A051159 -id:A051159 - cp's OEIS frontend

A034852 Rows of (Pascal's triangle - Losanitsch's triangle) (n >= 0, k >= 0).

0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 2, 2, 2, 0, 0, 2, 4, 4, 2, 0, 0, 3, 6, 10, 6, 3, 0, 0, 3, 9, 16, 16, 9, 3, 0, 0, 4, 12, 28, 32, 28, 12, 4, 0, 0, 4, 16, 40, 60, 60, 40, 16, 4, 0, 0, 5, 20, 60, 100, 126, 100, 60, 20, 5, 0, 0, 5, 25, 80, 160, 226, 226, 160, 80, 25, 5, 0, 0, 6, 30, 110, 240

Offset: 0

N. J. A. Sloane

Also number of linear unbranched n-4-catafusenes of C_{2v} symmetry.
Number of n-bead black-white reversible strings with k black beads; also binary grids; string is not palindromic. - Yosu Yurramendi, Aug 08 2008
The first seven columns are A004526, A002620, A006584, A032091, A032092, A032093, A032094. Row sums give essentially A032085. - Yosu Yurramendi, Aug 08 2008
From Álvar Ibeas, Jun 01 2020: (Start)
T(n, k) is the sum of odd-degree coefficients of the Gaussian polynomial [n, k]_q. The area below a NE lattice path between (0,0) and (k, n-k) is even for A034851(n, k) paths and odd for T(n, k) of them.
For a (non-reversible) string of k black and n-k white beads, consider the minimum number of bead transpositions needed to place the black ones to the left and the white ones to the right (in other words, the number of inversions of the permutation obtained by labeling the black beads by integers 1,...,k and the white ones by k+1,...,n, in the same order they take on the string). It is even for A034851(n, k) strings and odd for T(n, k) cases.
(End)

			Triangle begins:
  0;
  0 0;
  0 1 0;
  0 1 1 0;
  0 2 2 2 0;
  0 2 4 4 2 0;
  ...

Reinhard Zumkeller, Rows n=0..150 of triangle, flattened
Johann Cigler, Some remarks on Rogers-Szegö polynomials and Losanitsch's triangle, arXiv:1711.03340 [math.CO], 2017.
S. J. Cyvin, B. N. Cyvin, and J. Brunvoll, Unbranched catacondensed polygonal systems containing hexagons and tetragons, Croatica Chem. Acta, 69 (1996), 757-774.
S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber. 30 (1897), 1917-1926.
S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber. 30 (1897), 1917-1926. (Annotated scanned copy)
N. J. A. Sloane, Classic Sequences

Cf. A007318, A034851, A051159.

Essentially the same as A034877.

a034852 n k = a034852_tabl !! n !! k
a034852_row n = a034852_tabl !! n
a034852_tabl = zipWith (zipWith (-)) a007318_tabl a034851_tabl
-- Reinhard Zumkeller, Mar 24 2012

nmax = 12; t[n_?EvenQ, k_?EvenQ] := (Binomial[n, k] - Binomial[n/2, k/2])/ 2; t[n_?EvenQ, k_?OddQ] := Binomial[n, k]/2; t[n_?OddQ, k_?EvenQ] := (Binomial[n, k] - Binomial[(n-1)/2, k/2])/2; t[n_?OddQ, k_?OddQ] := (Binomial[n, k] - Binomial[(n-1)/2, (k-1)/2])/2; Flatten[ Table[t[n, k], {n, 0, nmax}, {k, 0, n}]] (* Jean-François Alcover, Nov 15 2011, after Yosu Yurramendi *)

Equals (A007318-A051159)/2. - Yosu Yurramendi, Aug 08 2008

T(n, k) = T(n - 1, k - 1) + T(n - 1, k); except when n is even and k odd, in which case T(n, k) = A034851(n, k) = T(n - 1, k - 1) + A034841(n - 1, k) = A034841(n - 1, k - 1) + T(n - 1, k) = C(n, k) / 2. - Álvar Ibeas, Jun 01 2020

More terms from James Sellers, May 04 2000

A179181 'PE(n,k)' triangle read by rows. PE(n,k) is the number of k-palindromes of n up to cyclic equivalence.

Original entry on oeis.org

1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 2, 0, 1, 1, 1, 2, 1, 1, 1, 1, 0, 3, 0, 3, 0, 1, 1, 1, 3, 2, 3, 2, 1, 1, 1, 0, 4, 0, 6, 0, 4, 0, 1, 1, 1, 4, 2, 6, 4, 4, 2, 1, 1, 1, 0, 5, 0, 10, 0, 10, 0, 5, 0, 1, 1, 1, 5, 3, 10, 6, 10, 5, 5, 3, 1, 1

Offset: 1

John P. McSorley, Jun 30 2010

A k-composition of n is an ordered collection of k positive integers (parts) which sum to n.
Two k-compositions of n are cyclically equivalent if one can be obtained from the other by a cyclic permutation of its parts.
A k-palindrome of n is a k-composition of n which is a palindrome.
Let PE(n,k) denote the number of k-palindromes of n up to cyclic equivalence.
This sequence is the 'PE(n,k)' triangle read by rows.

			The triangle begins
  1
  1,1
  1,0,1
  1,1,1,1
  1,0,2,0,1
  1,1,2,1,1,1
  1,0,3,0,3,0,1
  1,1,3,2,3,2,1,1
  1,0,4,0,6,0,4,0,1
  1,1,4,2,6,4,4,2,1,1
For example, row 8 is 1,1,3,2,3,2,1,1.
We have PE(8,3)=3 because there are 3 3-palindromes of 8, namely: 161, 242, and 323, and none are cyclically equivalent to the others.
We have PE(8,4)=2 because there are 3 4-palindromes of 8, namely: 3113, 1331, and 2222, but 3113 and 1331 are cyclically equivalent.

John P. McSorley: Counting k-compositions with palindromic and related structures. Preprint, 2010.

Andrew Howroyd, Table of n, a(n) for n = 1..1275

If we ignore cyclic equivalence then we have sequence A051159.
The row sums of the 'PE(n, k)' triangle give sequence A056503.
Cf. A179317, A179519.

T[n_, k_] := (3 - (-1)^k)/4*Sum[MoebiusMu[d]*QBinomial[n/d - 1, k/d - 1, -1], {d, Divisors@ GCD[n, k]}]; Table[DivisorSum[GCD[n, k], T[n/#, k/#] &], {n, 12}, {k, n}] // Flatten (* Michael De Vlieger, Oct 31 2021, after Jean-François Alcover at A179317 *)

p(n, k) = if(n%2==1&&k%2==0, 0, binomial((n-1)\2, (k-1)\2));
APE(n, k) = if(k%2,1,1/2) * sumdiv(gcd(n,k), d, moebius(d) * p(n/d, k/d));
T(n, k) = sumdiv(gcd(n,k), d, APE(n/d, k/d));
for(n=1, 10, for(k=1, n, print1(T(n,k), ", ")); print) \\ Andrew Howroyd, Oct 07 2017

PE(n, k) = Sum_{d | gcd(n,k)} A179317(n/d, k/d). - Andrew Howroyd, Oct 07 2017

Terms a(56) and beyond from Andrew Howroyd, Oct 07 2017

A136482 Triangle read by rows: T(n,k) = 2*A007318(n,k) - A034851(n,k) (i.e., twice Pascal's triangle - the Losanitch triangle).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 4, 4, 1, 1, 6, 8, 6, 1, 1, 7, 14, 14, 7, 1, 1, 9, 21, 30, 21, 9, 1, 1, 10, 30, 51, 51, 30, 10, 1, 1, 12, 40, 84, 102, 84, 40, 12, 1, 1, 13, 52, 124, 186, 186, 124, 52, 13, 1, 1, 15, 65, 180, 310, 378, 310, 180, 65, 15, 1, 1, 16, 80, 245, 490, 688, 688, 490, 245

Offset: 0

Gary W. Adamson, Dec 31 2007

Row sums are apparently in A135098. - R. J. Mathar, May 01 2008

			Row n=3 is 2*(1,3,3,1) - (1,2,2,1) = (1,4,4,1).

Cf. A007318, A034851, A136489.

A007318 := proc(n,k) binomial(n,k) ; end: A051159 := proc(n,k) binomial(n mod 2, k mod 2)*binomial(floor(n/2),floor(k/2)) ; end: A034851 := proc(n,k) (A007318(n,k)+A051159(n,k))/2 ; end: A136482 := proc(n,k) 2*A007318(n,k)-A034851(n,k) ; end: for n from 0 to 13 do for k from 0 to n do printf("%d,",A136482(n,k)) ; od: od: # R. J. Mathar, May 01 2008

Edited and corrected by R. J. Mathar, May 01 2008

A136489 Triangle T(n, k) = 3A007318(n, k) - 2A034851(n, k).

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 5, 5, 1, 1, 8, 10, 8, 1, 1, 9, 18, 18, 9, 1, 1, 12, 27, 40, 27, 12, 1, 1, 13, 39, 67, 67, 39, 13, 1, 1, 16, 52, 112, 134, 112, 52, 16, 1, 1, 17, 68, 164, 246, 246, 164, 68, 17, 1, 1, 20, 85, 240, 410, 504, 410, 240, 85, 20, 1

Offset: 0

Gary W. Adamson, Jan 01 2008

			First few rows of the triangle are:
  1;
  1,   1;
  1,   4,   1;
  1,   5,   5,   1;
  1,   8,  10,   8,   1;
  1,   9,  18,  18,   9,   1;
  1,  12,  27,  40,  27,  12,   1;
  1,  13,  39,  67,  67,  39,  13,   1;
  1,  16,  52, 112, 134, 112,  52,  16,   1;
  1,  17,  68, 164, 246, 246, 164,  68,  17,   1;
  ...

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A034851, A042948, A077957, A122746 (row sums).

A136489:= func< n,k | 2*Binomial(n,k) - Binomial(n mod 2, k mod 2)*Binomial(Floor(n/2), Floor(k/2)) >;
[A136489(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Aug 01 2023

T[n_, k_]:= 2*Binomial[n,k] -Binomial[Mod[n,2], Mod[k,2]]*Binomial[Floor[n/2], Floor[k/2]];
Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Aug 01 2023 *)

def A136489(n,k): return 2*binomial(n,k) - binomial(n%2, k%2)*binomial(n//2, k//2)
flatten([[A136489(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Aug 01 2023

T(n, k) = 3*A007318(n, k) - 2*A034851(n, k).
Sum_{k=0..n} T(n, k) = A122746(n).
From G. C. Greubel, Aug 01 2023: (Start)
T(n, k) = 2*A007318(n, k) - A051159(n, k).
T(n, k) = T(n-1, k) + T(n-1, k-1) if k is even.
T(n, n-k) = T(n, k).
T(n, n-1) = A042948(n).
Sum_{k=0..n} (-1)^k * T(n, k) = 2*[n=0] - A077957(n). (End)

A239632 Number of parts in all palindromic compositions of n.

Original entry on oeis.org

0, 1, 3, 4, 10, 12, 28, 32, 72, 80, 176, 192, 416, 448, 960, 1024, 2176, 2304, 4864, 5120, 10752, 11264, 23552, 24576, 51200, 53248, 110592, 114688, 237568, 245760, 507904, 524288, 1081344, 1114112, 2293760, 2359296, 4849664, 4980736, 10223616, 10485760, 21495808, 22020096, 45088768, 46137344

Offset: 0

Geoffrey Critzer, Mar 22 2014

			a(5)=12 because we have: 5, 1+3+1, 2+1+2, 1+1+1+1+1 with a total of 12 parts.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,4,0,-4).

Cf. A051159.

nn=30; r=Solve[p==y/(1-x) - y + 1 + y^2*x^2/(1-x^2)*p, p]; CoefficientList[Series[D[p/.r,y]/.y->1, {x,0,nn}], x]
CoefficientList[Series[(x + 3 x^2 - 2 x^4)/(1 - 2 x^2)^2, {x, 0, 40}], x] (* Vincenzo Librandi, Mar 23 2014 *)

G.f.: (x + 3*x^2 - 2*x^4)/(1 - 2*x^2)^2.
a(n) = Sum_{k=1..n} A051159(n,k)*k.
a(n) = 4*a(n-2) - 4*a(n-4) for n > 3. - Giovanni Resta, Mar 23 2014
a(2k) = (2k+1)*2^(k-1) for k>0, a(2k+1) = (2k+2)*2^(k-1) for k>=0. - Gregory L. Simay, Dec 05 2022
E.g.f.: (2*(1 + x)*cosh(sqrt(2)*x) + sqrt(2)*(1 + 2*x)*sinh(sqrt(2)*x) - 2)/4. - Stefano Spezia, Apr 25 2024

A121908 S-D transform of Catalan numbers A000108.

Original entry on oeis.org

1, 2, 3, 9, 19, 72, 181, 752, 2051, 8902, 25417, 113249, 333101, 1510888, 4538219, 20853973, 63626003, 295288350, 911918665, 4265460227, 13300767273, 62608960656, 196778953279, 931129725342, 2945833819213, 14000655099890, 44541071348599, 212484364171847

Offset: 0

Philippe Deléham, Sep 01 2006

nonn

			1 1 2 5 14 42 132 ... (A000108)
2 1 7 9 56 90 ...
3 6 16 47 146 ...
9 10 63 99 ...
19 53 162 ...
72 109 ...
181 ...
Row 1 : A000108
Row 2 : 1+1=2, 2-1=1, 5+2=7, 14-5=9, 42+14=56, 132-42=90, ...
Row 3 : 1+2=3, 7-1=6, 9+7=16, 56-9=47, 90+56=146, ...
Row 4 : 6+3=9, 16-6=10, 47+16=63, 146-47=99, ...
Row 5 : 10+9=19, 63-10=53, 99+63=162, ...
Row 6 : 53+19=72, 162-53=109, ...
Row 7 : 109+72=181, ...
First diagonal of this triangular array form this sequence.

Alois P. Heinz, Table of n, a(n) for n = 0..1000

a:= proc(n) option remember; `if`(n<6, [1, 2, 3, 9, 19, 72][n+1],
      ((16*n^2+72*n-153)*n *a(n-1)
       +(304*n^4-1276*n^3+1213*n^2+487*n-754) *a(n-2)
       -(288*n^3-768*n^2-294*n+1424) *a(n-3)
       -(560*n^4-3772*n^3+6497*n^2+1253*n-4558) *a(n-4)
       +17*(n-4)*(16*n^2-8*n-29) *a(n-5)
       +17*(n-5)*(n-4)*(16*n^2-4*n-13) *a(n-6)) /
      (n*(n+1)*(16*n^2-36*n+7)))
    end:
seq(a(n), n=0..40);  # Alois P. Heinz, Jul 12 2014

T[n_, k_] := Binomial[Mod[n, 2], Mod[k, 2]] Binomial[Quotient[n, 2], Quotient[k, 2]];
a[n_] := Sum[T[n, k] CatalanNumber[k], {k, 0, n}];
a /@ Range[0, 40] (* Jean-François Alcover, Nov 19 2020 *)

a(n) = Sum_{k=0..n} A051159(n,k) * A000108(k).

Recurrence: see Maple program.

More terms from Alois P. Heinz, Jul 12 2014

A121909 S-D transform of factorial numbers A000142.

Original entry on oeis.org

1, 2, 3, 10, 29, 162, 799

Offset: 0

Philippe Deléham, Sep 01 2006

nonn

			Row 1 : 1, 1, 2, 6, 24, 120, ...(A000142)
Row 2 : 1+1=2, 2-1=1, 6+2=8, 24-6=18, 120+24=144, ...
Row 3 : 1+2=3, 9-1=7, 18+8=26, 144-18=126, ...
Row 4 : 7+3=10, 26-7=19, 126+26=152, ...
Row 5 : 19+10=29, 152-19=133, ...
Row 6 : 133+29=162, ...
First term of each row form this sequence.

a(n)=Sum{k, 0<=k<=n}A051159(n,k)*A000142(k).

A171145 The sequence of coefficients of a polynomial recursion: p(x,n)=If[Mod[n, 2] == 0, (x + 1)p(x, n - 1), (x^2 + nx + 1)^Floor[n/2]].

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 4, 4, 1, 1, 10, 27, 10, 1, 1, 11, 37, 37, 11, 1, 1, 21, 150, 385, 150, 21, 1, 1, 22, 171, 535, 535, 171, 22, 1, 1, 36, 490, 3024, 7539, 3024, 490, 36, 1, 1, 37, 526, 3514, 10563, 10563, 3514, 526, 37, 1, 1, 55, 1215, 13530, 76845, 188001, 76845

Offset: 1

Roger L. Bagula and Gary W. Adamson, Dec 04 2009

Row sums are:
{1, 2, 5, 10, 49, 98, 729, 1458, 14641, 29282, 371293, 742586,...}.
The modulo 2 of this appears to be a staggered Sierpinski-type fractal.

			{1},
{1, 1},
{1, 3, 1},
{1, 4, 4, 1},
{1, 10, 27, 10, 1},
{1, 11, 37, 37, 11, 1},
{1, 21, 150, 385, 150, 21, 1},
{1, 22, 171, 535, 535, 171, 22, 1},
{1, 36, 490, 3024, 7539, 3024, 490, 36, 1},
{1, 37, 526, 3514, 10563, 10563, 3514, 526, 37, 1},
{1, 55, 1215, 13530, 76845, 188001, 76845, 13530, 1215, 55, 1},
{1, 56, 1270, 14745, 90375, 264846, 264846, 90375, 14745, 1270, 56, 1}

Cf. A051159, A169623, A007318, A171142, A171143

Clear[p, n, x, a]
p[x, 1] := 1;
p[x_, n_] := p[x, n] = If[Mod[n, 2] == 0, (x + 1)*p[x, n - 1], (x^2 + n*x + 1)^Floor[n/2]];
a = Table[CoefficientList[p[x, n], x], {n, 1, 12}];
Flatten[a]

p(x,n)=If[Mod[n, 2] == 0, (x + 1)*p(x, n - 1), (x^2 + n*x + 1)^Floor[n/2]]

A171146 The sequence of coefficients of a polynomial recursion: p(x,n)=If[Mod[n, 2] == 0, (x + 1)p(x, n - 1), (x^2 + (2n - 1)*x + 1)^Floor[n/2]] ( correction).

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 6, 6, 1, 1, 18, 83, 18, 1, 1, 19, 101, 101, 19, 1, 1, 39, 510, 2275, 510, 39, 1, 1, 40, 549, 2785, 2785, 549, 40, 1, 1, 68, 1738, 19856, 86995, 19856, 1738, 68, 1, 1, 69, 1806, 21594, 106851, 106851, 21594, 1806, 69, 1, 1, 105, 4415, 93030, 985645

Offset: 1

Roger L. Bagula and Gary W. Adamson, Dec 04 2009

Row sums are:

{1, 2, 7, 14, 121, 242, 3375, 6750, 130321, 260642, 6436343, 12872686...}.

			{1},
{1, 1},
{1, 5, 1},
{1, 6, 6, 1},
{1, 18, 83, 18, 1},
{1, 19, 101, 101, 19, 1},
{1, 39, 510, 2275, 510, 39, 1},
{1, 40, 549, 2785, 2785, 549, 40, 1},
{1, 68, 1738, 19856, 86995, 19856, 1738, 68, 1},
{1, 69, 1806, 21594, 106851, 106851, 21594, 1806, 69, 1},
{1, 105, 4415, 93030, 985645, 4269951, 985645, 93030, 4415, 105, 1},
{1, 106, 4520, 97445, 1078675, 5255596, 5255596, 1078675, 97445, 4520, 106, 1}

Cf. A051159 , A169623, A007318, A171142, A171143

Clear[p, n, x, a]
p[x, 1] := 1;
p[x_, n_] := p[x, n] = If[Mod[n, 2] == 0, (x + 1)*p[x, n - 1], (x^2 + (2*n - 1)*x + 1)^Floor[n/2]];
a = Table[CoefficientList[p[x, n], x], {n, 1, 12}];
Flatten[a]

p(x,n)=If[Mod[n, 2] == 0, (x + 1)*p(x, n - 1), (x^2 + (2*n - 1)*x + 1)^Floor[n/2]]

A171147 The sequence of coefficients of a polynomial recursion: p(x,n)=If[Mod[n, 2] == 0, (x + 1)p(x, n - 1), (x^2 + (2n)*x + 1)^Floor[n/2]].

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 7, 7, 1, 1, 20, 102, 20, 1, 1, 21, 122, 122, 21, 1, 1, 42, 591, 2828, 591, 42, 1, 1, 43, 633, 3419, 3419, 633, 43, 1, 1, 72, 1948, 23544, 108870, 23544, 1948, 72, 1, 1, 73, 2020, 25492, 132414, 132414, 25492, 2020, 73, 1, 1, 110, 4845, 106920

Offset: 1

Roger L. Bagula and Gary W. Adamson, Dec 04 2009

Row sums are:

{1, 2, 8, 16, 144, 288, 4096, 8192, 160000, 320000, 7962624, 15925248...}.

			{1},
{1, 1},
{1, 6, 1},
{1, 7, 7, 1},
{1, 20, 102, 20, 1},
{1, 21, 122, 122, 21, 1},
{1, 42, 591, 2828, 591, 42, 1},
{1, 43, 633, 3419, 3419, 633, 43, 1},
{1, 72, 1948, 23544, 108870, 23544, 1948, 72, 1},
{1, 73, 2020, 25492, 132414, 132414, 25492, 2020, 73, 1},
{1, 110, 4845, 106920, 1185810, 5367252, 1185810, 106920, 4845, 110, 1},
{1, 111, 4955, 111765, 1292730, 6553062, 6553062, 1292730, 111765, 4955, 111, 1}

Cf. A051159 , A169623, A007318, A171142, A171143

Clear[p, n, x, a]
p[x, 1] := 1;
p[x_, n_] := p[x, n] = If[Mod[n, 2] == 0, (x + 1)*p[x, n - 1], (x^2 + (2*n)*x + 1)^Floor[n/2]];
a = Table[CoefficientList[p[x, n], x], {n, 1, 12}];
Flatten[a]

p(x,n)=If[Mod[n, 2] == 0, (x + 1)*p(x, n - 1), (x^2 + (2*n)*x + 1)^Floor[n/2]]

cp's OEIS Frontend

A034852 Rows of (Pascal's triangle - Losanitsch's triangle) (n >= 0, k >= 0).

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A179181 'PE(n,k)' triangle read by rows. PE(n,k) is the number of k-palindromes of n up to cyclic equivalence.

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A136482 Triangle read by rows: T(n,k) = 2*A007318(n,k) - A034851(n,k) (i.e., twice Pascal's triangle - the Losanitch triangle).

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A136489 Triangle T(n, k) = 3*A007318(n, k) - 2*A034851(n, k).

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A239632 Number of parts in all palindromic compositions of n.

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A121908 S-D transform of Catalan numbers A000108.

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A121909 S-D transform of factorial numbers A000142.

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A171145 The sequence of coefficients of a polynomial recursion: p(x,n)=If[Mod[n, 2] == 0, (x + 1)*p(x, n - 1), (x^2 + n*x + 1)^Floor[n/2]].

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A136489 Triangle T(n, k) = 3A007318(n, k) - 2A034851(n, k).

A171145 The sequence of coefficients of a polynomial recursion: p(x,n)=If[Mod[n, 2] == 0, (x + 1)p(x, n - 1), (x^2 + nx + 1)^Floor[n/2]].

A171146 The sequence of coefficients of a polynomial recursion: p(x,n)=If[Mod[n, 2] == 0, (x + 1)p(x, n - 1), (x^2 + (2n - 1)*x + 1)^Floor[n/2]] ( correction).

A171147 The sequence of coefficients of a polynomial recursion: p(x,n)=If[Mod[n, 2] == 0, (x + 1)p(x, n - 1), (x^2 + (2n)*x + 1)^Floor[n/2]].