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 1-4 of 4 results.

A107238 A Chebyshev transform of number triangle A107230.

Original entry on oeis.org

1, 1, 1, 0, 2, 1, 0, 3, 3, 1, 0, 4, 8, 4, 1, 0, 5, 15, 15, 5, 1, 0, 6, 27, 36, 24, 6, 1, 0, 7, 42, 84, 70, 35, 7, 1, 0, 8, 64, 160, 200, 120, 48, 8, 1, 0, 9, 90, 300, 450, 405, 189, 63, 9, 1, 0, 10, 125, 500, 1000, 1050, 735, 280, 80, 10, 1, 0, 11, 165, 825, 1925, 2695, 2156, 1232, 396
Offset: 0

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Author

Paul Barry, May 14 2005

Keywords

Comments

Product of the number triangle A107230 by the Riordan array ((1-x^2)/(1+x^2),x/(1+x^2)). First column if C(1,n), second column is n (A001477), third column is essentially A034828.

Examples

			Triangle begins
1;
1,1;
0,2,1;
0,3,3,1;
0,4,8,4,1;
0,5,15,15,5,1;
		

Formula

Number triangle T(n, k)=sum{j=0..floor(n/2), (n/(n-j))(-1)^j*C(n-j, j)*A107230(n-2j, k)} (with T(0, n)=0^n).

A132894 Number of (1,0) steps in all paths of length n with steps U=(1,1), D=(1,-1) and H=(1,0), starting at (0,0), staying weakly above the x-axis (i.e., in all length-n left factors of Motzkin paths).

Original entry on oeis.org

0, 1, 4, 15, 52, 175, 576, 1869, 6000, 19107, 60460, 190333, 596652, 1863745, 5804176, 18028755, 55873872, 172818243, 533589660, 1644921789, 5063762220, 15568666029, 47811348816, 146675181975, 449538774048, 1376564658525
Offset: 0

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Author

Emeric Deutsch, Oct 07 2007

Keywords

Comments

Number of peaks (i.e., UDs) in all paths of length n+1 with steps U=(1,1), D=(1,-1) and H=(1,0), starting at (0,0), staying weakly above the x-axis (i.e., in all length n+1 left factors of Motzkin paths). Example: a(2)=4 because in the 13 (=A005773(4)) length-3 left factors of Motzkin paths, namely HHH, HHU, H(UD), HUH, HUU, (UD)H, (UD)U, UHD, UHH, UHU, U(UD), UUH and UUU, we have altogether 4 peaks (shown between parentheses).
This could be called the Motzkin transform of A077043 because the substitution x -> x*A001006(x) in the independent variable of the g.f. of A077043 yields the g.f. of this sequence here. - R. J. Mathar, Nov 10 2008

Examples

			a(2) = 4 because in the 5 (=A005773(3)) length-2 left factors of Motzkin paths, namely HH, HU, UD, UH and UU, we have altogether 4 H steps.
G.f. = x + 4*x^2 + 15*x^3 + 52*x^4 + 175*x^5 + 576*x^6 + 1869*x^7 + 6000*x^8 + ...
		

Crossrefs

Column k=1 of A328347.

Programs

  • Maple
    a := n -> add(k*binomial(n, k)*binomial(n-k, floor((n-k)/2)), k=0..n): seq(a(n), n=0..25);
    # second Maple program:
    a:= proc(n) a(n):=`if`(n<2, n, 2*n/(n-1)*a(n-1)+3*a(n-2)) end:
    seq(a(n), n=0..40);  # Alois P. Heinz, Jul 15 2013
  • Mathematica
    a[n_] := n*Hypergeometric2F1[3/2, 1-n, 2, 4]; Table[ a[n] // Abs, {n, 0, 25}] (* Jean-François Alcover, Jul 10 2013 *)
    a[ n_] := If[ n < 0, 0, -(-1)^n n Hypergeometric2F1[ 3/2, 1 - n, 2, 4]]; (* Michael Somos, Aug 06 2014 *)
  • Sage
    A132894 = lambda n: (-1)^(n+1)*jacobi_P(n-1,1,-n+1/2,-7)
    [Integer(A132894(n).n(40),16) for n in range(26)] # Peter Luschny, Sep 23 2014

Formula

a(n) = Sum_{k=0..n} k*A107230(n,k).
a(n) = Sum_{k=0..floor((n+1)/2)} k*A132893(n+1,k).
a(n) = Sum_{k=0..n} k*C(n,k)*C(n-k, floor((n-k)/2)).
G.f.: z/((1-3*z)*sqrt(1-2*z-3*z^2)).
a(n) = Sum_{k=0..n} k*C(n,k)*C(2*k,k)*(-1)^(n-k). - Wadim Zudilin, Oct 11 2010
E.g.f.: exp(x)*x*(BesselI(0, 2*x) + BesselI(1, 2*x)). - Peter Luschny, Aug 25 2012
a(n) = 2*n/(n-1)*a(n-1) + 3*a(n-2) for n>=2, a(n) = n for n<2. a(n) = n*A005773(n). - Alois P. Heinz, Jul 15 2013
a(n) ~ 3^(n-1/2)*sqrt(n/Pi). - Vaclav Kotesovec, Oct 08 2013
a(n) = (-1)^(n+1)*JacobiP(n-1,1,-n+1/2,-7). - Peter Luschny, Sep 23 2014

A107231 a(n) = C(n+2,2)*C(n,floor(n/2)).

Original entry on oeis.org

1, 3, 12, 30, 90, 210, 560, 1260, 3150, 6930, 16632, 36036, 84084, 180180, 411840, 875160, 1969110, 4157010, 9237800, 19399380, 42678636, 89237148, 194699232, 405623400, 878850700, 1825305300, 3931426800, 8143669800, 17450721000
Offset: 0

Views

Author

Paul Barry, May 13 2005

Keywords

Comments

Third column of A107230. Related to the generalized pentagonal numbers A001318. The sequence 0,0,1,3,12,... is an inverse Chebyshev transform of 0,0,1,3,8,... (see A034828). This transform maps a g.f. g(x) to (1/sqrt(1-4x^2))g(c(x^2)). Thus A001318, as first differences of A034828, can be expressed in terms of A107231.

Crossrefs

Programs

  • Mathematica
    Table[Binomial[n + 2, 2]*Binomial[n, Floor[n/2]], {n,0,50}] (* G. C. Greubel, Jun 13 2017 *)
  • PARI
    for(n=0,50, print1(binomial(n+2,2)*binomial(n,n\2), ", ")) \\ G. C. Greubel, Jun 13 2017

Formula

G.f.: (1+x)*(1-sqrt(1-4*x^2))^3*(sqrt(1-4*x^2)-4*x^2+1)^2/(8*x^4*(1-4*x^2)^(5/2)*(sqrt(1-4*x^2)+2*x-1)^2).
a(n) = Sum_{k=0..floor((n+2)/2)} binomial(n+2, k)*A034828(n+2-2*k). [corrected by Jason Yuen, Sep 02 2024]
Conjecture: n*a(n) +(n-4)*a(n-1) +2*(-2*n-5)*a(n-2) -4*n*a(n-3)=0. - R. J. Mathar, Nov 24 2012
G.f.: (1+x)/((1+2*x)^(3/2)*(1-2*x)^(5/2)). - Vladimir Reshetnikov, Aug 01 2018
Sum_{n>=0} 1/a(n) = Pi^2/9 - 2*Pi/sqrt(3) + 4. - Amiram Eldar, Sep 03 2024

A171814 Triangle T : T(n,k)= A007318(n,k)*A001700(n-k).

Original entry on oeis.org

1, 3, 1, 10, 6, 1, 35, 30, 9, 1, 126, 140, 60, 12, 1, 462, 630, 350, 100, 15, 1, 1716, 2772, 1890, 700, 150, 18, 1, 6435, 12012, 9702, 4410, 1225, 210, 21, 1, 24310, 51480, 48048, 25872, 8820, 1960, 280, 24, 1
Offset: 0

Views

Author

Philippe Deléham, Dec 19 2009

Keywords

Examples

			Triangle begins:
     1;
     3,    1;
    10,    6,    1;
    35,   30,    9,   1;
   126,  140,   60,  12,   1;
   462,  630,  350, 100,  15,  1;
  1716, 2772, 1890, 700, 150, 18, 1;
  ...
		

Crossrefs

Programs

  • Mathematica
    T[n_,k_]:=n!SeriesCoefficient[Exp[2*x]*(BesselI[0,2*x]+BesselI[1,2*x])*x^k / k!,{x,0,n}]; Table[T[n,k],{n,0,8},{k,0,n}]//Flatten (* Stefano Spezia, Dec 23 2023 *)

Formula

Sum_{k, 0<=k<=n} T(n,k)*x^k = A168491(n), A099323(n+1), A001405(n), A005773(n+1), A001700(n), A026378(n+1), A005573(n), A122898(n) for x = -4, -3, -2, -1, 0, 1, 2, 3 respectively.
Conjectural g.f.: 1/(2*t)*( sqrt( (1 - x*t)/(1 - (4 + x)*t) ) - 1 ) = 1 + (3 + x)*t + (10 + 6*x + x^2)*t^2 + .... - Peter Bala, Nov 10 2013
E.g.f. of column k: exp(2*x)*(BesselI(0,2*x)+BesselI(1,2*x))*x^k / k!. - Mélika Tebni, Dec 23 2023
Showing 1-4 of 4 results.