cp's OEIS Frontend

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.

Showing 1-10 of 12 results. Next

A085478 Triangle read by rows: T(n, k) = binomial(n + k, 2*k).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 6, 5, 1, 1, 10, 15, 7, 1, 1, 15, 35, 28, 9, 1, 1, 21, 70, 84, 45, 11, 1, 1, 28, 126, 210, 165, 66, 13, 1, 1, 36, 210, 462, 495, 286, 91, 15, 1, 1, 45, 330, 924, 1287, 1001, 455, 120, 17, 1, 1, 55, 495, 1716, 3003, 3003, 1820, 680, 153, 19, 1
Offset: 0

Views

Author

Philippe Deléham, Aug 14 2003

Keywords

Comments

Coefficient array for Morgan-Voyce polynomial b(n,x). A053122 (unsigned) is the coefficient array for B(n,x). Reversal of A054142. - Paul Barry, Jan 19 2004
This triangle is formed from even-numbered rows of triangle A011973 read in reverse order. - Philippe Deléham, Feb 16 2004
T(n,k) is the number of nondecreasing Dyck paths of semilength n+1, having k+1 peaks. T(n,k) is the number of nondecreasing Dyck paths of semilength n+1, having k peaks at height >= 2. T(n,k) is the number of directed column-convex polyominoes of area n+1, having k+1 columns. - Emeric Deutsch, May 31 2004
Riordan array (1/(1-x), x/(1-x)^2). - Paul Barry, May 09 2005
The triangular matrix a(n,k) = (-1)^(n+k)*T(n,k) is the matrix inverse of A039599. - Philippe Deléham, May 26 2005
The n-th row gives absolute values of coefficients of reciprocal of g.f. of bottom-line of n-wave sequence. - Floor van Lamoen (fvlamoen(AT)planet.nl), Sep 24 2006
Unsigned version of A129818. - Philippe Deléham, Oct 25 2007
T(n, k) is also the number of idempotent order-preserving full transformations (of an n-chain) of height k >=1 (height(alpha) = |Im(alpha)|) and of waist n (waist(alpha) = max(Im(alpha))). - Abdullahi Umar, Oct 02 2008
A085478 is jointly generated with A078812 as a triangular array of coefficients of polynomials u(n,x): initially, u(1,x) = v(1,x) = 1; for n>1, u(n,x) = u(n-1,x)+x*v(n-1)x and v(n,x) = u(n-1,x)+(x+1)*v(n-1,x). See the Mathematica section. - Clark Kimberling, Feb 25 2012
Per Kimberling's recursion relations, see A102426. - Tom Copeland, Jan 19 2016
Subtriangle of the triangle given by (0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 26 2012
T(n,k) is also the number of compositions (ordered partitions) of 2*n+1 into 2*k+1 parts which are all odd. Proof: The o.g.f. of column k, x^k/(1-x)^(2*k+1) for k >= 0, is the o.g.f. of the odd-indexed members of the sequence with o.g.f. (x/(1-x^2))^(2*k+1) (bisection, odd part). Thus T(n,k) is obtained from the sum of the multinomial numbers A048996 for the partitions of 2*n+1 into 2*k+1 parts, all of which are odd. E.g., T(3,1) = 3 + 3 from the numbers for the partitions [1,1,5] and [1,3,3], namely 3!/(2!*1!) and 3!/(1!*2!), respectively. The number triangle with the number of these partitions as entries is A152157. - Wolfdieter Lang, Jul 09 2012
The matrix elements of the inverse are T^(-1)(n,k) = (-1)^(n+k)*A039599(n,k). - R. J. Mathar, Mar 12 2013
T(n,k) = A258993(n+1,k) for k = 0..n-1. - Reinhard Zumkeller, Jun 22 2015
The n-th row polynomial in descending powers of x is the n-th Taylor polynomial of the algebraic function F(x)*G(x)^n about 0, where F(x) = (1 + sqrt(1 + 4*x))/(2*sqrt(1 + 4*x)) and G(x) = ((1 + sqrt(1 + 4*x))/2)^2. For example, for n = 4, (1 + sqrt(1 + 4*x))/(2*sqrt(1 + 4*x)) * ((1 + sqrt(1 + 4*x))/2)^8 = (x^4 + 10*x^3 + 15*x^2 + 7*x + 1) + O(x^5). - Peter Bala, Feb 23 2018
Row n also gives the coefficients of the characteristc polynomial of the tridiagonal n X n matrix M_n given in A332602: Phi(n, x) := Det(M_n - x*1_n) = Sum_{k=0..n} T(n, k)*(-x)^k, for n >= 0, with Phi(0, x) := 1. - Wolfdieter Lang, Mar 25 2020
It appears that the largest root of the n-th degree polynomial is equal to the sum of the distinct diagonals of a (2*n+1)-gon including the edge, 1. The largest root of x^3 - 6*x^2 + 5*x - 1 is 5.048917... = the sum of (1 + 1.80193... + 2.24697...). Alternatively, the largest root of the n-th degree polynomial is equal to the square of sigma(2*n+1). Check: 5.048917... is the square of sigma(7), 2.24697.... Given N = 2*n+1, sigma(N) (N odd) can be defined as 1/(2*sin(Pi/(2*N))). Relating to the 9-gon, the largest root of x^4 - 10*x^3 + 15*x^2 - 7*x + 1 is 8.290859..., = the sum of (1 + 1.879385... + 2.532088... + 2.879385...), and is the square of sigma(9), 2.879385... Refer to A231187 for a further clarification of sigma(7). - Gary W. Adamson, Jun 28 2022
For n >=1, the n-th row is given by the coefficients of the minimal polynomial of -4*sin(Pi/(4*n + 2))^2. - Eric W. Weisstein, Jul 12 2023
Denoting this lower triangular array by L, then L * diag(binomial(2*k,k)^2) * transpose(L) is the LDU factorization of A143007, the square array of crystal ball sequences for the A_n X A_n lattices. - Peter Bala, Feb 06 2024
T(n, k) is the number of occurrences of the periodic substring (01)^k in the periodic string (01)^n (see Proposition 4.7 at page 7 in Fang). - Stefano Spezia, Jun 09 2024

Examples

			Triangle begins as:
  1;
  1    1;
  1    3    1;
  1    6    5    1;
  1   10   15    7    1;
  1   15   35   28    9    1;
  1   21   70   84   45   11    1;
  1   28  126  210  165   66   13    1;
  1   36  210  462  495  286   91   15    1;
  1   45  330  924 1287 1001  455  120   17    1;
  1   55  495 1716 3003 3003 1820  680  153   19    1;
...
From _Philippe Deléham_, Mar 26 2012: (Start)
(0, 1, 0, 1, 0, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, 0, ...) begins:
  1
  0, 1
  0, 1,  1
  0, 1,  3,   1
  0, 1,  6,   5,   1
  0, 1, 10,  15,   7,   1
  0, 1, 15,  35,  28,   9,  1
  0, 1, 21,  70,  84,  45, 11,  1
  0, 1, 28, 126, 210, 165, 66, 13, 1. (End)
		

Crossrefs

Programs

  • GAP
    Flat(List([0..12], n-> List([0..n], k-> Binomial(n+k, 2*k) ))); # G. C. Greubel, Aug 01 2019
  • Haskell
    a085478 n k = a085478_tabl !! n !! k
    a085478_row n = a085478_tabl !! n
    a085478_tabl = zipWith (zipWith a007318) a051162_tabl a025581_tabl
    -- Reinhard Zumkeller, Jun 22 2015
    
  • Magma
    [Binomial(n+k, 2*k): k in [0..n], n in [0..12]]; // G. C. Greubel, Aug 01 2019
    
  • Maple
    T := (n,k) -> binomial(n+k,2*k): seq(seq(T(n,k), k=0..n), n=0..11);
  • Mathematica
    (* First program *)
    u[1, x_]:= 1; v[1, x_]:= 1; z = 13;
    u[n_, x_]:= u[n-1, x] + x*v[n-1, x];
    v[n_, x_]:= u[n-1, x] + (x+1)*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[%]   (* A085478 *)
    Table[Expand[v[n, x]], {n, 1, z}]
    cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
    TableForm[cv]
    Flatten[%]   (* A078812 *) (*Clark Kimberling, Feb 25 2012 *)
    (* Second program *)
    Table[Binomial[n + k, 2 k], {n, 0, 12}, {k, 0, n}] // Flatten (* G. C. Greubel, Aug 01 2019 *)
    CoefficientList[Table[Fibonacci[2 n + 1, Sqrt[x]], {n, 0, 10}], x] // Flatten (* Eric W. Weisstein, Jul 03 2023 *)
    Join[{{1}}, CoefficientList[Table[MinimalPolynomial[-4 Sin[Pi/(4 n + 2)]^2, x], {n, 20}], x]] (* Eric W. Weisstein, Jul 12 2023 *)
  • PARI
    T(n,k) = binomial(n+k,n-k)
    
  • Sage
    [[binomial(n+k,2*k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Aug 01 2019
    

Formula

T(n, k) = (n+k)!/((n-k)!*(2*k)!).
G.f.: (1-z)/((1-z)^2-tz). - Emeric Deutsch, May 31 2004
Row sums are A001519 (Fibonacci(2n+1)). Diagonal sums are A011782. Binomial transform of A026729 (product of lower triangular matrices). - Paul Barry, Jun 21 2004
T(n, 0) = 1, T(n, k) = 0 if n=0} T(n-1-j, k-1)*(j+1). T(0, 0) = 1, T(0, k) = 0 if k>0; T(n, k) = T(n-1, k-1) + T(n-1, k) + Sum_{j>=0} (-1)^j*T(n-1, k+j)*A000108(j). For the column k, g.f.: Sum_{n>=0} T(n, k)*x^n = (x^k) / (1-x)^(2*k+1). - Philippe Deléham, Feb 15 2004
Sum_{k=0..n} T(n,k)*x^(2*k) = A000012(n), A001519(n+1), A001653(n), A078922(n+1), A007805(n), A097835(n), A097315(n), A097838(n), A078988(n), A097841(n), A097727(n), A097843(n), A097730(n), A098244(n), A097733(n), A098247(n), A097736(n), A098250(n), A097739(n), A098253(n), A097742(n), A098256(n), A097767(n), A098259(n), A097770(n), A098262(n), A097773(n), A098292(n), A097776(n) for x=0,1,2,...,27,28 respectively. - Philippe Deléham, Dec 31 2007
T(2*n,n) = A005809(n). - Philippe Deléham, Sep 17 2009
A183160(n) = Sum_{k=0..n} T(n,k)*T(n,n-k). - Paul D. Hanna, Dec 27 2010
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k). - Philippe Deléham, Feb 06 2012
O.g.f. for column k: x^k/(1-x)^(2*k+1), k >= 0. [See the o.g.f. of the triangle above, and a comment on compositions. - Wolfdieter Lang, Jul 09 2012]
E.g.f.: (2/sqrt(x + 4))*sinh((1/2)*t*sqrt(x + 4))*cosh((1/2)*t*sqrt(x)) = t + (1 + x)*t^3/3! + (1 + 3*x + x^2)*t^5/5! + (1 + 6*x + 5*x^2 + x^3)*t^7/7! + .... Cf. A091042. - Peter Bala, Jul 29 2013
T(n, k) = A065941(n+3*k, 4*k) = A108299(n+3*k, 4*k) = A194005(n+3*k, 4*k). - Johannes W. Meijer, Sep 05 2013
Sum_{k=0..n} (-1)^k*T(n,k)*A000108(k) = A000007(n) for n >= 0. - Werner Schulte, Jul 12 2017
Sum_{k=0..floor(n/2)} T(n-k,k)*A000108(k) = A001006(n) for n >= 0. - Werner Schulte, Jul 12 2017
From Peter Bala, Jun 26 2025: (Start)
The n-th row polynomial b(n, x) = (-1)^n * U(2*n, (i/2)*sqrt(x)), where U(n,x) is the n-th Chebyshev polynomial of the second kind.
b(n, x) = (-1)^n * Dir(n, -1 - x/2), where Dir(n, x) is the n-th row polynomial of the triangle A244419.
b(n, -1 - x) is the n-th row polynomial of A098493. (End)

A046521 Array T(i,j) = binomial(-1/2-i,j)*(-4)^j, i,j >= 0 read by antidiagonals going down.

Original entry on oeis.org

1, 2, 1, 6, 6, 1, 20, 30, 10, 1, 70, 140, 70, 14, 1, 252, 630, 420, 126, 18, 1, 924, 2772, 2310, 924, 198, 22, 1, 3432, 12012, 12012, 6006, 1716, 286, 26, 1, 12870, 51480, 60060, 36036, 12870, 2860, 390, 30, 1, 48620, 218790, 291720, 204204, 87516, 24310
Offset: 0

Views

Author

Keywords

Comments

Or, a triangle related to A000984 (central binomial) and A000302 (powers of 4).
This is an example of a Riordan matrix. See the Shapiro et al. reference quoted under A053121 and Notes 1 and 2 of the Wolfdieter Lang reference, p. 306.
As a number triangle, this is the Riordan array (1/sqrt(1-4x),x/(1-4x)). - Paul Barry, May 30 2005
The A- and Z- sequences for this Riordan matrix are (see the Wolfdieter Lang link under A006232 for the D. G. Rogers, D. Merlini et al. and R. Sprugnoli references on Riordan A- and Z-sequences with a summary): A-sequence [1,4,0,0,0,...] and Z-sequence 4+2*A000108(n)*(-1)^(n+1)=[2, 2, -4, 10, -28, 84, -264, 858, -2860, 9724, -33592, 117572, -416024, 1485800, -5348880, 19389690, -70715340, 259289580, -955277400, 3534526380], n >= 0. The o.g.f. for the Z-sequence is 4-2*c(-x) with the Catalan number o.g.f. c(x). - Wolfdieter Lang, Jun 01 2007
As a triangle, T(2n,n) is A001448. Row sums are A046748. Diagonal sums are A176280. - Paul Barry, Apr 14 2010
From Wolfdieter Lang, Aug 10 2017: (Start)
The row polynomials R(n, x) of Riordan triangles R = (G(x), F(x)), with F(x)= x*Fhat(x), belong to the class of Boas-Buck polynomials (see the reference). Hence they satisfy the Boas-Buck identity (we use the notation of Rainville, Theorem 50, p. 141):
(E_x - n*1)*R(n, x) = -Sum_{k=0..n-1} (alpha(k)*1 + beta(k)*E_x)*R(n-1.k, x), for n >= 0, where E_x = x*d/dx (Euler operator). The Boas-Buck sequences are given by alpha(k) := [x^k] ((d/dx)log(G(x))) and beta(k) := [x^k] (d/dx)log(Fhat(x)).
This entails a recurrence for the sequence of column m of the Riordan triangle T, n > m >= 0: T(n, m) = (1/(n-m))*Sum_{k=m..n-1} (alpha(n-1-k) + m*beta(n-1-k))*T(k, m), with input T(m,m).
For the present case the Boas-Buck identity for the row polynomials is (E_x - n*1)*R(n, x) = -Sum_{k=0..n-1} 2^(2*k+1)*(1 + 2*E_x)*R(n-1-k, x), for n >= 0. For the ensuing recurrence for the columns m of the triangle T see the formula and example section. (End)
From Peter Bala, Mar 04 2018: (Start)
The following two remarks are particular cases of more general results for Riordan arrays of the form (f(x), x/(1 - k*x)).
1) Let R(n,x) denote the n-th row polynomial of this triangle. The polynomial R(n,4*x) has the e.g.f. Sum_{k = 0..n} T(n,k)*(4*x)/k!. The e.g.f. for the n-th diagonal of the triangle (starting at n = 0 for the main diagonal) equals exp(x) * the e.g.f. for the polynomial R(n,4*x). For example, when n = 3 we have exp(x)*(20 + 30*(4*x) + 10*(4*x)^2/2! + (4*x)^3/3!) = 20 + 140*x + 420*x^2/2! + 924*x^3/3! + 1716*x^4/4! + ....
2) Let P(n,x) = Sum_{k = 0..n} T(n,k)*x^(n-k) denote the n-th row polynomial in descending powers of x. P(n,x) is the n-th degree Taylor polynomial of (1 + 4*x)^(n-1/2) about 0. For example, for n = 4 we have (1 + 4*x)^(7/2) = 70*x^4 + 140*x^3 + 70*x^2 + 14*x + 1 + O(x^5).
Let C(x) = (1 - sqrt(1 - 4*x))/(2*x) denote the o.g.f. of the Catalan numbers A000108. The derivatives of C(x) are determined by the identity (-1)^n * x^n/n! * (d/dx)^n(C(x)) = 1/(2*x)*( 1 - P(n,-x)/(1 - 4*x)^(n-1/2) ), n = 0,1,2,.... See Lang 2002. Cf. A283150 and A283151. (End)

Examples

			Array begins:
  1,  2,   6,  20,   70, ...
  1,  6,  30, 140,  630, ...
  1, 10,  70, 420, 2310, ...
  1, 14, 126, 924, 6006, ...
Recurrence from A-sequence: 140 = a(4,1) = 20 + 4*30.
Recurrence from Z-sequence: 252 = a(5,0) = 2*70 + 2*140 - 4*70 + 10*14 - 28*1.
From _Paul Barry_, Apr 14 2010: (Start)
As a number triangle, T(n, m) begins:
n\k       0      1       2       3      4      5     6    7   8  9 10 ...
0:        1
1:        2      1
2:        6      6       1
3:       20     30      10       1
4:       70    140      70      14      1
5:      252    630     420     126     18      1
6:      924   2772    2310     924    198     22     1
7:     3432  12012   12012    6006   1716    286    26    1
8:    12870  51480   60060   36036  12870   2860   390   30   1
9:    48620 218790  291720  204204  87516  24310  4420  510  34  1
10:  184756 923780 1385670 1108536 554268 184756 41990 6460 646 38  1
... [Reformatted and extended by _Wolfdieter Lang_, Aug 10 2017]
Production matrix begins
      2, 1,
      2, 4, 1,
     -4, 0, 4, 1,
     10, 0, 0, 4, 1,
    -28, 0, 0, 0, 4, 1,
     84, 0, 0, 0, 0, 4, 1,
   -264, 0, 0, 0, 0, 0, 4, 1,
    858, 0, 0, 0, 0, 0, 0, 4, 1,
  -2860, 0, 0, 0, 0, 0, 0, 0, 4, 1 (End)
Boas-Buck recurrence for column m = 2, and n = 4: T(4, 2) = (2*(2*2+1)/2) * Sum_{k=2..3} 4^(3-k)*T(k, 2) = 5*(4*1 + 1*10) = 70. - _Wolfdieter Lang_, Aug 10 2017
From _Peter Bala_, Feb 15 2018: (Start)
With C(x) = (1 - sqrt( 1 - 4*x))/(2*x),
-x^3/3! * (d/dx)^3(C(x)) = 1/(2*x)*( 1 - (1 - 10*x + 30*x^2 - 20*x^3)/(1 - 4*x)^(5/2) ).
x^4/4! * (d/dx)^4(C(x)) = 1/(2*x)*( 1 - (1 - 14*x + 70*x^2 - 140*x^3 + 70*x^4 )/(1 - 4*x)^(7/2) ). (End)
		

References

  • Ralph P. Boas, jr. and R. Creighton Buck, Polynomial Expansions of analytic functions, Springer, 1958, pp. 17 - 21, (last sign in eq. (6.11) should be -).
  • Earl D. Rainville, Special Functions, The Macmillan Company, New York, 1960, ch. 8, sect. 76, 140 - 146.

Crossrefs

Columns include: A000984 (m=0), A002457 (m=1), A002802 (m=2), A020918 (m=3), A020920 (m=4), A020922 (m=5), A020924 (m=6), A020926 (m=7), A020928 (m=8), A020930 (m=9), A020932 (m=10).
Row sums: A046748.

Programs

  • GAP
    Flat(List([0..9],n->List([0..n],m->Binomial(2*n,n)*Binomial(n,m)/Binomial(2*m,m)))); # Muniru A Asiru, Jul 19 2018
    
  • Magma
    [Binomial(n+1,k+1)*Catalan(n)/Catalan(k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 28 2024
    
  • Mathematica
    t[i_, j_] := If[i < 0 || j < 0, 0, (2*i + 2*j)!*i!/(2*i)!/(i + j)!/j!]; Flatten[Reverse /@ Table[t[n, k - n] , {k, 0, 9}, {n, k, 0, -1}]][[1 ;; 51]] (* Jean-François Alcover, Jun 01 2011, after PARI prog. *)
  • PARI
    T(i,j)=if(i<0 || j<0,0,(2*i+2*j)!*i!/(2*i)!/(i+j)!/j!)
    
  • SageMath
    def A046521(n,k): return binomial(n+1, k+1)*catalan_number(n)/catalan_number(k)
    flatten([[A046521(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jul 28 2024

Formula

T(n, m) = binomial(2*n, n)*binomial(n, m)/binomial(2*m, m), n >= m >= 0.
G.f. for column m: ((x/(1-4*x))^m)/sqrt(1-4*x).
Recurrence from the A-sequence given above: a(n,m) = a(n-1,m-1) + 4*a(n-1,m), for n >= m >= 1.
Recurrence from the Z-sequence given above: a(n,0) = Sum_{j=0..n-1} Z(j)*a(n-1,j), n >= 1; a(0,0)=1.
As a number triangle, T(n,k) = C(2*n,n)*C(n,k)/C(2*k,k) = C(n-1/2,n-k)*4^(n-k). - Paul Barry, Apr 14 2010
From Peter Bala, Apr 11 2012: (Start):
One of three infinite families of integral factorial ratio sequences of height 1 (see Bober, Theorem 1.2). The other two are A007318 and A068555.
The triangular array equals exp(S), where the infinitesimal generator S has [2,6,10,14,18,...] on the main subdiagonal and zeros elsewhere.
Recurrence equation for the square array: T(n+1,k) = (k+1)/(4*n+2)*T(n,k+1). (End)
T(n,k) = 4^(n-k)*A006882(2*n - 1)/(A006882(2*n - 2*k)*A006882(2*k - 1)) = 4^(n-k)*(2*n - 1)!!/((2*n - 2*k)!*(2*k - 1)!!). - Peter Bala, Nov 07 2016
Boas-Buck recurrence for column m, m > n >= 0: T(n, m) = (2*(2*m+1)/(n-m))*Sum_{k=m..n-1} 4^(n-1-k)*T(k, m), with input T(n, n) = 1. See a comment above. - Wolfdieter Lang, Aug 10 2017
From Peter Bala, Aug 13 2021: (Start)
Analogous to the binomial transform we have the following sequence transformation formula: g(n) = Sum_{k = 0..n} T(n,k)*b^(n-k)*f(k) iff f(n) = Sum_{k = 0..n} (-1)^(n-k)*T(n,k)*b^(n-k)*g(k). See Prodinger, bottom of p. 413, with b replaced with 4*b, c = 1 and d = 1/2.
Equivalently, if F(x) = Sum_{n >= 0} f(n)*x^n and G(x) = Sum_{n >= 0} g(n)*x^n are a pair of formal power series then
G(x) = 1/sqrt(1 - 4*b*x) * F(x/(1 - 4*b*x)) iff F(x) = 1/sqrt(1 + 4*b*x) * G(x/(1 + 4*b*x)).
The m-th power of this array has entries m^(n-k)*T(n,k). (End)

A126075 Triangle T(n,k), 0 <= k <= n, read by rows, defined by: T(0,0)=1, T(n,k)=0 if k < 0 or if k > n, T(n,0) = 2*T(n-1,0) + T(n-1,1), T(n,k) = T(n-1,k-1) + T(n-1,k+1) for k >= 1.

Original entry on oeis.org

1, 2, 1, 5, 2, 1, 12, 6, 2, 1, 30, 14, 7, 2, 1, 74, 37, 16, 8, 2, 1, 185, 90, 45, 18, 9, 2, 1, 460, 230, 108, 54, 20, 10, 2, 1, 1150, 568, 284, 128, 64, 22, 11, 2, 1, 2868, 1434, 696, 348, 150, 75, 24, 12, 2, 1
Offset: 0

Views

Author

Philippe Deléham, Mar 02 2007

Keywords

Comments

Riordan array (c(x^2)/(1-2xc(x^2)),xc(x^2)) where c(x)=g.f. of Catalan numbers A000108. - Philippe Deléham, Mar 18 2007
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

Examples

			Triangle begins:
     1;
     2,    1;
     5,    2,   1;
    12,    6,   2,   1;
    30,   14,   7,   2,   1;
    74,   37,  16,   8,   2,  1;
   185,   90,  45,  18,   9,  2,  1;
   460,  230, 108,  54,  20, 10,  2,  1;
  1150,  568, 284, 128,  64, 22, 11,  2, 1;
  2868, 1434, 696, 348, 150, 75, 24, 12, 2, 1;
		

Crossrefs

Programs

  • Maple
    A126075 := proc (n, k)
    add( 2^(n-k-2*j)*binomial(n, j), j = 0..floor((n-k)/2) ) - add( 2^(n-k-2-2*j)*binomial(n, j), j = 0..floor((n-k-2)/2) )
    end proc:
    # display sequence in triangular form
    for n from 0 to 10 do seq(A126075(n, k), k = 0..n) end do;
    # Peter Bala, Feb 20 2018
  • Mathematica
    T[0, 0, x_, y_] := 1; T[n_, 0, x_, y_] := x*T[n - 1, 0, x, y] + T[n - 1, 1, x, y]; T[n_, k_, x_, y_] := T[n, k, x, y] = If[k < 0 || k > n, 0, T[n - 1, k - 1, x, y] + y*T[n - 1, k, x, y] + T[n - 1, k + 1, x, y]]; Table[T[n, k, 2, 0], {n, 0, 49}, {k, 0, n}] // Flatten  (* G. C. Greubel, Apr 21 2017 *)

Formula

Sum_{k=0..n} T(n,k) = A127358(n). T(n,0)=A054341(n).
Sum_{k=0..n} T(n,k)*(-k+1) = 2^n. - Philippe Deléham, Mar 25 2007
From Peter Bala, Feb 20 2018: (Start)
T(n,k) = Sum_{j = 0..floor((n-k)/2)} 2^(n-k-2*j)*binomial(n, j) - Sum_{j = 0..floor((n-k-2)/2)} 2^(n-k-2-2*j)*binomial(n, j), 0 <= k <= n. - Peter Bala, Feb 20 2018
The n-th row polynomial in descending powers of x is the n-th Taylor polynomial of the rational function (1 - x^2)/(1 - 2*x) * (1 + x^2)^n about 0. For example, for n = 4, (1 - x^2)/(1 - 2*x) * (1 + x^2)^4 = (30*x^4 + 14*x*3 + 7*x^2 + 2*x + 1) + O(x^5). (End)

A112857 Triangle T(n,k) read by rows: number of Green's R-classes in the semigroup of order-preserving partial transformations (of an n-element chain) consisting of elements of height k (height(alpha) = |Im(alpha)|).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 7, 5, 1, 1, 15, 17, 7, 1, 1, 31, 49, 31, 9, 1, 1, 63, 129, 111, 49, 11, 1, 1, 127, 321, 351, 209, 71, 13, 1, 1, 255, 769, 1023, 769, 351, 97, 15, 1, 1, 511, 1793, 2815, 2561, 1471, 545, 127, 17, 1, 1, 1023, 4097, 7423, 7937, 5503, 2561, 799, 161, 19, 1
Offset: 0

Views

Author

Abdullahi Umar, Aug 25 2008

Keywords

Comments

Sum of rows of T(n, k) is A007051; T(n,k) = |A118801(n,k)|.
Row-reversed variant of A119258. - R. J. Mathar, Jun 20 2011
Pairwise sums of row terms starting from the right yields triangle A038207. - Gary W. Adamson, Feb 06 2012
Riordan array (1/(1 - x), x/(1 - 2*x)). - Philippe Deléham, Jan 17 2014
Appears to coincide with the triangle T(n,m) (n >= 1, 1 <= m <= n) giving number of set partitions of [n], avoiding 1232, with m blocks [Crane, 2015]. See also A250118, A250119. - N. J. A. Sloane, Nov 25 2014
(A007318)^2 = A038207 = T*|A167374|. See A118801 for other relations to the Pascal matrix. - Tom Copeland, Nov 17 2016

Examples

			T(3,2) = 5 because in a regular semigroup of transformations the Green's R-classes coincide with convex partitions of subsets of {1,2,3} with convex classes (modulo the subsets): {1}, {2}/{1}, {3}/{2}, {3}/{1,2}, {3}/{1}, {2,3}
Triangle T(n,k) (with rows n >= 0 and columns k = 0..n) begins:
  1;
  1,    1;
  1,    3,    1;
  1,    7,    5,    1;
  1,   15,   17,    7,    1;
  1,   31,   49,   31,    9,    1;
  1,   63,  129,  111,   49,   11,    1;
  1,  127,  321,  351,  209,   71,   13,   1;
  1,  255,  769, 1023,  769,  351,   97,  15,   1;
  1,  511, 1793, 2815, 2561, 1471,  545, 127,  17,  1;
  1, 1023, 4097, 7423, 7937, 5503, 2561, 799, 161, 19, 1;
  ...
As to matrix M, top row of M^3 = (1, 7, 5, 1, 0, 0, 0, ...)
		

Crossrefs

Programs

  • Maple
    A112857 := proc(n,k) if k=0 or k=n then 1; elif k <0 or k>n then 0; else 2*procname(n-1,k)+procname(n-1,k-1) ; end if; end proc: # R. J. Mathar, Jun 20 2011
  • Mathematica
    Table[Abs[1 + (-1)^k*2^(n - k + 1)*Sum[ Binomial[n - 2 j - 2, k - 2 j - 1], {j, 0, Floor[k/2]}]] - 4 Boole[And[n == 1, k == 0]], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Nov 24 2016 *)

Formula

T(n,k) = Sum_{j = k..n} C(n,j)*C(j-1,k-1).
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) for n >= 2 and 1 <= k <= n-1 with T(n,0) = 1 = T(n,n) for n >= 0.
n-th row = top row of M^n, deleting the zeros, where M is an infinite square production matrix with (1,1,1,...) as the superdiagonal and (1,2,2,2,...) as the main diagonal. - Gary W. Adamson, Feb 06 2012
From Peter Bala, Mar 05 2018 (Start):
The following remarks are particular cases of more general results for Riordan arrays of the form (f(x), x/(1 - k*x)).
Let R(n,x) denote the n-th row polynomial of this triangle. The polynomial R(n,2*x) has the e.g.f. Sum_{k = 0..n} T(n,k)*(2*x)^k/k!. The e.g.f. for the n-th diagonal of the triangle (starting at n = 0 for the main diagonal) equals exp(x) * the e.g.f. for the polynomial R(n,2*x). For example, when n = 3 we have exp(x)*(1 + 7*(2*x) + 5*(2*x)^2/2! + (2*x)^3/3!) = 1 + 15*x + 49*x^2/2! + 111*x^3/3! + 209*x^4/4! + ....
Let P(n,x) = Sum_{k = 0..n} T(n,k)*x^(n-k) denote the n-th row polynomial in descending powers of x. Then P(n,x) is the n-th degree Taylor polynomial of the function (1 + 2*x)^n/(1 + x) about 0. For example, for n = 4 we have (1 + 2*x)^4/(1 + x) = x^4 + 15*x^3 + 17*x^2 + 7*x + 1 + O(x^5).
See A118801 for a signed version of this triangle and A145661 for a signed version of the row reversed triangle. (End)
Bivariate o.g.f.: Sum_{n,k>=0} T(n,k)*x^n*y^k = (1 - 2*x)/((1 - x)*(1 - 2*x - x*y)). - Petros Hadjicostas, Feb 14 2021
The matrix inverse of the Lucas triangle A029635 is -T(n, k)/(-2)^(n-k+1). - Peter Luschny, Dec 22 2024

A113139 Number triangle, equal to half of Delannoy square array A008288.

Original entry on oeis.org

1, 3, 1, 13, 5, 1, 63, 25, 7, 1, 321, 129, 41, 9, 1, 1683, 681, 231, 61, 11, 1, 8989, 3653, 1289, 377, 85, 13, 1, 48639, 19825, 7183, 2241, 575, 113, 15, 1, 265729, 108545, 40081, 13073, 3649, 833, 145, 17, 1, 1462563, 598417, 224143, 75517, 22363, 5641
Offset: 0

Views

Author

Paul Barry, Oct 15 2005

Keywords

Comments

Row sums are A047781(n+1). Diagonal sums are A113140. Inverse is A113141.

Examples

			Triangle begins
     1;
     3,    1;
    13,    5,    1;
    63,   25,    7,   1;
   321,  129,   41,   9,  1;
  1683,  681,  231,  61, 11,  1;
  8989, 3653, 1289, 377, 85, 13, 1;
  ...
A113139 as a square array = A110171 * A008288:
  / 1   1   1   1 ... \   / 1         \ / 1 1  1  1 ...\
  | 3   5   7   9 ... |   | 2  1       || 1 3  5  7 ...|
  |13  25  41  61 ... | = | 8  4 1     || 1 5 13 25 ...|
  |63 129 231 377 ... |   |38 18 6 1   || 1 7 25 63 .. |
  |...                |   |...         || 1...         |
- _Peter Bala_, Dec 09 2015
		

Crossrefs

A001850 (column 0), A002002 (column 1), A026002 (column 2), A190666 (column 3), A047781 (row sums), A113140 (diagonal sums), A113141 (matrix inverse). Cf. A006318, A008288, A110171.

Programs

  • Maple
    T := (n,k) -> (-1)^(n-k)*hypergeom([n+1, -n+k], [1], 2):
    seq(seq(simplify(T(n,k)),k=0..n),n=0..8); # Peter Luschny, Mar 02 2017
  • Mathematica
    Table[Sum[Binomial[n - k, j] Binomial[n + j, k + j], {j, 0, n}], {n, 0, 9}, {k, 0, n}] // Flatten (* Michael De Vlieger, Dec 09 2015 *)

Formula

T(n, k) = Sum_{j=0..n} C(n-k, j)*C(n+j, k+j).
T(n, k) = Sum_{j=0..n} C(n, j)*C(n-k, j-k)*2^(n-j).
From Peter Bala, Dec 09 2015: (Start)
T(n,k) = A008288(n - k, n).
O.g.f.: 2/( sqrt(x^2 - 6*x + 1)*(t*sqrt(x^2 - 6*x + 1) + t*x - t + 2) ) = 1 + (3 + t)*x + (13 + 5*t + t^2)*x^2 + ....
Riordan array (f(x), x*g(x)), where f(x) = 1/sqrt(1 - 6*x + x^2) is the o.g.f. for the central Delannoy numbers, A001850, and g(x) = 1/x* revert( x*(1 - x)/(1 + x) ) = 1 + 2*x + 6*x^2 + 22*x^3 + 90*x^4 + 394*x^5 + ... is the o.g.f. for the large Schroder numbers, A006318.
Read as a square array, this is the generalized Riordan array (f(x), g(x)) in the sense of the Bala link, which factorizes as (1 + x*g'(x)/g(x), x*g(x)) * (1/(1 - x), (1 + x)/(1 - x)) = A110171 * A008288. See the example below. (End)
T(n,k) = (-1)^(n-k)*hypergeom([n+1, -n+k], [1], 2). - Peter Luschny, Mar 02 2017
From Peter Bala, Feb 16 2020: (Start)
T(n,k) = P(n-k, k, 0, 3), where P(n, alpha, beta, x) is the n-th Jacobi polynomial with parameters alpha and beta.
T(n,k) = binomial(n,k) * hypergeom( [n + 1, k - n], [k + 1], -1 ).
The n-th row polynomial in descending powers of x is the n-th Taylor polynomial of the rational function (1 + x)^n/(1 - x)^(n+1) about 0. For example, for n = 4, (1 + x)^4/(1 - x)^5 = 1 + 9*x + 41*x^2 + 129*x^3 + 321*x^4 + O(x^5). Cf. A110171. (End)

A112554 Riordan array (c(x^2)^2, x*c(x^2)), c(x) the g.f. of A000108.

Original entry on oeis.org

1, 0, 1, 2, 0, 1, 0, 3, 0, 1, 5, 0, 4, 0, 1, 0, 9, 0, 5, 0, 1, 14, 0, 14, 0, 6, 0, 1, 0, 28, 0, 20, 0, 7, 0, 1, 42, 0, 48, 0, 27, 0, 8, 0, 1, 0, 90, 0, 75, 0, 35, 0, 9, 0, 1, 132, 0, 165, 0, 110, 0, 44, 0, 10, 0, 1, 0, 297, 0, 275, 0, 154, 0, 54, 0, 11, 0, 1, 429, 0, 572, 0, 429, 0, 208, 0, 65, 0, 12, 0, 1
Offset: 0

Views

Author

Paul Barry, Sep 13 2005

Keywords

Comments

Inverse of A112552.
The n-th row polynomial (in descending powers of x) is equal to the n-th degree Taylor polynomial of the polynomial function (1 - x^4)*(1 + x^2)^n about 0. For example, when n = 6, (1 - x^4)*(1 + x^2)^6 = 1 + 6*x^2 + 14*x^4 + 14*x^6 + O(x^8). - Peter Bala, Feb 19 2018

Examples

			Triangle begins
   1;
   0, 1;
   2, 0,  1;
   0, 3,  0, 1;
   5, 0,  4, 0, 1;
   0, 9,  0, 5, 0, 1;
  14, 0, 14, 0, 6, 0, 1;
		

Crossrefs

Row sums A037952, matrix inverse A112552.
Cf. A000108, A037952 (row sums), A112552, A112553.

Programs

  • Magma
    A112554:= func< n,k | ((1+(-1)^(n-k))/2)*(Binomial(n, Floor((n-k)/2)) - Binomial(n, Floor((n-k-4)/2))) >;
    [A112554(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Jan 13 2022
  • Maple
    seq(seq((1 + (-1)^(n-k))/2*( binomial(n, floor((n - k)/2)) - binomial(n, floor((n - k - 4)/2 )) ), k = 0..n), n = 0..10); # Peter Bala, Feb 19 2018
  • Mathematica
    T[n_, k_] := (1 + (-1)^(n-k))/2 (Binomial[n, Floor[(n-k)/2]] - Binomial[n, Floor[(n-k-4)/2]]);
    Table[T[n, k], {n, 0, 12}, {k, 0, n}] (* Jean-François Alcover, Jun 13 2019 *)
  • Sage
    # Algorithm of L. Seidel (1877)
    # Prints the first n rows of a signed version of the triangle.
    def Signed_A112554_triangle(n) :
        D = [0]*(n+4); D[1] = 1
        b = False; h = 2
        for i in range(2*n+2) :
            if b :
                for k in range(h,0,-1) : D[k] += D[k-1]
                h += 1
            else :
                for k in range(1,h, 1) : D[k] -= D[k+1]
            b = not b
            if b and i > 0 : print([D[z] for z in (2..h-1)])
    Signed_A112554_triangle(13) # Peter Luschny, May 01 2012
    

Formula

Sum_{k=0..n} T(n, k) = binomial(n+1, floor(n/2)) = A037952(n+1).
T(n, k) = ((1 + (-1)^(n-k))/2)*binomial(n, floor((n-k)/2)) - binomial(n, floor((n-k-4)/2 )). - Peter Bala, Feb 19 2018

A139377 A Jacobsthal-Catalan triangle.

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 5, 6, 3, 1, 11, 15, 10, 4, 1, 21, 41, 30, 15, 5, 1, 43, 113, 92, 51, 21, 6, 1, 85, 327, 284, 171, 79, 28, 7, 1, 171, 982, 897, 570, 286, 115, 36, 8, 1, 341, 3066, 2895, 1913, 1016, 446, 160, 45, 9, 1
Offset: 0

Views

Author

Paul Barry, Apr 15 2008

Keywords

Comments

First column is A001045(n+1). Second column is A139379. Row sums are A139379(n+1).
Diagonal sums are A135582. Inverse of the Riordan array (1-x-x^2+4x^3-2x^4,x(1-x)).

Examples

			Triangle begins
    1;
    1,   1;
    3,   2,   1;
    5,   6,   3,   1;
   11,  15,  10,   4,   1;
   21,  41,  30,  15,   5,   1;
   43, 113,  92,  51,  21,   6,   1;
   85, 327, 284, 171,  79,  28,   7,   1;
  171, 982, 897, 570, 286, 115,  36,   8,   1;
The production matrix for this array is
   1, 1,
   2, 1, 1,
  -2, 1, 1, 1,
   0, 1, 1, 1, 1,
   0, 1, 1, 1, 1, 1,
   0, 1, 1, 1, 1, 1, 1,
   0, 1, 1, 1, 1, 1, 1
		

Crossrefs

Cf. A001045 (first column), A139379 (second column and row sums), A135582 (sums along shallow diagonals).

Programs

  • Maple
    #define auxiliary sequence
    with(combinat):
    b := proc (n)
      (2/3)*(1+I)^(n-1) + (2/3)*(1-I)^(n-1) - (4/3)*(1+I)^(n-2)-(4/3)*(1-I)^(n-2) + (1/3)*(-1)^n*fibonacci(n+1) + (2/3)*(-1)^n*fibonacci(n);
    end proc:
    A139377 := proc (n, k)
      add(b(j)*binomial(2*n-k-j, n), j = 0..n-k);
    end proc:
    #display sequence as a triangle
    for n from 0 to 10 do
      seq(A139377(n, k), k = 0..n);
    end do; # Peter Bala, Feb 20 2018

Formula

Riordan array (1/(1-x-2x^2), xc(x)) where c(x) is the g.f. of A000108
From Peter Bala, Feb 20 2018: (Start)
Define a(n) = floor(2^(n+2)/3) - floor(2^(n+1)/3) = A001045(n+1). Then T(n,0) = a(n) and T(n,k) = Sum_{j = 0..n-k} a(j)*k/(2*n-k-2*j)*binomial(2*n-k-2*j,n-k-j) for 1 <= k <= n.
Define b(n) = (2/3)*(1+i)^(n-1) + (2/3)*(1-i)^(n-1) - (4/3)*(1+i)^(n-2) - (4/3)*(1-i)^(n-2) + (1/3)*(-1)^n*Fibonacci(n+1) + (2/3)*(-1)^n*Fibonacci(n). Then T(n,k) = Sum_{j = 0..n-k} b(j)*binomial(2*n-k-j,n) for 0 <= k <= n.
The n-th row polynomial in descending powers of x is the n-th Taylor polynomial of the rational function (1 - 2*x)/((1 - x)*(1 + x - x^2)*(1 - 2*x + 2*x^2)) * 1/(1 - x)^n about 0. For example, for n = 4, (1 - 2*x)/((1 - x)*(1 + x - x^2)*(1 - 2*x + 2*x^2)) * 1/(1 - x)^4 = (11*x^4 + 15*x^3 + 10*x^2 + 4*x + 1) + O(x^5). (End)

A264773 Triangle T(n,k) = binomial(4*n - 3*k, 3*n - 2*k), 0 <= k <= n.

Original entry on oeis.org

1, 4, 1, 28, 5, 1, 220, 36, 6, 1, 1820, 286, 45, 7, 1, 15504, 2380, 364, 55, 8, 1, 134596, 20349, 3060, 455, 66, 9, 1, 1184040, 177100, 26334, 3876, 560, 78, 10, 1, 10518300, 1560780, 230230, 33649, 4845, 680, 91, 11, 1, 94143280, 13884156, 2035800, 296010, 42504, 5985, 816, 105, 12, 1
Offset: 0

Views

Author

Peter Bala, Nov 30 2015

Keywords

Comments

Riordan array (f(x),x*g(x)), where g(x) = 1 + x + 4*x^2 + 22*x^3 + 140*x^4 + ... is the o.g.f. for A002293 and f(x) = g(x)/(4 - 3*g(x)) = 1 + 4*x + 28*x^2 + 220*x^3 + 1820*x^4 + ... is the o.g.f. for A005810.
More generally, if (R(n,k))n,k>=0 is a proper Riordan array and m is a nonnegative integer and a > b are integers then the array with (n,k)-th element R((m + 1)*n - a*k, m*n - b*k) is also a Riordan array (not necessarily proper). Here we take R as Pascal's triangle and m = a = 3 and b = 2. See A092392, A264772, A264774 and A113139 for further examples.

Examples

			Triangle begins
  n\k |       0      1     2    3   4   5   6   7
------+-----------------------------------------------
   0  |       1
   1  |       4      1
   2  |      28      5     1
   3  |     220     36     6    1
   4  |    1820    286    45    7   1
   5  |   15504   2380   364   55   8   1
   6  |  134596  20349  3060  455  66   9   1
   7  | 1184040 177100 26334 3876 560  78  10   1
...
		

Crossrefs

A005810 (column 0), A052203 (column 1), A257633 (column 2), A224274 (column 3), A004331 (column 4). Cf. A002293, A007318, A092392 (C(2n-k,n)), A119301 (C(3n-k,n-k)), A264772, A264774.

Programs

  • Magma
    /* As triangle: */ [[Binomial(4*n-3*k, 3*n-2*k): k in [0..n]]: n in [0.. 10]]; // Vincenzo Librandi, Dec 02 2015
  • Maple
    A264773:= proc(n,k) binomial(4*n - 3*k, 3*n - 2*k); end proc:
    seq(seq(A264773(n,k), k = 0..n), n = 0..10);
  • Mathematica
    A264773[n_,k_] := Binomial[4*n - 3*k, n - k];
    Table[A264773[n, k], {n, 0, 10}, {k, 0, n}] (* Paolo Xausa, Feb 06 2024 *)

Formula

T(n,k) = binomial(4*n - 3*k, n - k).
O.g.f.: f(x)/(1 - t*x*g(x)), where f(x) = Sum_{n >= 0} binomial(4*n,n)*x^n and g(x) = Sum_{n >= 0} 1/(3*n + 1)*binomial(4*n,n)*x^n.

A264774 Triangle T(n,k) = binomial(5*n - 4*k, 4*n - 3*k), 0 <= k <= n.

Original entry on oeis.org

1, 5, 1, 45, 6, 1, 455, 55, 7, 1, 4845, 560, 66, 8, 1, 53130, 5985, 680, 78, 9, 1, 593775, 65780, 7315, 816, 91, 10, 1, 6724520, 736281, 80730, 8855, 969, 105, 11, 1, 76904685, 8347680, 906192, 98280, 10626, 1140, 120, 12, 1, 886163135, 95548245, 10295472, 1107568, 118755, 12650, 1330, 136, 13, 1
Offset: 0

Views

Author

Peter Bala, Nov 30 2015

Keywords

Comments

Riordan array (f(x),x*g(x)), where g(x) = 1 + x + 5*x^2 + 35*x^3 + 285*x^4 + ... is the o.g.f. for A002294 and f(x) = g(x)/(5 - 4*g(x)) = 1 + 5*x + 45*x^2 + 455*x^3 + 4845*x^4 + ... is the o.g.f. for A001449.
More generally, if (R(n,k))n,k>=0 is a proper Riordan array and m is a nonnegative integer and a > b are integers then the array with (n,k)-th element R((m + 1)*n - a*k, m*n - b*k) is also a Riordan array (not necessarily proper). Here we take R as Pascal's triangle and m = a = 4 and b = 3. See A092392, A264772, A264773 and A113139 for further examples.

Examples

			Triangle begins
  n\k |       0      1     2    3   4   5   6   7
------+---------------------------------------------
   0  |       1
   1  |       5      1
   2  |      45      6     1
   3  |     455     55     7    1
   4  |    4845    560    66    8   1
   5  |   53130   5985   680   78   9   1
   6  |  593775  65780  7315  816  91  10   1
   7  | 6724520 736281 80730 8855 969 105  11  1
...
		

Crossrefs

Cf. A001449 (column 0), A079589(column 1). Cf. A002294, A007318, A092392 (C(2n-k,n)), A113139, A119301 (C(3n-k,n-k)), A264772, A264773.

Programs

  • Magma
    /* As triangle */ [[Binomial(5*n-4*k, 4*n-3*k): k in [0..n]]: n in [0.. 10]]; // Vincenzo Librandi, Dec 02 2015
  • Maple
    A264774:= proc(n,k) binomial(5*n - 4*k, 4*n - 3*k); end proc:
    seq(seq(A264774(n,k), k = 0..n), n = 0..10);
  • Mathematica
    Table[Binomial[5 n - 4 k, 4 n - 3 k], {n, 0, 9}, {k, 0, n}] // Flatten (* Michael De Vlieger, Dec 01 2015 *)

Formula

T(n,k) = binomial(5*n - 4*k, n - k).
O.g.f.: f(x)/(1 - t*x*g(x)), where f(x) = Sum_{n >= 0} binomial(5*n,n)*x^n and g(x) = Sum_{n >= 0} 1/(4*n + 1)*binomial(5*n,n)*x^n.

A283150 Riordan array (1/(1-9x)^(1/3), x/(9x-1)).

Original entry on oeis.org

1, 3, -1, 18, -12, 1, 126, -126, 21, -1, 945, -1260, 315, -30, 1, 7371, -12285, 4095, -585, 39, -1, 58968, -117936, 49140, -9360, 936, -48, 1, 480168, -1120392, 560196, -133380, 17784, -1368, 57, -1, 3961386, -10563696, 6162156, -1760616, 293436, -30096, 1881, -66, 1, 33011550, -99034650, 66023100
Offset: 0

Views

Author

Tom Richardson, Mar 01 2017

Keywords

Comments

Triangle read by rows. This is an example of a Riordan group involution. Dual Riordan array of A283151. With A283151 and A248324, forms doubly infinite Riordan array. For b and c the sequences A283151 and A248324, respectively, and i,j >= 0, the doubly infinite array with d(i,j) = a(i,j), d(-j,-i) = b(i,j), d(i,-j) = c(j,i), and d(-i,j) = 0 (except d(0,0)=1) is a doubly infinite Riordan array.
Matrix inverse of a(m,n) is a(m,n). - Werner Schulte, Aug 05 2017

Examples

			The triangle begins
        1;
        3,        -1;
       18,       -12,       1;
      126,      -126,      21,       -1;
      945,     -1260,     315,      -30,      1;
     7371,    -12285,    4095,     -585,     39,     -1;
    58968,   -117936,   49140,    -9360,    936,    -48,    1;
   480168,  -1120392,  560196,  -133380,  17784,  -1368,   57,  -1;
  3961386, -10563696, 6162156, -1760616, 293436, -30096, 1881, -66, 1;
		

Crossrefs

Programs

  • Maple
    T := (n, k) -> (-1)^k*binomial(n - 2/3, n - k)*9^(n - k):
    for n from 0 to 6 do seq(T(n, k), k = 0..n) od; # Peter Luschny, Sep 03 2021
  • PARI
    a(m,n) = binomial(-n-1/3, m-n)*(-1)^m*9^(m-n);
    tabl(nn) = for(n=0, nn, for (k=0, n, print1(a(n, k), ", ")); print); \\ Michel Marcus, Aug 07 2017

Formula

a(m,n) = binomial(-n-1/3, m-n)*(-1)^m*9^(m-n).
G.f.: (1-9x)^(2/3)/(xy-9x+1).
Recurrence: a(m,n) = a(m, n-1)*(n-1-m)/(9*n-6) for 0 < n <= m. - Werner Schulte, Aug 05 2017
From Peter Bala, Mar 05 2018 (Start):
Let P(n,x) = Sum_{k = 0..n} T(n,k)*x^(n-k) denote the n-th row polynomial in descending powers of x. Then (-1)^n*P(n,x) is the n-th degree Taylor polynomial of (1 - 9*x)^(n-2/3) about 0. For example, for n = 4 we have (1 - 9*x)^(10/3) = 945*x^4 - 1260*x^3 + 315*x^2 - 30*x + 1 + O(x^5).
Let R(n,x) denote the n-th row polynomial of this triangle. The polynomial R(n,9*x) has the e.g.f. Sum_{k = 0..n} T(n,k)*(9*x)^k/k!. The e.g.f. for the n-th diagonal of the triangle (starting at n = 0 for the main diagonal) equals exp(-x) * the e.g.f. for the polynomial R(n,9*x). For example, when n = 3 we have exp(-x)*(126 - 126*(9*x) + 21*(9*x)^2/2! - (9*x)^3/3!) = 126 - 1260*x + 4095*x^2/2! - 9360*x^3/3! + 17784*x^4/4! - ....
Let F(x) = (1 - ( 1 - 9*x)^(2/3))/(3*x) denote the o.g.f. of A155579. The derivatives of F(x) are related to the row polynomials P(n,x) by the identity x^n/n! * (d/dx)^n(F(x)) = 1/(3*x)*( (-1)^n - P(n,x)/(1 - 9*x)^(n-2/3) ), n = 0,1,2,.... Cf. A283151 and A046521. (End)
From Peter Bala, Aug 18 2021: (Start)
T(n,k) = (-1)^k*binomial(n-2/3, n-k)*9^(n-k).
Analogous to the binomial transform we have the following sequence transformation formula: g(n) = Sum_{k = 0..n} T(n,k)*b^(n-k)*f(k) iff f(n) = Sum_{k = 0..n} T(n,k)*b^(n-k)*g(k). See Prodinger, bottom of p. 413, with b replaced with 9*b, c = -1 and d = 1/3.
Equivalently, if F(x) = Sum_{n >= 0} f(n)*x^n and G(x) = Sum_{n >= 0} g(n)*x^n are a pair of formal power series then
G(x) = (1/(1 - 9*b*x)^(1/3)) * F(x/(1 - 9*b*x)) iff F(x) = (1/(1 + 9*b*x)^(1/3)) * G(x/(1 + 9*b*x)).
The infinitesimal generator of the unsigned array has the sequence (9*n+3) n>=0 on the main subdiagonal and zeros elsewhere. The m-th power of the unsigned array has entries m^(n-k)*|T(n,k)|. (End)

Extensions

Offset corrected by Werner Schulte, Aug 05 2017
Showing 1-10 of 12 results. Next