A189980 -id:A189980 - cp's OEIS frontend

A194005 Triangle of the coefficients of an (n+1)-th order differential equation associated with A103631.

1, 1, 1, 1, 2, 1, 1, 3, 2, 1, 1, 4, 3, 3, 1, 1, 5, 4, 6, 3, 1, 1, 6, 5, 10, 6, 4, 1, 1, 7, 6, 15, 10, 10, 4, 1, 1, 8, 7, 21, 15, 20, 10, 5, 1, 1, 9, 8, 28, 21, 35, 20, 15, 5, 1, 1, 10, 9, 36, 28, 56, 35, 35, 15, 6, 1, 1, 11, 10, 45, 36, 84, 56, 70, 35, 21, 6, 1

Offset: 0

Johannes W. Meijer and A. Hirschberg (a.hirschberg(AT)tue.nl), Aug 11 2011

This triangle is a companion to Parks' triangle A103631.
The coefficients of triangle A103631(n,k) appear in appendix 2 of Park’s remarkable article “A new proof of the Routh-Hurwitz stability criterion using the second method of Liapunov” if we assume that the b(n) coefficients are all equal to 1, see the second Maple program.
The a(n,k) coefficients of the triangle given above are related to the coefficients of a linear (n+1)-th order differential equation for the case b(n)=1, see the examples.
a(n,k) is also the number of symmetric binary strings of odd length n with Hamming weight k>0 and no consecutive 1's. - Christian Barrientos and Sarah Minion, Feb 27 2018

			For the 5th-order linear differential equation the coefficients a(k) are: a(0) = 1, a(1) = a(4,0) = 1, a(2) = a(4,1) = 4, a(3) = a(4,2) = 3, a(4) = a(4,3) = 3 and a(5) = a(4,4) = 1.
The corresponding Hurwitz matrices A(k) are, see Parks: A(5) = Matrix([[a(1),a(0),0,0,0], [a(3),a(2),a(1),a(0),0], [a(5),a(4),a(3),a(2),a(1)], [0,0,a(5),a(4),a(3)], [0,0,0,0,a(5)]]), A(4) = Matrix([[a(1),a(0),0,0], [a(3),a(2),a(1),a(0)], [a(5),a(4),a(3),a(2)], [0,0,a(5),a(4)]]), A(3) = Matrix([[a(1),a(0),0], [a(3),a(2),a(1)], [a(5),a(4),a(3)]]), A(2) = Matrix([[a(1),a(0)], [a(3),a(2)]]) and A(1) = Matrix([[a(1)]]).
The values of b(k) are, see Parks: b(1) = d(1), b(2) = d(2)/d(1), b(3) = d(3)/(d(1)*d(2)), b(4) = d(1)*d(4)/(d(2)*d(3)) and b(5) = d(2)*d(5)/(d(3)*d(4)).
These a(k) values lead to d(k) = 1 and subsequently to b(k) = 1 and this confirms our initial assumption, see the comments.
'
Triangle starts:
  [0] 1;
  [1] 1, 1;
  [2] 1, 2, 1;
  [3] 1, 3, 2,  1;
  [4] 1, 4, 3,  3,  1;
  [5] 1, 5, 4,  6,  3,  1;
  [6] 1, 6, 5, 10,  6,  4,  1;
  [7] 1, 7, 6, 15, 10, 10,  4,  1;
  [8] 1, 8, 7, 21, 15, 20, 10,  5, 1;
  [9] 1, 9, 8, 28, 21, 35, 20, 15, 5, 1;

Reinhard Zumkeller, Rows n = 0..150 of triangle, flattened
Henry W. Gould, A Variant of Pascal's Triangle, The Fibonacci Quarterly, Vol. 3, Nr. 4, Dec. 1965, pp. 257-271, with corrections.
P.C. Parks, A new proof of the Routh-Hurwitz stability criterion using the second method of Liapunov , Math. Proc. of the Cambridge Philosophical Society, Vol. 58, Issue 04 (1962) p. 694-702.
Chris Zheng, Jeffrey Zheng, Triangular Numbers and Their Inherent Properties, Variant Construction from Theoretical Foundation to Applications, Springer, Singapore, 51-65.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A065941 and A103631.
Triangle sums (see A180662): A000071 (row sums; alt row sums), A075427 (Kn22), A000079 (Kn3), A109222(n+1)-1 (Kn4), A000045 (Fi1), A034943 (Ca3), A001519 (Gi3), A000930 (Ze3)
Interesting diagonals: T(n,n-4) = A189976(n+5) and T(n,n-5) = A189980(n+6)
Cf. A052509.

a194005 n k = a194005_tabl !! n !! k
a194005_row n = a194005_tabl !! n
a194005_tabl = [1] : [1,1] : f [1] [1,1] where
   f row' row = rs : f row rs where
     rs = zipWith (+) ([0,1] ++ row') (row ++ [0])
-- Reinhard Zumkeller, Nov 22 2012

A194005 := proc(n, k): binomial(floor((2*n+1-k)/2), n-k) end:
for n from 0 to 11 do seq(A194005(n, k), k=0..n) od;
seq(seq(A194005(n,k), k=0..n), n=0..11);
nmax:=11: for n from 0 to nmax+1 do b(n):=1 od:
A103631 := proc(n,k) option remember: local j: if k=0 and n=0 then b(1)
elif k=0 and n>=1 then 0 elif k=1 then b(n+1) elif k=2 then b(1)*b(n+1)
elif k>=3 then expand(b(n+1)*add(procname(j,k-2), j=k-2..n-2)) fi: end:
for n from 0 to nmax do for k from 0 to n do
A194005(n,k):= add(A103631(n1,k), n1=k..n) od: od:
seq(seq(A194005(n,k),k=0..n), n=0..nmax);

Flatten[Table[Binomial[Floor[(2n+1-k)/2],n-k],{n,0,20},{k,0,n}]] (* Harvey P. Dale, Apr 15 2012 *)

T(n,k) = binomial(floor((2*n+1-k)/2), n-k).
T(n,k) = sum(A103631(n1,k), n1=k..n), 0<=k<=n and n>=0.
T(n,k) = sum(binomial(floor((2*n1-k-1)/2), n1-k), n1=k..n).
T(n,0) = T(n,n) = 1, T(n,k) = T(n-2,k-2) + T(n-1,k), 0 < k < n. - Reinhard Zumkeller, Nov 23 2012

A244419 Coefficient triangle of polynomials related to the Dirichlet kernel. Rising powers. Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2)).

Original entry on oeis.org

1, 1, 2, -1, 2, 4, -1, -4, 4, 8, 1, -4, -12, 8, 16, 1, 6, -12, -32, 16, 32, -1, 6, 24, -32, -80, 32, 64, -1, -8, 24, 80, -80, -192, 64, 128, 1, -8, -40, 80, 240, -192, -448, 128, 256, 1, 10, -40, -160, 240, 672, -448, -1024, 256, 512, -1, 10, 60, -160, -560, 672, 1792, -1024, -2304, 512, 1024

Offset: 0

Wolfdieter Lang, Jul 29 2014

This is the row reversed version of A180870. See also A157751 and A228565.
The Dirichlet kernel is D(n,x) = Sum_{k=-n..n} exp(i*k*x) = 1 + 2*Sum_{k=1..n} T(n,x) = S(n, 2*y) + S(n-1, 2*y) = S(2*n, sqrt(2*(1+y))) with y = cos(x), n >= 0, with the Chebyshev polynomials T (A053120) and S (A049310). This triangle T(n, k) gives in row n the coefficients of the polynomial Dir(n,y) = D(n,x=arccos(y)) = Sum_{m=0..n} T(n,m)*y^m. See A180870, especially the Peter Bala comments and formulas.
This is the Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2)) due to the o.g.f. for Dir(n,y) given by (1+z)/(1 - 2*y*z + z^2) = G(z)/(1 - y*F(z)) with G(z) = (1+z)/(1+z^2) and F(z) = 2*z/(1+z^2) (see the Peter Bala formula under A180870). For Riordan triangles and references see the W. Lang link 'Sheffer a- and z- sequences' under A006232.
The A- and Z- sequences of this Riordan triangle are (see the mentioned W. Lang link in the preceding comment also for the references): The A-sequence has o.g.f. 1+sqrt(1-x^2) and is given by A(2*k+1) = 0 and A(2*k) [2, -1/2, -1/8, -1/16, -5/128, -7/256, -21/1024, -33/2048, -429/32768, -715/65536, ...], k >= 0. (See A098597 and A046161.)
  The Z-sequence has o.g.f. sqrt((1-x)/(1+x)) and is given by
  [1, -1, 1/2, -1/2, 3/8, -3/8, 5/16, -5/16, 35/128, -35/128, ...]. (See A001790 and A046161.)
The column sequences are A057077, 2*(A004526 with even numbers signed), 4*A008805 (signed), 8*A058187 (signed), 16*A189976 (signed), 32*A189980 (signed) for m = 0, 1, ..., 5.
The row sums give A005408 (from the o.g.f. due to the Riordan property), and the alternating row sums give A033999.
The row polynomials Dir(n, x), n >= 0, give solutions to the diophantine equation (a + 1)*X^2 - (a - 1)*Y^2 = 2 by virtue of the identity (a + 1)*Dir(n, -a)^2 - (a - 1)*Dir(n, a)^2 = 2, which is easily proved inductively using the recurrence Dir(n, a) = (1 + a)*(-1)^(n-1)*Dir(n-1, -a) + a*Dir(n-1, a) given below by Wolfdieter Lang. - Peter Bala, May 08 2025

			The triangle T(n,m) begins:
  n\m  0   1   2    3    4    5    6     7     8    9    10 ...
  0:   1
  1:   1   2
  2:  -1   2   4
  3:  -1  -4   4    8
  4:   1  -4 -12    8   16
  5:   1   6 -12  -32   16   32
  6:  -1   6  24  -32  -80   32   64
  7:  -1  -8  24   80  -80 -192   64   128
  8:   1  -8 -40   80  240 -192 -448   128   256
  9:   1  10 -40 -160  240  672 -448 -1024   256  512
  10: -1  10  60 -160 -560  672 1792 -1024 -2304  512  1024
  ...
Example for A-sequence recurrence: T(3,1) = Sum_{j=0..2} A(j)*T(2,j) = 2*(-1) + 0*2 + (-1/2)*4 = -4. Example for Z-sequence recurrence: T(4,0) = Sum_{j=0..3} Z(j)*T(3,j) = 1*(-1) + (-1)*(-4) + (1/2)*4 + (-1/2)*8 = +1. (For the A- and Z-sequences see a comment above.)
Example for the alternate recurrence: T(4,2) = 2*T(3,1) - T(3,2) = 2*(-4) - 4 = -12. T(4,3) = 0*T(3,2) + T(3,3) = T(3,3) = 8. - _Wolfdieter Lang_, Jul 30 2014

Wikipedia, Dirichlet kernel.

Dir(n, x) : A005408 (x = 1), A002878 (x = 3/2), A001834 (x = 2), A030221 (x = 5/2), A002315 (x = 3), A033890 (x = 7/2), A057080 (x = 4), A057081 (x = 9/2), A054320 (x = 5), A077416 (x = 6), A028230 (x = 7), A159678 (x = 8), A049629 (x = 9), A083043 (x = 10),
(-1)^n * Dir(n, x): A122367 (x = -3/2); A079935 (x = -2), A004253 (x = -5/2), A001653 (x = -3), A049685 (x = -7/2), A070997 (x = -4), A070998 (x = -9/2), A072256(n+1) (x = -5).
Cf. A180870 (row reversed), A157751, A228565, A049310, A053120, A057077, A004526, A008805, A058187, A189976, A189980, A005408, A033999.

T[n_, k_] := T[n, k] = Which[k == 0, (-1)^Quotient[n, 2], (0 <= n && n < k) || (n == -1 && k == 1), 0, True, 2 T[n-1, k-1] - T[n-2, k]];
Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 28 2019, from Sage *)

def T(n, k):
    if k == 0: return (-1)^(n//2)
    if (0 <= n and n < k) or (n == -1 and k == 1): return 0
    return 2*T(n-1, k-1) - T(n-2, k)
for n in range(11): [T(n,k) for k in (0..n)] # Peter Luschny, Jul 29 2014

T(n, m) = [y^m] Dir(n,y) for n >= m >= 0 and 0 otherwise, with the polynomials Dir(y) defined in a comment above.
T(n, m) = 2^m*(S(n,m) + S(n-1,m)) with the entries S(n,m) of A049310 given there explicitly.
O.g.f. for polynomials Dir(y) see a comment above (Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2))).
O.g.f. for column m: ((1 + x)/(1 + x^2))*(2*x/(1 + x^2))^m, m >= 0, (Riordan property).
Recurrence for the polynomials: Dir(n, y) = 2*y*Dir(n-1, y) - Dir(n-2, y), n >= 1, with input D(-1, y) = -1 and D(0, y) = 1.
Triangle three-term recurrence: T(n,m) = 2*T(n-1,m-1) - T(n-2,m) for n >= m >= 1 with T(n,m) = 0 if 0 <= n < m, T(0,0) = 1, T(-1,1) = 0 and T(n,0) = A057077(n) = (-1)^(floor(n/2)).
From Wolfdieter Lang, Jul 30 2014: (Start)
In analogy to A157751 one can derive a recurrence for the row polynomials Dir(n, y) = Sum_{m=0..n} T(n,m)*y^m also using a negative argument but only one recursive step: Dir(n,y) = (1+y)*(-1)^(n-1)*Dir(n-1,-y) + y*Dir(n-1,y), n >= 1, Dir(0,y) = 1 (Dir(-1,y) = -1). See also A180870 from where this formula can be derived by row reversion.
This entails another triangle recurrence T(n,m) = (1 + (-1)^(n-m))*T(n-1,m-1) - (-1)^(n-m)*T(n-1,m), for n >= m >= 1 with T(n,m) = 0 if n < m and T(n,0) = (-1)^floor(n/2). (End)
From Peter Bala, Aug 14 2022: (Start)
The row polynomials Dir(n,x), n >= 0, are related to the Chebyshev polynomials of the first kind T(n,x) by the binomial transform as follows:
(2^n)*(x - 1)^(n+1)*Dir(n,x) = (-1) * Sum_{k = 0..2*n+1} binomial(2*n+1,k)*T(k,-x).
Note that Sum_{k = 0..2*n} binomial(2*n,k)*T(k,x) = (2^n)*(1 + x)^n*T(n,x). (End)
From Peter Bala, May 04 2025: (Start)
For n >= 1, the n-th row polynomial Dir(n, x) = (-1)^n * (U(n, -x) - U(n-1, -x)) = U(2*n, sqrt((1+x)/2)), where U(n, x) denotes the n-th Chebyshev polynomial of the second kind.
For n >= 1 and x < 1, Dir(n, x) = (-1)^n * sqrt(2/(1 - x )) * T(2*n+1, sqrt((1 - x)/2)), where T(n, x) denotes the n-th Chebyshev polynomial of the first kind.
Dir(n, x)^2 - 2*x*Dir(n, x)*Dir(n+1, x) + Dir(n+1, x)^2 = 2*(1 + x).
Dir(n, x) = (-1)^n * R(n, -2*(x+1)), where R(n, x) is the n-th row polynomial of the triangle A085478.
Dir(n, x) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n+k, 2*k) * (2*x + 2)^k. (End)

A157751 Triangle of coefficients of polynomials F(n,x) in descending powers of x generated by F(n,x)=(x+1)*F(n-1,x)+F(n-1,-x), with initial F(0,x)=1.

Original entry on oeis.org

1, 1, 2, 1, 2, 4, 1, 4, 4, 8, 1, 4, 12, 8, 16, 1, 6, 12, 32, 16, 32, 1, 6, 24, 32, 80, 32, 64, 1, 8, 24, 80, 80, 192, 64, 128, 1, 8, 40, 80, 240, 192, 448, 128, 256, 1, 10, 40, 160, 240, 672, 448, 1024, 256, 512, 1, 10, 60, 160, 560, 672, 1792, 1024, 2304, 512, 1024, 1, 12, 60, 280, 560, 1792, 1792, 4608, 2304, 5120, 1024, 2048

Offset: 0

Clark Kimberling, Mar 05 2009

Conjecture 1. If n>1 is even then F(n,x) has no real roots.
Conjecture 2. If n>0 is odd then F(n,x) has exactly one real root, r,
and if n>4 then 0 < -r < n.
Conjectures 1 and 2 are true. [From Alain Thiery (Alain.Thiery(AT)math.u-bordeaux1.fr), May 14 2010]
Cayley (1876) states "We, in fact, find  1 + sin u = 1 + x,  1 - sin 3u = (1 + x)(1 - 2x)^2,  1 + sin 5u = (1 + x)(1 + 2x - 4x^2)^2,  1 - sin 7u = (1 + x)(1 - 4x - 4x^2 + 8x^3)^2, &c.". - Michael Somos, Jun 19 2012
Appears to be the unsigned row reverse of A180870 and A228565. - Peter Bala, Feb 17 2014
From Wolfdieter Lang, Jul 29 2014: (Start)
This triangle is the Riordan triangle ((1+z)/(1-z^2), 2*z/(1-z^2)). For Riordan triangles see the W. Lang link 'Sheffer a-and z-sequences' under A006232, also for references. The o.g.f. given by Peter Bala in the formula section refers to the row reversed triangle. The usual information on this triangle, like o.g.f. for the columns, the row sums, the alternating row sums, the recurrences using A- and Z-sequences, etc. follows from this Riordan property. The Riordan proof follows from the given o.g.f. by Peter Bala, call it Grev(x,z), by row reversion: G(x,z) = Grev(1/x,x*z) = (1+z)/(1- 2*x*z - z^2) = G(z)*(1/(1 - x*F(z))) with G(z) = (1+z)/(1-z^2) and F(z) = z*2/(1-z^2). See A244419 for the discussion for a signed version of this triangle.
(End)

			Rows 0 to 8:
1
1 2
1 2 4
1 4 4 8
1 4 12 8 16
1 6 12 32 16 32
1 6 24 32 80 32 64
1 8 24 80 80 192 64 128
1 8 40 80 240 192 448 128 256
(Row 8) = (1, 4*2, 10*4, 10*8, 15*16, 6*32, 7*64, 1*128, 1*256).
First few polynomials:
F(0,x)=1, F(1,x)=x+2, F(2,x)=x^2+2*x+4, F(3,x)=x^3+4*x^2+4*x+8.
The row polynomials R(n,x) start: 1, 1 + 2*x = x*F(1,1/x), 1 + 2*x + 4^x^2 = x^2*F(2,1/x), ...  - _Wolfdieter Lang_, Jul 29 2014

A. Cayley, On an Expression for 1 +- sin(2p+1)u in Terms of sin u, Messenger of Mathematics, 5 (1876), pp. 7-8 = Mathematical Papers Vol. 10, n. 630, pp. 1-2.

G. C. Greubel, Rows n=0..100 of triangle, flattened

Cf. A140070, A140071, A180870, A228565, A078057, A123335, A000012, 2*A004526, 4*A008805, 8*A058187, 16*A189976, 32*A189980.

T[n_, 0]:= 1; T[n_, n_]:= 2^n; T[n_, k_]:= T[n, k] = T[n-1, k] + (1 + (-1)^(n-k))*T[n-1, k-1]; Table[T[n, k], {n, 0, 15}, {k, 0, n}] (* G. C. Greubel, Sep 24 2018 *)

t(n,k) = if(k==0, 1, if(k==n, 2^n, t(n-1,k) + (1+(-1)^(n-k))*t(n-1,k-1)));
for(n=0,15, for(k=0,n, print1(t(n,k), ", "))) \\ G. C. Greubel, Sep 24 2018

Count the top row as row 0 and let C(n,k) denote the usual binomial
coefficient. For row 2n, define p(0)=C(n,0), p(1)=C(n,1), p(2)=C(n+1,2),
p(3)=C(n+1,3), p(4)=C(n+2,4), p(5)=c(n+2,5),..., until reaching two final
1's: p(2n-1)=C(2n-1,2n-1) and p(2n)=C(2n,2n). Then the k-th number in row
2n is p(k)*2^k. For row 2n+1, define q(0)=C(n,0), q(1)=C(n+1,1),
q(2)=C(n+1,2), q(3)=C(n+2,3),..., until reaching q(2n+1)=C(2n+1,2n+1).
Then the k-th number in row 2n+1 is q(k)*2^k.
From Peter Bala, Jan 17 2014: (Start)
Working with an offset of 0, the o.g.f. is (1 + x*z)/(1 - 2*z - x^2*z^2) = 1 + (x + 2)*z + (x^2 + 2*x + 4)*z^2 + ....
Recurrence equation: T(n,k) = 2*T(n-1,k-1) + T(n-2,k-2) with T(n,0) = 1.
The polynomials G(n,x) defined by G(0,x) = 1 and G(n,x) = x*F(n-1,x) for n >= 1 satisfy G(n,x) = (x + 1)*G(n-1,x) - G(n-1,-x). Cf. A140070 and A140071. (End)
From Wolfdieter Lang, Jul 29 2014: (Start)
O.g.f. for the row polynomials (rising powers of x) R(n,x) = x^n*F(n,1/x): (1+z)/(1 - 2*x*z - z^2). Riordan triangle ((1+z)/(1-z^2), 2*z/(1-z^2)). See a comment above.
Recurrence for the row polynomials R(n,x) = (1+x)*R(n-1,x) - (-1)^n*x*R(n-1,-x), n >= 1, R(0,x) = 1.
R(n,x) = Ftilde(n,2*x) + Ftilde(n-1,2*x) with the monic Fibonacci polynomials Ftilde(n,x) given in A168561.
Recurrence for the triangle: R(n,m) = R(n-1,m) + (1 + (-1)^(n-m))*R(n-1,m-1), n >= m >= 1, R(n,m) = 0 if n < m, R(n,0) = 1.
O.g.f. column sequences ((1+x)/(1-x^2))*(2*x/(1-x^2))^m, m >= 0. See A000012, 2*A004526, 4*A008805, 8*A058187, 16*A189976, 32*A189980, ...
Row sums A078057. Alternating row sums A123335.
(End)

Offset corrected to 0. Cf.s added, keyword easy added by Wolfdieter Lang, Jul 29 2014

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A194005 Triangle of the coefficients of an (n+1)-th order differential equation associated with A103631.

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A244419 Coefficient triangle of polynomials related to the Dirichlet kernel. Rising powers. Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2)).

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A157751 Triangle of coefficients of polynomials F(n,x) in descending powers of x generated by F(n,x)=(x+1)*F(n-1,x)+F(n-1,-x), with initial F(0,x)=1.

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