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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A049310 Triangle of coefficients of Chebyshev's S(n,x) := U(n,x/2) polynomials (exponents in increasing order).

Original entry on oeis.org

1, 0, 1, -1, 0, 1, 0, -2, 0, 1, 1, 0, -3, 0, 1, 0, 3, 0, -4, 0, 1, -1, 0, 6, 0, -5, 0, 1, 0, -4, 0, 10, 0, -6, 0, 1, 1, 0, -10, 0, 15, 0, -7, 0, 1, 0, 5, 0, -20, 0, 21, 0, -8, 0, 1, -1, 0, 15, 0, -35, 0, 28, 0, -9, 0, 1, 0, -6, 0, 35, 0, -56, 0, 36, 0, -10, 0, 1, 1, 0, -21, 0, 70, 0, -84, 0
Offset: 0

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Comments

G.f. for row polynomials S(n,x) (signed triangle): 1/(1-x*z+z^2). Unsigned triangle |a(n,m)| has Fibonacci polynomials F(n+1,x) as row polynomials with g.f. 1/(1-x*z-z^2). |a(n,m)| triangle has rows of Pascal's triangle A007318 in the even-numbered diagonals (odd-numbered ones have only 0's).
Row sums (unsigned triangle) A000045(n+1) (Fibonacci). Row sums (signed triangle) S(n,1) sequence = periodic(1,1,0,-1,-1,0) = A010892.
Alternating row sums A049347(n) = S(n,-1) = periodic(1,-1,0). - Wolfdieter Lang, Nov 04 2011
S(n,x) is the characteristic polynomial of the adjacency matrix of the n-path. - Michael Somos, Jun 24 2002
S(n,x) is also the matching polynomial of the n-path. - Eric W. Weisstein, Apr 10 2017
|T(n,k)| = number of compositions of n+1 into k+1 odd parts. Example: |T(7,3)| = 10 because we have (1,1,3,3), (1,3,1,3), (1,3,3,1), (3,1,1,3), (3,1,3,1), (3,3,1,1), (1,1,1,5), (1,1,5,1), (1,5,1,1) and (5,1,1,1). - Emeric Deutsch, Apr 09 2005
S(n,x)= R(n,x) + S(n-2,x), n >= 2, S(-1,x)=0, S(0,x)=1, R(n,x):=2*T(n,x/2) = Sum_{m=0..n} A127672(n,m)*x^m (monic integer Chebyshev T-Polynomials). This is the rewritten so-called trace of the transfer matrix formula for the T-polynomials. - Wolfdieter Lang, Dec 02 2010
In a regular N-gon inscribed in a unit circle, the side length is d(N,1) = 2*sin(Pi/N). The length ratio R(N,k):=d(N,k)/d(N,1) for the (k-1)-th diagonal, with k from {2,3,...,floor(N/2)}, N >= 4, equals S(k-1,x) = sin(k*Pi/N)/sin(Pi/N) with x=rho(N):=R(N,2) = 2*cos(Pi/N). Example: N=7 (heptagon), rho=R(7,2), sigma:=R(N,3) = S(2,rho) = rho^2 - 1. Motivated by the quoted paper by P. Steinbach. - Wolfdieter Lang, Dec 02 2010
From Wolfdieter Lang, Jul 12 2011: (Start)
In q- or basic analysis, q-numbers are [n]_q := S(n-1,q+1/q) = (q^n-(1/q)^n)/(q-1/q), with the row polynomials S(n,x), n >= 0.
The zeros of the row polynomials S(n-1,x) are (from those of Chebyshev U-polynomials):
x(n-1;k) = +- t(k,rho(n)), k = 1..ceiling((n-1)/2), n >= 2, with t(n,x) the row polynomials of A127672 and rho(n):= 2*cos(Pi/n). The simple vanishing zero for even n appears here as +0 and -0.
Factorization of the row polynomials S(n-1,x), x >= 1, in terms of the minimal polynomials of cos(2 Pi/2), called Psi(n,x), with coefficients given by A181875/A181876:
S(n-1,x) = (2^(n-1))*Product_{n>=1}(Psi(d,x/2), 2 < d | 2n).
(From the rewritten eq. (3) of the Watkins and Zeitlin reference, given under A181872.) [See the W. Lang ArXiv link, Proposition 9, eq. (62). - Wolfdieter Lang, Apr 14 2018]
(End)
The discriminants of the S(n,x) polynomials are found in A127670. - Wolfdieter Lang, Aug 03 2011
This is an example for a subclass of Riordan convolution arrays (lower triangular matrices) called Bell arrays. See the L. W. Shapiro et al. reference under A007318. If a Riordan array is named (G(z),F(z)) with F(z)=z*Fhat(z), the o.g.f. for the row polynomials is G(z)/(1-x*z*Fhat(z)), and it becomes a Bell array if G(z)=Fhat(z). For the present Bell type triangle G(z)=1/(1+z^2) (see the o.g.f. comment above). This leads to the o.g.f. for the column no. k, k >= 0, x^k/(1+x^2)^(k+1) (see the formula section), the one for the row sums and for the alternating row sums (see comments above). The Riordan (Bell) A- and Z-sequences (defined in a W. Lang link under A006232, with references) have o.g.f.s 1-x*c(x^2) and -x*c(x^2), with the o.g.f. of the Catalan numbers A000108. Together they lead to a recurrence given in the formula section. - Wolfdieter Lang, Nov 04 2011
The determinant of the N x N matrix S(N,[x[1], ..., x[N]]) with elements S(m-1,x[n]), for n, m = 1, 2, ..., N, and for any x[n], is identical with the determinant of V(N,[x[1], ..., x[N]]) with elements x[n]^(m-1) (a Vandermondian, which equals Product_{1 <= i < j<= N} (x[j] - x[i])). This is a special instance of a theorem valid for any N >= 1 and any monic polynomial system p(m,x), m>=0, with p(0,x) = 1. For this theorem see the Vein-Dale reference, p. 59. Thanks to L. Edson Jeffery for an email asking for a proof of the non-singularity of the matrix S(N,[x[1], ...., x[N]]) if and only if the x[j], j = 1..N, are pairwise distinct. - Wolfdieter Lang, Aug 26 2013
These S polynomials also appear in the context of modular forms. The rescaled Hecke operator T*n = n^((1-k)/2)*T_n acting on modular forms of weight k satisfies T*(p^n) = S(n, T*p), for each prime p and positive integer n. See the Koecher-Krieg reference, p. 223. - _Wolfdieter Lang, Jan 22 2016
For a shifted o.g.f. (mod signs), its compositional inverse, and connections to Motzkin and Fibonacci polynomials, non-crossing partitions and other combinatorial structures, see A097610. - Tom Copeland, Jan 23 2016
From M. Sinan Kul, Jan 30 2016; edited by Wolfdieter Lang, Jan 31 2016 and Feb 01 2016: (Start)
Solutions of the Diophantine equation u^2 + v^2 - k*u*v = 1 for integer k given by (u(k,n), v(k,n)) = (S(n,k), S(n-1,k)) because of the Cassini-Simson identity: S(n,x)^2 - S(n+1,x)*S(n-1, x) = 1, after use of the S-recurrence. Note that S(-n, x) = -S(-n-2, x), n >= 1, and the periodicity of some S(n, k) sequences.
Hence another way to obtain the row polynomials would be to take powers of the matrix [x, -1; 1,0]: S(n, x) = (([x, -1; 1, 0])^n)[1,1], n >= 0.
See also a Feb 01 2016 comment on A115139 for a well-known S(n, x) sum formula.
Then we have with the present T triangle
A039834(n) = -i^(n+1)*T(n-1, k) where i is the imaginary unit and n >= 0.
A051286(n) = Sum_{i=0..n} T(n,i)^2 (see the Philippe Deléham, Nov 21 2005 formula),
A181545(n) = Sum_{i=0..n+1} abs(T(n,i)^3),
A181546(n) = Sum_{i=0..n+1} T(n,i)^4,
A181547(n) = Sum_{i=0..n+1} abs(T(n,i)^5).
S(n, 0) = A056594(n), and for k = 1..10 the sequences S(n-1, k) with offset n = 0 are A128834, A001477, A001906, A001353, A004254, A001109, A004187, A001090, A018913, A004189.
(End)
For more on the Diophantine equation presented by Kul, see the Ismail paper. - Tom Copeland, Jan 31 2016
The o.g.f. for the Legendre polynomials L(n,x) is 1 / sqrt(1- 2x*z + z^2), and squaring it gives the o.g.f. of U(n,x), A053117, so Sum_{k=0..n} L(k,x/2) L(n-k,x/2) = S(n,x). This gives S(n,x) = L(n/2,x/2)^2 + 2*Sum_{k=0..n/2-1} L(k,x/2) L(n-k,x/2) for n even and S(n,x) = 2*Sum_{k=0..(n-1)/2} L(k,x/2) L(n-k,x/2) for odd n. For a connection to elliptic curves and modular forms, see A053117. For the normalized Legendre polynomials, see A100258. For other properties and relations to other polynomials, see Allouche et al. - Tom Copeland, Feb 04 2016
LG(x,h1,h2) = -log(1 - h1*x + h2*x^2) = Sum_{n>0} F(n,-h1,h2,0,..,0) x^n/n is a log series generator of the bivariate row polynomials of A127672 with A127672(0,0) = 0 and where F(n,b1,b2,..,bn) are the Faber polynomials of A263916. Exp(LG(x,h1,h2)) = 1 / (1 - h1*x + h2*x^2 ) is the o.g.f. of the bivariate row polynomials of this entry. - Tom Copeland, Feb 15 2016 (Instances of the bivariate o.g.f. for this entry are on pp. 5 and 18 of Sunada. - Tom Copeland, Jan 18 2021)
For distinct odd primes p and q the Legendre symbol can be written as Legendre(q,p) = Product_{k=1..P} S(q-1, 2*cos(2*Pi*k/p)), with P = (p-1)/2. See the Lemmermeyer reference, eq. (8.1) on p. 236. Using the zeros of S(q-1, x) (see above) one has S(q-1, x) = Product_{l=1..Q} (x^2 - (2*cos(Pi*l/q))^2), with Q = (q-1)/2. Thus S(q-1, 2*cos(2*Pi*k/p)) = ((-4)^Q)*Product_{l=1..Q} (sin^2(2*Pi*k/p) - sin^2(Pi*l/q)) = ((-4)^Q)*Product_{m=1..Q} (sin^2(2*Pi*k/p) - sin^2(2*Pi*m/q)). For the proof of the last equality see a W. Lang comment on the triangle A057059 for n = Q and an obvious function f. This leads to Eisenstein's proof of the quadratic reciprocity law Legendre(q,p) = ((-1)^(P*Q)) * Legendre(p,q), See the Lemmermeyer reference, pp. 236-237. - Wolfdieter Lang, Aug 28 2016
For connections to generalized Fibonacci polynomials, compare their generating function on p. 5 of the Amdeberhan et al. link with the o.g.f. given above for the bivariate row polynomials of this entry. - Tom Copeland, Jan 08 2017
The formula for Ramanujan's tau function (see A000594) for prime powers is tau(p^k) = p^(11*k/2)*S(k, p^(-11/2)*tau(p)) for k >= 1, and p = A000040(n), n >= 1. See the Hardy reference, p. 164, eqs. (10.3.4) and (10.3.6) rewritten in terms of S. - Wolfdieter Lang, Jan 27 2017
From Wolfdieter Lang, May 08 2017: (Start)
The number of zeros Z(n) of the S(n, x) polynomials in the open interval (-1,+1) is 2*b(n) for even n >= 0 and 1 + 2*b(n) for odd n >= 1, where b(n) = floor(n/2) - floor((n+1)/3). This b(n) is the number of integers k in the interval (n+1)/3 < k <= floor(n/2). See a comment on the zeros of S(n, x) above, and b(n) = A008615(n-2), n >= 0. The numbers Z(n) have been proposed (with a conjecture related to A008611) by Michel Lagneau, as the number of zeros of Fibonacci polynomials on the imaginary axis (-I,+I), with I=sqrt(-1). They are Z(n) = A008611(n-1), n >= 0, with A008611(-1) = 0. Also Z(n) = A194960(n-4), n >= 0. Proof using the A008611 version. A194960 follows from this.
In general the number of zeros Z(a;n) of S(n, x) for n >= 0 in the open interval (-a,+a) for a from the interval (0,2) (x >= 2 never has zeros, and a=0 is trivial: Z(0;n) = 0) is with b(a;n) = floor(n//2) - floor((n+1)*arccos(a/2)/Pi), as above Z(a;n) = 2*b(a;n) for even n >= 0 and 1 + 2*b(a;n) for odd n >= 1. For the closed interval [-a,+a] Z(0;n) = 1 and for a from (0,1) one uses for Z(a;n) the values b(a;n) = floor(n/2) - ceiling((n+1)*arccos(a/2)/Pi) + 1. (End)
The Riordan row polynomials S(n, x) (Chebyshev S) belong to the Boas-Buck class (see a comment and references in A046521), hence they satisfy the Boas-Buck identity: (E_x - n*1)*S(n, x) = (E_x + 1)*Sum_{p=0..n-1} (1 - (-1)^p)*(-1)^((p+1)/2)*S(n-1-p, x), for n >= 0, where E_x = x*d/dx (Euler operator). For the triangle T(n, k) this entails a recurrence for the sequence of column k, given in the formula section. - Wolfdieter Lang, Aug 11 2017
The e.g.f. E(x,t) := Sum_{n>=0} (t^n/n!)*S(n,x) for the row polynomials is obtained via inverse Laplace transformation from the above given o.g.f. as E(x,t) = ((1/xm)*exp(t/xm) - (1/xp)*exp(t/xp) )/(xp - xm) with xp = (x + sqrt(x^2-4))/2 and xm = (x - sqrt(x^2-4))/2. - Wolfdieter Lang, Nov 08 2017
From Wolfdieter Lang, Apr 12 2018: (Start)
Factorization of row polynomials S(n, x), for n >= 1, in terms of C polynomials (not Chebyshev C) with coefficients given in A187360. This is obtained from the factorization into Psi polynomials (see the Jul 12 2011 comment above) but written in terms of minimal polynomials of 2*cos(2*Pi/n) with coefficients in A232624:
S(2*k, x) = Product_{2 <= d | (2*k+1)} C(d, x)*(-1)^deg(d)*C(d, -x), with deg(d) = A055034(d) the degree of C(d, x).
S(2*k+1, x) = Product_{2 <= d | 2*(k+1)} C(d, x) * Product_{3 <= 2*d + 1 | (k+1)} (-1)^(deg(2*d+1))*C(2*d+1, -x).
Note that (-1)^(deg(2*d+1))*C(2*d+1, -x)*C(2*d+1, x) pairs always appear.
The number of C factors of S(2*k, x), for k >= 0, is 2*(tau(2*k+1) - 1) = 2*(A099774(k+1) - 1) = 2*A095374(k), and for S(2*k+1, x), for k >= 0, it is tau(2*(k+1)) + tau_{odd}(k+1) - 2 = A302707(k), with tau(2*k+1) = A099774(k+1), tau(n) = A000005 and tau(2*(k+1)) = A099777(k+1).
For the reverse problem, the factorization of C polynomials into S polynomials, see A255237. (End)
The S polynomials with general initial conditions S(a,b;n,x) = x*S(a,b;n-1,x) - S(a,b;n-2,x), for n >= 1, with S(a,b;-1,x) = a and S(a,b;0,x) = b are S(a,b;n,x) = b*S(n, x) - a*S(n-1, x), for n >= -1. Recall that S(-2, x) = -1 and S(-1, x) = 0. The o.g.f. is G(a,b;z,x) = (b - a*z)/(1 - x*z + z^2). - Wolfdieter Lang, Oct 18 2019
Also the convolution triangle of A101455. - Peter Luschny, Oct 06 2022
From Wolfdieter Lang, Apr 26 2023: (Start)
Multi-section of S-polynomials: S(m*n+k, x) = S(m+k, x)*S(n-1, R(m, x)) - S(k, x)*S(n-2, R(m, x)), with R(n, x) = S(n, x) - S(n-2, x) (see A127672), S(-2, x) = -1, and S(-1, x) = 0, for n >= 0, m >= 1, and k = 0, 1, ..., m-1.
O.g.f. of {S(m*n+k, y)}_{n>=0}: G(m,k,y,x) = (S(k, y) - (S(k, y)*R(m, y) - S(m+k, y))*x)/(1 - R(m,y)*x + x^2).
See eqs. (40) and (49), with r = x or y and s =-1, of the G. Detlefs and W. Lang link at A034807. (End)
S(n, x) for complex n and complex x: S(n, x) = ((-i/2)/sqrt(1 - (x/2)^2))*(q(x/2)*exp(+n*log(q(x/2))) - (1/q(x/2))*exp(-n*log(q(x/2)))), with q(x) = x + sqrt(1 - x^2)*i. Here log(z) = |z| + Arg(z)*i, with Arg(z) from [-Pi,+Pi) (principal branch). This satisfies the recurrence relation for S because it is derived from the Binet - de Moivre formula for S. Examples: S(n/m, 0) = cos((n/m)*Pi/4), for n >= 0 and m >= 1. S(n*i, 0) = (1/2)*(1 + exp(n*Pi))*exp(-(n/2)*Pi), for n >= 0. S(1+i, 2+i) = 0.6397424847... + 1.0355669490...*i. Thanks to Roberto Alfano for asking a question leading to this formula. - Wolfdieter Lang, Jun 05 2023
Lim_{n->oo} S(n, x)/S(n-1, x) = r(x) = (x - sqrt(x^2 -4))/2, for |x| >= 2. For x = +-2, this limit is +-1. - Wolfdieter Lang, Nov 15 2023

Examples

			The triangle T(n, k) begins:
  n\k  0  1   2   3   4   5   6    7   8   9  10  11
  0:   1
  1:   0  1
  2:  -1  0   1
  3:   0 -2   0   1
  4:   1  0  -3   0   1
  5:   0  3   0  -4   0   1
  6:  -1  0   6   0  -5   0   1
  7:   0 -4   0  10   0  -6   0    1
  8:   1  0 -10   0  15   0  -7    0   1
  9:   0  5   0 -20   0  21   0   -8   0   1
  10: -1  0  15   0 -35   0  28    0  -9   0   1
  11:  0 -6   0  35   0 -56   0   36   0 -10   0   1
  ... Reformatted and extended by _Wolfdieter Lang_, Oct 24 2012
For more rows see the link.
E.g., fourth row {0,-2,0,1} corresponds to polynomial S(3,x)= -2*x + x^3.
From _Wolfdieter Lang_, Jul 12 2011: (Start)
Zeros of S(3,x) with rho(4)= 2*cos(Pi/4) = sqrt(2):
  +- t(1,sqrt(2)) = +- sqrt(2) and
  +- t(2,sqrt(2)) = +- 0.
Factorization of S(3,x) in terms of Psi polynomials:
S(3,x) = (2^3)*Psi(4,x/2)*Psi(8,x/2) = x*(x^2-2).
(End)
From _Wolfdieter Lang_, Nov 04 2011: (Start)
A- and Z- sequence recurrence:
T(4,0) = - (C(0)*T(3,1) + C(1)*T(3,3)) = -(-2 + 1) = +1,
T(5,3) = -3 - 1*1 = -4.
(End)
Boas-Buck recurrence for column k = 2, n = 6: S(6, 2) = (3/4)*(0 - 2* S(4 ,2) + 0 + 2*S(2, 2)) = (3/4)*(-2*(-3) + 2) = 6. - _Wolfdieter Lang_, Aug 11 2017
From _Wolfdieter Lang_, Apr 12 2018: (Start)
Factorization into C polynomials (see the Apr 12 2018 comment):
S(4, x) = 1 - 3*x^2 + x^4 = (-1 + x + x^2)*(-1 - x + x^2) = (-C(5, -x)) * C(5, x); the number of factors is 2 = 2*A095374(2).
S(5, x) = 3*x - 4*x^3 + x^5 = x*(-1 + x)*(1 + x)*(-3 + x^2) = C(2, x)*C(3, x)*(-C(3, -x))*C(6, x); the number of factors is 4 = A302707(2). (End)
		

References

  • G. H. Hardy, Ramanujan: twelve lectures on subjects suggested by his life and work, AMS Chelsea Publishing, Providence, Rhode Island, 2002, p. 164.
  • Max Koecher and Aloys Krieg, Elliptische Funktionen und Modulformen, 2. Auflage, Springer, 2007, p. 223.
  • Franz Lemmermeyer, Reciprocity Laws. From Euler to Eisenstein, Springer, 2000.
  • D. S. Mitrinovic, Analytic Inequalities, Springer-Verlag, 1970; p. 232, Sect. 3.3.38.
  • Theodore J. Rivlin, Chebyshev polynomials: from approximation theory to algebra and number theory, 2. ed., Wiley, New York, 1990, pp. 60 - 61.
  • R. Vein and P. Dale, Determinants and Their Applications in Mathematical Physics, Springer, 1999.

Crossrefs

Cf. A000005, A000217, A000292, A000332, A000389, A001227, A007318, A008611, A008615, A101455, A010892, A011973, A053112 (without zeros), A053117, A053119 (reflection), A053121 (inverse triangle), A055034, A097610, A099774, A099777, A100258, A112552 (first column clipped), A127672, A168561 (absolute values), A187360. A194960, A232624, A255237.
Triangles of coefficients of Chebyshev's S(n,x+k) for k = 5, 4, 3, 2, 1, 0, -1, -2, -3, -4, -5: A207824, A207823, A125662, A078812, A101950, A049310, A104562, A053122, A207815, A159764, A123967.

Programs

  • Magma
    A049310:= func< n,k | ((n+k) mod 2) eq 0 select (-1)^(Floor((n+k)/2)+k)*Binomial(Floor((n+k)/2), k) else 0 >;
    [A049310(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Jul 25 2022
  • Maple
    A049310 := proc(n,k): binomial((n+k)/2,(n-k)/2)*cos(Pi*(n-k)/2)*(1+(-1)^(n-k))/2 end: seq(seq(A049310(n,k), k=0..n),n=0..11); # Johannes W. Meijer, Aug 08 2011
    # Uses function PMatrix from A357368. Adds a row above and a column to the left.
    PMatrix(10, n -> ifelse(irem(n, 2) = 0, 0, (-1)^iquo(n-1, 2))); # Peter Luschny, Oct 06 2022
  • Mathematica
    t[n_, k_] /; EvenQ[n+k] = ((-1)^((n+k)/2+k))*Binomial[(n+k)/2, k]; t[n_, k_] /; OddQ[n+k] = 0; Flatten[Table[t[n, k], {n, 0, 12}, {k, 0, n}]][[;; 86]] (* Jean-François Alcover, Jul 05 2011 *)
    Table[Coefficient[(-I)^n Fibonacci[n + 1, - I x], x, k], {n, 0, 10}, {k, 0, n}] //Flatten (* Clark Kimberling, Aug 02 2011; corrected by Eric W. Weisstein, Apr 06 2017 *)
    CoefficientList[ChebyshevU[Range[0, 10], -x/2], x] // Flatten (* Eric W. Weisstein, Apr 06 2017 *)
    CoefficientList[Table[(-I)^n Fibonacci[n + 1, -I x], {n, 0, 10}], x] // Flatten (* Eric W. Weisstein, Apr 06 2017 *)
  • PARI
    {T(n, k) = if( k<0 || k>n || (n + k)%2, 0, (-1)^((n + k)/2 + k) * binomial((n + k)/2, k))} /* Michael Somos, Jun 24 2002 */
    
  • SageMath
    @CachedFunction
    def A049310(n,k):
        if n< 0: return 0
        if n==0: return 1 if k == 0 else 0
        return A049310(n-1,k-1) - A049310(n-2,k)
    for n in (0..9): [A049310(n,k) for k in (0..n)] # Peter Luschny, Nov 20 2012
    

Formula

T(n,k) := 0 if n < k or n+k odd, otherwise ((-1)^((n+k)/2+k))*binomial((n+k)/2, k); T(n, k) = -T(n-2, k)+T(n-1, k-1), T(n, -1) := 0 =: T(-1, k), T(0, 0)=1, T(n, k)= 0 if n < k or n+k odd; g.f. k-th column: (1 / (1 + x^2)^(k + 1)) * x^k. - Michael Somos, Jun 24 2002
T(n,k) = binomial((n+k)/2, (n-k)/2)*cos(Pi*(n-k)/2)*(1+(-1)^(n-k))/2. - Paul Barry, Aug 28 2005
Sum_{k=0..n} T(n,k)^2 = A051286(n). - Philippe Deléham, Nov 21 2005
Recurrence for the (unsigned) Fibonacci polynomials: F(1)=1, F(2)=x; for n > 2, F(n) = x*F(n-1) + F(n-2).
From Wolfdieter Lang, Nov 04 2011: (Start)
The Riordan A- and Z-sequences, given in a comment above, lead together to the recurrence:
T(n,k) = 0 if n < k, if k=0 then T(0,0)=1 and
T(n,0)= -Sum_{i=0..floor((n-1)/2)} C(i)*T(n-1,2*i+1), otherwise T(n,k) = T(n-1,k-1) - Sum_{i=1..floor((n-k)/2)} C(i)*T(n-1,k-1+2*i), with the Catalan numbers C(n)=A000108(n).
(End)
The row polynomials satisfy also S(n,x) = 2*(T(n+2, x/2) - T(n, x/2))/(x^2-4) with the Chebyshev T-polynomials. Proof: Use the trace formula 2*T(n, x/2) = S(n, x) - S(n-2, x) (see the Dec 02 2010 comment above) and the S-recurrence several times. This is a formula which expresses the S- in terms of the T-polynomials. - Wolfdieter Lang, Aug 07 2014
From Tom Copeland, Dec 06 2015: (Start)
The non-vanishing, unsigned subdiagonals Diag_(2n) contain the elements D(n,k) = Sum_{j=0..k} D(n-1,j) = (k+1) (k+2) ... (k+n) / n! = binomial(n+k,n), so the o.g.f. for the subdiagonal is (1-x)^(-(n+1)). E.g., Diag_4 contains D(2,3) = D(1,0) + D(1,1) + D(1,2) + D(1,3) = 1 + 2 + 3 + 4 = 10 = binomial(5,2). Diag_4 is shifted A000217; Diag_6, shifted A000292: Diag_8, shifted A000332; and Diag_10, A000389.
The non-vanishing antidiagonals are signed rows of the Pascal triangle A007318.
For a reversed, unsigned version with the zeros removed, see A011973. (End)
The Boas-Buck recurrence (see a comment above) for the sequence of column k is: S(n, k) = ((k+1)/(n-k))*Sum_{p=0..n-1-k} (1 - (-1)^p)*(-1)^((p+1)/2) * S(n-1-p, k), for n > k >= 0 and input S(k, k) = 1. - Wolfdieter Lang, Aug 11 2017
The m-th row consecutive nonzero entries in order are (-1)^c*(c+b)!/c!b! with c = m/2, m/2-1, ..., 0 and b = m-2c if m is even and with c = (m-1)/2, (m-1)/2-1, ..., 0 with b = m-2c if m is odd. For the 8th row starting at a(36) the 5 consecutive nonzero entries in order are 1,-10,15,-7,1 given by c = 4,3,2,1,0 and b = 0,2,4,6,8. - Richard Turk, Aug 20 2017
O.g.f.: exp( Sum_{n >= 0} 2*T(n,x/2)*t^n/n ) = 1 + x*t + (-1 + x^2)*t^2 + (-2*x + x^3)*t^3 + (1 - 3*x^2 + x^4)*t^4 + ..., where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Aug 15 2022

A048993 Triangle of Stirling numbers of 2nd kind, S(n,k), n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 7, 6, 1, 0, 1, 15, 25, 10, 1, 0, 1, 31, 90, 65, 15, 1, 0, 1, 63, 301, 350, 140, 21, 1, 0, 1, 127, 966, 1701, 1050, 266, 28, 1, 0, 1, 255, 3025, 7770, 6951, 2646, 462, 36, 1, 0, 1, 511, 9330, 34105, 42525, 22827, 5880, 750, 45, 1
Offset: 0

Views

Author

N. J. A. Sloane, Dec 11 1999

Keywords

Comments

Also known as Stirling set numbers.
S(n,k) enumerates partitions of an n-set into k nonempty subsets.
The o.g.f. for the sequence of diagonal k (k=0 for the main diagonal) is G(k,x) = ((x^k)/(1-x)^(2*k+1))*Sum_{m=0..k-1} A008517(k,m+1)*x^m. A008517 is the second-order Eulerian triangle. - Wolfdieter Lang, Oct 14 2005
From Philippe Deléham, Nov 14 2007: (Start)
Sum_{k=0..n} S(n,k)*x^k = B_n(x), where B_n(x) = Bell polynomials.
The first few Bell polynomials are:
B_0(x) = 1;
B_1(x) = 0 + x;
B_2(x) = 0 + x + x^2;
B_3(x) = 0 + x + 3x^2 + x^3;
B_4(x) = 0 + x + 7x^2 + 6x^3 + x^4;
B_5(x) = 0 + x + 15x^2 + 25x^3 + 10x^4 + x^5;
B_6(x) = 0 + x + 31x^2 + 90x^3 + 65x^4 + 15x^5 + x^6;
(End)
This is the Sheffer triangle (1, exp(x) - 1), an exponential (binomial) convolution triangle. The a-sequence is given by A006232/A006233 (Cauchy sequence). The z-sequence is the zero sequence. See the link under A006232 for the definition and use of these sequences. The row sums give A000110 (Bell), and the alternating row sums give A000587 (see the Philippe Deléham formulas and crossreferences below). - Wolfdieter Lang, Oct 16 2014
Also the inverse Bell transform of the factorial numbers (A000142). For the definition of the Bell transform see A264428 and for cross-references A265604. - Peter Luschny, Dec 31 2015
From Wolfdieter Lang, Feb 21 2017: (Start)
The transposed (trans) of this lower triagonal Sheffer matrix of the associated type S = (1, exp(x) - 1) (taken as N X N matrix for arbitrarily large N) provides the transition matrix from the basis {x^n/n!}, n >= 0, to the basis {y^n/n!}, n >= 0, with y^n/n! = Sum_{m>=n} S^{trans}(n, m) x^m/m! = Sum_{m>=0} x^m/m!*S(m, n).
The Sheffer transform with S = (g, f) of a sequence {a_n} to {b_n} for n >= 0, in matrix notation vec(b) = S vec(a), satisfies, with e.g.f.s A and B, B(x) = g(x)*A(f(x)) and B(x) = A(y(x)) identically, with vec(xhat) = S^{trans,-1} vec(yhat) in symbolic notation with vec(xhat)_n = x^n/n! (similarly for vec(yhat)).
(End)
Number of partitions of {1, 2, ..., n+1} into k+1 nonempty subsets such that no subset contains two adjacent numbers. - Thomas Anton, Sep 26 2022

Examples

			The triangle S(n,k) begins:
  n\k 0 1    2     3      4       5       6      7      8     9   10 11 12
  0:  1
  1:  0 1
  2:  0 1    1
  3:  0 1    3     1
  4:  0 1    7     6      1
  5:  0 1   15    25     10       1
  6:  0 1   31    90     65      15       1
  7:  0 1   63   301    350     140      21      1
  8:  0 1  127   966   1701    1050     266     28      1
  9:  0 1  255  3025   7770    6951    2646    462     36     1
 10:  0 1  511  9330  34105   42525   22827   5880    750    45    1
 11:  0 1 1023 28501 145750  246730  179487  63987  11880  1155   55  1
 12:  0 1 2047 86526 611501 1379400 1323652 627396 159027 22275 1705 66  1
 ... reformatted and extended - _Wolfdieter Lang_, Oct 16 2014
Completely symmetric function S(4, 2) = h^{(2)}_2 = 1^2 + 2^2 + 1^1*2^1 = 7; S(5, 2) = h^{(2)}_3 = 1^3 + 2^3 + 1^2*2^1 + 1^1*2^2 = 15. - _Wolfdieter Lang_, May 26 2017
From _Wolfdieter Lang_, Aug 11 2017: (Start)
Recurrence: S(5, 3) = S(4, 2) + 2*S(4, 3) = 7 + 3*6 = 25.
Boas-Buck recurrence for column m = 3, and n = 5: S(5, 3) = (3/2)*((5/2)*S(4, 3) + 10*Bernoulli(2)*S(3, 3)) = (3/2)*(15 + 10*(1/6)*1) = 25. (End)
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 835.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 310.
  • J. H. Conway and R. K. Guy, The Book of Numbers, Springer, p. 92.
  • F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 223.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 244.
  • J. Riordan, An Introduction to Combinatorial Analysis, p. 48.

Crossrefs

See especially A008277 which is the main entry for this triangle.
A000110(n) = sum(S(n, k)) k=0..n, n >= 0. Cf. A085693.
Cf. A084938, A106800 (mirror image), A138378, A213061 (mod 2).

Programs

  • Haskell
    a048993 n k = a048993_tabl !! n !! k
    a048993_row n = a048993_tabl !! n
    a048993_tabl = iterate (\row ->
       [0] ++ (zipWith (+) row $ zipWith (*) [1..] $ tail row) ++ [1]) [1]
    -- Reinhard Zumkeller, Mar 26 2012
  • Maple
    for n from 0 to 10 do seq(Stirling2(n,k),k=0..n) od; # yields sequence in triangular form # Emeric Deutsch, Nov 01 2006
  • Mathematica
    t[n_, k_] := StirlingS2[n, k]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Robert G. Wilson v *)
  • Maxima
    create_list(stirling2(n,k),n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */
    
  • PARI
    for(n=0, 22, for(k=0, n, print1(stirling(n, k, 2), ", ")); print()); \\ Joerg Arndt, Apr 21 2013
    

Formula

S(n, k) = k*S(n-1, k) + S(n-1, k-1), n > 0; S(0, k) = 0, k > 0; S(0, 0) = 1.
Equals [0, 1, 0, 2, 0, 3, 0, 4, 0, 5, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] where DELTA is Deléham's operator defined in A084938.
Sum_{k = 0..n} x^k*S(n, k) = A213170(n), A000587(n), A000007(n), A000110(n), A001861(n), A027710(n), A078944(n), A144180(n), A144223(n), A144263(n) respectively for x = -2, -1, 0, 1, 2, 3, 4, 5, 6, 7. - Philippe Deléham, May 09 2004, Feb 16 2013
S(n, k) = Sum_{i=0..k} (-1)^(k+i)binomial(k, i)i^n/k!. - Paul Barry, Aug 05 2004
Sum_{k=0..n} k*S(n,k) = B(n+1)-B(n), where B(q) are the Bell numbers (A000110). - Emeric Deutsch, Nov 01 2006
Equals the inverse binomial transform of A008277. - Gary W. Adamson, Jan 29 2008
G.f.: 1/(1-xy/(1-x/(1-xy/(1-2x/(1-xy/1-3x/(1-xy/(1-4x/(1-xy/(1-5x/(1-... (continued fraction equivalent to Deléham DELTA construction). - Paul Barry, Dec 06 2009
G.f.: 1/Q(0), where Q(k) = 1 - (y+k)*x - (k+1)*y*x^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 09 2013
Inverse of padded A008275 (padded just as A048993 = padded A008277). - Tom Copeland, Apr 25 2014
E.g.f. for the row polynomials s(n,x) = Sum_{k=0..n} S(n,k)*x^k is exp(x*(exp(z)-1)) (Sheffer property). E.g.f. for the k-th column sequence with k leading zeros is ((exp(x)-1)^k)/k! (Sheffer property). - Wolfdieter Lang, Oct 16 2014
G.f. for column k: x^k/Product_{j=1..k} (1-j*x), k >= 0 (with the empty product for k = 0 put to 1). See Abramowitz-Stegun, p. 824, 24.1.4 B. - Wolfdieter Lang, May 26 2017
Boas-Buck recurrence for column sequence m: S(n, k) = (k/(n - k))*(n*S(n-1, k)/2 + Sum_{p=k..n-2} (-1)^(n-p)*binomial(n,p)*Bernoulli(n-p)*S(p, k)), for n > k >= 0, with input T(k,k) = 1. See a comment and references in A282629. An example is given below. - Wolfdieter Lang, Aug 11 2017
The n-th row polynomial has the form x o x o ... o x (n factors), where o denotes the white diamond multiplication operator defined in Bala - see Example E4. - Peter Bala, Jan 07 2018
Sum_{k=1..n} k*S(n,k) = A138378(n). - Alois P. Heinz, Jan 07 2022
S(n,k) = Sum_{j=k..n} (-1)^(j-k)*A059297(n,j)*A354794(j,k). - Mélika Tebni, Jan 27 2023

A027641 Numerator of Bernoulli number B_n.

Original entry on oeis.org

1, -1, 1, 0, -1, 0, 1, 0, -1, 0, 5, 0, -691, 0, 7, 0, -3617, 0, 43867, 0, -174611, 0, 854513, 0, -236364091, 0, 8553103, 0, -23749461029, 0, 8615841276005, 0, -7709321041217, 0, 2577687858367, 0, -26315271553053477373, 0, 2929993913841559, 0, -261082718496449122051
Offset: 0

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Author

Keywords

Comments

a(n)/A027642(n) (Bernoulli numbers) provide the a-sequence for the Sheffer matrix A094816 (coefficients of orthogonal Poisson-Charlier polynomials). See the W. Lang link under A006232 for a- and z-sequences for Sheffer matrices. The corresponding z-sequence is given by the rationals A130189(n)/A130190(n).
Harvey (2008) describes a new algorithm for computing Bernoulli numbers. His method is to compute B(k) modulo p for many small primes p and then reconstruct B(k) via the Chinese Remainder Theorem. The time complexity is O(k^2 log(k)^(2+eps)). The algorithm is especially well-suited to parallelization. - Jonathan Vos Post, Jul 09 2008
Regard the Bernoulli numbers as forming a vector = B_n, and the variant starting (1, 1/2, 1/6, 0, -1/30, ...), (i.e., the first 1/2 has sign +) as forming a vector Bv_n. The relationship between the Pascal triangle matrix, B_n, and Bv_n is as follows: The binomial transform of B_n = Bv_n. B_n is unchanged when multiplied by the Pascal matrix with rows signed (+-+-, ...), i.e., (1; -1,-1; 1,2,1; ...). Bv_n is unchanged when multiplied by the Pascal matrix with columns signed (+-+-, ...), i.e., (1; 1,-1; 1,-2,1; 1,-3,3,-1; ...). - Gary W. Adamson, Jun 29 2012
The sequence of the Bernoulli numbers B_n = a(n)/A027642(n) is the inverse binomial transform of the sequence {A164555(n)/A027642(n)}, illustrated by the fact that they appear as top row and left column in A190339. - Paul Curtz, May 13 2016
Named by de Moivre (1773; "the numbers of Mr. James Bernoulli") after the Swiss mathematician Jacob Bernoulli (1655-1705). - Amiram Eldar, Oct 02 2023

Examples

			B_n sequence begins 1, -1/2, 1/6, 0, -1/30, 0, 1/42, 0, -1/30, 0, 5/66, 0, -691/2730, 0, 7/6, 0, -3617/510, ...
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 810.
  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 49.
  • Harold T. Davis, Tables of the Mathematical Functions. Vols. 1 and 2, 2nd ed., 1963, Vol. 3 (with V. J. Fisher), 1962; Principia Press of Trinity Univ., San Antonio, TX, Vol. 2, p. 230.
  • Harold M. Edwards, Riemann's Zeta Function, Academic Press, NY, 1974; see p. 11.
  • Steven R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.6.1.
  • Herman H. Goldstine, A History of Numerical Analysis, Springer-Verlag, 1977; Section 2.6.
  • L. M. Milne-Thompson, Calculus of Finite Differences, 1951, p. 137.
  • Hans Rademacher, Topics in Analytic Number Theory, Springer, 1973, Chap. 1.

Crossrefs

This is the main entry for the Bernoulli numbers and has all the references, links and formulas. Sequences A027642 (the denominators of B_n) and A000367/A002445 = B_{2n} are also important!
A refinement is A194587.

Programs

  • Magma
    [Numerator(Bernoulli(n)): n in [0..40]]; // Vincenzo Librandi, Mar 17 2014
    
  • Maple
    B := n -> add((-1)^m*m!*Stirling2(n, m)/(m+1), m=0..n);
    B := n -> bernoulli(n);
    seq(numer(bernoulli(n)), n=0..40); # Zerinvary Lajos, Apr 08 2009
  • Mathematica
    Table[ Numerator[ BernoulliB[ n]], {n, 0, 40}] (* Robert G. Wilson v, Oct 11 2004 *)
    Numerator[ Range[0, 40]! CoefficientList[ Series[x/(E^x - 1), {x, 0, 40}], x]]
    Numerator[CoefficientList[Series[PolyGamma[1, 1/x]/x - x, {x, 0, 40}, Assumptions -> x > 0], x]] (* Vladimir Reshetnikov, Apr 24 2013 *)
  • Maxima
    B(n):=(-1)^((n))*sum((stirling1(n,k)*stirling2(n+k,n))/binomial(n+k,k),k,0,n);
    makelist(num(B(n)),n,0,20); /* Vladimir Kruchinin, Mar 16 2013 */
    
  • PARI
    a(n)=numerator(bernfrac(n))
    
  • Python
    from sympy import bernoulli
    from fractions import Fraction
    [bernoulli(i).as_numer_denom()[0] for i in range(51)]  # Indranil Ghosh, Mar 18 2017
    
  • Python
    from sympy import bernoulli
    def A027641(n): return bernoulli(n).p
    print([A027641(n) for n in range(80)])  # M. F. Hasler, Jun 11 2019
  • SageMath
    [bernoulli(n).numerator() for n in range(41)]  # Peter Luschny, Feb 19 2016
    
  • SageMath
    # Alternatively:
    def A027641_list(len):
        f, R, C = 1, [1], [1]+[0]*(len-1)
        for n in (1..len-1):
            f *= n
            for k in range(n, 0, -1):
                C[k] = C[k-1] / (k+1)
            C[0] = -sum(C[k] for k in (1..n))
            R.append((C[0]*f).numerator())
        return R
    A027641_list(41)  # Peter Luschny, Feb 20 2016
    

Formula

E.g.f: x/(exp(x) - 1); take numerators.
Recurrence: B^n = (1+B)^n, n >= 2 (interpreting B^j as B_j).
B_{2n}/(2n)! = 2*(-1)^(n-1)*(2*Pi)^(-2n) Sum_{k>=1} 1/k^(2n) (gives asymptotics) - Rademacher, p. 16, Eq. (9.1). In particular, B_{2*n} ~ (-1)^(n-1)*2*(2*n)!/(2*Pi)^(2*n).
Sum_{i=1..n-1} i^k = ((n+B)^(k+1)-B^(k+1))/(k+1) (interpreting B^j as B_j).
B_{n-1} = - Sum_{r=1..n} (-1)^r binomial(n, r) r^(-1) Sum_{k=1..r} k^(n-1). More concisely, B_n = 1 - (1-C)^(n+1), where C^r is replaced by the arithmetic mean of the first r n-th powers of natural numbers in the expansion of the right-hand side. [Bergmann]
Sum_{i>=1} 1/i^(2k) = zeta(2k) = (2*Pi)^(2k)*|B_{2k}|/(2*(2k)!).
B_{2n} = (-1)^(m-1)/2^(2m+1) * Integral{-inf..inf, [d^(m-1)/dx^(m-1) sech(x)^2 ]^2 dx} (see Grosset/Veselov).
Let B(s,z) = -2^(1-s)(i/Pi)^s s! PolyLog(s,exp(-2*i*Pi/z)). Then B(2n,1) = B_{2n} for n >= 1. Similarly the numbers B(2n+1,1), which might be called Co-Bernoulli numbers, can be considered, and it is remarkable that Leonhard Euler in 1755 already calculated B(3,1) and B(5,1) (Opera Omnia, Ser. 1, Vol. 10, p. 351). (Cf. the Luschny reference for a discussion.) - Peter Luschny, May 02 2009
The B_n sequence is the left column of the inverse of triangle A074909, the "beheaded" Pascal's triangle. - Gary W. Adamson, Mar 05 2012
From Sergei N. Gladkovskii, Dec 04 2012: (Start)
E.g.f. E(x)= 2 - x/(tan(x) + sec(x) - 1)= Sum_{n>=0} a(n)*x^n/n!, a(n)=|B(n)|, where B(n) is Bernoulli number B_n.
E(x)= 2 + x - B(0), where B(k)= 4*k+1 + x/(2 + x/(4*k+3 - x/(2 - x/B(k+1)))); (continued fraction, 4-step). (End)
E.g.f.: x/(exp(x)-1)= U(0); U(k)= 2*k+1 - x(2*k+1)/(x + (2*k+2)/(1 + x/U(k+1))); (continued fraction). - Sergei N. Gladkovskii, Dec 05 2012
E.g.f.: 2*(x-1)/(x*Q(0)-2) where Q(k) = 1 + 2*x*(k+1)/((2*k+1)*(2*k+3) - x*(2*k+1)*(2*k+3)^2/(x*(2*k+3) + 4*(k+1)*(k+2)/Q(k+1))); (recursively defined continued fraction). - Sergei N. Gladkovskii, Feb 26 2013
a(n) = numerator(B(n)), B(n) = (-1)^n*Sum_{k=0..n} Stirling1(n,k) * Stirling2(n+k,n) / binomial(n+k,k). - Vladimir Kruchinin, Mar 16 2013
E.g.f.: x/(exp(x)-1) = E(0) where E(k) = 2*k+1 - x/(2 + x/E(k+1)); (continued fraction). - Sergei N. Gladkovskii, Mar 16 2013
G.f. for Bernoulli(n) = a(n)/A027642(n): psi_1(1/x)/x - x, where psi_n(z) is the polygamma function, psi_n(z) = (d/dz)^(n+1) log(Gamma(z)). - Vladimir Reshetnikov, Apr 24 2013
E.g.f.: 2*E(0) - 2*x, where E(k)= x + (k+1)/(1 + 1/(1 - x/E(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 10 2013
B_n = Sum_{m=0..n} (-1)^m *A131689(n, m)/(m + 1), n >= 0. See one of the Maple programs. - Wolfdieter Lang, May 05 2017
a(n) = numerator((-1)^n*A155585(n-1)*n/(4^n-2^n)), for n>=1. - Mats Granvik, Nov 26 2017
From Artur Jasinski, Dec 30 2020: (Start)
a(n) = numerator(-2*cos(Pi*n/2)*Gamma(n+1)*zeta(n)/(2*Pi)^n), for n=0 and n>1.
a(n) = numerator(-n*zeta(1-n)), for n=0 and n>1. (End)
a(n) = numerator(Sum_{k=0..n-1} (-1)^(k-1)*k!*Stirling2(n-1,k) / ((k+1)*(k+2))), for n>0 (see Jha link). - Bill McEachen, Jul 17 2025

A048994 Triangle of Stirling numbers of first kind, s(n,k), n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, -1, 1, 0, 2, -3, 1, 0, -6, 11, -6, 1, 0, 24, -50, 35, -10, 1, 0, -120, 274, -225, 85, -15, 1, 0, 720, -1764, 1624, -735, 175, -21, 1, 0, -5040, 13068, -13132, 6769, -1960, 322, -28, 1, 0, 40320, -109584, 118124, -67284, 22449, -4536, 546, -36, 1, 0, -362880, 1026576, -1172700, 723680, -269325, 63273, -9450, 870, -45, 1
Offset: 0

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Author

Keywords

Comments

The unsigned numbers are also called Stirling cycle numbers: |s(n,k)| = number of permutations of n objects with exactly k cycles.
Mirror image of the triangle A054654. - Philippe Deléham, Dec 30 2006
Also the triangle gives coefficients T(n,k) of x^k in the expansion of C(x,n) = (a(k)*x^k)/n!. - Mokhtar Mohamed, Dec 04 2012
From Wolfdieter Lang, Nov 14 2018: (Start)
This is the Sheffer triangle of Jabotinsky type (1, log(1 + x)). See the e.g.f. of the triangle below.
This is the inverse Sheffer triangle of the Stirling2 Sheffer triangle A008275.
The a-sequence of this Sheffer triangle (see a W. Lang link in A006232)
is from the e.g.f. A(x) = x/(exp(x) -1) a(n) = Bernoulli(n) = A027641(n)/A027642(n), for n >= 0. The z-sequence vanishes.
The Boas-Buck sequence for the recurrences of columns has o.g.f. B(x) = Sum_{n>=0} b(n)*x^n = 1/((1 + x)*log(1 + x)) - 1/x. b(n) = (-1)^(n+1)*A002208(n+1)/A002209(n+1), b = {-1/2, 5/12, -3/8, 251/720, -95/288, 19087/60480,...}. For the Boas-Buck recurrence of Riordan and Sheffer triangles see the Aug 10 2017 remark in A046521, adapted to the Sheffer case, also for two references. See the recurrence and example below. (End)
Let G(n,m,k) be the number of simple labeled graphs on [n] with m edges and k components. Then T(n,k) = Sum (-1)^m*G(n,m,k). See the Read link below. Equivalently, T(n,k) = Sum mu(0,p) where the sum is over all set partitions p of [n] containing k blocks and mu is the Moebius function in the incidence algebra associated to the set partition lattice on [n]. - Geoffrey Critzer, May 11 2024

Examples

			Triangle begins:
  n\k 0     1       2       3      4      5      6    7    8   9 ...
  0   1
  1   0     1
  2   0    -1       1
  3   0     2      -3       1
  4   0    -6      11      -6      1
  5   0    24     -50      35    -10      1
  6   0  -120     274    -225     85    -15      1
  7   0   720   -1764    1624   -735    175    -21    1
  8   0 -5040   13068  -13132   6769  -1960    322  -28    1
  9   0 40320 -109584  118124 -67284  22449  -4536  546  -36   1
  ... - _Wolfdieter Lang_, Aug 22 2012
------------------------------------------------------------------
From _Wolfdieter Lang_, Nov 14 2018: (Start)
Recurrence: s(5,2)= s(4, 1) - 4*s(4, 2) = -6 - 4*11 = -50.
Recurrence from the a- and z-sequences: s(6, 3) = 2*(1*1*(-50) + 3*(-1/2)*35 + 6*(1/6)*(-10) + 10*0*1) = -225.
Boas-Buck recurrence for column k = 3, with b = {-1/2, 5/12, -3/8, ...}:
s(6, 3) = 6!*((-3/8)*1/3! + (5/12)*(-6)/4! + (-1/2)*35/5!) = -225. (End)
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 833.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974; Chapter V, also p. 310.
  • J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, NY, 1996, p. 93.
  • F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 226.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 245.
  • J. Riordan, An Introduction to Combinatorial Analysis, p. 48.

Crossrefs

See especially A008275 which is the main entry for this triangle. A132393 is an unsigned version, and A008276 is another version.
A000142(n) = Sum_{k=0..n} |s(n, k)| for n >= 0.
Row sums give A019590(n+1).

Programs

  • Haskell
    a048994 n k = a048994_tabl !! n !! k
    a048994_row n = a048994_tabl !! n
    a048994_tabl = map fst $ iterate (\(row, i) ->
    (zipWith (-) ([0] ++ row) $ map (* i) (row ++ [0]), i + 1)) ([1], 0)
    -- Reinhard Zumkeller, Mar 18 2013
  • Maple
    A048994:= proc(n,k) combinat[stirling1](n,k) end: # R. J. Mathar, Feb 23 2009
    seq(print(seq(coeff(expand(k!*binomial(x,k)),x,i),i=0..k)),k=0..9); # Peter Luschny, Jul 13 2009
    A048994_row := proc(n) local k; seq(coeff(expand(pochhammer(x-n+1,n)), x,k), k=0..n) end: # Peter Luschny, Dec 30 2010
  • Mathematica
    Table[StirlingS1[n, m], {n, 0, 9}, {m, 0, n}] (* Peter Luschny, Dec 30 2010 *)
  • Maxima
    create_list(stirling1(n,k),n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */
    
  • PARI
    a(n,k) = if(k<0 || k>n,0, if(n==0,1,(n-1)*a(n-1,k)+a(n-1,k-1)))
    
  • PARI
    trg(nn)=for (n=0, nn-1, for (k=0, n, print1(stirling(n,k,1), ", ");); print();); \\ Michel Marcus, Jan 19 2015
    

Formula

s(n, k) = A008275(n,k) for n >= 1, k = 1..n; column k = 0 is {1, repeat(0)}.
s(n, k) = s(n-1, k-1) - (n-1)*s(n-1, k), n, k >= 1; s(n, 0) = s(0, k) = 0; s(0, 0) = 1.
The unsigned numbers a(n, k)=|s(n, k)| satisfy a(n, k)=a(n-1, k-1)+(n-1)*a(n-1, k), n, k >= 1; a(n, 0) = a(0, k) = 0; a(0, 0) = 1.
Triangle (signed) = [0, -1, -1, -2, -2, -3, -3, -4, -4, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, ...]; Triangle(unsigned) = [0, 1, 1, 2, 2, 3, 3, 4, 4, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, ...]; where DELTA is Deléham's operator defined in A084938.
Sum_{k=0..n} (-m)^(n-k)*s(n, k) = A000142(n), A001147(n), A007559(n), A007696(n), ... for m = 1, 2, 3, 4, ... .- Philippe Deléham, Oct 29 2005
A008275*A007318 as infinite lower triangular matrices. - Gerald McGarvey, Aug 20 2009
T(n,k) = n!*[x^k]([t^n]exp(x*log(1+t))). - Peter Luschny, Dec 30 2010, updated Jun 07 2020
From Wolfdieter Lang, Nov 14 2018: (Start)
Recurrence from the Sheffer a-sequence (see a comment above): s(n, k) = (n/k)*Sum_{j=0..n-k} binomial(k-1+j, j)*Bernoulli(j)*s(n-1, k-1+j), for n >= 1 and k >= 1, with s(n, 0) = 0 if n >= 1, and s(0,0) = 1.
Boas-Buck type recurrence for column k: s(n, k) = (n!*k/(n - k))*Sum_{j=k..n-1} b(n-1-j)*s(j, k)/j!, for n >= 1 and k = 0..n-1, with input s(n, n) = 1. For sequence b see the Boas-Buck comment above. (End)
T(n,k) = Sum_{j=k..n} (-1)^(n-j)*A271705(n,j)*A216294(j,k). - Mélika Tebni, Feb 23 2023

Extensions

Offset corrected by R. J. Mathar, Feb 23 2009
Formula corrected by Philippe Deléham, Sep 10 2009

A088218 Total number of leaves in all rooted ordered trees with n edges.

Original entry on oeis.org

1, 1, 3, 10, 35, 126, 462, 1716, 6435, 24310, 92378, 352716, 1352078, 5200300, 20058300, 77558760, 300540195, 1166803110, 4537567650, 17672631900, 68923264410, 269128937220, 1052049481860, 4116715363800, 16123801841550, 63205303218876, 247959266474052
Offset: 0

Views

Author

Michael Somos, Sep 24 2003

Keywords

Comments

Essentially the same as A001700, which has more information.
Note that the unique rooted tree with no edges has no leaves, so a(0)=1 is by convention. - Michael Somos, Jul 30 2011
Number of ordered partitions of n into n parts, allowing zeros (cf. A097070) is binomial(2*n-1,n) = a(n) = essentially A001700. - Vladeta Jovovic, Sep 15 2004
Hankel transform is A000027; example: Det([1,1,3,10;1,3,10,35;3,10,35,126; 10,35,126,462]) = 4. - Philippe Deléham, Apr 13 2007
a(n) is the number of functions f:[n]->[n] such that for all x,y in [n] if xA045992(n). - Geoffrey Critzer, Apr 02 2009
Hankel transform of the aeration of this sequence is A000027 doubled: 1,1,2,2,3,3,... - Paul Barry, Sep 26 2009
The Fi1 and Fi2 triangle sums of A039599 are given by the terms of this sequence. For the definitions of these triangle sums see A180662. - Johannes W. Meijer, Apr 20 2011
Alternating row sums of Riordan triangle A094527. See the Philippe Deléham formula. - Wolfdieter Lang, Nov 22 2012
(-2)*a(n) is the Z-sequence for the Riordan triangle A110162. For the notion of Z- and A-sequences for Riordan arrays see the W. Lang link under A006232 with details and references. - Wolfdieter Lang, Nov 22 2012
From Gus Wiseman, Jun 27 2021: (Start)
Also the number of integer compositions of 2n with alternating (or reverse-alternating) sum 0 (ranked by A344619). This is equivalent to Ran Pan's comment at A001700. For example, the a(0) = 1 through a(3) = 10 compositions are:
() (11) (22) (33)
(121) (132)
(1111) (231)
(1122)
(1221)
(2112)
(2211)
(11121)
(12111)
(111111)
For n > 0, a(n) is also the number of integer compositions of 2n with alternating sum 2.
(End)
Number of terms in the expansion of (x_1+x_2+...+x_n)^n. - César Eliud Lozada, Jan 08 2022

Examples

			G.f. = 1 + x + 3*x^2 + 10*x^3 + 35*x^4 + 126*x^5 + 462*x^6 + 1716*x^7 + ...
The five rooted ordered trees with 3 edges have 10 leaves.
..x........................
..o..x.x..x......x.........
..o...o...o.x..x.o..x.x.x..
..r...r....r....r.....r....
		

References

  • L. W. Shapiro and C. J. Wang, Generating identities via 2 X 2 matrices, Congressus Numerantium, 205 (2010), 33-46.

Crossrefs

Same as A001700 modulo initial term and offset.
First differences are A024718.
Main diagonal of A071919 and of A305161.
A signed version is A110556.
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
A003242 counts anti-run compositions.
A025047 counts wiggly compositions (ascend: A025048, descend: A025049).
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A106356 counts compositions by number of maximal anti-runs.
A124754 gives the alternating sum of standard compositions.
A345197 counts compositions by sum, length, and alternating sum.
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0: counted by A088218 (this sequence), ranked by A344619/A344619.
- k = 1: counted by A000984, ranked by A345909/A345911.
- k = -1: counted by A001791, ranked by A345910/A345912.
- k = 2: counted by A088218 (this sequence), ranked by A345925/A345922.
- k = -2: counted by A002054, ranked by A345924/A345923.
- k >= 0: counted by A116406, ranked by A345913/A345914.
- k <= 0: counted by A058622(n-1), ranked by A345915/A345916.
- k > 0: counted by A027306, ranked by A345917/A345918.
- k < 0: counted by A294175, ranked by A345919/A345920.
- k != 0: counted by A058622, ranked by A345921/A345921.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd: counted by A000302, ranked by A053738/A053738.

Programs

  • Magma
    [Binomial(2*n-1, n): n in [0..30]]; // Vincenzo Librandi, Aug 07 2014
  • Maple
    seq(binomial(2*n-1, n),n=0..24); # Peter Luschny, Sep 22 2014
  • Mathematica
    a[ n_] := SeriesCoefficient[(1 - x)^-n, {x, 0, n}];
    c = (1 - (1 - 4 x)^(1/2))/(2 x);CoefficientList[Series[1/(1-(c-1)),{x,0,20}],x] (* Geoffrey Critzer, Dec 02 2010 *)
    Table[Binomial[2 n - 1, n], {n, 0, 20}] (* Vincenzo Librandi, Aug 07 2014 *)
    a[ n_] := If[ n < 0, 0, With[ {m = 2 n}, m! SeriesCoefficient[ (1 + BesselI[0, 2 x]) / 2, {x, 0, m}]]]; (* Michael Somos, Nov 22 2014 *)
  • PARI
    {a(n) = sum( i=0, n, binomial(n+i-2,i))};
    
  • PARI
    {a(n) = if( n<0, 0, polcoeff( (1 + 1 / sqrt(1 - 4*x + x * O(x^n))) / 2, n))};
    
  • PARI
    {a(n) = if( n<0, 0, polcoeff( 1 / (1 - x + x * O(x^n))^n, n))};
    
  • PARI
    {a(n) = if( n<0, 0, binomial( 2*n - 1, n))};
    
  • PARI
    {a(n) = if( n<1, n==0, polcoeff( subst((1 - x) / (1 - 2*x), x, serreverse( x - x^2 + x * O(x^n))), n))};
    
  • Sage
    def A088218(n):
        return rising_factorial(n,n)/falling_factorial(n,n)
    [A088218(n) for n in (0..24)]  # Peter Luschny, Nov 21 2012
    

Formula

G.f.: (1 + 1 / sqrt(1 - 4*x)) / 2.
a(n) = binomial(2*n - 1, n).
a(n) = (n+1)*A000108(n)/2, n>=1. - B. Dubalski (dubalski(AT)atr.bydgoszcz.pl), Feb 05 2002 (in A060150)
a(n) = (0^n + C(2n, n))/2. - Paul Barry, May 21 2004
a(n) is the coefficient of x^n in 1 / (1 - x)^n and also the sum of the first n coefficients of 1 / (1 - x)^n. Given B(x) with the property that the coefficient of x^n in B(x)^n equals the sum of the first n coefficients of B(x)^n, then B(x) = B(0) / (1 - x).
G.f.: 1 / (2 - C(x)) = (1 - x*C(x))/sqrt(1-4*x) where C(x) is g.f. for Catalan numbers A000108. Second equation added by Wolfdieter Lang, Nov 22 2012.
From Paul Barry, Nov 02 2004: (Start)
a(n) = Sum_{k=0..n} binomial(2*n, k)*cos((n-k)*Pi);
a(n) = Sum_{k=0..n} binomial(n, (n-k)/2)*(1+(-1)^(n-k))*cos(k*Pi/2)/2 (with interpolated zeros);
a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)*cos((n-2*k)*Pi/2) (with interpolated zeros); (End)
a(n) = A110556(n)*(-1)^n, central terms in triangle A110555. - Reinhard Zumkeller, Jul 27 2005
a(n) = Sum_{0<=k<=n} A094527(n,k)*(-1)^k. - Philippe Deléham, Mar 14 2007
From Paul Barry, Mar 29 2010: (Start)
G.f.: 1/(1-x/(1-2x/(1-(1/2)x/(1-(3/2)x/(1-(2/3)x/(1-(4/3)x/(1-(3/4)x/(1-(5/4)x/(1-... (continued fraction);
E.g.f.: (of aerated sequence) (1 + Bessel_I(0, 2*x))/2. (End)
a(n + 1) = A001700(n). a(n) = A024718(n) - A024718(n - 1).
E.g.f.: E(x) = 1+x/(G(0)-2*x) ; G(k) = (k+1)^2+2*x*(2*k+1)-2*x*(2*k+3)*((k+1)^2)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec 21 2011
a(n) = Sum_{k=0..n}(-1)^k*binomial(2*n,n+k). - Mircea Merca, Jan 28 2012
a(n) = rf(n,n)/ff(n,n), where rf is the rising factorial and ff the falling factorial. - Peter Luschny, Nov 21 2012
D-finite with recurrence: n*a(n) +2*(-2*n+1)*a(n-1) = 0. - R. J. Mathar, Dec 04 2012
a(n) = hypergeom([1-n,-n],[1],1). - Peter Luschny, Sep 22 2014
G.f.: 1 + x/W(0), where W(k) = 4*k+1 - (4*k+3)*x/(1 - (4*k+1)*x/(4*k+3 - (4*k+5)*x/(1 - (4*k+3)*x/W(k+1) ))) ; (continued fraction). - Sergei N. Gladkovskii, Nov 13 2014
a(n) = A000984(n) + A001791(n). - Gus Wiseman, Jun 28 2021
E.g.f.: (1 + exp(2*x) * BesselI(0,2*x)) / 2. - Ilya Gutkovskiy, Nov 03 2021
From Amiram Eldar, Mar 12 2023: (Start)
Sum_{n>=0} 1/a(n) = 5/3 + 4*Pi/(9*sqrt(3)).
Sum_{n>=0} (-1)^n/a(n) = 3/5 - 8*log(phi)/(5*sqrt(5)), where phi is the golden ratio (A001622). (End)
a(n) ~ 2^(2*n-1)/sqrt(n*Pi). - Stefano Spezia, Apr 17 2024

A008288 Square array of Delannoy numbers D(i,j) (i >= 0, j >= 0) read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 5, 5, 1, 1, 7, 13, 7, 1, 1, 9, 25, 25, 9, 1, 1, 11, 41, 63, 41, 11, 1, 1, 13, 61, 129, 129, 61, 13, 1, 1, 15, 85, 231, 321, 231, 85, 15, 1, 1, 17, 113, 377, 681, 681, 377, 113, 17, 1, 1, 19, 145, 575, 1289, 1683, 1289, 575, 145, 19, 1, 1, 21, 181, 833, 2241, 3653, 3653
Offset: 0

Views

Author

Keywords

Comments

In the Formula section, some contributors use T(n,k) = D(n-k, k) (for 0 <= k <= n), which is the triangular version of the square array (D(n,k): n,k >= 0). Conversely, D(n,k) = T(n+k,k) for n,k >= 0. - Petros Hadjicostas, Aug 05 2020
Also called the tribonacci triangle [Alladi and Hoggatt (1977)]. - N. J. A. Sloane, Mar 23 2014
D(n,k) is the number of lattice paths from (0,0) to (n,k) using steps (1,0), (0,1), (1,1). - Joerg Arndt, Jul 01 2011 [Corrected by N. J. A. Sloane, May 30 2020]
Or, triangle read by rows of coefficients of polynomials P[n](x) defined by P[0] = 1, P[1] = x+1; for n >= 2, P[n] = (x+1)*P[n-1] + x*P[n-2].
D(n, k) is the number of k-matchings of a comb-like graph with n+k teeth. Example: D(1, 3) = 7 because the graph consisting of a horizontal path ABCD and the teeth Aa, Bb, Cc, Dd has seven 3-matchings: four triples of three teeth and the three triples {Aa, Bb, CD}, {Aa, Dd, BC}, {Cc, Dd, AB}. Also D(3, 1)=7, the 1-matchings of the same graph being the seven edges: {AB}, {BC}, {CD}, {Aa}, {Bb}, {Cc}, {Dd}. - Emeric Deutsch, Jul 01 2002
Sum of n-th antidiagonal of the array D is A000129(n+1). - Reinhard Zumkeller, Dec 03 2004 [Edited by Petros Hadjicostas, Aug 05 2020 so that the counting of antidiagonals of D starts at n = 0. That is, the sum of the terms in the n-th row of the triangles T is A000129(n+1).]
The A-sequence for this Riordan type triangle (see one of Paul Barry's comments under Formula) is A112478 and the Z-sequence the trivial: {1, 0, 0, 0, ...}. See the W. Lang link under A006232 for Sheffer a- and z-sequences where also Riordan A- and Z-sequences are explained. This leads to the recurrence for the triangle given below. - Wolfdieter Lang, Jan 21 2008
The triangle or chess sums, see A180662 for their definitions, link the Delannoy numbers with twelve different sequences, see the crossrefs. All sums come in pairs due to the symmetrical nature of this triangle. The knight sums Kn14 and Kn15 have been added. It is remarkable that all knight sums are related to the tribonacci numbers, that is, A000073 and A001590, but none of the others. - Johannes W. Meijer, Sep 22 2010
This sequence, A008288, is jointly generated with A035607 as an array of coefficients of polynomials u(n,x): initially, u(1,x) = v(1,x) = 1; for n > 1, u(n,x) = x*u(n-1,x) + v(n-1) and v(n,x) = 2*x*u(n-1,x) + v(n-1,x). See the Mathematica section. - Clark Kimberling, Mar 09 2012
Row n, for n > 0, of Roger L. Bagula's triangle in the Example section 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 = sqrt(x) + 1/sqrt(x); see A230000. - Clark Kimberling, Nov 13 2013
In an n-dimensional hypercube lattice, D(n,k) gives the number of nodes situated at a Minkowski (Manhattan) distance of k from a given node. In cellular automata theory, the cells at Manhattan distance k are called the von Neumann neighborhood of radius k. For k=1, see A005843. - Dmitry Zaitsev, Dec 10 2015
These numbers appear as the coefficients of series relating spherical and bispherical harmonics, in the solutions of Laplace's equation in 3D. [Majic 2019, Eq. 22] - Matt Majic, Nov 24 2019
From Peter Bala, Feb 19 2020: (Start)
The following remarks assume an offset of 1 in the row and column indices of the triangle.
The sequence of row polynomials T(n,x), beginning with T(1,x) = x, T(2,x) = x + x^2, T(3,x) = x + 3*x^2 + x^3, ..., is a strong divisibility sequence of polynomials in the ring Z[x]; that is, for all positive integers n and m, poly_gcd(T(n,x), T(m,x)) = T(gcd(n, m), x) - apply Norfleet (2005), Theorem 3. Consequently, the sequence (T(n,x): n >= 1) is a divisibility sequence in the polynomial ring Z[x]; that is, if n divides m then T(n,x) divides T(m,x) in Z[x].
Let S(x) = 1 + 2*x + 6*x^2 + 22*x^3 + ... denote the o.g.f. for the large Schröder numbers A006318. The power series (x*S(x))^n, n = 2, 3, 4, ..., can be expressed as a linear combination with polynomial coefficients of S(x) and 1: (x*S(x))^n = T(n-1,-x) - T(n,-x)*S(x). The result can be extended to negative integer n if we define T(0,x) = 0 and T(-n,x) = (-1)^(n+1) * T(n,x)/x^n. Cf. A115139.
[In the previous two paragraphs, D(n,x) was replaced with T(n,x) because the contributor is referring to the rows of the triangle T(n,k), not the rows of the array D(n,k). - Petros Hadjicostas, Aug 05 2020] (End)
Named after the French amateur mathematician Henri-Auguste Delannoy (1833-1915). - Amiram Eldar, Apr 15 2021
D(i,j) = D(j,i). With this and Dmitry Zaitsev's Dec 10 2015 comment, D(i,j) can be considered the number of points at L1 distance <= i in Z^j or the number of points at L1 distance <= j in Z^i from any given point. The rows and columns of D(i,j) are the crystal ball sequences on cubic lattices. See the first example below. The n-th term in the k-th crystal ball sequence can be considered the number of points at distance <= n from any point in a k-dimensional cubic lattice, or the number of points at distance <= k from any point in an n-dimensional cubic lattice. - Shel Kaphan, Jan 01 2023 and Jan 07 2023
Dimensions of hom spaces Hom(R^{(i)}, R^{(j)}) in the Delannoy category attached to the oligomorphic group of order preserving self-bijections of the real line. - Noah Snyder, Mar 22 2023

Examples

			The square array D(i,j) (i >= 0, j >= 0) begins:
  1, 1,  1,   1,   1,   1,    1,    1,    1,    1, ... = A000012
  1, 3,  5,   7,   9,  11,   13,   15,   17,   19, ... = A005408
  1, 5, 13,  25,  41,  61,   85,  113,  145,  181, ... = A001844
  1, 7, 25,  63, 129, 231,  377,  575,  833, 1159, ... = A001845
  1, 9, 41, 129, 321, 681, 1289, 2241, 3649, 5641, ... = A001846
  ...
For D(2,5) = 61, which is seen above in the row labeled A001844, we calculate the sum (9 + 11 + 41) of the 3 nearest terms above and/or to the left. - _Peter Munn_, Jan 01 2023
D(2,5) = 61 can also be obtained from the row labeled A005408 using a recurrence mentioned in the formula section:  D(2,5) = D(1,5) + 2*Sum_{k=0..4} D(1,k), so D(2,5) = 11 + 2*(1+3+5+7+9) = 11 + 2*25. - _Shel Kaphan_, Jan 01 2023
As a triangular array (on its side) this begins:
   0,   0,   0,   0,   1,   0,  11,   0, ...
   0,   0,   0,   1,   0,   9,   0,  61, ...
   0,   0,   1,   0,   7,   0,  41,   0, ...
   0,   1,   0,   5,   0,  25,   0, 129, ...
   1,   0,   3,   0,  13,   0,  63,   0, ...
   0,   1,   0,   5,   0,  25,   0, 129, ...
   0,   0,   1,   0,   7,   0,  41,   0, ...
   0,   0,   0,   1,   0,   9,   0,  61, ...
   0,   0,   0,   0,   1,   0,  11,   0, ...
   [Edited by _Shel Kaphan_, Jan 01 2023]
From _Roger L. Bagula_, Dec 09 2008: (Start)
As a triangle T(n,k) (with rows n >= 0 and columns k = 0..n), this begins:
   1;
   1,  1;
   1,  3,   1;
   1,  5,   5,   1;
   1,  7,  13,   7,    1;
   1,  9,  25,  25,    9,    1;
   1, 11,  41,  63,   41,   11,    1;
   1, 13,  61, 129,  129,   61,   13,   1;
   1, 15,  85, 231,  321,  231,   85,  15,   1;
   1, 17, 113, 377,  681,  681,  377, 113,  17,  1;
   1, 19, 145, 575, 1289, 1683, 1289, 575, 145, 19, 1;
   ... (End)
Triangle T(n,k) recurrence: 63 = T(6,3) = 25 + 13 + 25 = T(5,2) + T(4,2) + T(5,3).
Triangle T(n,k) recurrence with A-sequence A112478: 63 = T(6,3) = 1*25 + 2*25 - 2*9 + 6*1 (T entries from row n = 5 only). [Here the formula T(n,k) = Sum_{j=0..n-k} A112478(j) * T(n-1, k-1+j) is used with n = 6 and k = 3; i.e., T(6,3) = Sum_{j=0..3} A111478(j) * T(5, 2+j). - _Petros Hadjicostas_, Aug 05 2020]
From _Philippe Deléham_, Mar 29 2012: (Start)
Subtriangle of the triangle given by (1, 0, 1, -1, 0, 0, 0, ...) DELTA (0, 1, 0, 0, 0, ...) where DELTA is the operator defined in A084938:
   1;
   1,  0;
   1,  1,  0;
   1,  3,  1,  0;
   1,  5,  5,  1,  0;
   1,  7, 13,  7,  1,  0;
   1,  9, 25, 25,  9,  1, 0;
   1, 11, 41, 63, 41, 11, 1, 0;
   ...
Subtriangle of the triangle given by (0, 1, 0, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, 0, ...) where DELTA is the operator defined in A084938:
   1;
   0, 1;
   0, 1,  1;
   0, 1,  3,  1;
   0, 1,  5,  5,  1;
   0, 1,  7, 13,  7,  1;
   0, 1,  9, 25, 25,  9,  1;
   0, 1, 11, 41, 63, 41, 11, 1;
   ... (End)
		

References

  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 593.
  • Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 81.
  • L. Moser and W. Zayachkowski, Lattice paths with diagonal steps, Scripta Mathematica, 26 (1963), 223-229.
  • G. Picou, Note #2235, L'Intermédiaire des Mathématiciens, 8 (1901), page 281. - N. J. A. Sloane, Mar 02 2022
  • D. B. West, Combinatorial Mathematics, Cambridge, 2021, p. 28.

Crossrefs

Sums of antidiagonals: A000129 (Pell numbers).
Main diagonal: A001850 (central Delannoy numbers), which has further information and references.
A002002, A026002, and A190666 are +-k-diagonals for k=1, 2, 3 resp. - Shel Kaphan, Jan 01 2023
See also A027618.
Cf. A059446.
Has same main diagonal as A064861. Different from A100936.
Read mod small primes: A211312, A211313, A211314, A211315.
Triangle sums (see the comments): A000129 (Row1); A056594 (Row2); A000073 (Kn11 & Kn21); A089068 (Kn12 & Kn22); A180668 (Kn13 & Kn23); A180669 (Kn14 & Kn24); A180670 (Kn15 & Kn25); A099463 (Kn3 & Kn4); A116404 (Fi1 & Fi2); A006498 (Ca1 & Ca2); A006498(3*n) (Ca3 & Ca4); A079972 (Gi1 & Gi2); A079972(4*n) (Gi3 & Gi4); A079973(3*n) (Ze1 & Ze2); A079973(2*n) (Ze3 & Ze4).
Cf. A102413, A128966. (D(n,1)) = A005843. Cf. A115139.

Programs

  • Haskell
    a008288 n k = a008288_tabl !! n !! k
    a008288_row n = a008288_tabl !! n
    a008288_tabl = map fst $ iterate
        (\(us, vs) -> (vs, zipWith (+) ([0] ++ us ++ [0]) $
                           zipWith (+) ([0] ++ vs) (vs ++ [0]))) ([1], [1, 1])
    -- Reinhard Zumkeller, Jul 21 2013
    
  • Maple
    A008288 := proc(n, k) option remember; if k = 0 then 1 elif n=k then 1 else procname(n-1, k-1) + procname(n-2, k-1) + procname(n-1, k) end if; end proc: seq(seq(A008288(n,k),k=0..n), n=0..10); # triangular indices n and k
    P[0]:=1; P[1]:=x+1; for n from 2 to 12 do P[n]:=expand((x+1)*P[n-1]+x*P[n-2]); lprint(P[n]); lprint(seriestolist(series(P[n],x,200))); end do:
  • Mathematica
    (* Next, A008288 jointly generated with A035607 *)
    u[1, x_] := 1; v[1, x_] := 1; z = 16;
    u[n_, x_] := x*u[n - 1, x] + v[n - 1, x];
    v[n_, x_] := 2 x*u[n - 1, x] + v[n - 1, x];
    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[%]    (* A008288 *)
    Table[Expand[v[n, x]], {n, 1, z}]
    cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
    TableForm[cv]
    Flatten[%]    (* A035607 *)
    (* Clark Kimberling, Mar 09 2012 *)
    d[n_, k_] := Binomial[n+k, k]*Hypergeometric2F1[-k, -n, -n-k, -1]; A008288 = Flatten[Table[d[n-k, k], {n, 0, 12}, {k, 0, n}]] (* Jean-François Alcover, Apr 05 2012, after 3rd formula *)
  • Python
    from functools import cache
    @cache
    def delannoy_row(n: int) -> list[int]:
        if n == 0: return [1]
        if n == 1: return [1, 1]
        rov = delannoy_row(n - 2)
        row = delannoy_row(n - 1) + [1]
        for k in range(n - 1, 0, -1):
            row[k] += row[k - 1] + rov[k - 1]
        return row
    for n in range(10): print(delannoy_row(n))  # Peter Luschny, Jul 30 2023
  • Sage
    for k in range(8):  # seen as an array, read row by row
        a = lambda n: hypergeometric([-n, -k], [1], 2)
        print([simplify(a(n)) for n in range(11)]) # Peter Luschny, Nov 19 2014
    

Formula

D(n, 0) = 1 = D(0, n) for n >= 0; D(n, k) = D(n, k-1) + D(n-1, k-1) + D(n-1, k).
Bivariate o.g.f.: Sum_{n >= 0, k >= 0} D(n, k)*x^n*y^k = 1/(1 - x - y - x*y).
D(n, k) = Sum_{d = 0..min(n,k)} binomial(k, d)*binomial(n+k-d, k) = Sum_{d=0..min(n,k)} 2^d*binomial(n, d)*binomial(k, d). [Edited by Petros Hadjicostas, Aug 05 2020]
Seen as a triangle read by rows: T(n, 0) = T(n, n) = 1 for n >= 0 and T(n, k) = T(n-1, k-1) + T(n-2, k-1) + T(n-1, k), 0 < k < n and n > 1. - Reinhard Zumkeller, Dec 03 2004
Read as a number triangle, this is the Riordan array (1/(1-x), x(1+x)/(1-x)) with T(n, k) = Sum_{j=0..n-k} C(n-k, j) * C(k, j) * 2^j. - Paul Barry, Jul 18 2005
T(n,k) = Sum_{j=0..n-k} C(k,j)*C(n-j,k). - Paul Barry, May 21 2006
Let y^k(n) be the number of Khalimsky-continuous functions f from [0,n-1] to Z such that f(0) = 0 and f(n-1) = k. Then y^k(n) = D(i,j) for i = (1/2)*(n-1-k) and j = (1/2)*(n-1+k) where n-1+k belongs to 2Z. - Shiva Samieinia (shiva(AT)math.su.se), Oct 08 2007
Recurrence for triangle from A-sequence (see the Wolfdieter Lang comment above): T(n,k) = Sum_{j=0..n-k} A112478(j) * T(n-1, k-1+j), n >= 1, k >= 1. [For k > n, the sum is empty, in which case T(n,k) = 0.]
From Peter Bala, Jul 17 2008: (Start)
The n-th row of the square array is the crystal ball sequence for the product lattice A_1 x ... x A_1 (n copies). A035607 is the table of the associated coordination sequences for these lattices.
The polynomial p_n(x) := Sum {k = 0..n} 2^k * C(n,k) * C(x,k) = Sum_{k = 0..n} C(n,k) * C(x+k,n), whose values [p_n(0), p_n(1), p_n(2), ... ] give the n-th row of the square array, is the Ehrhart polynomial of the n-dimensional cross polytope (the hyperoctahedron) [Bump et al. (2000), Theorem 6].
The first few values are p_0(x) = 1, p_1(x) = 2*x + 1, p_2(x) = 2*x^2 + 2*x + 1 and p_3(x) = (4*x^3 + 6*x^2 + 8*x + 3)/3.
The reciprocity law p_n(m) = p_m(n) reflects the symmetry of the table.
The polynomial p_n(x) is the unique polynomial solution of the difference equation (x+1)*f(x+1) - x*f(x-1) = (2*n+1)*f(x), normalized so that f(0) = 1.
These polynomials have their zeros on the vertical line Re x = -1/2 in the complex plane; that is, the polynomials p_n(x-1), n = 1,2,3,..., satisfy a Riemann hypothesis [Bump et al. (2000), Theorem 4]. The o.g.f. for the p_n(x) is (1 + t)^x/(1 - t)^(x + 1) = 1 + (2*x + 1)*t + (2*x^2 + 2*x + 1)*t^2 + ... .
The square array of Delannoy numbers has a close connection with the constant log(2). The entries in the n-th row of the array occur in the series acceleration formula log(2) = (1 - 1/2 + 1/3 - ... + (-1)^(n+1)/n) + (-1)^n * Sum_{k>=1} (-1)^(k+1)/(k*D(n,k-1)*D(n,k)). [T(n,k) was replaced with D(n,k) in the formula to agree with the beginning of the paragraph. - Petros Hadjicostas, Aug 05 2020]
For example, the fourth row of the table (n = 3) gives the series log(2) = 1 - 1/2 + 1/3 - 1/(1*1*7) + 1/(2*7*25) - 1/(3*25*63) + 1/(4*63*129) - ... . See A142979 for further details.
Also the main diagonal entries (the central Delannoy numbers) give the series acceleration formula Sum_{n>=1} 1/(n*D(n-1,n-1)*D(n,n)) = (1/2)*log(2), a result due to Burnside. [T(n,n) was replaced here with D(n,n) to agree with the previous paragraphs. - Petros Hadjicostas, Aug 05 2020]
Similar relations hold between log(2) and the crystal ball sequences of the C_n lattices A142992. For corresponding results for the constants zeta(2) and zeta(3), involving the crystal ball sequences for root lattices of type A_n and A_n x A_n, see A108625 and A143007 respectively. (End)
From Peter Bala, Oct 28 2008: (Start)
Hilbert transform of Pascal's triangle A007318 (see A145905 for the definition of this term).
D(n+a,n) = P_n(a,0;3) for all integer a such that a >= -n, where P_n(a,0;x) is the Jacobi polynomial with parameters (a,0) [Hetyei]. The related formula A(n,k) = P_k(0,n-k;3) defines the table of asymmetric Delannoy numbers, essentially A049600. (End)
Seen as a triangle read by rows: T(n, k) = Hyper2F1([k-n, -k], [1], 2). - Peter Luschny, Aug 02 2014, Oct 13 2024.
From Peter Bala, Jun 25 2015: (Start)
O.g.f. for triangle T(n,k): A(z,t) = 1/(1 - (1 + t)*z - t*z^2) = 1 + (1 + t)*z + (1 + 3*t + t^2)*z^2 + (1 + 5*t + 5*t^2 + t^3)*z^3 + ....
1 + z*d/dz(A(z,t))/A(z,t) is the o.g.f. for A102413. (End)
E.g.f. for the n-th subdiagonal of T(n,k), n >= 0, equals exp(x)*P(n,x), where P(n,x) is the polynomial Sum_{k = 0..n} binomial(n,k)*(2*x)^k/k!. For example, the e.g.f. for the second subdiagonal is exp(x)*(1 + 4*x + 4*x^2/2) = 1 + 5*x + 13*x^2/2! + 25*x^3/3! + 41*x^4/4! + 61*x^5/5! + .... - Peter Bala, Mar 05 2017 [The n-th subdiagonal of triangle T(n,k) is the n-th row of array D(n,k).]
Let a_i(n) be multiplicative with a_i(p^e) = D(i, e), p prime and e >= 0, then Sum_{n > 0} a_i(n)/n^s = (zeta(s))^(2*i+1)/(zeta(2*s))^i for i >= 0. - Werner Schulte, Feb 14 2018
Seen as a triangle read by rows: T(n,k) = Sum_{i=0..k} binomial(n-i, i) * binomial(n-2*i, k-i) for 0 <= k <= n. - Werner Schulte, Jan 09 2019
Univariate generating function: Sum_{k >= 0} D(n,k)*z^k = (1 + z)^n/(1 - z)^(n+1). [Dziemianczuk (2013), Eq. 5.3] - Matt Majic, Nov 24 2019
(n+1)*D(n+1,k) = (2*k+1)*D(n,k) + n*D(n-1,k). [Majic (2019), Eq. 22] - Matt Majic, Nov 24 2019
For i, j >= 1, D(i,j) = D(i,j-1) + 2*Sum_{k=0..i-1} D(k,j-1), or, because D(i,j) = D(j,i), D(i,j) = D(i-1,j) + 2*Sum_{k=0..j-1} D(i-1,k). - Shel Kaphan, Jan 01 2023
Sum_{k=0..n} T(n,k)^2 = A026933(n). - R. J. Mathar, Nov 07 2023
Let S(x) = (1 - x - (1 - 6*x + x^2)^(1/2))/(2*x) denote the g.f. of the sequence of large Schröder numbers A006318. Read as a lower triangular array, the signed n-th row polynomial R(n, -x) = 1/sqrt(1 - 6*x + x^2) *( 1/S(x)^(n+1) + (x*S(x))^(n+1) ). For example, R(4, -x) = 1 - 7*x + 13*x^2 - 7*x^3 + x^4 = 1/sqrt(1 - 6*x + x^2) * ( 1/S(x)^5 + (x*S(x))^5 ). Cf. A102413. - Peter Bala, Aug 01 2024

Extensions

Expanded description from Clark Kimberling, Jun 15 1997
Additional references from Sylviane R. Schwer (schwer(AT)lipn.univ-paris13.fr), Nov 28 2001
Changed the notation to make the formulas more precise. - N. J. A. Sloane, Jul 01 2002

A164555 Numerators of the "original" Bernoulli numbers; also the numerators of the Bernoulli polynomials at x=1.

Original entry on oeis.org

1, 1, 1, 0, -1, 0, 1, 0, -1, 0, 5, 0, -691, 0, 7, 0, -3617, 0, 43867, 0, -174611, 0, 854513, 0, -236364091, 0, 8553103, 0, -23749461029, 0, 8615841276005, 0, -7709321041217, 0, 2577687858367, 0, -26315271553053477373, 0, 2929993913841559, 0, -261082718496449122051
Offset: 0

Views

Author

Paul Curtz, Aug 15 2009

Keywords

Comments

Apart from a sign flip in a(1), the same as A027641.
a(n) is also the numerator of the n-th term of the binomial transform of the sequence of Bernoulli numbers, i.e., of the sequence of fractions A027641(n)/A027642(n).
a(n)/A027642(n) with e.g.f. x/(1-exp(-x)) is the a-sequence for the Sheffer matrix A094645, see the W. Lang link under A006232 for Sheffer a- and z-sequences. - Wolfdieter Lang, Jun 20 2011
a(n)/A027642(n) also give the row sums of the rational triangle of the coefficients of the Bernoulli polynomials A053382/A053383 (falling powers) or A196838/A196839 (rising powers). - Wolfdieter Lang, Oct 25 2011
Given M = the beheaded Pascal's triangle, A074909; with B_n as a vector V, with numerators shown: (1, 1, 1, ...). Then M*V = [1, 2, 3, 4, 5, ...]. If the sign in a(1) is negative in V, then M*V = [1, 0, 0, 0, ...]. - Gary W. Adamson, Mar 09 2012
One might interpret the term "'original' Bernoulli numbers" as numbers given by the e.g.f. x/(1-exp(-x)). - Peter Luschny, Jun 17 2012
Let B(n) = a(n)/A027642(n) then B(n) = Integral_{x=0..1} F_n(x) where F_n(x) are the signed Fubini polynomials F_n(x) = Sum_{k=0..n} (-1)^n*Stirling2(n,k)*k!*(-x)^k (see illustration). - Peter Luschny, Jan 09 2017

Examples

			From _Peter Luschny_, Aug 13 2017: (Start)
Integral_{x=0..1} 1 = 1,
Integral_{x=0..1} x = 1/2,
Integral_{x=0..1} 2*x^2 - x = 1/6,
Integral_{x=0..1} 6*x^3 - 6*x^2 + x = 0,
Integral_{x=0..1} 24*x^4 - 36*x^3 + 14*x^2 - x = -1/30,
Integral_{x=0..1} 120*x^5 - 240*x^4 + 150*x^3 - 30*x^2 + x = 0,
...
Integral_{x=0..1} Sum_{k=0..n} (-1)^n*Stirling2(n,k)*k!*(-x)^k = Bernoulli(n). (End)
		

References

  • Jacob Bernoulli, Ars Conjectandi, Basel: Thurneysen Brothers, 1713. See page 97.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 106-108.

Crossrefs

Programs

  • Haskell
    a164555 n = a164555_list !! n
    a164555_list = 1 : map (numerator . sum) (zipWith (zipWith (%))
       (zipWith (map . (*)) (tail a000142_list) a242179_tabf) a106831_tabf)
    -- Reinhard Zumkeller, Jul 04 2014
    
  • Maple
    A164555 := proc(n) if n <= 2 then 1; else numer(bernoulli(n)) ; fi; end: # R. J. Mathar, Aug 26 2009
    seq(numer(n!*coeff(series(t/(1-exp(-t)),t,n+2),t,n)),n=0..40); # Peter Luschny, Jun 17 2012
  • Mathematica
    CoefficientList[ Series[ x/(1 - Exp[-x]), {x, 0, 40}], x]*Range[0, 40]! // Numerator (* Jean-François Alcover, Mar 04 2013 *)
    Table[Numerator[BernoulliB[n,1]], {n, 0, 40}] (* Vaclav Kotesovec, Jan 03 2021 *)
  • Sage
    a = lambda n: bernoulli_polynomial(1,n).numerator()
    [a(n) for n in (0..40)] # Peter Luschny, Jan 09 2017

Formula

a(n) = numerator(B(n)) with B(n) = Sum_{k=0..n} (-1)^(n-k) * C(n+1, k+1) * S(n+k, k) / C(n+k, k) and S the Stirling set numbers. - Peter Luschny, Jun 25 2016
a(n) = numerator(n*EulerPolynomial(n-1, 1)/(2*(2^n-1))) for n>=1. - Peter Luschny, Sep 01 2017
From Artur Jasinski, Jan 01 2021: (Start)
a(n) = numerator(-2*cos(Pi*n/2)*Gamma(n+1)*zeta(n)/(2*Pi)^n) for n != 1.
a(n) = numerator(-n*zeta(1-n)) for n >= 1. In the case n = 0 take the limit. (End)

Extensions

Edited and extended by R. J. Mathar, Sep 03 2009
Name extended by Peter Luschny, Jan 09 2017

A132393 Triangle of unsigned Stirling numbers of the first kind (see A048994), read by rows, T(n,k) for 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 3, 1, 0, 6, 11, 6, 1, 0, 24, 50, 35, 10, 1, 0, 120, 274, 225, 85, 15, 1, 0, 720, 1764, 1624, 735, 175, 21, 1, 0, 5040, 13068, 13132, 6769, 1960, 322, 28, 1, 0, 40320, 109584, 118124, 67284, 22449, 4536, 546, 36, 1, 0, 362880, 1026576, 1172700, 723680, 269325, 63273, 9450, 870, 45, 1
Offset: 0

Views

Author

Philippe Deléham, Nov 10 2007, Oct 15 2008, Oct 17 2008

Keywords

Comments

Another name: Triangle of signless Stirling numbers of the first kind.
Triangle T(n,k), 0<=k<=n, read by rows given by [0,1,1,2,2,3,3,4,4,5,5,...] DELTA [1,0,1,0,1,0,1,0,1,...] where DELTA is the operator defined in A084938.
A094645*A007318 as infinite lower triangular matrices.
Row sums are the factorial numbers. - Roger L. Bagula, Apr 18 2008
Exponential Riordan array [1/(1-x), log(1/(1-x))]. - Ralf Stephan, Feb 07 2014
Also the Bell transform of the factorial numbers (A000142). For the definition of the Bell transform see A264428 and for cross-references A265606. - Peter Luschny, Dec 31 2015
This is the lower triagonal Sheffer matrix of the associated or Jabotinsky type |S1| = (1, -log(1-x)) (see the W. Lang link under A006232 for the notation and references). This implies the e.g.f.s given below. |S1| is the transition matrix from the monomial basis {x^n} to the rising factorial basis {risefac(x,n)}, n >= 0. - Wolfdieter Lang, Feb 21 2017
T(n, k), for n >= k >= 1, is also the total volume of the n-k dimensional cell (polytope) built from the n-k orthogonal vectors of pairwise different lengths chosen from the set {1, 2, ..., n-1}. See the elementary symmetric function formula for T(n, k) and an example below. - Wolfdieter Lang, May 28 2017
From Wolfdieter Lang, Jul 20 2017: (Start)
The compositional inverse w.r.t. x of y = y(t;x) = x*(1 - t(-log(1-x)/x)) = x + t*log(1-x) is x = x(t;y) = ED(y,t) := Sum_{d>=0} D(d,t)*y^(d+1)/(d+1)!, the e.g.f. of the o.g.f.s D(d,t) = Sum_{m>=0} T(d+m, m)*t^m of the diagonal sequences of the present triangle. See the P. Bala link for a proof (there d = n-1, n >= 1, is the label for the diagonals).
This inversion gives D(d,t) = P(d, t)/(1-t)^(2*d+1), with the numerator polynomials P(d, t) = Sum_{m=0..d} A288874(d, m)*t^m. See an example below. See also the P. Bala formula in A112007. (End)
For n > 0, T(n,k) is the number of permutations of the integers from 1 to n which have k visible digits when viewed from a specific end, in the sense that a higher value hides a lower one in a subsequent position. - Ian Duff, Jul 12 2019

Examples

			Triangle T(n,k) begins:
  1;
  0,    1;
  0,    1,     1;
  0,    2,     3,     1;
  0,    6,    11,     6,    1;
  0,   24,    50,    35,   10,    1;
  0,  120,   274,   225,   85,   15,   1;
  0,  720,  1764,  1624,  735,  175,  21,  1;
  0, 5040, 13068, 13132, 6769, 1960, 322, 28, 1;
  ...
---------------------------------------------------
Production matrix is
  0, 1
  0, 1, 1
  0, 1, 2,  1
  0, 1, 3,  3,  1
  0, 1, 4,  6,  4,  1
  0, 1, 5, 10, 10,  5,  1
  0, 1, 6, 15, 20, 15,  6, 1
  0, 1, 7, 21, 35, 35, 21, 7, 1
  ...
From _Wolfdieter Lang_, May 09 2017: (Start)
Three term recurrence: 50 = T(5, 2) = 1*6 + (5-1)*11 = 50.
Recurrence from the Sheffer a-sequence [1, 1/2, 1/6, 0, ...]: 50 = T(5, 2) = (5/2)*(binomial(1, 1)*1*6 + binomial(2, 1)*(1/2)*11 + binomial(3, 1)*(1/6)*6 + 0) = 50. The vanishing z-sequence produces the k=0 column from T(0, 0) = 1. (End)
Elementary symmetric function T(4, 2) = sigma^{(3)}_2 = 1*2 + 1*3 + 2*3 = 11. Here the cells (polytopes) are 3 rectangles with total area 11. - _Wolfdieter Lang_, May 28 2017
O.g.f.s of diagonals: d=2 (third diagonal) [0, 6, 50, ...] has D(2,t) = P(2, t)/(1-t)^5, with P(2, t) = 2 + t, the n = 2 row of A288874. - _Wolfdieter Lang_, Jul 20 2017
Boas-Buck recurrence for column k = 2 and n = 5: T(5, 2) = (5!*2/3)*((3/8)*T(2,2)/2! + (5/12)*T(3,2)/3! + (1/2)*T(4,2)/4!) = (5!*2/3)*(3/16 + (5/12)*3/3! + (1/2)*11/4!) = 50. The beta sequence begins: {1/2, 5/12, 3/8, ...}. - _Wolfdieter Lang_, Aug 11 2017
		

References

  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, pages 31, 187, 441, 996.
  • R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd. ed., Table 259, p. 259.
  • Steve Roman, The Umbral Calculus, Dover Publications, New York (1984), pp. 149-150

Crossrefs

Essentially a duplicate of A048994. Cf. A008275, A008277, A112007, A130534, A288874, A354795.

Programs

  • Haskell
    a132393 n k = a132393_tabl !! n !! k
    a132393_row n = a132393_tabl !! n
    a132393_tabl = map (map abs) a048994_tabl
    -- Reinhard Zumkeller, Nov 06 2013
    
  • Maple
    a132393_row := proc(n) local k; seq(coeff(expand(pochhammer (x,n)),x,k),k=0..n) end: # Peter Luschny, Nov 28 2010
  • Mathematica
    p[t_] = 1/(1 - t)^x; Table[ ExpandAll[(n!)SeriesCoefficient[ Series[p[t], {t, 0, 30}], n]], {n, 0, 10}]; a = Table[(n!)* CoefficientList[SeriesCoefficient[ Series[p[t], {t, 0, 30}], n], x], {n, 0, 10}]; Flatten[a] (* Roger L. Bagula, Apr 18 2008 *)
    Flatten[Table[Abs[StirlingS1[n,i]],{n,0,10},{i,0,n}]] (* Harvey P. Dale, Feb 04 2014 *)
  • Maxima
    create_list(abs(stirling1(n,k)),n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */
    
  • PARI
    column(n,k) = my(v1, v2); v1 = vector(n-1, i, 0); v2 = vector(n, i, 0); v2[1] = 1; for(i=1, n-1, v1[i] = (i+k)*(i+k-1)/2*v2[i]; for(j=1, i-1, v1[j] *= (i-j)*(i+k)/(i-j+2)); v2[i+1] = vecsum(v1)/i); v2 \\ generates n first elements of the k-th column starting from the first nonzero element. - Mikhail Kurkov, Mar 05 2025

Formula

T(n,k) = T(n-1,k-1)+(n-1)*T(n-1,k), n,k>=1; T(n,0)=T(0,k); T(0,0)=1.
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A000142(n), A001147(n), A007559(n), A007696(n), A008548(n), A008542(n), A045754(n), A045755(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8 respectively. - Philippe Deléham, Nov 13 2007
Expand 1/(1-t)^x = Sum_{n>=0}p(x,n)*t^n/n!; then the coefficients of the p(x,n) produce the triangle. - Roger L. Bagula, Apr 18 2008
Sum_{k=0..n} T(n,k)*2^k*x^(n-k) = A000142(n+1), A000165(n), A008544(n), A001813(n), A047055(n), A047657(n), A084947(n), A084948(n), A084949(n) for x = 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively. - Philippe Deléham, Sep 18 2008
a(n) = Sum_{k=0..n} T(n,k)*3^k*x^(n-k) = A001710(n+2), A001147(n+1), A032031(n), A008545(n), A047056(n), A011781(n), A144739(n), A144756(n), A144758(n) for x=1,2,3,4,5,6,7,8,9,respectively. - Philippe Deléham, Sep 20 2008
Sum_{k=0..n} T(n,k)*4^k*x^(n-k) = A001715(n+3), A002866(n+1), A007559(n+1), A047053(n), A008546(n), A049308(n), A144827(n), A144828(n), A144829(n) for x=1,2,3,4,5,6,7,8,9 respectively. - Philippe Deléham, Sep 21 2008
Sum_{k=0..n} x^k*T(n,k) = x*(1+x)*(2+x)*...*(n-1+x), n>=1. - Philippe Deléham, Oct 17 2008
From Wolfdieter Lang, Feb 21 2017: (Start)
E.g.f. k-th column: (-log(1 - x))^k, k >= 0.
E.g.f. triangle (see the Apr 18 2008 Baluga comment): exp(-x*log(1-z)).
E.g.f. a-sequence: x/(1 - exp(-x)). See A164555/A027642. The e.g.f. for the z-sequence is 0. (End)
From Wolfdieter Lang, May 28 2017: (Start)
The row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k, for n >= 0, are R(n, x) = risefac(x,n-1) := Product_{j=0..n-1} x+j, with the empty product for n=0 put to 1. See the Feb 21 2017 comment above. This implies:
T(n, k) = sigma^{(n-1)}_(n-k), for n >= k >= 1, with the elementary symmetric functions sigma^{(n-1)}_m of degree m in the n-1 symbols 1, 2, ..., n-1, with binomial(n-1, m) terms. See an example below.(End)
Boas-Buck type recurrence for column sequence k: T(n, k) = (n!*k/(n - k)) * Sum_{p=k..n-1} beta(n-1-p)*T(p, k)/p!, for n > k >= 0, with input T(k, k) = 1, and beta(k) = A002208(k+1)/A002209(k+1). See a comment and references in A286718. - Wolfdieter Lang, Aug 11 2017
T(n,k) = Sum_{j=k..n} j^(j-k)*binomial(j-1, k-1)*A354795(n,j) for n > 0. - Mélika Tebni, Mar 02 2023
n-th row polynomial: n!*Sum_{k = 0..2*n} (-1)^k*binomial(-x, k)*binomial(-x, 2*n-k) = n!*Sum_{k = 0..2*n} (-1)^k*binomial(1-x, k)*binomial(-x, 2*n-k). - Peter Bala, Mar 31 2024
From Mikhail Kurkov, Mar 05 2025: (Start)
For a general proof of the formulas below via generating functions, see Mathematics Stack Exchange link.
Recursion for the n-th row (independently of other rows): T(n,k) = 1/(n-k)*Sum_{j=2..n-k+1} binomial(-k,j)*T(n,k+j-1)*(-1)^j for 1 <= k < n with T(n,n) = 1.
Recursion for the k-th column (independently of other columns): T(n,k) = 1/(n-k)*Sum_{j=2..n-k+1} (j-2)!*binomial(n,j)*T(n-j+1,k) for 1 <= k < n with T(n,n) = 1 (see Fedor Petrov link). (End)

A008290 Triangle T(n,k) of rencontres numbers (number of permutations of n elements with k fixed points).

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 2, 3, 0, 1, 9, 8, 6, 0, 1, 44, 45, 20, 10, 0, 1, 265, 264, 135, 40, 15, 0, 1, 1854, 1855, 924, 315, 70, 21, 0, 1, 14833, 14832, 7420, 2464, 630, 112, 28, 0, 1, 133496, 133497, 66744, 22260, 5544, 1134, 168, 36, 0, 1, 1334961, 1334960, 667485, 222480, 55650, 11088, 1890, 240, 45, 0, 1
Offset: 0

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Author

Keywords

Comments

This is a binomial convolution triangle (Sheffer triangle) of the Appell type: (exp(-x)/(1-x),x), i.e., the e.g.f. of column k is (exp(-x)/(1-x))*(x^k/k!). See the e.g.f. given by V. Jovovic below. - Wolfdieter Lang, Jan 21 2008
The formula T(n,k) = binomial(n,k)*A000166(n-k), with the derangements numbers (subfactorials) A000166 (see also the Charalambides reference) shows the Appell type of this triangle. - Wolfdieter Lang, Jan 21 2008
T(n,k) is the number of permutations of {1,2,...,n} having k pairs of consecutive right-to-left minima (0 is considered a right-to-left minimum for each permutation). Example: T(4,2)=6 because we have 1243, 1423, 4123, 1324, 3124 and 2134; for example, 1324 has right-to-left minima in positions 0-1,3-4 and 2134 has right-to-left minima in positions 0,2-3-4, the consecutive ones being joined by "-". - Emeric Deutsch, Mar 29 2008
T is an example of the group of matrices outlined in the table in A132382--the associated matrix for the sequence aC(0,1). - Tom Copeland, Sep 10 2008
A refinement of this triangle is given by A036039. - Tom Copeland, Nov 06 2012
This triangle equals (A211229(2*n,2*k)) n,k >= 0. - Peter Bala, Dec 17 2014

Examples

			exp((y-1)*x)/(1-x) = 1 + y*x + (1/2!)*(1+y^2)*x^2 + (1/3!)*(2 + 3*y + y^3)*x^3 + (1/4!)*(9 + 8*y + 6*y^2 + y^4)*x^4 + (1/5!)*(44 + 45*y + 20*y^2 + 10*y^3 + y^5)*x^5 + ...
Triangle begins:
       1
       0      1
       1      0     1
       2      3     0     1
       9      8     6     0    1
      44     45    20    10    0    1
     265    264   135    40   15    0   1
    1854   1855   924   315   70   21   0  1
   14833  14832  7420  2464  630  112  28  0 1
  133496 133497 66744 22260 5544 1134 168 36 0 1
...
From _Peter Bala_, Feb 13 2017: (Start)
The infinitesimal generator has integer entries given by binomial(n,k)*(n-k-1)! for n >= 2 and 0 <= k <= n-2 and begins
   0
   0  0
   1  0  0
   2  3  0  0
   6  8  6  0 0
  24 30 20 10 0 0
...
It is essentially A238363 (unsigned and omitting the main diagonal), A211603 (with different offset) and appears to be A092271, again without the main diagonal. (End)
		

References

  • Ch. A. Charalambides, Enumerative Combinatorics, Chapman & Hall/CRC, Boca Raton, Florida, 2002, p. 173, Table 5.2 (without row n=0 and column k=0).
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 194.
  • Arnold Kaufmann, Introduction à la combinatorique en vue des applications, Dunod, Paris, 1968. See p. 92.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 65.

Crossrefs

Mirror of triangle A098825.
Cf. A080955.
Cf. A000012, A000142 (row sums), A000354.
Cf. A170942. Sub-triangle of A211229.
T(2n,n) gives A281262.

Programs

  • Haskell
    a008290 n k = a008290_tabl !! n !! k
    a008290_row n = a008290_tabl !! n
    a008290_tabl = map reverse a098825_tabl
    -- Reinhard Zumkeller, Dec 16 2013
  • Maple
    T:= proc(n,k) T(n, k):= `if`(k=0, `if`(n<2, 1-n, (n-1)*
          (T(n-1, 0)+T(n-2, 0))), binomial(n, k)*T(n-k, 0))
        end:
    seq(seq(T(n, k), k=0..n), n=0..12);  # Alois P. Heinz, Mar 15 2013
  • Mathematica
    a[0] = 1; a[1] = 0; a[n_] := Round[n!/E] /; n >= 1 size = 8; Table[Binomial[n, k]a[n - k], {n, 0, size}, {k, 0, n}] // TableForm (* Harlan J. Brothers, Mar 19 2007 *)
    T[n_, k_] := Subfactorial[n-k]*Binomial[n, k]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 12 2017 *)
    T[n_, k_] := If[n<1, Boole[n==0 && k==0], T[n, k] = T[n-1, k-1] + T[n-1, k]*(n-1-k) + T[n-1, k+1]*(k+1)]; (* Michael Somos, Sep 13 2024 *)
    T[0, 0]:=1; T[n_, 0]:=T[n, 0]=n  T[n-1, 0]+(-1)^n; T[n_, k_]:=T[n, k]=n/k T[n-1, k-1];
    Flatten@Table[T[n, k], {n, 0, 9}, {k, 0, n}] (* Oliver Seipel, Nov 26 2024 *)
  • PARI
    {T(n, k) = if(k<0 || k>n, 0, n!/k! * sum(i=0, n-k, (-1)^i/i!))}; /* Michael Somos, Apr 26 2000 */
    

Formula

T(n, k) = T(n-1, k)*n + binomial(n, k)*(-1)^(n-k) = T(n, k-1)/k + binomial(n, k)*(-1)^(n-k)/(n-k+1) = T(n-1, k-1)*n/k = T(n-k, 0)*binomial(n, k) = A000166(n-k)*binomial(n,k) [with T(0, 0) = 1]; so T(n, n) = 1, T(n, n-1) = 0, T(n, n-2) = n*(n-1)/2 for n >= 0.
Sum_{k=0..n} T(n, k) = Sum_{k=0..n} k * T(n, k) = n! for all n > 0, n, k integers. - Wouter Meeussen, May 29 2001
From Vladeta Jovovic, Aug 12 2002: (Start)
O.g.f. for k-th column: (1/k!)*Sum_{i>=k} i!*x^i/(1+x)^(i+1).
O.g.f. for k-th row: k!*Sum_{i=0..k} (-1)^i/i!*(1-x)^i. (End)
E.g.f.: exp((y-1)*x)/(1-x). - Vladeta Jovovic, Aug 18 2002
E.g.f. for number of permutations with exactly k fixed points is x^k/(k!*exp(x)*(1-x)). - Vladeta Jovovic, Aug 25 2002
Sum_{k=0..n} T(n, k)*x^k is the permanent of the n X n matrix with x's on the diagonal and 1's elsewhere; for x = 0, 1, 2, 3, 4, 5, 6 see A000166, A000142, A000522, A010842, A053486, A053487, A080954. - Philippe Deléham, Dec 12 2003; for x = 1+i see A009551 and A009102. - John M. Campbell, Oct 11 2011
T(n, k) = Sum_{j=0..n} A008290(n, j)*k^(n-j) is the permanent of the n X n matrix with 1's on the diagonal and k's elsewhere; for k = 0, 1, 2 see A000012, A000142, A000354. - Philippe Deléham, Dec 13 2003
T(n,k) = Sum_{j=0..n} (-1)^(j-k)*binomial(j,k)*n!/j!. - Paul Barry, May 25 2006
T(n,k) = (n!/k!)*Sum_{j=0..n-k} ((-1)^j)/j!, 0 <= k <= n. From the Appell type of the triangle and the subfactorial formula.
T(n,0) = n*Sum_{j=0..n-1} (j/(j+1))*T(n-1,j), T(0,0)=1. From the z-sequence of this Sheffer triangle z(j)=j/(j+1) with e.g.f. (1-exp(x)*(1-x))/x. See the W. Lang link under A006232 for Sheffer a- and z-sequences. - Wolfdieter Lang, Jan 21 2008
T(n,k) = (n/k)*T(n-1,k-1) for k >= 1. See above. From the a-sequence of this Sheffer triangle a(0)=1, a(n)=0, n >= 1 with e.g.f. 1. See the W. Lang link under A006232 for Sheffer a- and z-sequences. - Wolfdieter Lang, Jan 21 2008
From Henk P. J. van Wijk, Oct 29 2012: (Start)
T(n,k) = T(n-1,k)*(n-1-k) + T(n-1,k+1)*(k+1) for k=0 and
T(n,k) = T(n-1,k-1) + T(n-1,k)*(n-1-k) + T(n-1,k+1)*(k+1) for k>=1.
(End)
T(n,k) = A098825(n,n-k). - Reinhard Zumkeller, Dec 16 2013
Sum_{k=0..n} k^2 * T(n, k) = 2*n! if n > 1. - Michael Somos, Jun 06 2017
From Tom Copeland, Jul 26 2017: (Start)
The lowering and raising operators of this Appell sequence of polynomials P(n,x) are L = d/dx and R = x + d/dL log[exp(-L)/(1-L)] = x-1 + 1/(1-L) = x + L + L^2 - ... such that L P(n,x) = n P(n-1,x) and R P(n,x) = P(n+1,x).
P(n,x) = (1-L)^(-1) exp(-L) x^n = (1+L+L^2+...)(x-1)^n = n! Sum_{k=0..n} (x-1)^k / k!.
The formalism of A133314 applies to the pair of entries A008290 and A055137.
The polynomials of this pair P_n(x) and Q_n(x) are umbral compositional inverses; i.e., P_n(Q.(x)) = x^n = Q_n(P.(x)), where, e.g., (Q.(x))^n = Q_n(x).
For more on the infinitesimal generator, noted by Bala below, see A238385. (End)
Sum_{k=0..n} k^m * T(n,k) = A000110(m)*n! if n >= m. - Zhujun Zhang, May 24 2019
Sum_{k=0..n} (k+1) * T(n,k) = A098558(n). - Alois P. Heinz, Mar 11 2022
From Alois P. Heinz, May 20 2023: (Start)
Sum_{k=0..n} (-1)^k * T(n,k) = A000023(n).
Sum_{k=0..n} (-1)^k * k * T(n,k) = A335111(n). (End)
T(n,k) = A145224(n,k)+A145225(n,k), refined by even and odd perms. - R. J. Mathar, Jul 06 2023

Extensions

Comments and more terms from Michael Somos, Apr 26 2000 and Christian G. Bower, Apr 26 2000

A053121 Catalan triangle (with 0's) read by rows.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 0, 2, 0, 1, 2, 0, 3, 0, 1, 0, 5, 0, 4, 0, 1, 5, 0, 9, 0, 5, 0, 1, 0, 14, 0, 14, 0, 6, 0, 1, 14, 0, 28, 0, 20, 0, 7, 0, 1, 0, 42, 0, 48, 0, 27, 0, 8, 0, 1, 42, 0, 90, 0, 75, 0, 35, 0, 9, 0, 1, 0, 132, 0, 165, 0, 110, 0, 44, 0, 10, 0, 1, 132, 0, 297, 0, 275, 0, 154, 0, 54, 0, 11, 0
Offset: 0

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Author

Keywords

Comments

Inverse lower triangular matrix of A049310(n,m) (coefficients of Chebyshev's S polynomials).
Walks with a wall: triangle of number of n-step walks from (0,0) to (n,m) where each step goes from (a,b) to (a+1,b+1) or (a+1,b-1) and the path stays in the nonnegative quadrant.
T(n,m) is the number of left factors of Dyck paths of length n ending at height m. Example: T(4,2)=3 because we have UDUU, UUDU, and UUUD, where U=(1,1) and D=(1,-1). (This is basically a different formulation of the previous - walks with a wall - property.) - Emeric Deutsch, Jun 16 2011
"The Catalan triangle is formed in the same manner as Pascal's triangle, except that no number may appear on the left of the vertical bar." [Conway and Smith]
G.f. for row polynomials p(n,x) := Sum_{m=0..n} (a(n,m)*x^m): c(z^2)/(1-x*z*c(z^2)). Row sums (x=1): A001405 (central binomial).
In the language of the Shapiro et al. reference such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to the Bell-subgroup of the Riordan-group. The g.f. Ginv(x) of the m=0 column of the inverse of a given Bell-matrix (here A049310) is obtained from its g.f. of the m=0 column (here G(x)=1/(1+x^2)) by Ginv(x)=(f^{(-1)}(x))/x, with f(x) := x*G(x) and f^{(-1)}is the compositional inverse function of f (here one finds, with Ginv(0)=1, c(x^2)). See the Shapiro et al. reference.
Number of involutions of {1,2,...,n} that avoid the patterns 132 and have exactly k fixed points. Example: T(4,2)=3 because we have 2134, 4231 and 3214. Number of involutions of {1,2,...,n} that avoid the patterns 321 and have exactly k fixed points. Example: T(4,2)=3 because we have 1243, 1324 and 2134. Number of involutions of {1,2,...,n} that avoid the patterns 213 and have exactly k fixed points. Example: T(4,2)=3 because we have 1243, 1432 and 4231. - Emeric Deutsch, Oct 12 2006
This triangle belongs to the family of triangles defined by: T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=x*T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+y*T(n-1,k)+T(n-1,k+1) for k>=1 . Other triangles arise by choosing different values for (x,y): (0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970; (1,0) -> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877; (1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598; (2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954; (3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791; (4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. - Philippe Deléham, Sep 25 2007
Riordan array (c(x^2),xc(x^2)), where c(x) is the g.f. of Catalan numbers A000108. - Philippe Deléham, Nov 25 2007
A053121^2 = triangle A145973. Convolved with A001405 = triangle A153585. - Gary W. Adamson, Dec 28 2008
By columns without the zeros, n-th row = A000108 convolved with itself n times; equivalent to A = (1 + x + 2x^2 + 5x^3 + 14x^4 + ...), then n-th row = coefficients of A^(n+1). - Gary W. Adamson, May 13 2009
Triangle read by rows,product of A130595 and A064189 considered as infinite lower triangular arrays; A053121 = A130595*A064189 = B^(-1)*A097609*B where B = A007318. - Philippe Deléham, Dec 07 2009
From Mark Dols, Aug 17 2010: (Start)
As an upper right triangle, rows represent powers of 5-sqrt(24):
5 - sqrt(24)^1 = 0.101020514...
5 - sqrt(24)^2 = 0.010205144...
5 - sqrt(24)^3 = 0.001030928...
(Divided by sqrt(96) these powers give a decimal representation of the columns of A007318, with 1/sqrt(96) being the middle column.) (End)
T(n,k) is the number of dispersed Dyck paths of length n (i.e., Motzkin paths of length n with no (1,0) steps at positive heights) having k (1,0)-steps. Example: T(5,3)=4 because, denoting U=(1,1), D=(1,-1), H=(1,0), we have HHHUD, HHUDH, HUDHH, and UDHHH. - Emeric Deutsch, Jun 01 2011
Let S(N,x) denote the N-th Chebyshev S-polynomial in x (see A049310, cf. [W. Lang]). Then x^n = sum_{k=0..n} T(n,k)*S(k,x). - L. Edson Jeffery, Sep 06 2012
This triangle a(n,m) appears also in the (unreduced) formula for the powers rho(N)^n for the algebraic number over the rationals rho(N) = 2*cos(Pi/N) = R(N, 2), the smallest diagonal/side ratio R in the regular N-gon:
rho(N)^n = sum(a(n,m)*R(N,m+1),m=0..n), n>=0, identical in N >= 1. R(N,j) = S(j-1, x=rho(N)) (Chebyshev S (A049310)). See a comment on this under A039599 (even powers) and A039598 (odd powers). Proof: see the Sep 06 2012 comment by L. Edson Jeffery, which follows from T(n,k) (called here a(n,k)) being the inverse of the Riordan triangle A049310. - Wolfdieter Lang, Sep 21 2013
The so-called A-sequence for this Riordan triangle of the Bell type (c(x^2), x*c(x^2)) (see comments above) is A(x) = 1 + x^2. This proves the recurrence given in the formula section by Henry Bottomley for a(n, m) = a(n-1, m-1) + a(n-1, m+1) for n>=1 and m>=1, with inputs. The Z-sequence for this Riordan triangle is Z(x) = x which proves the recurrence a(n,0) = a(n-1,1), n>=1, a(0,0) = 1. For A- and Z-sequences for Riordan triangles see the W. Lang link under A006232. - Wolfdieter Lang, Sep 22 2013
Rows of the triangle describe decompositions of tensor powers of the standard (2-dimensional) representation of the Lie algebra sl(2) into irreducibles. Thus a(n,m) is the multiplicity of the m-th ((m+1)-dimensional) irreducible representation in the n-th tensor power of the standard one. - Mamuka Jibladze, May 26 2015
The Riordan row polynomials p(n, x) belong to the Boas-Buck class (see a comment and references in A046521), hence they satisfy the Boas-Buck identity: (E_x - n*1)*p(n, x) = (E_x + 1)*Sum_{j=0..n-1} (1/2)*(1 - (-1)^j)*binomial(j+1, (j+1)/2)*p(n-1-j, x), for n >= 0, where E_x = x*d/dx (Euler operator). For the triangle a(n, m) this entails a recurrence for the sequence of column m, given in the formula section. - Wolfdieter Lang, Aug 11 2017
From Roger Ford, Jan 22 2018: (Start)
For row n, the nonzero values represent the odd components (loops) formed by n+1 nonintersecting arches above and below the x-axis with the following constraints: The top has floor((n+3)/2) starting arches at position 1 and the next consecutive odd positions. All other starting top arches are in even positions. The bottom arches are a rainbow of arches. If the component=1 then the arch configuration is a semimeander solution.
Examples: For row 3 {0, 2, 0, 1} there are 3 arch configurations: 2 arch configurations have a component=1; 1 has a component=3. c=components, U=top arch starting in odd position, u=top arch starting in an even position, d=ending top arch:
.
top UuUdUddd c=3 top UdUuUddd c=1 top UdUdUudd c=1
/\ /\
//\\ / \
// \\ / /\ \ /\
// \\ / / \ \ / \
///\ /\\\ /\ / / /\ \ \ /\ /\ / /\ \
\\\ \/ /// \ \ \ \/ / / / \ \ \ \/ / / /
\\\ /// \ \ \ / / / \ \ \ / / /
\\\/// \ \ \/ / / \ \ \/ / /
\\// \ \ / / \ \ / /
\/ \ \/ / \ \/ /
\ / \ /
\/ \/
For row 4 {2, 0, 3, 0, 1} there are 6 arch configurations: 2 have a component=1; 3 have a component=3: 1 has a component=1. (End)

Examples

			Triangle a(n,m) begins:
  n\m  0   1   2   3   4   5   6  7  8  9 10 ...
  0:   1
  1:   0   1
  2:   1   0   1
  3:   0   2   0   1
  4:   2   0   3   0   1
  5:   0   5   0   4   0   1
  6:   5   0   9   0   5   0   1
  7:   0  14   0  14   0   6   0  1
  8:  14   0  28   0  20   0   7  0  1
  9:   0  42   0  48   0  27   0  8  0  1
  10: 42   0  90   0  75   0  35  0  9  0  1
  ... (Reformatted by _Wolfdieter Lang_, Sep 20 2013)
E.g., the fourth row corresponds to the polynomial p(3,x)= 2*x + x^3.
From _Paul Barry_, May 29 2009: (Start)
Production matrix is
  0, 1,
  1, 0, 1,
  0, 1, 0, 1,
  0, 0, 1, 0, 1,
  0, 0, 0, 1, 0, 1,
  0, 0, 0, 0, 1, 0, 1,
  0, 0, 0, 0, 0, 1, 0, 1,
  0, 0, 0, 0, 0, 0, 1, 0, 1,
  0, 0, 0, 0, 0, 0, 0, 1, 0, 1 (End)
Boas-Buck recurrence for column k = 2, n = 6: a(6, 2) = (3/4)*(0 + 2*a(4 ,2) + 0 + 6*a(2, 2)) = (3/4)*(2*3 + 6) = 9. - _Wolfdieter Lang_, Aug 11 2017
		

References

  • J. H. Conway and D. A. Smith, On Quaternions and Octonions, A K Peters, Ltd., Natick, MA, 2003. See p. 60. MR1957212 (2004a:17002)
  • A. Nkwanta, Lattice paths and RNA secondary structures, in African Americans in Mathematics, ed. N. Dean, Amer. Math. Soc., 1997, pp. 137-147.

Crossrefs

Cf. A008315, A049310, A000108, A001405 (row sums), A145973, A153585, A108786, A037952. Another version: A008313. A039598 and A039599 without zeros, and odd and even numbered rows.
Variant without zero-diagonals: A033184 and with rows reversed: A009766.

Programs

  • Haskell
    a053121 n k = a053121_tabl !! n !! k
    a053121_row n = a053121_tabl !! n
    a053121_tabl = iterate
       (\row -> zipWith (+) ([0] ++ row) (tail row ++ [0,0])) [1]
    -- Reinhard Zumkeller, Feb 24 2012
    
  • Maple
    T:=proc(n,k): if n+k mod 2 = 0 then (k+1)*binomial(n+1,(n-k)/2)/(n+1) else 0 fi end: for n from 0 to 13 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form; Emeric Deutsch, Oct 12 2006
    F:=proc(l,p) if ((l-p) mod 2) = 1 then 0 else (p+1)*l!/( ( (l-p)/2 )! * ( (l+p)/2 +1)! ); fi; end;
    r:=n->[seq( F(n,p),p=0..n)]; [seq(r(n),n=0..15)]; # N. J. A. Sloane, Jan 29 2011
    A053121 := proc(n,k) option remember; `if`(k>n or k<0,0,`if`(n=k,1,
    procname(n-1,k-1)+procname(n-1,k+1))) end proc:
    seq(print(seq(A053121(n,k), k=0..n)),n=0..12); # Peter Luschny, May 01 2011
  • Mathematica
    a[n_, m_] /; n < m || OddQ[n-m] = 0; a[n_, m_] = (m+1) Binomial[n+1, (n-m)/2]/(n+1); Flatten[Table[a[n, m], {n, 0, 12}, {m, 0, n}]] [[1 ;; 90]] (* Jean-François Alcover, May 18 2011 *)
    T[0, 0] := 1; T[n_, k_]/;0<=k<=n := T[n, k] = T[n-1, k-1]+T[n-1, k+1]; T[n_, k_] := 0; Flatten@Table[T[n, k], {n, 0, 12}, {k, 0, n}] (* Oliver Seipel, Dec 31 2024 *)
  • PARI
    T(n, m)=if(nCharles R Greathouse IV, Mar 09 2016
  • Sage
    def A053121_triangle(dim):
        M = matrix(ZZ,dim,dim)
        for n in (0..dim-1): M[n,n] = 1
        for n in (1..dim-1):
            for k in (0..n-1):
                M[n,k] = M[n-1,k-1] + M[n-1,k+1]
        return M
    A053121_triangle(13) # Peter Luschny, Sep 19 2012
    

Formula

a(n, m) := 0 if n
a(n, m) = (4*(n-1)*a(n-2, m) + 2*(m+1)*a(n-1, m-1))/(n+m+2), a(n, m)=0 if n
G.f. for m-th column: c(x^2)*(x*c(x^2))^m, where c(x) = g.f. for Catalan numbers A000108.
G.f.: G(t,z) = c(z^2)/(1 - t*z*c(z^2)), where c(z) = (1 - sqrt(1-4*z))/(2*z) is the g.f. for the Catalan numbers (A000108). - Emeric Deutsch, Jun 16 2011
a(n, m) = a(n-1, m-1) + a(n-1, m+1) if n > 0 and m >= 0, a(0, 0)=1, a(0, m)=0 if m > 0, a(n, m)=0 if m < 0. - Henry Bottomley, Jan 25 2001
Sum_{k>=0} T(m,k)^2 = A000108(m). - Paul D. Hanna, Apr 23 2005
Sum_{k>=0} T(m, k)*T(n, k) = 0 if m+n is odd; Sum_{k>=0} T(m, k)*T(n, k) = A000108((m+n)/2) if m+n is even. - Philippe Deléham, May 26 2005
T(n,k)=sum{i=0..n, (-1)^(n-i)*C(n,i)*sum{j=0..i, C(i,j)*(C(i-j,j+k)-C(i-j,j+k+2))}}; Column k has e.g.f. BesselI(k,2x)-BesselI(k+2,2x). - Paul Barry, Feb 16 2006
Sum_{k=0..n} T(n,k)*(k+1) = 2^n. - Philippe Deléham, Mar 22 2007
Sum_{j>=0} T(n,j)*binomial(j,k) = A054336(n,k). - Philippe Deléham, Mar 30 2007
T(2*n+1,2*k+1) = A039598(n,k), T(2*n,2*k) = A039599(n,k). - Philippe Deléham, Apr 16 2007
Sum_{k=0..n} T(n,k)^x = A000027(n+1), A001405(n), A000108(n), A003161(n), A129123(n) for x = 0,1,2,3,4 respectively. - Philippe Deléham, Nov 22 2009
Sum_{k=0..n} T(n,k)*x^k = A126930(n), A126120(n), A001405(n), A054341(n), A126931(n) for x = -1, 0, 1, 2, 3 respectively. - Philippe Deléham, Nov 28 2009
Sum_{k=0..n} T(n,k)*A000045(k+1) = A098615(n). - Philippe Deléham, Feb 03 2012
Recurrence for row polynomials C(n, x) := Sum_{m=0..n} a(n, m)*x^m = x*Sum_{k=0..n} Chat(k)*C(n-1-k, x), n >= 0, with C(-1, 1/x) = 1/x and Chat(k) = A000108(k/2) if n is even and 0 otherwise. From the o.g.f. of the row polynomials: G(z; x) := Sum_{n >= 0} C(n, x)*z^n = c(z^2)*(1 + x*z*G(z, x)), with the o.g.f. c of A000108. - Ahmet Zahid KÜÇÜK and Wolfdieter Lang, Aug 23 2015
The Boas-Buck recurrence (see a comment above) for the sequence of column m is: a(n, m) = ((m+1)/(n-m))*Sum_{j=0..n-1-m} (1/2)*(1 - (-1)^j)*binomial(j+1, (j+1)/2)* a(n-1-j, k), for n > m >= 0 and input a(m, m) = 1. - Wolfdieter Lang, Aug 11 2017
Sum_{m=1..n} a(n,m) = A037952(n). - R. J. Mathar, Sep 23 2021

Extensions

Edited by N. J. A. Sloane, Jan 29 2011
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