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 1-6 of 6 results.

A008459 Square the entries of Pascal's triangle.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 9, 9, 1, 1, 16, 36, 16, 1, 1, 25, 100, 100, 25, 1, 1, 36, 225, 400, 225, 36, 1, 1, 49, 441, 1225, 1225, 441, 49, 1, 1, 64, 784, 3136, 4900, 3136, 784, 64, 1, 1, 81, 1296, 7056, 15876, 15876, 7056, 1296, 81, 1, 1, 100, 2025, 14400, 44100, 63504, 44100, 14400, 2025, 100, 1
Offset: 0

Views

Author

Keywords

Comments

Number of lattice paths from (0, 0) to (n, n) with steps (1, 0) and (0, 1), having k right turns. - Emeric Deutsch, Nov 23 2003
Product of A007318 and A105868. - Paul Barry, Nov 15 2005
Number of partitions that fit in an n X n box with Durfee square k. - Franklin T. Adams-Watters, Feb 20 2006
From Peter Bala, Oct 23 2008: (Start)
Narayana numbers of type B. Row n of this triangle is the h-vector of the simplicial complex dual to an associahedron of type B_n (a cyclohedron) [Fomin & Reading, p. 60]. See A063007 for the corresponding f-vectors for associahedra of type B_n. See A001263 for the h-vectors for associahedra of type A_n. The Hilbert transform of this triangular array is A108625 (see A145905 for the definition of this term).
Let A_n be the root lattice generated as a monoid by {e_i - e_j: 0 <= i, j <= n + 1}. Let P(A_n) be the polytope formed by the convex hull of this generating set. Then the rows of this array are the h-vectors of a unimodular triangulation of P(A_n) [Ardila et al.]. A063007 is the corresponding array of f-vectors for these type A_n polytopes. See A086645 for the array of h-vectors for type C_n polytopes and A108558 for the array of h-vectors associated with type D_n polytopes.
(End)
The n-th row consists of the coefficients of the polynomial P_n(t) = Integral_{s = 0..2*Pi} (1 + t^2 - 2*t*cos(s))^n/Pi/2 ds. For example, when n = 3, we get P_3(t) = t^6 + 9*t^4 + 9*t^2 + 1; the coefficients are 1, 9, 9, 1. - Theodore Kolokolnikov, Oct 26 2010
Let E(y) = Sum_{n >= 0} y^n/n!^2 = BesselJ(0, 2*sqrt(-y)). Then this triangle is the generalized Riordan array (E(y), y) with respect to the sequence n!^2 as defined in Wang and Wang. - Peter Bala, Jul 24 2013
From Colin Defant, Sep 16 2018: (Start)
Let s denote West's stack-sorting map. T(n,k) is the number of permutations pi of [n+1] with k descents such that s(pi) avoids the patterns 132, 231, and 321. T(n,k) is also the number of permutations pi of [n+1] with k descents such that s(pi) avoids the patterns 132, 312, and 321.
T(n,k) is the number of permutations of [n+1] with k descents that avoid the patterns 1342, 3142, 3412, and 3421. (End)
The number of convex polyominoes whose smallest bounding rectangle has size (k+1)*(n+1-k) and which contain the lower left corner of the bounding rectangle (directed convex polyominoes). - Günter Rote, Feb 27 2019
Let P be the poset [n] X [n] ordered by the product order. T(n,k) is the number of antichains in P containing exactly k elements. Cf. A063746. - Geoffrey Critzer, Mar 28 2020

Examples

			Pascal's triangle begins
  1
  1  1
  1  2   1
  1  3   3   1
  1  4   6   4   1
  1  5  10  10   5   1
  1  6  15  20  15   6   1
  1  7  21  35  35  21   7   1
...
so the present triangle begins
  1
  1   1
  1   4    1
  1   9    9     1
  1  16   36    16     1
  1  25  100   100    25    1
  1  36  225   400   225   36   1
  1  49  441  1225  1225  441  49   1
...
		

References

  • T. K. Petersen, Eulerian Numbers, Birkhauser, 2015, Chapter 12.
  • J. Riordan, An introduction to combinatorial analysis, Dover Publications, Mineola, NY, 2002, page 191, Problem 15. MR1949650
  • P. G. Tait, On the Linear Differential Equation of the Second Order, Proceedings of the Royal Society of Edinburgh, 9 (1876), 93-98 (see p. 97) [From Tom Copeland, Sep 09 2010, vol number corrected Sep 10 2010]

Crossrefs

Row sums are in A000984. Columns 0-3 are A000012, A000290, A000537, A001249.
Family of polynomials (see A062145): this sequence (c=1), A132813 (c=2), A062196 (c=3), A062145 (c=4), A062264 (c=5), A062190 (c=6).
Cf. A007318, A055133, A116647, A001263, A086645, A063007, A108558, A108625 (Hilbert transform), A145903, A181543, A086645 (logarithmic derivative), A105868 (inverse binomial transform), A093118.

Programs

  • GAP
    Flat(List([0..10],n->List([0..n],k->Binomial(n,k)^2))); # Muniru A Asiru, Mar 30 2018
    
  • Magma
    /* As triangle */ [[Binomial(n, k)^2: k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Dec 15 2016
    
  • Maple
    seq(seq(binomial(n, k)^2, k=0..n), n=0..10);
  • Mathematica
    Table[Binomial[n, k]^2, {n, 0, 11}, {k, 0, n}]//Flatten (* Alonso del Arte, Dec 08 2013 *)
  • Maxima
    create_list(binomial(n,k)^2,n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */
    
  • Maxima
    T(n,k):=if n=k then 1 else if k=0 then 1 else T(n-1,k)*(n+k)/(n-k)+T(n-1,k-1); /* Vladimir Kruchinin, Oct 18 2014 */
    
  • Maxima
    A(x,y):=1/sqrt(1-2*x-2*x*y+x^2-2*x^2*y+x^2*y^2);
    taylor(x*A(x,y)+x*y*A(x,y)+sqrt(1+4*x^2*y*A(x,y)^2),x,0,7,y,0,7); /* Vladimir Kruchinin, Oct 23 2020 */
    
  • PARI
    {T(n, k) = if( k<0 || k>n, 0, binomial(n, k)^2)}; /* Michael Somos, May 03 2004 */
    
  • PARI
    {T(n,k)=polcoeff(polcoeff(sum(m=0,n,(2*m)!/m!^2*x^(2*m)*y^m/(1-x-x*y+x*O(x^n))^(2*m+1)),n,x),k,y)} \\ Paul D. Hanna, Oct 31 2010
    
  • Python
    def A008459(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),n-comb(r+1,2))**2 # Chai Wah Wu, Nov 12 2024

Formula

T(n,k) = A007318(n,k)^2. - Sean A. Irvine, Mar 29 2018
E.g.f.: exp((1+y)*x)*BesselI(0, 2*sqrt(y)*x). - Vladeta Jovovic, Nov 17 2003
G.f.: 1/sqrt(1-2*x-2*x*y+x^2-2*x^2*y+x^2*y^2); g.f. for row n: (1-t)^n P_n[(1+t)/(1-t)] where the P_n's are the Legendre polynomials. - Emeric Deutsch, Nov 23 2003 [The original version of the bivariate g.f. has been modified with the roles of x and y interchanged so that now x corresponds to n and y to k. - Petros Hadjicostas, Oct 22 2017]
G.f. for column k is Sum_{j = 0..k} C(k, j)^2*x^(k+j)/(1 - x)^(2*k+1). - Paul Barry, Nov 15 2005
Column k has g.f. (x^k)*Legendre_P(k, (1+x)/(1-x))/(1 - x)^(k+1) = (x^k)*Sum_{j = 0..k} C(k, j)^2*x^j/(1 - x)^(2*k+1). - Paul Barry, Nov 19 2005
Let E be the operator D*x*D, where D denotes the derivative operator d/dx. Then (1/n!^2) * E^n(1/(1 - x)) = (row n generating polynomial)/(1 - x)^(2*n+1) = Sum_{k >= 0} binomial(n+k, k)^2*x^k. For example, when n = 3 we have (1/3!)^2*E^3(1/(1 - x)) = (1 + 9*x + 9*x^2 + x^3)/(1 - x)^7 = (1/3!)^2 * Sum_{k >= 0} ((k+1)*(k+2)*(k+3))^2*x^k. - Peter Bala, Oct 23 2008
G.f.: A(x, y) = Sum_{n >= 0} (2*n)!/n!^2 * x^(2*n)*y^n/(1 - x - x*y)^(2*n+1). - Paul D. Hanna, Oct 31 2010
From Peter Bala, Jul 24 2013: (Start)
Let E(y) = Sum_{n >= 0} y^n/n!^2 = BesselJ(0, 2*sqrt(-y)). Generating function: E(y)*E(x*y) = 1 + (1 + x)*y + (1 + 4*x + x^2)*y^2/2!^2 + (1 + 9*x + 9*x^2 + x^3)*y^3/3!^2 + .... Cf. the unsigned version of A021009 with generating function exp(y)*E(x*y).
The n-th power of this array has the generating function E(y)^n*E(x*y). In particular, the matrix inverse A055133 has the generating function E(x*y)/E(y). (End)
T(n,k) = T(n-1,k)*(n+k)/(n-k) + T(n-1,k-1), T(n,0) = T(n,n) = 1. - Vladimir Kruchinin, Oct 18 2014
Observe that the recurrence T(n,k) = T(n-1,k)*(n+k)/(n-k) - T(n-1,k-1), for n >= 2 and 1 <= k < n, with boundary conditions T(n,0) = T(n,n) = 1 gives Pascal's triangle A007318. - Peter Bala, Dec 21 2014
n-th row polynomial R(n, x) = [z^n] (1 + (1 + x)*z + x*z^2)^n. Note that 1/n*[z^(n-1)] (1 + (1 + x)*z + x*z^2)^n gives the row polynomials of A001263. - Peter Bala, Jun 24 2015
Binomial transform of A105868. If G(x,t) = 1/sqrt(1 - 2*(1 + t)*x + (1 - t)^2*x^2) denotes the o.g.f. of this array then 1 + x*d/dx log(G(x,t)) = 1 + (1 + t)*x + (1 + 6*t + t^2)*x^2 + ... is the o.g.f. for A086645. - Peter Bala, Sep 06 2015
T(n,k) = Sum_{i=0..n} C(n-i,k)*C(n,i)*C(n+i,i)*(-1)^(n-i-k). - Vladimir Kruchinin, Jan 14 2018
G.f. satisfies A(x,y) = x*A(x,y)+x*y*A(x,y)+sqrt(1+4*x^2*y*A(x,y)^2). - Vladimir Kruchinin, Oct 23 2020
G.f. satisfies the differential equation y * d^2(A(x,y))/dy^2 - x^2 * d^2(x*A(x,y))/dx^2 + 2*x^2* A(x,y)^3 = 0. - Sergii Voloshyn, Mar 07 2025
T(n,k) = Sum_{i=0..n} C(2*n+1,i)*C(n+k-i,n)^2*(-1)^i. - Natalia L. Skirrow, Apr 14 2025

A000275 Coefficients of a Bessel function (reciprocal of J_0(z)); also pairs of permutations with rise/rise forbidden.

Original entry on oeis.org

1, 1, 3, 19, 211, 3651, 90921, 3081513, 136407699, 7642177651, 528579161353, 44237263696473, 4405990782649369, 515018848029036937, 69818743428262376523, 10865441556038181291819, 1923889742567310611949459, 384565973956329859109177427, 86180438505835750284241676121
Offset: 0

Views

Author

Keywords

Comments

a(n) has the Lucas property, namely a(n) is congruent to a(n_0)a(n_1)...a(n_k) modulo p for any prime p where n_0,n_1,... are the base p digits of n. (Carlitz via McIntosh)

Examples

			From _Peter Bala_, Aug 08 2011: (Start)
a(3) = 19: The 19 pairs of permutations in the group S_3 x S_3 with no common rises correspond to the zero entries in the table below.
  ======================================
   Number of common rises in S_3 x S_3
  ======================================
     | 123   132   213   231   312   321
  ======================================
  123|  2     1     1     1     1     0
  132|  1     1     0     1     0     0
  213|  1     0     1     0     1     0
  231|  1     1     0     1     0     0
  312|  1     0     1     0     1     0
  321|  0     0     0     0     0     0
(End)
G.f. = 1 + x + 3*x^2 + 19*x^3 + 211*x^4 + 3651*x^5 + 90921*x^6 + ...
		

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Row 2 of A212855.
Cf. A055133 (absolute value of column 0 of triangle), A192721 (column 1), A115368.
Column k=1 of A340986.

Programs

  • Maple
    A000275 := proc(n) sum(z^k/k!^2, k = 0..infinity);
    series(%^x, z=0, n+1): n!^2*coeff(%,z,n); add(abs(coeff(%,x,k)), k=0..n) end:
    seq(A000275(n), n=0..17); # Peter Luschny, May 27 2017
  • Mathematica
    a[0] = 1; a[n_] := a[n] = Sum[(-1)^(r+n+1)*Binomial[n, r]^2 a[r], {r, 0, n-1}]; Table[a[n], {n, 0, 17}] (* Jean-François Alcover, Aug 05 2013 *)
    CoefficientList[Series[1/BesselJ[0,Sqrt[4*x]], {x, 0, 20}], x]* Range[0, 20]!^2 (* Vaclav Kotesovec, Mar 02 2014 *)
    a[ n_] := If[ n < 0, 0, (n! 2^n)^2 SeriesCoefficient[ 1 / BesselJ[ 0, x], {x, 0, 2 n}]]; (* Michael Somos, Aug 20 2015 *)
  • PARI
    {a(n) = if( n<0, 0, n!^2 * 4^n * polcoeff( 1 / besselj(0, x + x * O(x^(2*n))), 2*n))}; /* Michael Somos, May 17 2004 */

Formula

a(n) = Sum_{r=0..n-1} (-1)^(r+n+1) binomial(n, r)^2 a(r), if n > 0.
Sum_{n>=0} a(n) * x^n / n!^2 = 1 / J_0(sqrt(4*x)). - _Michael Somos, May 17 2004
From Peter Bala, Aug 08 2011: (Start)
Conjectural formula: 1 = Sum_{n>=0} a(n)*x^n*Sum_{k>=0} binomial(n+k,k)^2*(-x)^k.
Apart from the initial term, first column of A192721. (End)
E.g.f.: 1/J_0(sqrt(4*x)) = 1 + x/Q(0), where Q(k) = (k+1)^2 - x + (k+1)^2*x/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec 06 2013
a(n) ~ c * (n!)^2 / r^n, where r = (1/4)*BesselJZero[0,1]^2 = 1.4457964907366961302939989396139517587678604516... and c = 1/(sqrt(r) * BesselJ(1, 2*sqrt(r))) = 1.60197469692804662664846689139151227422675123376219... - Vaclav Kotesovec, Mar 02 2014, updated Apr 01 2018

Extensions

More terms from Christian G. Bower, Apr 25 2000

A086646 Triangle, read by rows, of numbers T(n,k), 0 <= k <= n, given by T(n,k) = A000364(n-k)*binomial(2*n, 2*k).

Original entry on oeis.org

1, 1, 1, 5, 6, 1, 61, 75, 15, 1, 1385, 1708, 350, 28, 1, 50521, 62325, 12810, 1050, 45, 1, 2702765, 3334386, 685575, 56364, 2475, 66, 1, 199360981, 245951615, 50571521, 4159155, 183183, 5005, 91, 1, 19391512145, 23923317720, 4919032300, 404572168, 17824950, 488488, 9100, 120, 1
Offset: 0

Views

Author

Philippe Deléham, Jul 26 2003

Keywords

Comments

The elements of the matrix inverse are apparently given by T^(-1)(n,k) = (-1)^(n+k)*A086645(n,k). - R. J. Mathar, Mar 14 2013
Let E(y) = Sum_{n >= 0} y^n/(2*n)! = cosh(sqrt(y)). Then this triangle is the generalized Riordan array (1/E(-y), y) with respect to the sequence (2*n)! as defined in Wang and Wang. - Peter Bala, Aug 06 2013
Let P_n be the poset of even size subsets of [2n] ordered by inclusion. Then Sum_{k=0..n}(-1)^(n-k)*T(n,k)*x^k is the characteristic polynomial of P_n. - Geoffrey Critzer, Feb 24 2021

Examples

			Triangle begins:
      1;
      1,     1;
      5,     6,     1;
     61,    75,    15,    1;
   1385,  1708,   350,   28,  1;
  50521, 62325, 12810, 1050, 45, 1;
  ...
From _Peter Bala_, Aug 06 2013: (Start)
Polynomial  |        Real zeros to 5 decimal places
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
R(5,-x)     | 1, 9.18062, 13.91597
R(10,-x)    | 1, 9.00000, 25.03855,  37.95073
R(15,-x)    | 1, 9.00000, 25.00000,  49.00895, 71.83657
R(20,-x)    | 1, 9.00000, 25.00000,  49.00000, 81.00205, 114.87399
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
(End)
		

Crossrefs

Cf. A000281.
Cf. A000795 (row sums).
Cf. A055133, A086645 (unsigned matrix inverse), A103364, A104033.
T(2n,n) give |A214445(n)|.

Programs

  • Maple
    A086646 := proc(n,k)
        if k < 0 or k > n then
            0 ;
        else
            A000364(n-k)*binomial(2*n,2*k) ;
        end if;
    end proc: # R. J. Mathar, Mar 14 2013
  • Mathematica
    R[0, _] = 1;
    R[n_, x_] := R[n, x] = x^n - Sum[(-1)^(n-k) Binomial[2n, 2k] R[k, x], {k, 0, n-1}];
    Table[CoefficientList[R[n, x], x], {n, 0, 8}] // Flatten (* Jean-François Alcover, Dec 19 2019 *)
    T[0, 0] := 1; T[n_, 0] := -Sum[(-1)^k T[n, k], {k, 1, n}]; T[n_, k_]/;0Oliver Seipel, Jan 11 2025 *)

Formula

cosh(u*t)/cos(t) = Sum_{n>=0} S_2n(u)*(t^(2*n))*(1/(2*n)!). S_2n(u) = Sum_{k>=0} T(n,k)*u^(2*k). Sum_{k>=0} (-1)^k*T(n,k) = 0. Sum_{k>=0} T(n,k) = 2^n*A005647(n); A005647: Salie numbers.
Triangle T(n,k) read by rows; given by [1, 4, 9, 16, 25, 36, 49, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, ...] where DELTA is the operator defined in A084938.
Sum_{k=0..n} (-1)^k*T(n,k)*4^(n-k) = A000281(n). - Philippe Deléham, Jan 26 2004
Sum_{k=0..n} T(n,k)*(-4)^k*9^(n-k) = A002438(n+1). - Philippe Deléham, Aug 26 2005
Sum_{k=0..n} (-1)^k*9^(n-k)*T(n,k) = A000436(n). - Philippe Deléham, Oct 27 2006
From Peter Bala, Aug 06 2013: (Start)
Let E(y) = Sum_{n >= 0} y^n/(2*n)! = cosh(sqrt(y)). Generating function: E(x*y)/E(-y) = 1 + (1 + x)*y/2! + (5 + 6*x + x^2)*y^2/4! + (61 + 75*x + 15*x^2 + x^3)*y^3/6! + .... The n-th power of this array has a generating function E(x*y)/E(-y)^n. In particular, the matrix inverse is a signed version of A086645 with a generating function E(-y)*E(x*y).
Recurrence equation for the row polynomials: R(n,x) = x^n - Sum_{k = 0..n-1} (-1)^(n-k)*binomial(2*n,2*k)*R(k,x) with initial value R(0,x) = 1.
It appears that for arbitrary complex x we have lim_{n -> infinity} R(n,-x^2)/R(n,0) = cos(x*Pi/2). A stronger result than pointwise convergence may hold: the convergence may be uniform on compact subsets of the complex plane. This would explain the observation that the real zeros of the polynomials R(n,-x) seem to converge to the odd squares 1, 9, 25, ... as n increases. Some numerical examples are given below. Cf. A055133, A091042 and A103364.
R(n,-1) = 0; R(n,-9) = (-1)^n*2*4^n; R(n,-25) = (-1)^n*2*(16^n - 4^n);
R(n,-49) = (-1)^n*2*(36^n - 16^n + 4^n). (End)

A103364 Matrix inverse of the Narayana triangle A001263.

Original entry on oeis.org

1, -1, 1, 2, -3, 1, -7, 12, -6, 1, 39, -70, 40, -10, 1, -321, 585, -350, 100, -15, 1, 3681, -6741, 4095, -1225, 210, -21, 1, -56197, 103068, -62916, 19110, -3430, 392, -28, 1, 1102571, -2023092, 1236816, -377496, 68796, -8232, 672, -36, 1, -27036487, 49615695, -30346380, 9276120, -1698732, 206388
Offset: 1

Views

Author

Paul D. Hanna, Feb 02 2005

Keywords

Comments

The first column is A103365. The second column is A103366. Row sums are all zeros (for n > 1). Absolute row sums form A103367.
Let E(y) = Sum_{n >= 0} y^n/(n!*(n+1)!) = (1/sqrt(y))*BesselI(1,2*sqrt(y)). Then this triangle is the generalized Riordan array (1/E(y), y) with respect to the sequence n!*(n+1)! as defined in Wang and Wang. - Peter Bala, Aug 07 2013

Examples

			Rows begin:
        1;
       -1,        1;
        2,       -3,       1;
       -7,       12,      -6,       1;
       39,      -70,      40,     -10,     1;
     -321,      585,    -350,     100,   -15,     1;
     3681,    -6741,    4095,   -1225,   210,   -21,   1;
   -56197,   103068,  -62916,   19110, -3430,   392, -28,   1;
  1102571, -2023092, 1236816, -377496, 68796, -8232, 672, -36, 1;
  ...
From _Peter Bala_, Aug 09 2013: (Start)
The real zeros of the row polynomials R(n,x) appear to converge to zeros of E(alpha*x) as n increases, where alpha = -3.67049 26605 ... ( = -(A115369/2)^2).
Polynomial | Real zeros to 5 decimal places
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
R(5,x)     | 1, 3.57754, 3.81904
R(10,x)    | 1, 3.35230, 7.07532,  9.14395
R(15,x)    | 1, 3.35231, 7.04943, 12.09668, 15.96334
R(20,x)    | 1, 3.35231, 7.04943, 12.09107, 18.47845, 24.35255
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Function   |
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
E(alpha*x) | 1, 3.35231, 7.04943, 12.09107, 18.47720, 26.20778, ...
Note: The n-th zero of E(alpha*x) may be calculated in Maple 17 using the instruction evalf( BesselJZeros(1,n)/ BesselJZeros(1,1))^2 ). (End)
		

Crossrefs

Programs

  • Mathematica
    T[n_, 1]:= Last[Table[(-1)^(n - 1)*(CoefficientList[Series[x/BesselJ[1, 2*x], {x, 0, 1000}], x])[[k]]*(n)!*(n - 1)!, {k, 1, 2*n - 1, 2}]]
    T[n_, n_] := 1; T[2, 1] := -1; T[3, 1] := 2; T[n_, k_] := T[n, k] = T[n - 1, k - 1]*n*(n - 1)/(k*(k - 1)); Table[T[n, k], {n, 1, 50}, {k, 1, n}] // Flatten (* G. C. Greubel, Jan 04 2016 *)
    T[n_, n_] := 1; T[n_, 1]/;n>1 := T[n, 1] = -Sum[T[n, j], {j, 2, n}]; T[n_, k_]/;1Oliver Seipel, Jan 01 2025 *)
  • PARI
    T(n,k)=if(n
    				

Formula

From Peter Bala, Aug 07 2013: (Start)
Let E(y) = Sum_{n >= 0} y^n/(n!*(n+1)!) = (1/sqrt(y))* BesselI(1,2*sqrt(y)). Generating function: E(x*y)/E(y) = 1 + (-1 + x)*y/(1!*2!) + (2 - 3*x + x^2)*y^2/(2!*3!) + (-7 + 12*x - 6*x^2 + x^3)*y^3/(3!*4!) + .... The n-th power of this array has a generating function E(x*y)/E(y)^n. In particular, the matrix inverse A001263 has a generating function E(y)*E(x*y).
Recurrence equation for the row polynomials: R(n,x) = x^n - Sum_{k = 0..n-1} 1/(n-k+1)*binomial(n,k)*binomial(n+1,k+1) *R(k,x) with initial value R(0,x) = 1.
Let alpha denote the root of E(x) = 0 that is smallest in absolute magnitude. Numerically, alpha = -3.67049 26605 ... = -(A115369/2)^2. It appears that for arbitrary complex x we have lim_{n->oo} R(n,x)/R(n,0) = E(alpha*x). Cf. A055133, A086646 and A104033.
A stronger result than pointwise convergence may hold: the convergence may be uniform on compact subsets of the complex plane. This would explain the observation that the real zeros of the polynomials R(n,x) seem to converge to the zeros of E(alpha*x) as n increases. Some numerical examples are given below. (End)
From Werner Schulte, Jan 04 2017, corrected May 05 2025: (Start)
T(n,k) = T(n-1,k-1)*n*(n-1)/(k*(k-1)) for 1 < k <= n;
T(n,k) = T(n+1-k,1)*A001263(n,k) for 1 <= k <= n;
Sum_{k=1..n} T(n,k)*A000108(k) = 1 for n > 0. (End)

A104033 Triangle, read by rows, equal to the matrix inverse of triangle A103327, where A103327(n,k) = binomial(2*n+1,2*k+1).

Original entry on oeis.org

1, -3, 1, 25, -10, 1, -427, 175, -21, 1, 12465, -5124, 630, -36, 1, -555731, 228525, -28182, 1650, -55, 1, 35135945, -14449006, 1782495, -104676, 3575, -78, 1, -2990414715, 1229758075, -151714563, 8912475, -305305, 6825, -105, 1, 329655706465, -135565467080, 16724709820, -982532408
Offset: 0

Views

Author

Paul D. Hanna, Feb 28 2005

Keywords

Comments

Column 0 equals signed A009843 (expansion of x/cosh(x)). Row sums form signed A000182 (expansion of tanh(x)).
The matrix logarithm is L(n,k) = -(-1)^(n-k)*A000182(n-k)*A103327(n,k), where A000182 = tangent numbers.
Let E(y) = cosh(sqrt(y)) = 1 + 3*y/3! + 5*y^2/5! + 7*y^3/7! + .... so that 1/E(y) = 1 - 3*y/3! + 25*y^2/5! - 427*y^3/7! + .... Then this triangle is the generalized Riordan array (1/E(y), y) with respect to the sequence (2*n+1)! as defined in Wang and Wang. - Peter Bala, Aug 06 2013

Examples

			Rows begin:
1;
-3, 1;
25, -10, 1;
-427, 175, -21, 1;
12465, -5124, 630, -36, 1;
-555731 ,228525, -28182, 1650, -55, 1;
35135945, -14449006, 1782495, -104676, 3575, -78, 1;
-2990414715, 1229758075, -151714563, 8912475, -305305, 6825, -105, 1;
329655706465, -135565467080, 16724709820, -982532408, 33669350, -754936, 11900, -136, 1; ...
From _Peter Bala_, Aug 06 2013: (Start)
The real zeros of the row polynomials R(n,x) seem to converge to the even squares as n increases.
Polynomial |        Real zeros to 6 decimal places
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
R(5,x)     | 3.999986
R(10,x)    | 4.000000, 15.999978
R(15,x)    | 4.000000, 16.000000, 35.999992, 64.414273, 76.998346
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
(End)
		

Crossrefs

Programs

  • PARI
    {T(n,k) = if(n=j, binomial(2*m-1,2*j-1))))^-1)[n+1,k+1])}
    for(n=0,10,for(k=0,n, print1(T(n,k),", "));print(""))
    
  • PARI
    {T(n,k) = binomial(2*n+1,2*k+1) * polcoeff(1/cosh(x+x*O(x^(2*n))), 2*n-2*k) * (2*n-2*k)!}
    for(n=0,10,for(k=0,n, print1(T(n,k),", "));print(""))

Formula

Column k: Sum_{j=0..n} C(2*n+1, 2*j+1) * T(j, k) = 0 (n>k), or 1 (n=k).
Row n: Sum_{j=0..n} T(n, j) * C(2*j+1, 2*k+1) = 0 (k
Sum_{k=0..n} T(n, k) * 4^k = 1 for n >= 0.
T(n, k) = (-1)^(n-k)*A000364(n-k)*A103327(n, k), where A000364 = Euler numbers.
Sum_{k=0..n} (-1)^(n-k)*T(n, k) = A002084(n). - Philippe Deléham, Aug 27 2005
From Peter Bala, Aug 06 2013: (Start)
Generating function: 1/sqrt(x)*sinh(sqrt(x)*t)/cosh(t) = t + (-3 + x)*t^3/3! + (25 - 10*x + x^2)*t^5/5! + ....
Recurrence equation for the row polynomials: R(n,x) = x^n - Sum_{k = 0..n-1} binomial(2*n+1,2*k+1)*R(k,x) with initial value R(0,x) = 1.
It appears that for arbitrary nonzero complex x we have
lim_{n -> inf} R(n,x^2)/R(n,0) = (1/(Pi/2*x))*sin(Pi/2*x).
A stronger result than pointwise convergence may hold: the convergence may be uniform on compact subsets of the complex plane. This would explain the observation that the real zeros of the polynomials R(n,x) seem to converge to the even squares 4, 16, 36, ... as n increases. Some numerical examples are given below. Cf. A055133, A086646 and A103364.
If p = 2*n + 1 is a prime then all the entries in row n are divisible by p, apart from T(n,n) = 1. Thus the row sum is congruent to 1 modulo p.
Row sums R(n,1) = (-1)^n*A000182(n+1).
R(n,4) = 1; R(n,16) = (1/2)*( 3^(2*n+1) - 1 ) = A096053(n);
R(n,36) = (1/3)*( 5^(2*n+1) - 3^(2*n+1) + 1 );
R(n,64) = (1/4)*( 7^(2*n+1) - 5^(2*n+1) + 3^(2*n+1) - 1 ). (End)

Extensions

Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, Jun 08 2007

A216207 Matrix inverse of A181543 (cubes of entries of Pascal's triangle).

Original entry on oeis.org

1, -1, 1, 7, -8, 1, -163, 189, -27, 1, 8983, -10432, 1512, -64, 1, -966751, 1122875, -163000, 7000, -125, 1, 179781181, -208818216, 30317625, -1304000, 23625, -216, 1, -53090086057, 61664945083, -8953081011, 385146125, -6988625, 64827, -343, 1, 23402291822743, -27182124061184, 3946556485312, -169776943616, 3081169000, -28625408, 153664, -512, 1
Offset: 0

Author

R. J. Mathar, Mar 12 2013

Keywords

Comments

Is the leftmost column derived from A212856?

Examples

			1;
-1,1;
7,-8,1;
-163,189,-27,1;
8983,-10432,1512,-64,1;
-966751,1122875,-163000,7000,-125,1;
179781181,-208818216,30317625,-1304000,23625,-216,1;
		

Crossrefs

Cf. A055133.

Formula

Sum_{j=k..n} A181543(n,j)*T(j,k) = delta(n,k).
Showing 1-6 of 6 results.