle (A007318), read by rows:
1;
1*, 1;
1, 2*, 1;
1*, 3, 3*, 1;
1, 4*, 6, 4*, 1;
1*, 5, 10*, 10, 5*, 1;
1, 6*, 15, 20*, 15, 6*, 1;
1*, 7, 21*, 35, 35*, 21, 7*, 1;
1, 8*, 28, 56*, 70, 56*, 28, 8*, 1;
1*, 9, 36*, 84, 126*, 126, 84*, 36, 9*, 1;
Triangle T(n,k) begins:
1;
2;
1,    3;
4,    4;
1,   10,  5;
6,   20,  6;
1,   21,  35,   7;
8,   56,  56,   8;
1,   36, 126,  84,  9;
10, 120, 252, 120, 10;

Alois P. Heinz, Rows n = 1..200, flattened

Cf. A109446.

T:= (n, k)-> binomial(n, 2*k+1-irem(n, 2)):
seq(seq(T(n, k), k=0..ceil((n-2)/2)), n=1..20);  # Alois P. Heinz, Feb 07 2014

Flatten[ Table[ If[ OddQ[n - k], Binomial[n, k], {}], {n, 0, 15}, {k, 0, n}]] (* Robert G. Wilson v *)

More terms from Robert G. Wilson v, Aug 30 2005

Corrected offset by Alois P. Heinz, Feb 07 2014

A128100 Triangle read by rows: T(n,k) is the number of ways to tile a 2 X n rectangle with k pieces of 2 X 2 tiles and n-2k pieces of 1 X 2 tiles (0 <= k <= floor(n/2)).

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 5, 5, 1, 8, 10, 3, 13, 20, 9, 1, 21, 38, 22, 4, 34, 71, 51, 14, 1, 55, 130, 111, 40, 5, 89, 235, 233, 105, 20, 1, 144, 420, 474, 256, 65, 6, 233, 744, 942, 594, 190, 27, 1, 377, 1308, 1836, 1324, 511, 98, 7, 610, 2285, 3522, 2860, 1295, 315, 35, 1, 987, 3970

Offset: 0

Emeric Deutsch, Feb 18 2007

Row sums are the Jacobsthal numbers (A001045). Column 0 yields the Fibonacci numbers (A000045); the other columns yield convolved Fibonacci numbers (A001629, A001628, A001872, A001873, etc.). Sum_{k=0..floor(n/2)} k*T(n,k) = A073371(n-2).
Triangle T(n,k), with zeros omitted, given by (1, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Jan 24 2012
Riordan array (1/(1-x-x^2), x^2/(1-x-x^2)), with zeros omitted. - Philippe Deléham, Feb 06 2012
Diagonal sums are A000073(n+2) (tribonacci numbers). - Philippe Deléham, Feb 16 2014
Number of induced subgraphs of the Fibonacci cube Gamma(n-1) that are isomorphic to the hypercube Q_k. Example: row n=4 is 5, 5, 1; indeed, the Fibonacci cube Gamma(3) is a square with an additional pendant edge attached to one of its vertices; it has 5 vertices (i.e., Q_0's), 5 edges (i.e., Q_1's) and 1 square (i.e., Q_2). - Emeric Deutsch, Aug 12 2014
Row n gives the coefficients of the polynomial p(n,x) defined as the numerator of the rational function given by f(n,x) = 1 + (x + 1)/f(n-1,x), where f(x,0) = 1. Conjecture: for n > 2, p(n,x) is irreducible if and only if n is a (prime - 2). - Clark Kimberling, Oct 22 2014

			Triangle starts:
   1;
   1;
   2,  1;
   3,  2;
   5,  5,  1;
   8, 10,  3;
  13, 20,  9,  1;
  21, 38, 22,  4;
From _Philippe Deléham_, Jan 24 2012: (Start)
Triangle (1, 1, -1, 0, 0, ...) DELTA (0, 1, -1, 0, 0, 0, ...) begins:
   1;
   1,  0;
   2,  1,  0;
   3,  2,  0,  0;
   5,  5,  1,  0,  0;
   8, 10,  3,  0,  0,  0;
  13, 20,  9,  1,  0,  0,  0;
  21, 38, 22,  4,  0,  0,  0,  0; (End)
From _Clark Kimberling_, Oct 22 2014: (Start)
Here are the first 4 polynomials p(n,x) as in Comment and generated by Mathematica program:
  1
  2 +  x
  3 + 2x
  5 + 5x + x^2. (End)

C.-P. Chou and H. A. Witek, An Algorithm and FORTRAN Program for Automatic Computation of the Zhang-Zhang Polynomial of Benzenoids, MATCH: Commun. Math. Comput. Chem, 68 (2012) 3-30. See Eq. (9). - From _N. J. A. Sloane_, Dec 23 2012
S. Klavzar, M. Mollard, Cube polynomial of Fibonacci and Lucas cubes, preprint.
S. Klavzar, M. Mollard, Cube polynomial of Fibonacci and Lucas cubes, Acta Appl. Math. 117, 2012, 93-105. - _Emeric Deutsch_, Aug 12 2014

Cf. A001045, A000045, A001629, A001628, A001872, A001873, A073371.

Cf. A109466, A119473.

G:=1/(1-z-(1+t)*z^2): Gser:=simplify(series(G,z=0,19)): for n from 0 to 16 do P[n]:=sort(coeff(Gser,z,n)) od: for n from 0 to 16 do seq(coeff(P[n],t,j),j=0..floor(n/2)) od; # yields sequence in triangular form

p[x_, n_] := 1 + (x + 1)/p[x, n - 1]; p[x_, 1] = 1;
Numerator[Table[Factor[p[x, n]], {n, 1, 20}]]  (* Clark Kimberling, Oct 22 2014 *)

G.f.: 1/(1-z-(1+t)z^2).
Sum_{k=0..n} T(n,k)*x^k = A053404(n), A015447(n), A015446(n), A015445(n), A015443(n), A015442(n), A015441(n), A015440(n), A006131(n), A006130(n), A001045(n+1), A000045(n+1), A000012(n), A010892(n), A107920(n+1), A106852(n), A106853(n), A106854(n), A145934(n), A145976(n), A145978(n), A146078(n), A146080(n), A146083(n), A146084(n) for x = 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, -1, -2, -3, -4, -5, -6, -7, -8, -9, -10, -11, -12, and -13, respectively. - Philippe Deléham, Jan 24 2012
T(n,k) = T(n-1,k) + T(n-2,k) + T(n-2,k-1). - Philippe Deléham, Jan 24 2012
G.f.: T(0)/2, where T(k) = 1 + 1/(1 - (2*k+1+ x*(1+y))*x/((2*k+2+ x*(1+y))*x + 1/T(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Nov 06 2013
T(n,k) = Sum_{i=k..floor(n/2)} binomial(n-i,i)*binomial(i,k). See Corollary 3.3 in the Klavzar et al. link. - Emeric Deutsch, Aug 12 2014

A172349 Triangle t(n,k) read by rows: fibonomial ratios c(n)/(c(k)*c(n-k)) where c are partial products of a generalized Fibonacci sequence with multiplier m=4.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 5, 5, 1, 1, 9, 45, 9, 1, 1, 29, 261, 261, 29, 1, 1, 65, 1885, 3393, 1885, 65, 1, 1, 181, 11765, 68237, 68237, 11765, 181, 1, 1, 441, 79821, 1037673, 3343613, 1037673, 79821, 441, 1, 1, 1165, 513765, 18598293, 134321005, 134321005

Offset: 0

Roger L. Bagula and Gary W. Adamson, Feb 01 2010

Start from the generalized Fibonacci sequence A006131 and its partial products c(n) = 1, 1, 1, 5, 45, 1305, 84825, 15353325, 6770816325, 7888001018625... Then t(n,k) = c(n)/(c(k)*c(n-k)).

Row sums are 1, 2, 3, 12, 65, 582, 7295, 160368, 5579485, 306868458, 26280601275,...

			1;
1, 1;
1, 1, 1;
1, 5, 5, 1;
1, 9, 45, 9, 1;
1, 29, 261, 261, 29, 1;
1, 65, 1885, 3393, 1885, 65, 1;
1, 181, 11765, 68237, 68237, 11765, 181, 1;
1, 441, 79821, 1037673, 3343613, 1037673, 79821, 441, 1;
1, 1165, 513765, 18598293, 134321005, 134321005, 18598293, 513765, 1165, 1;
1, 2929, 3412285, 300963537, 6052711133, 13566421505, 6052711133, 300963537, 3412285, 2929, 1;

Cf. A010048 (m=1), A015109 (m=2), A172347 (m=3), A172350 (m=5).

Clear[f, c, a, t];
f[0, a_] := 0; f[1, a_] := 1;
f[n_, a_] := f[n, a] = f[n - 1, a] + a*f[n - 2, a];
c[n_, a_] := If[n == 0, 1, Product[f[i, a], {i, 1, n}]];
t[n_, m_, a_] := c[n, a]/(c[m, a]*c[n - m, a]);
Table[Table[Table[t[n, m, a], {m, 0, n}], {n, 0, 10}], {a, 1, 10}];
Table[Flatten[Table[Table[t[n, m, a], {m, 0, n}], {n, 0, 10}]], {a, 1, 10}]

A344236 Number of n-step walks from a universal vertex to the other on the diamond graph.

Original entry on oeis.org

0, 1, 2, 5, 14, 33, 90, 221, 582, 1465, 3794, 9653, 24830, 63441, 162762, 416525, 1067574, 2733673, 7003970, 17938661, 45954542, 117709185, 301527354, 772364093, 1978473510, 5067929881, 12981823922, 33253543445, 85180839134, 218195012913, 558918369450

Offset: 0

M. Eren Kesim, May 12 2021

a(n) is the number of n-step walks from vertex A to vertex C on the graph below.
B--C
| /|
|/ |
A--D

			Let A, B, C and D be the vertices of the diamond graph, where A and C are the universal vertices. Then, a(3) = 5 walks from A to C are: (A, B, A, C), (A, C, A, C), (A, C, B, C), (A, C, D, C), and (A, D, A, C).

Index entries for linear recurrences with constant coefficients, signature (0,5,4).

Cf. A006131, A344261.

f := proc(n) option remember; if n <= 2 then n; else 5*f(n - 2) + 4*f(n - 3); end if; end proc

LinearRecurrence[{0, 5, 4}, {0, 1, 2}, 30]

my(p=Mod('x,'x^2-'x-4)); a(n) = (vecsum(Vec(lift(p^n))) + n%2) >> 1; \\ Kevin Ryde, May 13 2021

def A344236_list(n):
    list = [0, 1, 2] + [0] * (n - 3)
    for i in range(3, n):
        list[i] = 5 * list[i - 2] + 4 * list[i - 3]
    return list
print(A344236_list(31)) # M. Eren Kesim, Jul 19 2021

a(n) = a(n-1) + 4*a(n-2) + (-1)^n for n > 1.
a(n) = A344261(n-1) + 2*a(n-2) + 2*A344261(n-2) for n > 1.
a(n) = A344261(n) - (-1)^n.
a(n) = A006131(n) - A344261(n).
a(n) = (A006131(n) - (-1)^n)/2.
a(n) = ((sqrt(17)-1)/(4*sqrt(17)))*((1-sqrt(17))/2)^n + ((sqrt(17)+1)/(4*sqrt(17)))*((1+sqrt(17))/2)^n - (1/2)*(-1)^n.
G.f.: (2*x^2 + x)/(-4*x^3 - 5*x^2 + 1).
a(n) = 5*a(n-2) + 4*a(n-3) for n > 2. - Stefano Spezia, May 13 2021

A344261 Number of n-step walks from one of the vertices with degree 3 to itself on the four-vertex diamond graph.

Original entry on oeis.org

1, 0, 3, 4, 15, 32, 91, 220, 583, 1464, 3795, 9652, 24831, 63440, 162763, 416524, 1067575, 2733672, 7003971, 17938660, 45954543, 117709184, 301527355, 772364092, 1978473511, 5067929880, 12981823923, 33253543444, 85180839135, 218195012912, 558918369451

Offset: 0

M. Eren Kesim, May 13 2021

a(n) is the number of n-step walks from vertex A to itself on the graph below.
B--C
| /|
|/ |
A--D

			Let A, B, C and D be the vertices of the four-vertex diamond graph, where A and C are the vertices with degree 3. Then, a(3) = 4 walks from A to itself are: (A, B, C, A), (A, C, B, A), (A, C, D, A) and (A, D, C, A).

Index entries for linear recurrences with constant coefficients, signature (0,5,4).

Cf. A006131, A344236.

f := proc(n) option remember; if n = 0 then 1; elif n = 1 then 0; elif n = 2 then 3; else 5*f(n - 2) + 4*f(n - 3); end if; end proc

LinearRecurrence[{0, 5, 4}, {1, 0, 3}, 30] (* Amiram Eldar, May 13 2021 *)

def A344261_list(n):
    list = [1, 0, 3] + [0] * (n - 3)
    for i in range(3, n):
        list[i] = 5 * list[i - 2] + 4 * list[i - 3]
    return list
print(A344261_list(31)) # M. Eren Kesim, Jul 19 2021

a(n) = a(n-1) + 4*a(n-2) - (-1)^n for n > 1.
a(n) = 5*a(n-2) + 4*a(n-3) for n > 2.
a(n) = A344236(n-1) + 2*a(n-2) + 2*A344236(n-2) for n > 1.
a(n) = A344236(n) + (-1)^n.
a(n) = A006131(n) - A344236(n).
a(n) = (A006131(n) + (-1)^n)/2.
a(n) = ((sqrt(17)-1)/(4*sqrt(17)))*((1-sqrt(17))/2)^n + ((sqrt(17)+1)/(4*sqrt(17)))*((1+sqrt(17))/2)^n + (1/2)*(-1)^n.
G.f.: (2*x^2 - 1)/(4*x^3 + 5*x^2 - 1).

A103280 Array read by antidiagonals, generated by the matrix M = [1,1,1;1,N,1;1,1,1].

Original entry on oeis.org

1, 1, 2, 1, 3, 6, 1, 4, 9, 16, 1, 5, 14, 27, 44, 1, 6, 21, 48, 81, 120, 1, 7, 30, 85, 164, 243, 328, 1, 8, 41, 144, 341, 560, 729, 896, 1, 9, 54, 231, 684, 1365, 1912, 2187, 2448, 1, 10, 69, 352, 1289, 3240, 5461, 6528, 6561, 6688, 1, 11, 86, 513, 2276, 7175, 15336, 21845

Offset: 0

Lambert Klasen (lambert.klasen(AT)gmx.net), Jan 27 2005

Consider the matrix M = [1,1,1;1,N,1;1,1,1];
Characteristic polynomial of M is x^3 + (-N - 2)*x^2 + (2*N - 2)*x.
Now (M^n)[1,2] is equivalent to the recursion a(1) = 1, a(2) = N+2, a(n) = (N+2)a(n-1)+(2N-2)a(n-2). (This also holds for negative N and fractional N.)
a(n+1)/a(n) converges to the upper root of the characteristic polynomial ((N + 2) + sqrt((N - 2)^2 + 8))/2 for n to infinity.
Columns of array follow the polynomials:
0
1
N + 2
N^2 + 2*N + 6
N^3 + 2*N^2 + 8*N + 16
N^4 + 2*N^3 + 10*N^2 + 24*N + 44
N^5 + 2*N^4 + 12*N^3 + 32*N^2 + 76*N + 120
N^6 + 2*N^5 + 14*N^4 + 40*N^3 + 112*N^2 + 232*N + 328
N^7 + 2*N^6 + 16*N^5 + 48*N^4 + 152*N^3 + 368*N^2 + 704*N + 896
N^8 + 2*N^7 + 18*N^6 + 56*N^5 + 196*N^4 + 528*N^3 + 1200*N^2 + 2112*N + 2448
etc.

			Array begins:
1,2,6,16,44,120,328,896,2448,6688,...
1,3,9,27,81,243,729,2187,6561,19683, ...
1,4,14,48,164,560,1912,6528,22288,76096,...
1,5,21,85,341,1365,5461,21845,87381,349525,...
1,6,30,144,684,3240,15336,72576,343440,1625184,...
1,7,41,231,1289,7175,39913,221991,1234633,6866503,...
...

Cf. A103279 (for (M^n)[1, 1]), A002605 (for N=0), A000244 (for N=1), A007070 (for N=2), A002450 (for N=3), A030192 (for N=4), A152268 (for N=5), A006131 (for N=-1), A000400 (bisection for N=-2), A015443 (for N=-3), A083102 (for N=-4).

T12(N, n) = if(n==1,1,if(n==2,N+2,(N+2)*T12(N,n-1)-(2*N-2)*T12(N,n-2)))
for(k=0,10,print1(k,": ");for(i=1,10,print1(T12(k,i),","));print())

T(N, 1)=1, T(N, 2)=N+2, T(N, n)=(N+2)*T(N, n-1)-(2*N-2)*T(N, n-2).

A176738 Expansion of 1 / ((1+x)(1-x-4x^2)). (5,4)-Padovan sequence.

Original entry on oeis.org

1, 0, 5, 4, 25, 40, 141, 300, 865, 2064, 5525, 13780, 35881, 91000, 234525, 598524, 1536625, 3930720, 10077221, 25800100, 66108985, 169309384, 433745325, 1110982860, 2845964161, 7289895600, 18673752245, 47833334644, 122528343625, 313861682200, 803975056701

Offset: 0

Wolfdieter Lang, Jul 14 2010

See A000931 (Padovan), and the W. Lang link given there.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,5,4).

Cf. A176737 ((4,3)-Padovan).

LinearRecurrence[{0,5,4},{1,0,5},40] (* Harvey P. Dale, May 27 2016 *)
f[n_] := Simplify[((-1)^(1 +n) + (2^(-1 -n)*((1 + Sqrt[17])^n*(-5 +3Sqrt[17]) + (1 -Sqrt[17])^n*(5 + 3Sqrt[17])))/Sqrt[17])/2]; Array[f, 31, 0] (* or *)
CoefficientList[Series[1/(1 -5x^2 -4x^3), {x, 0, 30}], x] (* or *)
RecurrenceTable[{a[n] == 5 a[n - 2] + 4 a[n - 3], a[0] == 1, a[1] == 0, a[2] == 5}, a, {n, 30}] (* Robert G. Wilson v, Dec 25 2017 *)

Vec(1 / ((1 + x)*(1 - x - 4*x^2)) + O(x^40)) \\ Colin Barker, Dec 25 2017

O.g.f.: 1/((1-x-4*x^2)*(1+x)) = ((3-4*x)/(1-x-4*x^2) -1/(1+x))/2.
a(n) = (3*b(n) - 4*b(n-1) - (-1)^n)/2, n>=0, with b(n):=A006131(n) ((1,4)-Fibonacci), b(-1):=0.
From Colin Barker, Dec 25 2017: (Start)
a(n) = ((-1)^(1+n) + (2^(-1-n)*((1+sqrt(17))^n*(-5+3*sqrt(17)) + (1-sqrt(17))^n*(5+3*sqrt(17)))) / sqrt(17)) / 2.
a(n) = 5*a(n-2) + 4*a(n-3) for n>2.
(End)

A189604 Number of n X 3 array permutations with each element not moving, or moving one space E, S or NW.

Original entry on oeis.org

1, 6, 20, 72, 256, 912, 3248, 11568, 41200, 146736, 522608, 1861296, 6629104, 23609904, 84087920, 299483568, 1066626544, 3798846768, 13529793392, 48187073712, 171620807920, 611236571184, 2176951329392, 7753327130544

Offset: 1

R. H. Hardin, Apr 24 2011

nonn

Column 3 of A189610.

Binomial transform of A006131 starting (1, 5, 9, 29, 65, ...). - Gary W. Adamson, Feb 19 2014

			Some solutions for 4 X 3:
.
   4  5  1    0  5  1    0  1  2    0  1  2
   0  3  2    7  4  2    3  4  5    3  4  5
   6  7  8    3  6  8    6 11  8   10  7  8
   9 10 11    9 10 11    9  7 10    6  9 11
.
   4  0  1    0  1  2    4  1  2
   7  3  2    3  8  5    0  3  5
  10 11  5    6  4  7    6  7  8
   6  9  8    9 10 11    9 10 11

R. H. Hardin, Table of n, a(n) for n = 1..200

Cf. A006131.

a[n_] := Sum[Sum[4^j Binomial[k-j+1, j], {j, 0, Quotient[k+1, 2]}]* Binomial[n-1, k], {k, 0, n-1}];
a /@ Range[1, 24] (* Jean-François Alcover, Sep 24 2019, after Gary W. Adamson *)

Empirical: a(n) = 3*a(n-1) + 2*a(n-2).

G.f.: (x+3*x^2)/(1-3*x-2*x^2). - Vladimir Kruchinin, May 13 2011

A231730 Triangular array read by rows: row n shows the coefficients of the polynomial u(n) = c(0) + c(1)x + ... + c(n)x^(n) which is the numerator of the n-th convergent of the continued fraction [k, k, k, ... ], where k = x + 1/2.

Original entry on oeis.org

1, 2, 5, 4, 4, 9, 22, 12, 8, 29, 56, 72, 32, 16, 65, 202, 232, 208, 80, 32, 181, 556, 924, 800, 560, 192, 64, 441, 1726, 2964, 3480, 2480, 1440, 448, 128, 1165, 4832, 10112, 12608, 11680, 7168, 3584, 1024, 256, 2929, 14066, 31632, 46752, 46816, 36288, 19712

Offset: 1

Clark Kimberling, Nov 13 2013

Sum of numbers in row n: A015521(n). Left edge: A006131. Right edge: powers of 2

			First 3 rows:
1 .... 2
5 .... 4 .... 4
9 .... 22 ... 12 ... 8
First 3 polynomials:  1 + 2*x, 5 + 4*x + 4*x^2, 9 + 22*x + 12*x^2 + 8*x^3.

Cf. A230000, A231731.

t[n_] := t[n] = Table[x + 1/2, {k, 0, n}];
b = Table[Factor[Convergents[t[n]]], {n, 0, 10}];
p[x_, n_] := p[x, n] = Last[Expand[Numerator[b]]][[n]];
u = Table[p[x, n], {n, 1, 10}]
v = CoefficientList[u, x]; Flatten[v]

A231774 Triangular array read by rows: row n shows the coefficients of the polynomial u(n) = c(0) + c(1)x + ... + c(n)x^(n) which is the numerator of the n-th convergent of the continued fraction [k, k, k, ... ], where k = (x + 1)/(x + 2).

Original entry on oeis.org

2, 1, 5, 6, 2, 9, 19, 13, 3, 29, 72, 69, 30, 5, 65, 213, 278, 182, 60, 8, 181, 682, 1084, 928, 451, 118, 13, 441, 1975, 3795, 4065, 2625, 1023, 223, 21, 1165, 5868, 13015, 16590, 13290, 6852, 2221, 414, 34, 2929, 16697, 42404, 63020, 60435, 38799, 16682

Offset: 1

Clark Kimberling, Nov 13 2013

Sum of numbers in row n: A002534(n). Left edge: A006131. Right edge: A000045 (Fibonacci numbers).

			First 3 rows:
2 ... 1
5 ... 6 .... 2
9 ... 19 ... 13 ... 3
First 3 polynomials:  2 + x, 5 + 6*x + 2*x^2, 9 + 19*x + 13*x^2 + 3*x^3.

Cf. A230000, A231775, A000045.

t[n_] := t[n] = Table[(x + 1)/(x + 2), {k, 0, n}];
b = Table[Factor[Convergents[t[n]]], {n, 0, 10}];
p[x_, n_] := p[x, n] = Last[Expand[Numerator[b]]][[n]];
u = Table[p[x, n], {n, 1, 10}]
v = CoefficientList[u, x]; Flatten[v]

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