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 13 results. Next

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)

Links

Crossrefs

Cf. A000364, A103327, A009843, A000182 (unsigned row sums), A055133, A086645, A086646, A096053, A103364.

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

A086645 Triangle read by rows: T(n, k) = binomial(2n, 2k).

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 15, 15, 1, 1, 28, 70, 28, 1, 1, 45, 210, 210, 45, 1, 1, 66, 495, 924, 495, 66, 1, 1, 91, 1001, 3003, 3003, 1001, 91, 1, 1, 120, 1820, 8008, 12870, 8008, 1820, 120, 1, 1, 153, 3060, 18564, 43758, 43758, 18564, 3060, 153, 1, 1, 190, 4845, 38760
Offset: 0

Views

Author

Philippe Deléham, Jul 26 2003

Keywords

Comments

Terms have the same parity as those of Pascal's triangle.
Coefficients of polynomials (1/2)*((1 + x^(1/2))^(2n) + (1 - x^(1/2))^(2n)).
Number of compositions of 2n having k parts greater than 1; example: T(3, 2) = 15 because we have 4+2, 2+4, 3+2+1, 3+1+2, 2+3+1, 2+1+3, 1+3+2, 1+2+3, 2+2+1+1, 2+1+2+1, 2+1+1+2, 1+2+2+1, 1+2+1+2, 1+1+2+2, 3+3. - Philippe Deléham, May 18 2005
Number of binary words of length 2n - 1 having k runs of consecutive 1's; example: T(3,2) = 15 because we have 00101, 01001, 01010, 01011, 01101, 10001, 10010, 10011, 10100, 10110, 10111, 11001, 11010, 11011, 11101. - Philippe Deléham, May 18 2005
Let M_n be the n X n matrix M_n(i, j) = T(i, j-1); then for n > 0, det(M_n) = A000364(n), Euler numbers; example: det([1, 1, 0, 0; 1, 6, 1, 0; 1, 15, 15, 1; 1, 28, 70, 28 ]) = 1385 = A000364(4). - Philippe Deléham, Sep 04 2005
Equals ConvOffsStoT transform of the hexagonal numbers, A000384: (1, 6, 15, 28, 45, ...); e.g., ConvOffs transform of (1, 6, 15, 28) = (1, 28, 70, 28, 1). - Gary W. Adamson, Apr 22 2008
From Peter Bala, Oct 23 2008: (Start)
Let C_n be the root lattice generated as a monoid by {+-2*e_i: 1 <= i <= n; +-e_i +- e_j: 1 <= i not equal to j <= n}. Let P(C_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(C_n) [Ardila et al.]. See A127674 for (a signed version of) the corresponding array of f-vectors for these type C_n polytopes. See A008459 for the array of h-vectors for type A_n polytopes and A108558 for the array of h-vectors associated with type D_n polytopes.
The Hilbert transform of this triangle is A142992 (see A145905 for the definition of this term).
(End)
Diagonal sums: A108479. - Philippe Deléham, Sep 08 2009
Coefficients of Product_{k=1..n} (cot(k*Pi/(2n+1))^2 - x) = Sum_{k=0..n} (-1)^k*binomial(2n,2k)*x^k/(2n+1-2k). - David Ingerman (daviddavifree(AT)gmail.com), Mar 30 2010
Generalized Narayana triangle for 4^n (or cosh(2x)). - Paul Barry, Sep 28 2010
Coefficients of the matrix inverse appear to be T^(-1)(n,k) = (-1)^(n+k)*A086646(n,k). - R. J. Mathar, Mar 12 2013
Let E(y) = Sum_{n>=0} y^n/(2*n)! = cosh(sqrt(y)). Then this triangle is the generalized Riordan array (E(y), y) with respect to the sequence (2*n)! as defined in Wang and Wang. Cf. A103327. - Peter Bala, Aug 06 2013
Row 6, (1,66,495,924,495,66,1), plays a role in expansions of powers of the Dedekind eta function. See the Chan link, p. 534, and A034839. - Tom Copeland, Dec 12 2016

Examples

			From _Peter Bala_, Oct 23 2008: (Start)
The triangle begins
n\k|..0.....1.....2.....3.....4.....5.....6
===========================================
0..|..1
1..|..1.....1
2..|..1.....6.....1
3..|..1....15....15.....1
4..|..1....28....70....28.....1
5..|..1....45...210...210....45.....1
6..|..1....66...495...924...495....66.....1
...
(End)
From _Peter Bala_, Aug 06 2013: (Start)
Viewed as the generalized Riordan array (cosh(sqrt(y)), y) with respect to the sequence (2*n)! the column generating functions begin
1st col: cosh(sqrt(y)) = 1 + y/2! + y^2/4! + y^3/6! + y^4/8! + ....
2nd col: 1/2!*y*cosh(sqrt(y)) = y/2! + 6*y^2/4! + 15*y^3/6! + 28*y^4/8! + ....
3rd col: 1/4!*y^2*cosh(sqrt(y)) = y^2/4! + 15*y^3/6! + 70*y^4/8! + 210*y^5/10! + .... (End)

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 224.

Links

Crossrefs

Cf. A098158, A000384, A081294.
Cf. A008459, A108558, A127674, A142992. - Peter Bala, Oct 23 2008
Cf. A103327 (binomial(2n+1, 2k+1)), A103328 (binomial(2n, 2k+1)), A091042 (binomial(2n+1, 2k)). -Wolfdieter Lang, Jan 06 2013
Cf. A086646 (unsigned matrix inverse), A103327.
Cf. A034839.

Programs

  • Magma
    /* As triangle: */ [[Binomial(2*n, 2*k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Dec 14 2016
  • Maple
    A086645:=(n,k)->binomial(2*n,2*k): seq(seq(A086645(n,k),k=0..n),n=0..12);
  • Mathematica
    Table[Binomial[2 n, 2 k], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Dec 13 2016 *)
  • Maxima
    create_list(binomial(2*n,2*k),n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */
  • PARI
    {T(n, k) = binomial(2*n, 2*k)};
  • PARI
    {T(n, k) = sum( i=0, min(k, n-k), 4^i * binomial(n, 2*i) * binomial(n - 2*i, k-i))}; /* Michael Somos, May 26 2005 */

Formula

T(n, k) = (2*n)!/((2*(n-k))!*(2*k)!) row sums = A081294. COLUMNS: A000012, A000384
Sum_{k>=0} T(n, k)*A000364(k) = A000795(n) = (2^n)*A005647(n).
Sum_{k>=0} T(n, k)*2^k = A001541(n). Sum_{k>=0} T(n, k)*3^k = 2^n*A001075(n). Sum_{k>=0} T(n, k)*4^k = A083884(n). - Philippe Deléham, Feb 29 2004
O.g.f.: (1 - z*(1+x))/(x^2*z^2 - 2*x*z*(1+z) + (1-z)^2) = 1 + (1 + x)*z +(1 + 6*x + x^2)*z^2 + ... . - Peter Bala, Oct 23 2008
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A081294(n), A001541(n), A090965(n), A083884(n), A099140(n), A099141(n), A099142(n), A165224(n), A026244(n) for x = 0,1,2,3,4,5,6,7,8,9 respectively. - Philippe Deléham, Sep 08 2009
Product_{k=1..n} (cot(k*Pi/(2n+1))^2 - x) = Sum_{k=0..n} (-1)^k*binomial(2n,2k)*x^k/(2n+1-2k). - David Ingerman (daviddavifree(AT)gmail.com), Mar 30 2010
From Paul Barry, Sep 28 2010: (Start)
G.f.: 1/(1-x-x*y-4*x^2*y/(1-x-x*y)) = (1-x*(1+y))/(1-2*x*(1+y)+x^2*(1-y)^2);
E.g.f.: exp((1+y)*x)*cosh(2*sqrt(y)*x);
T(n,k) = Sum_{j=0..n} C(n,j)*C(n-j,2*(k-j))*4^(k-j). (End)
T(n,k) = 2*T(n-1,k) + 2*T(n-1,k-1) + 2*T(n-2,k-1) - T(n-2,k) - T(n-2,k-2), with T(0,0)=T(1,0)=T(1,1)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Nov 26 2013
From Peter Bala, Sep 22 2021: (Start)
n-th row polynomial R(n,x) = (1-x)^n*T(n,(1+x)/(1-x)), where T(n,x) is the n-th Chebyshev polynomial of the first kind. Cf. A008459.
R(n,x) = Sum_{k = 0..n} binomial(n,2*k)*(4*x)^k*(1 + x)^(n-2*k).
R(n,x) = n*Sum_{k = 0..n} (n+k-1)!/((n-k)!*(2*k)!)*(4*x)^k*(1-x)^(n-k) for n >= 1. (End)

A091042 Triangle of even numbered entries of odd numbered rows of Pascal's triangle A007318.

Original entry on oeis.org

1, 1, 3, 1, 10, 5, 1, 21, 35, 7, 1, 36, 126, 84, 9, 1, 55, 330, 462, 165, 11, 1, 78, 715, 1716, 1287, 286, 13, 1, 105, 1365, 5005, 6435, 3003, 455, 15, 1, 136, 2380, 12376, 24310, 19448, 6188, 680, 17, 1, 171, 3876, 27132, 75582, 92378, 50388, 11628, 969, 19, 1, 210, 5985, 54264, 203490, 352716, 293930, 116280, 20349, 1330, 21
Offset: 0

Views

Author

Wolfdieter Lang, Jan 23 2004

Keywords

Comments

The row polynomials Pe(n, x) := Sum_{m=0..n} a(n, m)*x^m appear as numerators of the generating functions for the even numbered column sequences of array A034870.
Elements have the same parity as those of Pascal's triangle.
All zeros of polynomial Pe(n, x) are negative. They are -tan^2(Pi/2*n+1), -tan^2(2*Pi/2*n+1), ..., -tan^2(n*Pi/2*n+1). Moreover, for m >= 1, Pe(m, -x^2) is the characteristic polynomial of the linear difference equation with constant coefficients for differences between multiples of 2*m+1 with even and odd digit sum in base 2*m in the interval [0,(2*m)^n). - Vladimir Shevelev and Peter J. C. Moses, May 22 2012
Row reverse of A103327. - Peter Bala, Jul 29 2013
The row polynomial Pe(d, x), multiplied by (2*d)!/d! = A001813(d), gives the numerator polynomial of the o.g.f. of the sequence of the diagonal d, for d >= 0, of the Sheffer triangle Lah[4,1] given in A048854. - Wolfdieter Lang, Oct 12 2017

Examples

			Triangle a(n, m) begins:
n\m  0   1    2     3      4      5      6      7     8    9  10 ...
0:   1
1:   1   3
2:   1  10    5
3:   1  21   35     7
4:   1  36  126    84      9
5:   1  55  330   462    165     11
6:   1  78  715  1716   1287    286     13
7:   1 105 1365  5005   6435   3003    455     15
8:   1 136 2380 12376  24310  19448   6188    680    17
9:   1 171 3876 27132  75582  92378  50388  11628   969   19
10:  1 210 5985 54264 203490 352716 293930 116280 20349 1330  21
... reformatted and extended. - _Wolfdieter Lang_, Oct 12 2017
From _Peter Bala_, Jan 30 2022: (Start)
(1/2)*(N^2 + N) = Sum_{j = 1..N} j.
(1/2)*(N^2 + N)^3 = Sum_{j = 1..N} j^3 + 3*Sum_{j = 1..N} j^5.
(1/2)*(N^2 + N)^5 = Sum_{j = 1..N} j^5 + 10*Sum_{j = 1..N} j^7 + 5*Sum_{j = 1..N} j^9.
(1/2)*(N^2 + N)^7 = Sum_{j = 1..N} j^7 + 21*Sum_{j = 1..N} j^9 + 35*Sum_{j = 1..N} j^11 + 7*Sum_{j = 1..N} j^13. (End)

References

  • A. M. Yaglom and I. M. Yaglom, An elementary proof of the Wallis, Leibniz and Euler formulas for pi. Uspekhi Matem. Nauk, VIII (1953), 181-187(in Russian).

Links

Crossrefs

Cf. A212500, A038754. A000302 (row sums), A085478, A103327 (row reverse), A048854, A103328.

Programs

  • GAP
    Flat(List([0..12], n-> List([0..n], k-> Binomial(2*n+1, 2*k) ))); # G. C. Greubel, Aug 01 2019
  • Magma
    [[Binomial(2*n+1,2*k): k in [0..n]]: n in [0..12]]; // G. C. Greubel, Aug 01 2019
  • Maple
    f := (x, t) -> cosh(sqrt(x)*t)*sinh(t); seq(seq(coeff(((2*n+1)!*coeff(series(f(x,t),t,2*n+2),t,2*n+1)),x,k),k=0..n),n=0..9); # Peter Luschny, Jul 29 2013
  • Mathematica
    T[n_, k_] /; 0 <= k <= n := T[n, k] = 2T[n-1, k] + 2T[n-1, k-1] + 2T[n-2, k-1] - T[n-2, k] - T[n-2, k-2]; T[0, 0] = T[1, 0] = 1; T[1, 1] = 3; T[, ] = 0;
Table[T[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* Jean-François Alcover, Jul 29 2018, after Philippe Deléham *)
Table[Binomial[2*n+1, 2*k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Aug 01 2019 *)
  • PARI
    T(n,k) = binomial(2*n+1, 2*k); \\ G. C. Greubel, Aug 01 2019
  • Python
    from math import comb, isqrt
def A091042(n): return comb((r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)))<<1|1,n-comb(r+1,2)<<1) # Chai Wah Wu, Apr 30 2025
  • Sage
    [[binomial(2*n+1, 2*k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Aug 01 2019

Formula

T(n, m) = binomial(2*n+1, 2*m) = A007318(2*n+1, 2*m), n >= m >= 0, otherwise 0.
From Peter Bala, Jul 29 2013: (Start)
E.g.f.: sinh(t)*cosh(sqrt(x)*t) = t + (1 + 3*x)*t^3/3! + (1 + 10*x + 5*x^2)*t^5/5! + (1 + 21*x + 35*x^2 + 7*x^3)*t^7/7! + ....
O.g.f.: A(x,t) = (1 + (x - 1)*t)/( (1 + (x - 1)*t)^2 - 4*t*x ) = 1 + (1 + 3*x)*t + (1 + 10*x + 5*x^2)*t^2 + ...
The function A( x/(x + 4), t*(x + 4)/4 ) = 1 + (1 + x)*t + (1 + 3*x + x^2)*t^2 + ... is the o.g.f. for A085478.
O.g.f. for n-th diagonal: ( Sum_{k = 0..n} binomial(2*n,2*k)*x^k )/(1 - x)^(2*n).
n-th row polynomial R(n,x) = (1/2)*( (1 + sqrt(x))^(2*n+1) - (sqrt(x) - 1)^(2*n+1) ).
Row sums A000302. (End)
T(n, k) = 2*T(n-1,k) + 2*T(n-1,k-1) + 2*T(n-2,k-1) - T(n-2,k) - T(n-2,k-2) with T(0,0)=T(1,0)=1, T(1,1)=3, T(n,k)=0 if k < 0 or if k > n. - Philippe Deléham, Nov 26 2013
From Peter Bala, Jan 31 2022: (Start)
Define S(r,N) = Sum_{j = 1..N} j^r. Then the following identity holds for n >= 0: (1/2)*(N^2 + N)^(2*n+1) = T(n,0)*S(2*n+1,N) + T(n,1)*S(2*n+3,N) + ... + T(n,n)* S(4*n+1,N). Some examples are given below. (End)

A002084 Sinh(x) / cos(x) = Sum_{n>=0} a(n)*x^(2n+1)/(2n+1)!.

Original entry on oeis.org

1, 4, 36, 624, 18256, 814144, 51475776, 4381112064, 482962852096, 66942218896384, 11394877025289216, 2336793875186479104, 568240131312188379136, 161669933656307658932224, 53204153193639888357113856, 20053432927718528320240287744
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Gandhi proves that a(n) == 1 (mod 2n+1) if 2n+1 is prime, that a(2n+1) == 4 (mod 10), and that a(2n+2) == 6 (mod 10). - Charles R Greathouse IV, Oct 16 2012

Examples

			x + 2/3*x^3 + 3/10*x^5 + 13/105*x^7 + 163/3240*x^9 + ...

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).

Links

Crossrefs

Cf. A002085.

Programs

  • Mathematica
    With[{nn=30},Take[CoefficientList[Series[Sinh[x]/Cos[x],{x,0,nn}],x] Range[0,nn-1]!,{2,-1,2}]] (* Harvey P. Dale, Jul 17 2012 *)
  • PARI
    a(n)=n++;my(v=Vec(1/cos(x+O(x^(2*n+1)))));v=vector(n,i,v[2*i-1]*(2*i-2)!);sum(g=1,n,binomial(2*n-1,2*g-2)*v[g]) \\ Charles R Greathouse IV, Oct 16 2012
  • PARI
    list(n)=n++;my(v=Vec(1/cos(x+O(x^(2*n+1)))));v=vector(n,i,v[2*i-1]*(2*i-2)!);vector(n,k,sum(g=1,k,binomial(2*k-1,2*g-2)*v[g])) \\ Charles R Greathouse IV, Oct 16 2012
  • Sage
    # Generalized algorithm of L. Seidel (1877)
def A002084_list(n) :
    R = []; A = {-1:0, 0:0}
    k = 0; e = 1
    for i in range(2*n) :
        Am = 1 if e == -1 else 0
        A[k + e] = 0
        e = -e
        for j in (0..i) :
            Am += A[k]
            A[k] = Am
            k += e
        if e == 1 : R.append(A[i//2])
    return R
A002084_list(10) # Peter Luschny, Jun 02 2012

Formula

E.g.f.: sinh(x)/cos(x) = Sum_{n>=0} a(n)*x^(2n+1)/(2n+1)!.
a(n) = Sum_{k=0..n} binomial(2n+1, 2k+1)*A000364(n-k) = Sum_{k=0..n} A103327(n, k)*A000324(n-k) = Sum_{k=0..n} (-1)^(n-k)*A104033(n, k). - Philippe Deléham, Aug 27 2005
a(n) ~ sinh(Pi/2) * 2^(2*n + 3) * (2*n + 1)! / Pi^(2*n+2). - Vaclav Kotesovec, Jul 05 2020

Extensions

a(13)-a(15) from Andrew Howroyd, Feb 05 2018

A103328 Triangle T(n, k) read by rows: binomial(2n, 2k+1).

Original entry on oeis.org

0, 2, 0, 4, 4, 0, 6, 20, 6, 0, 8, 56, 56, 8, 0, 10, 120, 252, 120, 10, 0, 12, 220, 792, 792, 220, 12, 0, 14, 364, 2002, 3432, 2002, 364, 14, 0, 16, 560, 4368, 11440, 11440, 4368, 560, 16, 0, 18, 816, 8568, 31824, 48620, 31824, 8568, 816, 18, 0, 20, 1140, 15504
Offset: 0

Views

Author

Ralf Stephan, Feb 06 2005

Keywords

Comments

A subset of Pascal's triangle A007318 with only even elements.

Examples

			Triangle begins
   0;
   2,   0;
   4,   4,    0;
   6,  20,    6,     0;
   8,  56,   56,     8,     0;
  10, 120,  252,   120,    10,    0;
  12, 220,  792,   792,   220,   12,   0;
  14, 364, 2002,  3432,  2002,  364,  14,  0;
  16, 560, 4368, 11440, 11440, 4368, 560, 16, 0;
  ...
From _Peter Bala_, Jan 30 2022: (Start)
(1/2)*(N^2 + N)^2 = 2*Sum_{j = 1..N} j^3.
(1/2)*(N^2 + N)^4 = 4*Sum_{j = 1..N} j^5 + 4*Sum_{j = 1..N} j^7.
(1/2)*(N^2 + N)^6 = 6*Sum_{j = 1..N} j^7 + 20*Sum_{j = 1..N} j^9 + 6*Sum_{j = 1..N} j^11.
(1/2)*(N^2 + N)^8 = 8*Sum_{j = 1..N} j^9 + 56*Sum_{j = 1..N} j^11 + 56*Sum_{j = 1..N} j^13 + 8*Sum_{j = 1..N} j^15. (End)

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 224.

Crossrefs

Cf. A091042, A086645, A103327.

Formula

From Peter Bala, Jan 31 2022: (Start)
T(n,k) = 2*T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k) + 2*T(n-2,k-1) - T(n-2,k-2), with T(0,0) = 0, T(1,0) = 2 and T(n,k) = 0 if k < 0 or if k > n-1.
n-th row polynomial R(n,x) = (1/(2*sqrt(x)))*( (1 + sqrt(x))^(2*n) - (1 - sqrt(x))^(2*n) ).
O.g.f.: A(x,t) = 2*t/(1 - 2*(x + 1)*t + (x - 1)^2*t^2) = 2*t + (4 + 4*x)*t^2 + (6 + 20*x + 6*x^2)*t^3 + ....
G.f.: (1/sqrt(x))*sinh(t)*sinh(sqrt(x)*t) = 2*t^2/2! + (4 + 4*x)*t^4/4! + (6 + 20*x^2 + 6*x^3)*t^6/6! + ....
O.g.f. for n-th diagonal: ( Sum_{k = 0..n} binomial(2*n,2*k+1)*x^k )/(1 - x)^(2*n) = 1/(2*sqrt(x))*((1 - sqrt(x))^(-2*n) - (1 + sqrt(x))^(-2*n)).
With a different offset, 2/(x-4)*A(x/(x-4), t*(x-4)/4) = t/(1 + t*(2 - x) + t^2) is a g.f. of A053122.
Define S(r,N) = Sum_{j = 1..N} j^r. Then the following identity holds for n >= 1:
(1/2)*(N^2 + N)^(2*n) = T(n,0)*S(2*n+1,N) + T(n,1)*S(2*n+3,N) + ... + T(n,n-1)*S(4*n-1,N). Some examples are given below. (End)

A109446 Binomial coefficients C(n,k) with n-k even, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 6, 1, 5, 10, 1, 1, 15, 15, 1, 7, 35, 21, 1, 1, 28, 70, 28, 1, 9, 84, 126, 36, 1, 1, 45, 210, 210, 45, 1, 11, 165, 462, 330, 55, 1, 1, 66, 495, 924, 495, 66, 1, 13, 286, 1287, 1716, 715, 78, 1, 1, 91, 1001, 3003, 3003, 1001, 91, 1, 15, 455, 3003, 6435, 5005
Offset: 0

Views

Author

Philippe Deléham, Aug 27 2005

Keywords

Comments

Binomial(n,2(n-k-1)) is also the number of permutations avoiding both 123 and 132 with k descents, i.e., positions with w[i]>w[i+1]. - Lara Pudwell, Dec 19 2018

Examples

			Starred terms in Pascal's triangle (A007318), read by rows:
  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*;
  1*,  8, 28*,  56, 70*,   56, 28*,   8, 1*;
  1,  9*,  36, 84*, 126, 126*,  84, 36*,  9, 1*;
Rows in A086645 (1; 1, 1; 1, 6, 1; ...) interspersed with rows in A103327 (1; 3, 1; 5, 10, 1; ...).
1; 1; 1, 1; 3, 1; 1, 6, 1; 5, 10, 1; 1, 15, 15, 1; 7, 35, 21, 1; ....

Links

Crossrefs

Cf. A109447. See A054142 for another version.

Programs

  • Maple
    T:= (n, k)-> binomial(n, 2*k+irem(n, 2)):
seq(seq(T(n, k), k=0..floor(n/2)), n=0..20);  # Alois P. Heinz, Feb 07 2014
  • Mathematica
    Flatten[ Table[ If[ EvenQ[n - k], Binomial[n, k], {}], {n, 0, 15}, {k, 0, n}]] (* Robert G. Wilson v *)

Extensions

More terms from Robert G. Wilson v, Aug 30 2005

A292219 Triangle read by rows. A generalization of unsigned Lah numbers, called L[4,3].

Original entry on oeis.org

1, 6, 1, 60, 20, 1, 840, 420, 42, 1, 15120, 10080, 1512, 72, 1, 332640, 277200, 55440, 3960, 110, 1, 8648640, 8648640, 2162160, 205920, 8580, 156, 1, 259459200, 302702400, 90810720, 10810800, 600600, 16380, 210, 1, 8821612800, 11762150400, 4116752640, 588107520, 40840800, 1485120, 28560, 272, 1
Offset: 0

Views

Author

Wolfdieter Lang, Sep 23 2017

Keywords

Comments

For the general L[d,a] triangles see A286724, also for references.
This is the generalized signless Lah number triangle L[4,3], the Sheffer triangle ((1 - 4*t)^(-3/2), t/(1 - 4*t)). It is defined as transition matrix
risefac[4,3](x, n) = Sum_{m=0..n} L[4,3](n, m)*fallfac[4,3](x, m), where risefac[4,3](x, n) := Product_{0..n-1} (x + (3 + 4*j)) for n >= 1 and risefac[4,3](x, 0) := 1, and fallfac[4,3](x, n):= Product_{0..n-1} (x - (3 + 4*j)) for n >= 1 and fallfac[4,3](x, 0) := 1.
In matrix notation: L[4,3] = S1phat[4,3]*S2hat[4,3] with the unsigned scaled Stirling1 and the scaled Stirling2 generalizations A225471 and A225469, respectively.
The a- and z-sequences for this Sheffer matrix have e.g.f.s Ea(t) = 1 + 4*t and Ez(t) = (1 + 4*t)*(1 - (1 + 4*t)^(-3/2))/t, respectively. That is, a = {1, 4, repeat(0)} and z(n) = 2*A292221(n). See the W. Lang link on a- and z-sequences there.
The inverse matrix T^(-1) = L^(-1)[4,3] is Sheffer ((1 + 4*t)^(-3/2), t/(1 + 4*t)). This means that T^(-1)(n, m) = (-1)^(n-m)*T(n, m).
fallfac[4,3](x, n) = Sum_{m=0..n} (-1)^(n-m)*T(n, m)*risefac[4,3](x, m), n >= 0.
Diagonal sequences have o.g.f. G(d, x) = A001813(d)*Sum_{m=0..d} A103327(d, m)*x^m/(1 - x)^(2*d + 1), for d >= 0 (d=0 main diagonal). G(d, x) generates {A001813(d)*binomial(2*(m + d) + 1, 2*d)}{m >= 0}. See the second W. Lang link on how to compute o.g.f.s of diagonal sequences of general Sheffer triangles. - _Wolfdieter Lang, Oct 12 2017

Examples

			The triangle T(n, m) begins:
  n\m          0           1          2         3        4       5     6   7  8
  0:           1
  1:           6           1
  2:          60          20          1
  3:         840         420         42         1
  4:       15120       10080       1512        72        1
  5:      332640      277200      55440      3960      110       1
  6:     8648640     8648640    2162160    205920     8580     156     1
  7:   259459200   302702400   90810720  10810800   600600   16380   210   1
  8:  8821612800 11762150400 4116752640 588107520 40840800 1485120 28560 272  1
  ...
Recurrence from a-sequence: T(4, 2) = (4/2)*T(3, 1) + 4*4*T(3, 2) = 2*420 + 16*42 = 1512.
Recurrence from z-sequence: T(4, 0) = 4*(z(0)*T(3, 0) + z(1)*T(3, 1) + z(2)*T(3, 2)+ z(3)*T(3, 3)) = 4*(6*840 - 6*420 + 40*42 -420*1) = 15120.
Meixner type identity for n = 2: (D_x - 4*(D_x)^2)*(60 + 20*x + 1*x^2 ) = (20 + 2*x) - 4*2 = 2*(6 + x).
Sheffer recurrence for R(3, x): [(6 + x) + 8*(3 + x)*D_x + 16*x*(D_x)^2] (60 + 20*x + 1*x^2) = (6 + x)*(60 + 20*x + x^2) + 8*(3 + x)*(20 + 2*x) + 16*2*x = 840 + 420*x + 42*x^2 + x^3 = R(3, x).
Boas-Buck recurrence for column m = 2 with n = 4: T(4, 2) = (2*4!/2)*(3 + 2*2)*(1*42/3! + 4*1/2!) = 1512.
Diagonal sequence d = 2: {60, 420, 1512, ...} has o.g.f. 12*(5 + 10*x + x^2)/(1 - x)^5 (see A001813(2) and row n=2 of A103327) generating {12*binomial(2*(m + 2) + 1, 4)}_{m >= 0}. - _Wolfdieter Lang_, Oct 12 2017

References

  • Steven Roman, The Umbral Calculus, Academic press, Orlando, London, 1984, p. 50.

Links

Crossrefs

Cf. A225469, A225471, A271703 L[1,0], A286724 L[2,1], A290596 L[3,1], A290597 L[3,2], A048854 L[4,1], A292221, A103327,
Diagonal sequences: A000012, 2*A014105(m+1), 12*A053126(m+4), 120*A053128(m+6), A053130(n+8), ... - Wolfdieter Lang, Oct 12 2017

Formula

T(n, m) = L[4,3](n,m) = Sum_{k=m..n} A225471(n, k)*A225469(k, m), 0 <= m <= n.
E.g.f. of row polynomials R(n, x) := Sum_{m=0..n} T(n, m)*x^m:
(1 - 4*t)^(-3/2)*exp(x*t/(1 - 4*t)) (this is the e.g.f. for the triangle).
E.g.f. of column m: (1 - 4*t)^(-3/2)*(t/(1 - 4*t))^m/m!, m >= 0.
Three term recurrence for column entries k >= 1: T(n, m) = (n/m)*T(n-1, m-1) + 4*n*T(n-1, m) with T(n, m) = 0 for n < m, and for the column m = 0: T(n, 0) = n*Sum_{j=0}^(n-1) z(j)*T(n-1, j), n >= 1, T(0, 0) = 0, from the a-sequence {1, 4 repeat(0)} and z(j) = 2*A292221(j) (see above).
Four term recurrence: T(n, m) = T(n-1, m-1) + 2*(4*n - 1)*T(n-1, m) - 8*(n-1)*(2*n - 1)*T(n-2, m), n >= m >= 0, with T(0, 0) =1, T(-1, m) = 0, T(n, -1) = 0 and T(n, m) = 0 if n < m.
Meixner type identity for (monic) row polynomials: (D_x/(1 + 4*D_x)) * R(n, x) = n * R(n-1, x), n >= 1, with R(0, x) = 1 and D_x = d/dx. That is, Sum_{k=0..n-1} (-4)^k*{D_x)^(k+1)*R(n, x) = n*R(n-1, x), n >= 1.
General recurrence for Sheffer row polynomials (see the Roman reference, p. 50, Corollary 3.7.2, rewritten for the present Sheffer notation):
R(n, x) = [(6 + x)*1 + 8*(3 + x)*D_x + 16*x*(D_x)^2]*R(n-1, x), n >= 1, with R(0, x) = 1.
Boas-Buck recurrence for column m (see a comment in A286724 with references): T(n, m) = (2*n!/(n-m))*(3 + 2*m)*Sum_{p=0..n-1-m} 4^p*T(n-1-p, m)/(n-1-p)!, for n > m >= 0, with input T(m, m) = 1.
Explicit form (from the o.g.f.s of diagonal sequences): ((2*(n-m))!/(n-m)!)*binomial(2*n + 1, 2*(n-m)), n >= m >= 0, and vanishing for n < m. - Wolfdieter Lang, Oct 12 2017

A123162 Triangle read by rows: T(n,k) = binomial(2*n - 1, 2*k - 1) for 0 < k <= n and T(n,0) = 1.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 5, 10, 1, 1, 7, 35, 21, 1, 1, 9, 84, 126, 36, 1, 1, 11, 165, 462, 330, 55, 1, 1, 13, 286, 1287, 1716, 715, 78, 1, 1, 15, 455, 3003, 6435, 5005, 1365, 105, 1, 1, 17, 680, 6188, 19448, 24310, 12376, 2380, 136, 1, 1, 19, 969, 11628, 50388, 92378, 75582, 27132, 3876, 171, 1
Offset: 0

Views

Author

Roger L. Bagula, Oct 02 2006

Keywords

Examples

			Triangle begins:
     1;
     1,  1;
     1,  3,   1;
     1,  5,  10,    1;
     1,  7,  35,   21,    1;
     1,  9,  84,  126,   36,    1;
     1, 11, 165,  462,  330,   55,    1;
     1, 13, 286, 1287, 1716,  715,   78,  1;
     1, 15, 455, 3003, 6435, 5005, 1365, 105, 1;
     ...

Links

Crossrefs

Cf. A002458, A007318, A034867, A103327, A122366, A123166, A259557.

Programs

  • GAP
    Flat(Concatenation([1],List([1..10],n->Concatenation([1],List([1..n],m->Binomial(2*n-1,2*m-1)))))); # Muniru A Asiru, Oct 11 2018
  • Magma
    A123162:= func< n,k | k eq 0 select 1 else Binomial(2*n-1, 2*k-1) >;
[A123162(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 18 2023
  • Mathematica
    T[n_, k_]= If [k==0, 1, Binomial[2*n-1, 2*k-1]];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten
  • Maxima
    T(n, k) := if k = 0 then 1 else binomial(2*n - 1, 2*k - 1)$
tabl(nn) := for n:0 thru nn do print(makelist(T(n, k), k, 0, n))$ /* Franck Maminirina Ramaharo, Oct 10 2018 */
  • SageMath
    def A123162(n,k): return binomial(2*n-1, 2*k-1) + int(k==0)
flatten([[A123162(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jun 01 2022

Formula

From Paul Barry, May 26 2008: (Start)
T(n,k) = binomial(2*n - 1, 2*k - 1) + 0^k.
Column k has g.f. (x^k/(1 - x)^(2*k + 0^k))*Sum_{j=0..k} binomial(2*k, 2*j)*x^j. (End)
From Franck Maminirina Ramaharo, Oct 10 2018: (Start)
Row n = coefficients in the expansion of ((x + sqrt(x))*(sqrt(x) - 1)^(2*n) + (x - sqrt(x))*(sqrt(x) + 1)^(2*n) + 2*x - 2)/(2*x - 2).
G.f.: (1 - (2 + x)*y + (1 - 2*x)*y^2 - (x - x^2)*y^3)/(1 - (3 + 2*x)*y + (3 + x^2)*y^2 - (1 - 2*x + x^2)*y^3).
E.g.f.: ((x + sqrt(x))*exp(y*(sqrt(x) - 1)^2) + (x - sqrt(x))*exp(y*(sqrt(x) + 1)^2) + (2*x - 2)*exp(y) - 2*x)/(2*x - 2). (End)
From G. C. Greubel, Jul 18 2023: (Start)
Sum_{k=0..n} T(n,k) = A123166(n).
T(n, n-1) = (n-1)*T(n, 1), n >= 2.
T(2*n, n) = A259557(n).
T(2*n+1, n+1) = A002458(n). (End)

Extensions

Edited by N. J. A. Sloane, Oct 04 2006
Partially edited and offset corrected by Franck Maminirina Ramaharo, Oct 10 2018

A343051 A triangle T(n,k) read by rows which can be used to calculate the area of a regular polygon with sides having length 1, provided that the polygon has an odd number of sides.

Original entry on oeis.org

1, 16, 3, 256, 800, 125, 4096, 62720, 115248, 16807, 65536, 3096576, 23514624, 34012224, 4782969, 1048576, 118947840, 2518720512, 13605588480, 17148710480, 2357947691, 16777216, 3898605568, 185305595904, 2609720475648, 11485488551680, 13234415217504, 1792160394037
Offset: 0

Views

Author

Peter Armstrong Maley, May 16 2021

Keywords

Comments

The examples will demonstrate how this works.

Examples

			16*A^2 - 3 = 0, A = 0.433012... the area of an equilateral triangle with sides of length 1.
256*A^4 - 800*A^2 + 125 = 0, A = 1.720477..., the area of a regular pentagon with sides of length 1.
4096*A^6 - 62720*A^4 + 115248*A^2 - 16807 = 0: A = 3.63391244..., the area of a regular heptagon with sides of length 1.
16777216*A^12 - 3898605568*A^10 + 185305595904*A^8 - 2609720475648*A^6 + 11485488551680*A^4 - 13234415217504*A^2 + 1792160394037 = 0: A = 13.185768328323878..., the area of a regular 13-gon with sides of length 1.
This sequence can be expressed as a triangle:
      1;
     16,       3;
    256,     800,      125;
   4096,   62720,   115248,    16807;
  65536, 3096576, 23514624, 34012224, 4782969;
  ...

Links

Crossrefs

Cf. A103327.

Programs

  • Mathematica
    T[n_, k_] := Binomial[2n+1, 2k+1] (2n+1)^(2k-1) 16^(n-k);
Table[T[n, k], {n, 0, 6}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 29 2021, after Andrew Howroyd *)

Formula

This sequence can be generated from A103327. This example is for the pentagon: 5*L^4 - 10*L^2 + 1 = 0, L = (4/5)*A. Thus 256*A^4 - 800*A^2 + 125 = 0. In the case of a heptagon, L = (4/7)*A.
T(n,k) = binomial(2*n+1, 2*k+1)*(2*n+1)^(2*k-1)*16^(n-k). - Andrew Howroyd, May 23 2021

A380281 Triangle T(n, k) read by rows: T(n, k) = 2^n*binomial(2*n + 1, 2*k + 1) * Pochhammer(1/2, n - k) * Pochhammer(1/2, k).

Original entry on oeis.org

1, 3, 1, 15, 10, 3, 105, 105, 63, 15, 945, 1260, 1134, 540, 105, 10395, 17325, 20790, 14850, 5775, 945, 135135, 270270, 405405, 386100, 225225, 73710, 10395, 2027025, 4729725, 8513505, 10135125, 7882875, 3869775, 1091475, 135135, 34459425, 91891800, 192972780, 275675400, 268017750, 175429800, 74220300, 18378360, 2027025
Offset: 0

Views

Author

Thomas Scheuerle, Jan 18 2025

Keywords

Examples

			Triangle begins:
n\k      0 |      1 |      2 |      3 |      4 |       5 |      6 |     7 |
[0]       1,
[1]       3,       1
[2]      15,      10,       3
[3]     105,     105,      63,       15
[4]     945,    1260,    1134,      540,     105
[5]   10395,   17325,   20790,    14850,    5775,     945
[6]  135135,  270270,  405405,   386100,  225225,   73710,   10395
[7] 2027025, 4729725, 8513505, 10135125, 7882875, 3869775, 1091475, 135135

Links

Crossrefs

T(n, 1) = A001147(n+1), T(n, 2) = A000457(n-1), T(n, 3) = A001881(n+3)*3, T(n, n) = A001147(n).
Cf. A076729, (conj. row sums), A103327, A173424.

Programs

  • Maple
    T := (n,k) -> 2^n*binomial(2*n + 1, 2*k + 1)*pochhammer(1/2, n - k)*pochhammer(1/2, k):
for n from 0 to 7 do seq(T(n, k), k = 0..n) od;  # Peter Luschny, Jan 21 2025
  • Mathematica
    A380281[n_, k_] := (2*n - 1)!!*Binomial[n, k]*Binomial[2*n + 1, 2*k + 1]/Binomial[2*n, 2*k];
Table[A380281[n, k], {n, 0, 10}, {k, 0, n}] (* Paolo Xausa, Jan 22 2025 *)
  • PARI
    T(n, k) = Vec(O(x^(1+n))+(1+x)^(n+1)*hypergeom([1,1/2+n+1],3/2,-x)*(2*(n+1))!/(2^(n+1)*(n+1)!))[1+k]
  • PARI
    doublefact(n) = prod(i=0, (n-1)\2, n - 2*i )
T(n, k) = doublefact(2*n-1) * binomial(n, k) * binomial(2*n+1, 2*k+1) / binomial(2*n, 2*k)
  • SageMath
    rf = rising_factorial
def T(n, k): return 2^n*binomial(2*n+1, 2*k+1)*rf(1/2, n-k)*rf(1/2, k)
for n in range(9): print([T(n, k) for k in range(n+1)])  # Peter Luschny, Jan 21 2025

Formula

Coefficients for the series representation of Owen's T-function Ot(x, m) = atan(m)/(2*Pi) + Sum_{s>=0} (-1)^(s+1)*m*(Sum_{r=0..s} T(s, r)*m^(2*s))*x^(2+2*s)/(2*Pi*(2+2*s)!).
Ot(x, m) - atan(m)/(2*Pi) = -V(x, x*m), where V is Nicholson's V-function. V(h, q) = Integral_{x=0..h} Integral_{y=0..q*x/h} phi(x)*phi(y) dydx, where phi(x) is the standard normal density exp(-x^2/2)/sqrt(2*Pi).
G.f. of row n: ((1 + x)^(n+1)*Hypergeometric2F1[1, 1/2 + n + 1, 3/2, -x]*(2*(n+1))!)/(2^(n+1)*(n+1)!).
T(n, k) = A103327(n, k)*A173424(n, k).
T(n, k) = (2*n-1)!! * binomial(n, k) * binomial(2*n+1, 2*k+1) / binomial(2*n, 2*k).
Conjecture: Row sums are A076729.
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