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 1-5 of 5 results.

A100326 Triangle, read by rows, where row n equals the inverse binomial of column n of square array A100324, which lists the self-convolutions of SHIFT(A003169).

Original entry on oeis.org

1, 1, 1, 3, 4, 1, 14, 20, 7, 1, 79, 116, 46, 10, 1, 494, 736, 311, 81, 13, 1, 3294, 4952, 2174, 626, 125, 16, 1, 22952, 34716, 15634, 4798, 1088, 178, 19, 1, 165127, 250868, 115048, 36896, 9094, 1724, 240, 22, 1, 1217270, 1855520, 862607, 285689, 74687, 15629, 2561, 311, 25, 1
Offset: 0

Views

Author

Paul D. Hanna, Nov 17 2004

Keywords

Comments

The leftmost column equals A003169 shift one place right.
Each column k > 0 equals the convolution of the prior column and A003169.
Row sums form A100327.
The elements of the matrix inverse are T^(-1)(n,k) = (-1)^(n+k) * A158687(n,k). - R. J. Mathar, Mar 15 2013

Examples

			Rows begin:
        1;
        1,       1;
        3,       4,      1;
       14,      20,      7,      1;
       79,     116,     46,     10,     1;
      494,     736,    311,     81,    13,     1;
     3294,    4952,   2174,    626,   125,    16,    1;
    22952,   34716,  15634,   4798,  1088,   178,   19,   1;
   165127,  250868, 115048,  36896,  9094,  1724,  240,  22,  1;
  1217270, 1855520, 862607, 285689, 74687, 15629, 2561, 311, 25,  1;
  ...
First column forms A003169 shift right.
Binomial transform of row 3 forms column 3 of square A100324: BINOMIAL([14,20,7,1]) = [14,34,61,96,140,194,259,...].
Binomial transform of row 4 forms column 4 of square A100324: BINOMIAL([79,116,46,10,1]) = [79,195,357,575,860,1224,...].
		

Crossrefs

Cf. A003169, A100324, A100327 (row sums), A158687, A264717 (central terms).

Programs

  • Haskell
    import Data.List (transpose)
    a100326 n k = a100326_tabl !! n !! k
    a100326_row n = a100326_tabl !! n
    a100326_tabl = [1] : f [[1]] where
    f xss@(xs:_) = ys : f (ys : xss) where
    ys = y : map (sum . zipWith (*) (zs ++ [y])) (map reverse zss)
    y = sum $ zipWith (*) [1..] xs
    zss@((:zs):) = transpose $ reverse xss
    -- Reinhard Zumkeller, Nov 21 2015
    
  • Maple
    A100326 := proc(n,k)
        if k < 0 or k > n then
            0 ;
        elif n = 0 then
            1 ;
        elif k = 0 then
            A003169(n)
        else
            add(procname(i+1,0)*procname(n-i-1,k-1),i=0..n-k) ;
        end if;
    end proc: # R. J. Mathar, Mar 15 2013
  • Mathematica
    lim= 9; t[0, 0]=1; t[n_, 0]:= t[n, 0]= Sum[(k+1)*t[n-1,k], {k,0,n-1}]; t[n_, k_]:= t[n, k]= Sum[t[j+1, 0]*t[n-j-1, k-1], {j,0,n-k}]; Flatten[Table[t[n, k], {n,0,lim}, {k,0,n}]] (* Jean-François Alcover, Sep 20 2011 *)
  • PARI
    T(n,k)=if(n
    				
  • SageMath
    @CachedFunction
    def T(n,k): # T = A100326
        if (k<0 or k>n): return 0
        elif (k==n): return 1
        elif (k==0): return sum((j+1)*T(n-1,j) for j in range(n))
        else: return sum(T(j+1,0)*T(n-j-1,k-1) for j in range(n-k+1))
    flatten([[T(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jan 30 2023

Formula

T(n, 0) = A003169(n) = Sum_{k=0..n-1} (k+1)*T(n-1, k) for n>0, with T(0, 0)=1.
T(n, k) = Sum_{i=0..n-k} T(i+1, 0)*T(n-i-1, k-1) for n > 0.
T(2*n, n) = A264717(n).
Sum_{k=0..n} T(n, k) = A100327(n).
G.f.: A(x, y) = (1 + G(x))/(1 - y*G(x)), where G(x) is the g.f. of A003169.
From G. C. Greubel, Jan 30 2023: (Start)
Sum_{k=0..n} (-1)^k*T(n, k) = A000007(n).
Sum_{k=0..n-1} (-1)^k*T(n, k) = A033999(n). (End)

A100327 Row sums of triangle A100326, in which row n equals the inverse binomial of column n of square array A100324.

Original entry on oeis.org

1, 2, 8, 42, 252, 1636, 11188, 79386, 579020, 4314300, 32697920, 251284292, 1953579240, 15336931928, 121416356108, 968187827834, 7769449728780, 62696580696172, 508451657412496, 4141712433518956, 33872033298518728, 278014853384816184, 2289376313410678312
Offset: 0

Views

Author

Paul D. Hanna, Nov 17 2004

Keywords

Comments

Self-convolution yields A100328, which equals column 1 of triangle A100326 (omitting leading zero).

Crossrefs

Programs

  • Magma
    A100327:= func< n | n eq 0 select 1 else (2/n)*(&+[Binomial(n, k)*Binomial(2*n+k, k-1): k in [1..n]]) >;
    [A100327(n): n in [0..30]]; // G. C. Greubel, Jan 30 2023
    
  • Maple
    A100327 := n -> simplify(2^n*binomial(3*n,2*n)*hypergeom([-1-2*n,-n], [-3*n], 1/2)/ (n+1/2)): seq(A100327(n), n=0..22); # Peter Luschny, Jun 10 2017
  • Mathematica
    Flatten[{1,Table[Sum[2*Binomial[n,k]*Binomial[2n+k,k-1]/n,{k,1,n}],{n,1,20}]}] (* Vaclav Kotesovec, Oct 17 2012 *)
  • PARI
    a(n)=if(n==0,1,sum(k=0,n,2*binomial(n,k)*binomial(2*n+k,k-1)/n))
    
  • PARI
    a(n)=polcoeff((1/x)*serreverse(x*(1-x+sqrt(1-4*x +x^2*O(x^n)))/(2+x)),n)
    for(n=0,25,print1(a(n),", ")) \\ Paul D. Hanna, Nov 22 2012
    
  • SageMath
    def A100327(n): return 2^n*binomial(3*n,2*n)*simplify(hypergeometric([-1-2*n,-n], [-3*n],1/2)/(n+1/2))
    [A100327(n) for n in range(31)] # G. C. Greubel, Jan 30 2023

Formula

G.f.: (1/x)*Series_Reversion( x*(1-x + sqrt(1 - 4*x)) / (2+x) ). - Paul D. Hanna, Nov 22 2012
G.f. A(x) = (1+G(x))/(1-G(x)), also A(x)^2 = (1+G(x))*G(x)/x, where G(x) = x*(1+G(x))/(1-G(x))^2 is the g.f. of A003169.
a(n) = 2*A003168(n) for n>0 with a(0)=1.
a(n) = Sum_{k=1..n} 2*binomial(n, k)*binomial(2n+k, k-1)/n for n>0 with a(0)=1.
Recurrence: 20*n*(2*n+1)*a(n) = (371*n^2 - 395*n + 96)*a(n-1) - 6*(27*n^2 - 103*n + 96)*a(n-2) + 4*(n-3)*(2*n-5)*a(n-3). - Vaclav Kotesovec, Oct 17 2012
a(n) ~ sqrt(4046 + 1122*sqrt(17))*((71 + 17*sqrt(17))/16)^n/(136*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 17 2012
a(n) = 2^n*binomial(3*n,2*n)*hypergeometric([-1-2*n,-n], [-3*n],1/2)/(n+1/2). - Peter Luschny, Jun 10 2017

A100325 Antidiagonal sums of square array A100324, which lists the self-convolutions of SHIFT(A003169).

Original entry on oeis.org

1, 2, 6, 25, 130, 774, 5009, 34231, 242988, 1773767, 13229272, 100362848, 772016385, 6007208105, 47198747457, 373929821070, 2983774582206, 23958802697161, 193448157014605, 1569625544848531, 12791865082236462
Offset: 0

Views

Author

Paul D. Hanna, Nov 17 2004

Keywords

Crossrefs

Programs

  • Mathematica
    f[n_]:= f[n]= If[n<2, 1, If[n==2, 3, ((324*n^2 -708*n +360)*f[n-1] -(371*n^2 -1831*n +2250)*f[n-2] + (20*n^2 -130*n +210)*f[n-3])/(16*n*(2*n-1)) ]]; (* f = A003169 *)
    A[n_, k_]:= A[n, k]= If[n==0, f[k], If[k==0, 1, Sum[f[k-j]*A[n-1,j], {j,0,k}]]]; (* A = 100324 *)
    a[n_]:= a[n]= Sum[A[n-k,k], {k,0,n}]; (* a = A100325 *)
    Table[a[n], {n, 0, 40}] (* G. C. Greubel, Jan 31 2023 *)
  • PARI
    {a(n)=local(A=1+x+x*O(x^n));if(n==0,1, for(i=1,n,A=1+x*A/(2-A)^2); sum(k=0,n,polcoeff(A^(n-k+1),k)))}
    
  • SageMath
    @CachedFunction
    def f(n): # f = A003169
        if (n<2): return 1
        elif (n==2): return 3
        else: return ((324*n^2-708*n+360)*f(n-1) - (371*n^2-1831*n+2250)*f(n-2) + (20*n^2-130*n+210)*f(n-3))/(16*n*(2*n-1))
    @CachedFunction
    def A(n, k): # A = 100324
        if (n==0): return f(k)
        elif (k==0): return 1
        else: return sum( f(k-j)*A(n-1, j) for j in range(k+1) )
    def T(n,k): return A(n-k, k)
    def A100325(n): return sum( T(n,k) for k in range(n+1) )
    [A100325(n) for n in range(41)] # G. C. Greubel, Jan 31 2023

Formula

G.f. A(x) = (1+G003169(x))/(1-x-x*G003169(x)), where G003169(x) is the g.f. of A003169.
a(n) ~ (sqrt(3056686 + 12607266/sqrt(17)) * ((71 + 17*sqrt(17))/16)^n) / (10201 * sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Jan 31 2023

A100328 Column 1 of triangle A100326, in which row n equals the inverse binomial of column n of square array A100324, with leading zero omitted.

Original entry on oeis.org

1, 4, 20, 116, 736, 4952, 34716, 250868, 1855520, 13979192, 106901032, 827644424, 6474611984, 51100656544, 406400018092, 3253636464756, 26201323746880, 212093247874904, 1724793778005528, 14084738953391768, 115447965121881856
Offset: 0

Views

Author

Paul D. Hanna, Nov 17 2004

Keywords

Comments

Self-convolution of A100327, which equals the row sums of triangle A100326.

Crossrefs

Programs

  • PARI
    {a(n)=if(n==0,1,sum(j=0,n, if(j==0,1,sum(k=0,j,2*binomial(j,k)*binomial(2*j+k,k-1)/j))* if(n-j==0,1,sum(k=0,n-j,2*binomial(n-j,k)*binomial(2*n-2*j+k,k-1)/(n-j)))))}

Formula

G.f.: (1+G003169(x))*G003169(x)/x, where G003169(x) is the g.f. of A003169.
Recurrence: 4*(n+1)*(2*n+1)*(17*n^2 - 28*n + 12)*a(n) = (1207*n^4 - 1988*n^3 + 1013*n^2 - 124*n - 12)*a(n-1) - 2*(n-2)*(2*n-3)*(17*n^2 + 6*n + 1)*a(n-2). - Vaclav Kotesovec, Jul 05 2014
a(n) ~ sqrt(95+393/sqrt(17)) * ((71+17*sqrt(17))/16)^n / (4*sqrt(2*Pi) * n^(3/2)). - Vaclav Kotesovec, Jul 05 2014
From Peter Bala, Sep 08 2024: (Start)
a(n) = (2/n) * Sum_{k = 0..n} binomial(n+1, n-k-1)*binomial(2*n, k)*2^(n-k) for n >= 1.
a(n) = 4*Jacobi_P(n-1, 2, n+1, 3)/n for n >= 1. Cf. A003168. (End)

A003169 Number of 2-line arrays; or number of P-graphs with 2n edges.

Original entry on oeis.org

1, 3, 14, 79, 494, 3294, 22952, 165127, 1217270, 9146746, 69799476, 539464358, 4214095612, 33218794236, 263908187100, 2110912146295, 16985386737830, 137394914285538, 1116622717709012, 9113225693455362, 74659999210200292
Offset: 1

Views

Author

Keywords

Comments

First column of triangle A100326. - Paul D. Hanna, Nov 16 2004

References

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

Crossrefs

Programs

  • Haskell
    a003169 = flip a100326 0  -- Reinhard Zumkeller, Nov 21 2015
  • Maple
    a[0]:=0:a[1]:=1:a[2]:=3:for n from 3 to 30 do a[n]:=((324*n^2-708*n+360)*a[n-1] -(371*n^2-1831*n+2250)*a[n-2]+(20*n^2-130*n+210)*a[n-3])/(16*n*(2*n-1)) od:seq(a[n],n=1..25); # Emeric Deutsch, Jan 31 2005
  • Mathematica
    lim = 21; t[0, 0] = 1; t[n_, 0] := t[n, 0] = Sum[(k + 1)*t[n - 1, k], {k, 0, n - 1}]; t[n_, k_] := t[n, k] = Sum[t[j + 1, 0]*t[n - j - 1, k - 1], {j, 0, n - k}]; Table[ t[n, 0], {n, lim}] (* Jean-François Alcover, Sep 20 2011, after Paul D. Hanna's comment *)
  • PARI
    {a(n)=if(n==0,0,if(n==1,1,if(n==2,3,( (324*n^2-708*n+360)*a(n-1) -(371*n^2-1831*n+2250)*a(n-2)+(20*n^2-130*n+210)*a(n-3))/(16*n*(2*n-1)) )))} \\ Paul D. Hanna, Nov 16 2004
    
  • PARI
    {a(n)=local(A=x+x*O(x^n));if(n==1,1, for(i=1,n,A=x*(1+A)/(1-A)^2); polcoeff(A,n))}
    
  • PARI
    seq(n)=Vec(serreverse(x*(1 - x)^2/(1 + x) + O(x*x^n))) \\ Andrew Howroyd, Mar 07 2023
    

Formula

For formula see Read reference.
D-finite with recurrence a(n) = ( (324*n^2-708*n+360)*a(n-1) - (371*n^2-1831*n+2250)*a(n-2) + (20*n^2-130*n+210)*a(n-3) )/(16*n*(2*n-1)) for n>2, with a(0)=0, a(1)=1, a(2)=3. - Paul D. Hanna, Nov 16 2004
G.f. satisfies: A(x) = x*(1+A(x))/(1-A(x))^2 where A(0)=0. G.f. satisfies: (1+A(x))/(1-A(x)) = 2*G003168(x)-1, where G003168 is the g.f. of A003168. - Paul D. Hanna, Nov 16 2004
a(n) = (1/n)*Sum_{i=0..n-1} binomial(n,i)*binomial(3*n-i-2,n-i-1). - Vladeta Jovovic, Sep 13 2006
Appears to be (1/n)*Jacobi_P(n-1,1,n-1,3). If so then a(n) = (1/(2*n-1))*Sum_{k = 0..n-1} binomial(n-1,k)*binomial(2*n+k-1,k+1) = (1/n)*Sum_{k = 0..n} binomial(n,k)*binomial(2*n-2,n+k-1)*2^k. [Added Jun 11 2023: these are correct, and can be proved using the WZ algorithm.] - Peter Bala, Aug 01 2012
a(n) ~ sqrt(33/sqrt(17)-7) * ((71+17*sqrt(17))/16)^n / (4*sqrt(2*Pi)*n^(3/2)). - Vaclav Kotesovec, Aug 09 2013
The o.g.f. A(x) = 1 + 3*x + 14*x^2 + ... taken with offset 0, satisfies 1 + x*A'(x)/A(x) = 1 + 3*x + 19*x^2 + 138*x^3 + ..., the o.g.f. for A156894. - Peter Bala, Oct 05 2015
From Peter Bala, Jun 11 2023: (Start)
a(n) = (1/n)*Sum_{k = 0..n-1} binomial(n,k+1)*binomial(2*n+k-1,k) (Carlitz, equation 3.19).
4*n*(17*n - 29)*(2*n - 1)*a(n) = (1207*n^3 - 4473*n^2 + 5258*n - 1920)*a(n-1) - 2*(2*n - 5)*(17*n - 12)*(n - 2)*a(n-2) with a(1) = 1 and a(2) = 3. (End)

Extensions

More terms from Emeric Deutsch, Jan 31 2005
Showing 1-5 of 5 results.