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.

A128053 a(n)=A128056(n)/A128055(n).

Original entry on oeis.org

1, -3, -3, 7, 7, -9, -9, 47, 47, -123, -123, 161, 161, -843, -843, 2207, 2207, -2889, -2889, 15127, 15127, -39603, -39603, 51841, 51841, -271443, -271443, 710647, 710647, -930249, -930249, 4870847, 4870847, -12752043, -12752043
Offset: 0

Views

Author

Paul Barry, Feb 13 2007

Keywords

Crossrefs

Cf. A128052.

Formula

a(n)=(-1)^C(n+1,2)*(F(n-1)+F(n+1))(5/6-cos(pi*n/3)/3)(1+(-1)^n)/2+(F(n)+F(n+2))(5/6-cos(pi*(n+1)/3)/3)(1-(-1)^n)/2.
Empirical g.f.: -(x-1) * (x^10-2*x^9-5*x^8+2*x^7+9*x^6+9*x^4+2*x^3-5*x^2-2*x+1) / ((x^4+3*x^2+1)*(x^8-3*x^6+8*x^4-3*x^2+1)). - Colin Barker, Jun 27 2013

A097893 Partial sums of the central trinomial coefficients (A002426).

Original entry on oeis.org

1, 2, 5, 12, 31, 82, 223, 616, 1723, 4862, 13815, 39468, 113257, 326198, 942425, 2730032, 7926659, 23061590, 67214399, 196211252, 573590621, 1678941350, 4920076877, 14433305000, 42381641381, 124558477682, 366371703833
Offset: 0

Views

Author

Emeric Deutsch, Sep 03 2004

Keywords

Comments

a(n) is the number of peaks at odd height in all Motzkin paths of length n+2. Example: a(2)=5 counts the peaks shown between parentheses in the 9 Motzkin paths of length 4: HHHH, HH(UD), H(UD)H, HUHD, (UD)HH, (UD)(UD), UHDH, UHHD and UUDD.
Binomial transform of 1,1,2,2,6,6,20,20,70,70...... (A000984 doubled). It would appear that the Hankel transform of this sequence is a signed version of A128055, with sign pattern given by s(n)=(2/3-sqrt(3)/3)cos(5*Pi*n/6)-sin(5*Pi*n/6)/3+(sqrt(3)/3+2/3)*cos(Pi*n/6)-sin(Pi*n/6)/3-cos(Pi*n/2)/3+sin(Pi*n/2)/3. - Paul Barry, Jan 03 2008
Define triangle T(n,1) = T(n,n) = 1 and T(r,c) = T(r,c-1) + T(r-1,c-1) + T(r-2,c-1). Then the sum of the terms in row(n) is a(n+1). - J. M. Bergot, Apr 30 2013

Links

Crossrefs

Cf. A002426, A025178.

Programs

  • Haskell
    a097893 n = a097893_list !! n
a097893_list = scanl1 (+) a002426_list
-- Reinhard Zumkeller, Jan 22 2013
  • Maple
    ser:=series(1/(1-z)/sqrt(1-2*z-3*z^2),z=0,32): 1,seq(coeff(ser,z^n),n=1..31);
a := n -> (n+1)*hypergeom([1/2,(1-n)/2,-n/2],[1,3/2],4):
seq(simplify(a(n)), n=0..26); # Peter Luschny, Oct 29 2015
  • Mathematica
    Table[ Sum[ Binomial[n, k]*Binomial[k, n-k], {k, 0, n}], {n, 0, 26}] // Accumulate (* Jean-François Alcover, Jul 10 2013 *)
CoefficientList[Series[1/((1-x)*Sqrt[1-2*x-3*x^2]), {x, 0, 50}], x] (* G. C. Greubel, Dec 21 2017 *)
  • PARI
    a(n)=sum(i=0,n,sum(j=0,i,binomial(i,i-j)*binomial(j,i-j)))
  • PARI
    vector(30, n, n--; sum(k=0, n\2, binomial(n+1, 2*k+1)* binomial(2*k, k))) \\ Altug Alkan, Oct 29 2015
  • PARI
    x='x+O('x^30); Vec(1/((1-x)*sqrt(1-2*x-3*x^2))) \\ G. C. Greubel, Dec 21 2017
  • Python
    from math import comb
def A097893(n): return sum(comb(n+1,(k<<1)|1)*comb(k<<1,k) for k in range((n>>1)+1)) # Chai Wah Wu, Aug 14 2025

Formula

G.f.: 1/((1-z)*sqrt(1-2*z-3*z^2)).
a(n) = Sum_{0<=j<=i<=n} C(i, i-j)*C(j, i-j). - Benoit Cloitre, Oct 23 2004
a(n) = sum_{k=0..n} Sum_{j=0..n-k} C(k,j)C(n-k,j)C(2j,j). - Paul Barry, Jan 03 2008
Logarithm g.f. atan(x*M(x)), M(x) - o.g.f. for Motzkin numbers (A001006). - Vladimir Kruchinin_, Aug 11 2010
D-finite with recurrence -n*a(n) +(3*n-1)*a(n-1) +(n-2)*a(n-2) +3*(1-n)*a(n-3)=0. - R. J. Mathar, Nov 09 2012 [Since A002426(n) = a(n) - a(n-1), this third-order recurrence follows easily from the second-order recurrence given in A002426. - Peter Bala, Oct 28 2015]
G.f.: G(0)/(1-x), where G(k)= 1 + x*(2+3*x)*(4*k+1)/( 4*k+2 - x*(2+3*x)*(4*k+2)*(4*k+3)/(x*(2+3*x)*(4*k+3) + 4*(k+1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jun 29 2013
a(n) ~ 3^(n+3/2)/(4*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 23 2013
a(n) = Sum_{k = 0..floor(n/2)} binomial(n + 1,2*k + 1) *binomial(2*k,k). Cf. A025178. - Peter Bala, Oct 28 2015
a(n) = (n+1)*hypergeom([1/2,(1-n)/2,-n/2],[1,3/2],4). - Peter Luschny, Oct 29 2015
a(n) = (n+1)*Sum_{k=0..floor(n/2)} multinomial(n;n-2*k,k,k)/(2*k+1). - Chai Wah Wu, Aug 14 2025

A128056 Hankel transform of A128057.

Original entry on oeis.org

1, -3, -6, 28, 56, -288, -576, 3008, 6016, -31488, -62976, 329728, 659456, -3452928, -6905856, 36159488, 72318976, -378667008, -757334016, 3965452288, 7930904576, -41526755328, -83053510656, 434873827328
Offset: 0

Views

Author

Paul Barry, Feb 13 2007

Keywords

Comments

a(n)=2^A128054(n)*A128053(n). Hankel transform of A128058.

Formula

a(n)=A128055(n)*((cos(pi*n/2)-sin(pi*n/2))((F(n-1)+F(n+1))(5/6-cos(pi*n/3)/3)(1+(-1)^n)/2 +(F(n)+F(n+2))(5/6-cos(pi*(n+1)/3)/3)(1-(-1)^n)/2)).
Empirical g.f.: -(2*x-1)*(4*x^2-x+1) / (16*x^4+12*x^2+1). - Colin Barker, Jun 27 2013

A162533 a(n) = Sum_{k=0..n} binomial(n,2k)*A002426(k).

Original entry on oeis.org

1, 1, 2, 4, 10, 26, 68, 176, 454, 1174, 3052, 7976, 20932, 55108, 145448, 384704, 1019462, 2706214, 7194956, 19155896, 51065260, 136284236, 364097912, 973654240, 2605983772, 6980545276, 18712478072, 50196568144, 134739960904, 361892443592, 972537193168
Offset: 0

Views

Author

Paul Barry, Jul 05 2009

Keywords

Comments

Hankel transform is (-1)^binomial(n,2)*(-2)^A128054(n) (see A128055).

Links

Crossrefs

Cf. A027826.

Programs

  • Mathematica
    b[n_] := If[n < 0, 0, 3^n Hypergeometric2F1[1/2, -n, 1, 4/3]]; Table[Sum[Binomial[n, 2*k]*b[k], {k, 0, n}], {n, 0, 50}] (* or *) CoefficientList[Series[(1-x)/sqrt(1-4*x+4*x^2-4*x^4), {x, 0, 50}], x] (* G. C. Greubel, Feb 27 2017 *)
  • PARI
    x='x+O('x^50); Vec((1-x)/sqrt(1-4*x+4*x^2-4*x^4)) \\ G. C. Greubel, Feb 27 2017

Formula

G.f.: (1-x)/((1-x)^2-x^2-2x^4/((1-x)^2-x^2-x^4/((1-x)^2-x^2-x^4/(1-... (continued fraction).
G.f.: (1-x)/sqrt(1-4*x+4*x^2-4*x^4) = (1-x)/sqrt((1-2*x)^2-4*x^4) = (1-x)/sqrt((1-x-2*x^2)*(1-x+2*x^2)). - Paul Barry, Oct 13 2009
Conjecture: n*a(n) + (4-5*n)*a(n-1) + 2*(4*n-7)*a(n-2) + 4*(3-n)*a(n-3) + 4*(2-n)*a(n-4) + 4*(n-4)*a(n-5) = 0. - R. J. Mathar, Nov 16 2011
a(n) ~ 3^(1/4) * (1 + sqrt(3))^(n + 1/2) / (2^(3/2) * sqrt(Pi*n)). - Vaclav Kotesovec, Jun 08 2019
Showing 1-4 of 4 results.

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