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.

A051286 Whitney number of level n of the lattice of the ideals of the fence of order 2n.

Original entry on oeis.org

1, 1, 2, 5, 11, 26, 63, 153, 376, 931, 2317, 5794, 14545, 36631, 92512, 234205, 594169, 1510192, 3844787, 9802895, 25027296, 63972861, 163701327, 419316330, 1075049011, 2758543201, 7083830648, 18204064403, 46812088751, 120452857976
Offset: 0

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Author

Keywords

Comments

A Chebyshev transform of the central trinomial numbers A002426: image of 1/sqrt(1-2x-3x^2) under the mapping that takes g(x) to (1/(1+x^2))*g(x/(1+x^2)). - Paul Barry, Jan 31 2005
a(n) has same parity as Fibonacci(n+1) = A000045(n+1); see A107597. - Paul D. Hanna, May 22 2005
This is the second kind of Whitney numbers, which count elements, not to be confused with the first kind, which sum Mobius functions. - Thomas Zaslavsky, May 07 2008
From Paul Barry, Mar 31 2010: (Start)
Apply the Riordan array (1/(1-x+x^2),x/(1-x+x^2)) to the aerated central binomial coefficients with g.f. 1/sqrt(1-4x^2).
Hankel transform is A174882. (End)
a(n) is the number of lattice paths in L[n]. The members of L[n] are lattice paths of weight n that start at (0,0), end on the horizontal axis and whose steps are of the following four kinds: an (1,0)-step h with weight 1, an (1,0)-step H with weight 2, a (1,1)-step U with weight 2, and a (1,-1)-step D with weight 1. The weight of a path is the sum of the weights of its steps. Example: a(3)=5 because we have hhh, hH, Hh, UD, and DU; a(4)=11 because we have hhhh, hhH, hHh, Hhh, HH, hUD, UhD, UDh, hDU, DhU, and DUh (see the Bona-Knopfmacher reference).
Apparently the number of peakless grand Motzkin paths of length n. - David Scambler, Jul 04 2013
A bijection between L[n] (as defined above) and peakless grand Motzkin paths of length n is now given in arXiv:2002.12874. - Sergi Elizalde, Jul 14 2021
a(n) is also the number of unimodal bargraphs with a centered maximum (i.e., whose column heights are weakly increasing in the left half and weakly decreasing in the right half) and semiperimeter n+1. - Sergi Elizalde, Jul 14 2021
Diagonal of the rational function 1 / ((1 - x)*(1 - y) - (x*y)^2). - Ilya Gutkovskiy, Apr 23 2025
a(n) is the number of rooted ordered trees with node weights summing to n, where the root has weight 0, non-root node weights are in {1,2}, and no nodes have the same weight as their parent node. - John Tyler Rascoe, Jun 10 2025

Examples

			a(3) = 5 because the ideals of size 3 of the fence F(6) = { x1 < x2 > x3 < x4 > x5 < x6 } are x1*x3*x5, x1*x2*x3, x3*x4*x5, x1*x5*x6, x3*x5*x6.
		

Crossrefs

Cf. main diagonal of A125250, column k=2 of A384747.
Cf. A051291, A051292, A078698, A107597, A185828 (log), A174882 (Hankel transf.).

Programs

  • Maple
    seq( sum('binomial(i-k,k)*binomial(i-k,k)', 'k'=0..floor(i/2)), i=0..30 ); # Detlef Pauly (dettodet(AT)yahoo.de), Nov 09 2001
    # second Maple program:
    a:= proc(n) option remember; `if`(n<4, [1$2, 2, 5][n+1],
         ((2*n-1)*a(n-1)+(n-1)*a(n-2)+(2*n-3)*a(n-3)-(n-2)*a(n-4))/n)
        end:
    seq(a(n), n=0..35);  # Alois P. Heinz, Aug 11 2016
  • Mathematica
    Table[Sum[Binomial[n-k,k]^2,{k,0,Floor[n/2]}],{n,0,40}] (* Emanuele Munarini, Mar 01 2011; corrected by Harvey P. Dale, Sep 12 2012 *)
    CoefficientList[Series[1/Sqrt[1-2*x-x^2-2*x^3+x^4], {x, 0, 20}], x] (* Vaclav Kotesovec, Jan 05 2013 *)
    a[n_] := HypergeometricPFQ[ {(1-n)/2, (1-n)/2, -n/2, -n/2}, {1, -n, -n}, 16]; Table[a[n], {n, 0, 29}] (* Jean-François Alcover, Feb 26 2013 *)
  • Maxima
    makelist(sum(binomial(n-k,k)^2,k,0,floor(n/2)),n,0,40);  /* Emanuele Munarini, Mar 01 2011 */
    
  • PARI
    a(n)=polcoeff(1/sqrt((1+x+x^2)*(1-3*x+x^2)+x*O(x^n)),n)
    
  • PARI
    a(n)=sum(k=0,n,binomial(n-k,k)^2) /* Paul D. Hanna */
    
  • PARI
    {a(n)=polcoeff( exp(sum(m=1,n, sum(k=0,m, binomial(2*m,2*k)*x^k) *x^m/m) +x*O(x^n)), n)}  /* Paul D. Hanna, Mar 18 2011 */
    
  • PARI
    {a(n)=local(A=1); A=sum(m=0, n, x^m*sum(k=0, m, binomial(m, k)^2*x^k) +x*O(x^n)); polcoeff(A, n)} \\ Paul D. Hanna, Sep 05 2014
    
  • PARI
    {a(n)=local(A=1+x); A=sum(m=0, n, x^m*sum(k=0, n, binomial(m+k, k)^2*x^k) * (1-x)^(2*m+1) +x*O(x^n)); polcoeff(A, n)} \\ Paul D. Hanna, Sep 05 2014
    
  • PARI
    {a(n)=local(A=1+x); A=sum(m=0, n\2, x^(2*m) * sum(k=0, n, binomial(m+k, k)^2*x^k) +x*O(x^n)); polcoeff(A, n)} \\ Paul D. Hanna, Sep 05 2014
    
  • PARI
    {a(n)=local(A=1+x); A=sum(m=0, n\2, x^(2*m) * sum(k=0, m, binomial(m, k)^2*x^k) / (1-x +x*O(x^n))^(2*m+1) ); polcoeff(A, n)} \\ Paul D. Hanna, Sep 05 2014
    
  • Python
    from sympy import binomial
    def a(n): return sum(binomial(n - k, k)**2 for k in range(n//2 + 1))
    print([a(n) for n in range(31)]) # Indranil Ghosh, Apr 18 2017

Formula

G.f.: 1/sqrt(1 - 2*x - x^2 - 2*x^3 + x^4).
a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*(-1)^k*A002426(n-2k). - Paul Barry, Jan 31 2005
From Paul D. Hanna, May 22 2005: (Start)
a(n) = Sum_{k=0..n} C(n-k, k)^2.
Limit_{n->oo} a(n+1)/a(n) = (sqrt(5)+3)/2.
G.f.: 1/sqrt((1+x+x^2)*(1-3*x+x^2)). (End)
a(n) = Sum_{k=0..n} A049310(n, k)^2. - Philippe Deléham, Nov 21 2005
a(n) = Sum_{k=0..n} (C(k,k/2)*(1+(-1)^k)/2) * Sum_{j=0..n} (-1)^((n-j)/2)*C((n+j)/2,j)*((1+(-1)^(n-j))/2)*C(j,k). - Paul Barry, Mar 31 2010
G.f.: exp( Sum_{n>=1} (x^n/n)*Sum_{k=0..n} C(2n,2k)*x^k ). - Paul D. Hanna, Mar 18 2011
Logarithmic derivative equals A185828. - Paul D. Hanna, Mar 18 2011
D-finite with recurrence: n*a(n) - (2*n-1)*a(n-1) - (n-1)*a(n-2) - (2*n-3)*a(n-3) + (n-2)*a(n-4) = 0. - R. J. Mathar, Dec 17 2011
The g.f. A(x) satisfies the differential equation (1-2*x-x^2-2*x^3+x^4)*A'(x) = (1+x+3*x^2-2*x^3)*A(x), from which the recurrence conjectured by Mathar follows. - Emanuele Munarini, Dec 18 2017
a(n) ~ phi^(2*n + 2) / (2 * 5^(1/4) * sqrt(Pi*n)), where phi = A001622 = (1 + sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Jan 05 2013, simplified Dec 18 2017
From Paul D. Hanna, Sep 05 2014: (Start)
G.f.: Sum_{n>=0} x^n * Sum_{k=0..n} C(n,k)^2 * x^k.
G.f.: Sum_{n>=0} x^n *[Sum_{k>=0} C(n+k,k)^2 * x^k] * (1-x)^(2*n+1).
G.f.: Sum_{n>=0} x^(2*n) * [Sum_{k>=0} C(n+k,k)^2 * x^k].
G.f.: Sum_{n>=0} x^(2*n) * [Sum_{k=0..n} C(n,k)^2 * x^k] /(1-x)^(2n+1).
(End)

A098181 Two consecutive odd numbers separated by multiples of four, repeated twice, between them, written in increasing order.

Original entry on oeis.org

1, 3, 4, 4, 5, 7, 8, 8, 9, 11, 12, 12, 13, 15, 16, 16, 17, 19, 20, 20, 21, 23, 24, 24, 25, 27, 28, 28, 29, 31, 32, 32, 33, 35, 36, 36, 37, 39, 40, 40, 41, 43, 44, 44, 45, 47, 48, 48, 49, 51, 52, 52, 53, 55, 56, 56, 57, 59, 60, 60, 61, 63, 64, 64, 65, 67, 68, 68, 69, 71, 72, 72
Offset: 0

Views

Author

Paul Barry, Aug 30 2004

Keywords

Comments

Essentially partial sums of A007877.
a(n) is the number of odd coefficients of the q-binomial coefficient [n+2 choose 2]. (Easy to prove.) - Richard Stanley, Oct 12 2016

Examples

			G.f. = 1 + 3*x + 4*x^2 + 4*x^3 + 5*x^4 + 7*x^5 + 8*x^6 + 8*x^7 + 9*x^8 + ...
		

Crossrefs

Cf. A098180.

Programs

  • GAP
    a:=[1,3,4,4];; for n in [5..80] do a[n]:=2*a[n-1]-2*a[n-2]+2*a[n-3] -a[n-4]; od; a; # G. C. Greubel, May 22 2019
  • Magma
    R:=PowerSeriesRing(Integers(), 80); Coefficients(R!( (1+x)/((1-x)^2*(1+x^2)) )); // G. C. Greubel, May 22 2019
    
  • Maple
    A:=seq((2*n+3 - cos(Pi*n/2) + sin(Pi*n/2))/2, n=0..50); \\ Bernard Schott, Jun 07 2019
  • Mathematica
    Table[Floor[Binomial[n+3, 2]/2] -Floor[Binomial[n+1, 2]/2], {n, 0, 80}] (* or *) CoefficientList[Series[(1+x)/((1-x)^2*(1+x^2)), {x, 0, 80}], x] (* Michael De Vlieger, Oct 12 2016 *)
  • PARI
    {a(n) = n\4*4 + [1, 3, 4, 4][n%4+1]}; /* Michael Somos, Sep 11 2014 */
    
  • Sage
    ((1+x)/((1-x)^2*(1+x^2))).series(x, 80).coefficients(x, sparse=False) # G. C. Greubel, May 22 2019
    

Formula

G.f.: (1+x)/((1-x)^2*(1+x^2)).
a(n) = ( (2*n+3) - cos(Pi*n/2) + sin(Pi*n/2) )/2.
a(n) = 2*a(n-1) - 2*a(n-2) + 2*a(n-3) - a(n-4).
a(n) = floor(C(n+3, 2)/2)-floor(C(n+1, 2)/2). - Paul Barry, Jan 01 2005
a(4*n) = 4*n+1, a(4*n+1) = 4*n+3, a(4*n+2) = a(4*n+3) = 4*n+4. - Philippe Deléham, Apr 06 2007
Euler transform of length 4 sequence [ 3, -2, 0, 1]. - Michael Somos, Sep 11 2014
a(-3-n) = -a(n) for all n in Z. - Michael Somos, Sep 11 2014
a(n) = log_2(|A174882(n+2)|). [Barry] - R. J. Mathar, Aug 18 2017
a(n) = (2*n+3 - (-1)^ceiling(n/2))/2. - Wesley Ivan Hurt, Sep 29 2017

Extensions

Name edited by G. C. Greubel, Jun 06 2019

A181407 a(n) = (n-4)*(n-3)*2^(n-2).

Original entry on oeis.org

3, 3, 2, 0, 0, 16, 96, 384, 1280, 3840, 10752, 28672, 73728, 184320, 450560, 1081344, 2555904, 5963776, 13762560, 31457280, 71303168, 160432128, 358612992, 796917760, 1761607680, 3875536896, 8489271296, 18522046464, 40265318400, 87241523200, 188441690112
Offset: 0

Views

Author

Paul Curtz, Jan 28 2011

Keywords

Comments

Binomial transform of (3, 0, -1, followed by A005563).
The sequence and its successive differences are:
3, 3, 2, 0, 0, 16, 96, 384, a(n),
0, -1, -2, 0, 16, 80, 288, 896, A178987,
-1, -1, 2, 16, 64, 208, 608, 2688, -A127276,
0, 3, 14, 48, 144, 400, 1056, 2688, A176027,
3, 11, 34, 96, 256, 656, 1632, 3968, A084266(n+1)
8, 23, 62, 160, 400, 976, 2336, 5504,
15, 39, 98, 240, 576, 1360, 3168, 7296.
Division of the k-th column by abs(A174882(k)) gives
3, 3, 1, 0, 0, 1, 3, 3, 5, 15, 21, 14, A064038(n-3),
0, -1, -1, 0, 1, 5, 9, 7, 10, 27, 35, 22, A160050(n-3),
-1, -1, 1, 2, 4, 13, 19, 13, 17, 43, 53, 32, A176126,
0, 3, 7, 6, 9, 25, 33, 21, 26, 63, 75, 44, A178242,
3, 11, 17, 12, 16, 41, 51, 31, 37, 87, 101, 58,
8 23, 31, 20, 25, 61, 73, 43, 50, 115, 131, 74,
15, 39, 49, 30, 36, 85, 99, 57, 65, 147, 165, 92.

Crossrefs

Programs

  • GAP
    List([0..40], n-> (n-4)*(n-3)*2^(n-2)); # G. C. Greubel, Feb 21 2019
  • Magma
    [(n-4)*(n-3)*2^(n-2): n in [0..40] ]; // Vincenzo Librandi, Feb 01 2011
    
  • Mathematica
    Table[(n-4)*(n-3)*2^(n-2), {n,0,40}] (* G. C. Greubel, Feb 21 2019 *)
  • PARI
    vector(40, n, n--; (n-4)*(n-3)*2^(n-2)) \\ G. C. Greubel, Feb 21 2019
    
  • Sage
    [(n-4)*(n-3)*2^(n-2) for n in (0..40)] # G. C. Greubel, Feb 21 2019
    

Formula

a(n) = 16*A001788(n-4).
a(n+1) - a(n) = A178987(n).
G.f.: (3 - 15*x + 20*x^2) / (1-2*x)^3. - R. J. Mathar, Jan 30 2011
E.g.f.: (x^2 - 3*x + 3)*exp(2*x). - G. C. Greubel, Feb 21 2019

A177002 Period 4: repeat [1, 2, 4, 2].

Original entry on oeis.org

1, 2, 4, 2, 1, 2, 4, 2, 1, 2, 4, 2, 1, 2, 4, 2, 1, 2, 4, 2, 1, 2, 4, 2, 1, 2, 4, 2, 1, 2, 4, 2, 1, 2, 4, 2, 1, 2, 4, 2, 1, 2, 4, 2, 1, 2, 4, 2, 1, 2, 4, 2, 1, 2, 4, 2, 1, 2, 4, 2, 1, 2, 4, 2, 1, 2, 4, 2, 1, 2, 4, 2, 1, 2, 4, 2, 1, 2, 4, 2, 1, 2, 4, 2, 1, 2
Offset: 0

Views

Author

Paul Curtz, Dec 08 2010

Keywords

Comments

Also the decimal expansion of 138/1111 and the continued fractions of (5+3*sqrt(10))/10 or (6*sqrt(10)-10)/13. - R. J. Mathar, Dec 13 2010

Crossrefs

Cf. A174882.

Programs

Formula

a(n) = | A174882(n+1) / A174882(n) |.
G.f.: (1+2*x+4*x^2+2*x^3)/((1-x)*(1+x)*(x^2+1)). - R. J. Mathar, Dec 13 2010
a(n) = 2+(1+(-1)^n)*(1-3*I^n)/4. - Bruno Berselli, Mar 15 2011
a(n) = a(n-1) * a(n-3) / a(n-2) for n>2. - Bruno Berselli, Feb 04 2013
From Wesley Ivan Hurt, Jul 09 2016: (Start)
a(n) = a(n-4) for n>3.
a(n) = (9 + cos(n*Pi) - 6*cos(n*Pi/2))/4. (End)

A178628 A (1,1) Somos-4 sequence associated to the elliptic curve E: y^2 - x*y - y = x^3 + x^2 + x.

Original entry on oeis.org

1, 1, -1, -4, -3, 19, 67, -40, -1243, -4299, 25627, 334324, 627929, -29742841, -372632409, 1946165680, 128948361769, 1488182579081, -52394610324649, -2333568937567764, -5642424912729707, 3857844273728205019
Offset: 1

Views

Author

Paul Barry, May 31 2010

Keywords

Comments

a(n) is (-1)^C(n,2) times the Hankel transform of the sequence with g.f.
1/(1-x^2/(1-x^2/(1-4x^2/(1+(3/16)x^2/(1-(76/9)x^2/(1-(201/361)x^2/(1-... where
1,4,-3/16,76/9,201/361,... are the x-coordinates of the multiples of z=(0,0)
on E:y^2-xy-y=x^3+x^2+x.

Crossrefs

Programs

  • Magma
    I:=[1,1,-1,-4]; [n le 4 select I[n] else (Self(n-1)*Self(n-3) + Self(n-2)^2)/Self(n-4): n in [1..30]]; // G. C. Greubel, Sep 18 2018
    
  • Mathematica
    RecurrenceTable[{a[n] == (a[n-1]*a[n-3] +a[n-2]^2)/a[n-4], a[1] == 1, a[2] == 1, a[3] == -1, a[4] == -4}, a, {n,1,30}] (* G. C. Greubel, Sep 18 2018 *)
  • PARI
    a(n)=local(E,z);E=ellinit([ -1,1,-1,1,0]);z=ellpointtoz(E,[0,0]); round(ellsigma(E,n*z)/ellsigma(E,z)^(n^2))
    
  • PARI
    m=30; v=concat([1,1,-1,-4], vector(m-4)); for(n=5, m, v[n] = ( v[n-1]*v[n-3] +v[n-2]^2)/v[n-4]); v \\ G. C. Greubel, Sep 18 2018
    
  • PARI
    {a(n) = subst(elldivpol(ellinit([-1, 1, -1, 1, 0]), n), x ,0)}; /* Michael Somos, Jul 05 2024 */
    
  • SageMath
    @CachedFunction
    def a(n): # a = A178628
        if n<5: return (0,1,1,-1,-4)[n]
        else: return (a(n-1)*a(n-3) + a(n-2)^2)/a(n-4)
    [a(n) for n in range(1,41)] # G. C. Greubel, Jul 05 2024

Formula

a(n) = (a(n-1)*a(n-3) + a(n-2)^2)/a(n-4), n>4.
a(n) = -a(-n). a(n) = (-a(n-1)*a(n-4) +4*a(n-2)*a(n-3))/a(n-5) for all n in Z except n=5. - Michael Somos, Jul 05 2024

Extensions

Offset changed to 0. - Michael Somos, Jul 05 2024
Showing 1-5 of 5 results.