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.

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A167868 a(n) = 3^n * Sum_{k=0..n} binomial(2*k,k)^3 / 3^k.

Original entry on oeis.org

1, 11, 249, 8747, 369241, 17110731, 840221217, 42944901219, 2260581606657, 121714776747971, 6671749658197129, 371062413164972955, 20887218937200347281, 1187720356043817041843, 68124474120573747125529
Offset: 0

Views

Author

Alexander Adamchuk, Nov 14 2009

Keywords

Comments

The expression a(n) = B^n*Sum_{k=0..n} binomial(2*k,k)/B^k gives A006134 for B=1, A082590 (B=2), A132310 (B=3), A002457 (B=4), A144635 (B=5), A167713 (B=16).
The expression a(n) = B^n*Sum_{k=0..n} binomial(2*k,k)^3/B^k gives A079727 for B=1, A167867 (B=2), A167868 (B=3), A167869 (B=4), A167870 (B=16), A167871 (B=64).

Crossrefs

Programs

  • Mathematica
    Table[3^n Sum[Binomial[2k,k]^3/3^k,{k,0,n}],{n,0,20}] (* Vincenzo Librandi, Mar 26 2012 *)

Formula

a(n) = 3^n * Sum_{k=0..n} binomial(2*k,k)^3 / 3^k.
Recurrence: n^3*a(n) = (67*n^3 - 96*n^2 + 48*n - 8)*a(n-1) - 24*(2*n-1)^3*a(n-2). - Vaclav Kotesovec, Aug 13 2013
a(n) ~ 2^(6*n+6)/(61*(Pi*n)^(3/2)). - Vaclav Kotesovec, Aug 13 2013

Extensions

More terms from Sean A. Irvine, Apr 27 2010

A167869 a(n) = 4^n * Sum_{k=0..n} binomial(2*k,k)^3 / 4^k.

Original entry on oeis.org

1, 12, 264, 9056, 379224, 17519904, 858968640, 43860112128, 2307187351512, 124161781334048, 6803252453289408, 378260174003539200, 21287072393719585216, 1210206988807094340864, 69402141007670673363456
Offset: 0

Views

Author

Alexander Adamchuk, Nov 14 2009

Keywords

Comments

The expression a(n) = B^n*Sum_{k=0..n} binomial(2*k,k)/B^k gives A006134 for B=1, A082590 (B=2), A132310 (B=3), A002457 (B=4), A144635 (B=5), A167713 (B=16).
The expression a(n) = B^n*Sum_{k=0..n} binomial(2*k,k)^3/B^k gives A079727 for B=1, A167867 (B=2), A167868 (B=3), A167869 (B=4), A167870 (B=16), A167871 (B=64).

Crossrefs

Programs

  • Mathematica
    Table[4^n Sum[Binomial[2k,k]^3/4^k,{k,0,n}],{n,0,30}] (* Vincenzo Librandi, Mar 26 2012 *)

Formula

a(n) = 4^n * Sum_{k=0..n} binomial(2*k,k)^3 / 4^k.
Recurrence: n^3*a(n) = 4*(17*n^3 - 24*n^2 + 12*n - 2)*a(n-1) - 32*(2*n-1)^3*a(n-2). - Vaclav Kotesovec, Aug 13 2013
a(n) ~ 2^(6*n+4)/(15*(Pi*n)^(3/2)). - Vaclav Kotesovec, Aug 13 2013

Extensions

More terms from Sean A. Irvine, Apr 27 2010

A167870 a(n) = 16^n * Sum_{k=0..n} binomial(2*k,k)^3 / 16^k.

Original entry on oeis.org

1, 24, 600, 17600, 624600, 25996608, 1204834752, 59701593600, 3086972400600, 164324590337600, 8935798773354816, 494019944564058624, 27678350810730366400, 1567912312203901862400, 89647910047704725798400
Offset: 0

Views

Author

Alexander Adamchuk, Nov 14 2009

Keywords

Comments

The expression a(n) = B^n*Sum_{k=0..n} binomial(2*k,k)/B^k gives A006134 for B=1, A082590 (B=2), A132310 (B=3), A002457 (B=4), A144635 (B=5), A167713 (B=16).
The expression a(n) = B^n*Sum_{k=0..n} binomial(2*k,k)^3/B^k gives A079727 for B=1, A167867 (B=2), A167868 (B=3), A167869 (B=4), A167870 (B=16), A167871 (B=64).

Crossrefs

Programs

  • Mathematica
    Table[16^n Sum[Binomial[2k,k]^3/16^k,{k,0,n}],{n,0,20}] (* Harvey P. Dale, Jan 21 2012 *)

Formula

a(n) = 16^n * Sum_{k=0..n} binomial(2*k,k)^3 / 16^k.
Recurrence: n^3*a(n) = 8*(10*n^3 - 12*n^2 + 6*n - 1)*a(n-1) - 128*(2*n-1)^3*a(n-2). - Vaclav Kotesovec, Aug 13 2013
a(n) ~ 2^(6*n+2)/(3*(Pi*n)^(3/2)). - Vaclav Kotesovec, Aug 13 2013

Extensions

More terms from Sean A. Irvine, Apr 27 2010

A167871 a(n) = 64^n * Sum_{k=0..n} binomial(2*k,k)^3 / 64^k.

Original entry on oeis.org

1, 72, 4824, 316736, 20614104, 1335305664, 86248451520, 5560325134848, 357992555533272, 23026456586057408, 1479999826835627328, 95071036081670530560, 6104320340924619384256, 391801560518407856592384
Offset: 0

Views

Author

Alexander Adamchuk, Nov 14 2009

Keywords

Comments

The expression a(n) = B^n*Sum_{k=0..n} binomial(2*k,k)/B^k gives A006134 for B=1, A082590 (B=2), A132310 (B=3), A002457 (B=4), A144635 (B=5), A167713 (B=16).
The expression a(n) = B^n*Sum_{k=0..n} binomial(2*k,k)^3/B^k gives A079727 for B=1, A167867 (B=2), A167868 (B=3), A167869 (B=4), A167870 (B=16), A167871 (B=64).
p^2 divides all a(n) from n = (p-1)/2 to n = p-1 for prime p of the form p = 4k+3, p = {3,7,11,19,23,31,43,47,59,...} = A002145.
p^2 divides all a(n) from n = (2p-1 - (p-1)/2) to n = 2p-1 for prime p of the form p = 4k+3.
p^2 divides all a(n) from n = (3p-1 - (p-1)/2) to n = 3p-1 for prime p of the form p = 4k+3.
p^2 divides all a(n) from n = (p^2-1)/2 to n = p^2-1 for prime p of the form p = 4k+3.

References

  • W. Feller, An Introduction to Probability Theory and Its Applications, Vol. 1, 3rd ed., Wiley, 1968, p. 361.

Crossrefs

Programs

  • Mathematica
    Table[64^n Sum[Binomial[2k,k]^3/64^k,{k,0,n}],{n,0,20}] (* Vincenzo Librandi, Mar 26 2012 *)

Formula

a(n) = 64^n * Sum_{k=0..n} binomial(2*k,k)^3 / 64^k.
Recurrence: n^3*a(n) = 8*(4*n-1)*(4*n^2 - 2*n + 1)*a(n-1) - 512*(2*n-1)^3 *a(n-2). - Vaclav Kotesovec, Aug 14 2013
a(n) ~ 64^n*(Pi/GAMMA(3/4)^4 - 2/(Pi^(3/2)*sqrt(n))). - Vaclav Kotesovec, Aug 14 2013

Extensions

More terms from Sean A. Irvine, Apr 25 2010

A226302 a(n) = P_n(-1), where P_n(x) is a certain polynomial arising in the enumeration of tatami mat coverings.

Original entry on oeis.org

1, -1, 2, -4, 6, -14, 20, -48, 70, -166, 252, -584, 924, -2092, 3432, -7616, 12870, -28102, 48620, -104824, 184756, -394404, 705432, -1494240, 2704156, -5692636, 10400600, -21785872, 40116600, -83688344, 155117520, -322494208, 601080390, -1246068806, 2333606220
Offset: 2

Views

Author

N. J. A. Sloane, Jun 06 2013

Keywords

Comments

See Erickson-Ruskey for precise definition. The polynomials P_n(x) are described as "mysterious".
Bisections give A082590 and A000984.

Crossrefs

Programs

  • Maple
    A226302 := proc(n)
        if type(n,even) then
            A000984(n/2-1) ;
        else
            -A082590((n-3)/2) ;
        end if;
    end proc: # R. J. Mathar, Nov 06 2013
  • Mathematica
    Rest[Rest[CoefficientList[Series[x^2*(1/Sqrt[1-4*x^2] - x/((1-2*x^2)*Sqrt[1-4*x^2])), {x, 0, 30}], x]]] (* Vaclav Kotesovec, Jun 14 2015, after Sergei N. Gladkovskii *)
    max = 30; Clear[g]; g[max + 2] = 1; g[k_] := g[k] = 1 - 2*x^2 - 2*x*(1 - 2*x^2)^2*(2*k+1)/( 2*x*(1 - 2*x^2)*(2*k+1) - (k+1)/(1 - x/g[k+1] )); gf = 1 - x/g[0]; CoefficientList[Series[gf, {x, 0, max}], x] (* Vaclav Kotesovec, Jun 14 2015, after Sergei N. Gladkovskii *)
    a = DifferenceRoot[Function[{a, n}, {(-(6*n^2) + 2*n + 4)*a[n+2] + (n^2 + n - 2)*a[n+4] + 8*(n - 1)*n*a[n] - 4*n*a[n+1] + 2*n*a[n+3] == 0, a[2] == 1, a[3] == -1, a[4] == 2, a[5] == -4}]]; Table[a[n], {n, 2, 36}] (* Jean-François Alcover, Feb 23 2019 *)
  • PARI
    Vec(x^2*(1/sqrt(1-4*x^2) - x/((1-2*x^2)*sqrt(1-4*x^2))) + O(x^50)) \\ G. C. Greubel, Jan 29 2017

Formula

Conjecture: (-n+2)*a(n) +(-n+2)*a(n-1) +2*(3*n-11)*a(n-2) +2*(3*n-14)*a(n-3) +4*(-2*n+9)*a(n-4) +8*(-n+6)*a(n-5)=0. - R. J. Mathar, Nov 06 2013
G.f. (for offset 0): 1/sqrt(1-4*x^2) - x/((1-2*x^2)*sqrt(1-4*x^2)) = 1 - x/W(0), where W(k)= 1 - 2*x^2 - 2*x*(1 - 2*x^2)^2*(2*k+1)/( 2*x*(1 - 2*x^2)*(2*k+1) - (k+1)/(1 - x/W(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jun 13 2015
Recurrence (for n>5): (n-5)*(n-2)*a(n) = -2*(n-4)*a(n-1) + 2*(n-5)*(3*n-10)*a(n-2) + 4*(n-4)*a(n-3) - 8*(n-5)*(n-4)*a(n-4). - Vaclav Kotesovec, Jun 14 2015
a(n) ~ (-1)^n * 2^(n-3/2) / sqrt(Pi*n). - Vaclav Kotesovec, Jun 14 2015

A167860 Primes p dividing every A167859(m) from m=(p-1)/2 to m=(p-1).

Original entry on oeis.org

7, 47, 191, 383, 439, 1151, 1399, 2351, 2879, 3119, 3511, 3559, 4127, 5087, 5431, 6911, 8887, 9127, 9791, 9887, 12391, 13151, 14407, 15551, 16607, 19543, 20399, 21031, 21319, 21839, 23039, 25391, 26399, 28087, 28463, 28711, 29287, 33223, 39551, 43103, 44879, 46271
Offset: 1

Views

Author

Alexander Adamchuk, Nov 13 2009

Keywords

Comments

Apparently A167860 is a subset of primes of the form 8*k + 7 (A007522).
Every A167859(m) from m=(p-1)/2 to m=(p-1) is divisible by prime p belonging to A167860.
7^3 divides A167859(13) and 7^2 divides A167859(10)-A167859(13).
Every A167859(m) from m=(kp-1 - (p-1)/2) to m=(kp-1) is divisible by prime p from A167860.
Every A167859(m) from m=((p^2-1)/2) to m=(p^2-1) is divisible by prime p from A167860. For p=7 every A167859(m) from m=((p^3-1)/2) to m=(p^3-1) and from m=((p^4-1)/2) to m(p^4-1)is divisible by p^2.

Crossrefs

Programs

  • Maple
    A167859 := proc(n)
        option remember;
        if n <= 1 then
            add( (binomial(2*k, k)/2^k)^2, k=0..n) ;
            4^n*% ;
        else
            4*(5*n^2 - 4*n + 1)*procname(n-1) - 16*(2*n - 1)^2*procname(n-2) ;
            %/n^2 ;
        end if;
    end proc:
    isA167860 := proc(p)
        local m ;
        for m from (p-1)/2 to p-1 do
            if modp(A167859(m),p) > 0 then
                return false;
            end if;
        end do:
        true ;
    end proc:
    A167860 := proc(n)
        option remember ;
        if n = 0 then
            2;
        else
            p := nextprime(procname(n-1)) ;
            while not isA167860(p) do
                p := nextprime(p) ;
            end do ;
            return p;
        end if;
    end proc:
    seq(A167860(n),n=1..10) ; # R. J. Mathar, Jan 22 2025
  • PARI
    is(p) = if(isprime(p)&&p%2, my(m=Mod(1, p), s=m); for(k=1, p\2, s+=(m*=(2*k-1)/k)^2); !s, 0); \\ Jinyuan Wang, Jul 24 2022

Extensions

More terms from Jinyuan Wang, Jul 24 2022

A233481 Number of singletons (strong fixed points) in pair-partitions.

Original entry on oeis.org

0, 1, 4, 21, 144, 1245, 13140, 164745, 2399040, 39834585, 742940100, 15374360925, 349484058000, 8654336615925, 231842662751700, 6679510641428625, 205916703920928000, 6762863294018456625, 235719416966063530500, 8689887736412502745125
Offset: 0

Views

Author

Wojciech Bozejko, Dec 11 2013

Keywords

Comments

For h(V) = number of singletons (non-crossing chords) in the pair-partition of 2n-elementary set P_2(2n), let T(2n) = sum_{V in P_2(2n)} h(V).
Elements of the sequence a(n) = T(2n).
a(n) is the number of linear chord diagrams on 2n vertices with one marked chord such that none of the remaining n-1 chords cross the marked chord, see [Young]. - Donovan Young, Aug 11 2020

Crossrefs

A081054 counts pair-partitions of a fixed size without singletons, i.e., linear chord diagrams with 2n nodes and n arcs in which each arc crosses another arc.

Programs

  • Maple
    a := n -> 2*n*GAMMA(1/2+n)*hypergeom([1/2,-n+1],[3/2],-1)/sqrt(Pi);
    seq(simplify(a(n)), n = 0..19); # Peter Luschny, Dec 16 2013
    # Alternative:
    u := (z/2)^2: egf := 2*u*exp(u)*hypergeom([1/2], [3/2], u): ser := series(egf, z, 40): seq((2*n)!*coeff(ser, z, 2*n), n = 0..19); # Peter Luschny, Mar 14 2023
  • Mathematica
    Table[Sum[(2 k - 1)!! (2 n - 2 k - 1)!!, {k, 0, n - 1}], {n,0,30}] (* T. D. Noe, Dec 13 2013 *)
  • Sage
    def A233481():
        a, b, n = 0, 1, 1
        while True:
            yield a
            n += 1
            a, b = b, n*((3*n-4)*b/(n-1)-(2*n-3)*a)
    a = A233481(); [next(a) for i in range(17)]  # Peter Luschny, Dec 14 2013

Formula

a(n) = T_{2n} = n*sum_{k=0..(n-1)} (2k-1)!!*(2n-2k-1)!!, where (2n-1)!! = 1*3*5*...*(2n-1).
From Peter Luschny, Dec 16 2013: (Start)
E.g.f.: x/((1-x)*sqrt(1-2*x)).
a(n) = 2*n*Gamma(1/2+n)*2_F_1([1/2,-n+1],[3/2],-1)/sqrt(Pi), where 2_F_1 is the hypergeometric function.
a(n) = n*((3*n-4)*a(n-1)/(n-1)-(2*n-3)*a(n-2)) for n>1.
a(n) = n*A034430(n-1) for n>=1.
a(n+1)/(n+1)! = JacobiP(n, 1/2, -n-1, 3).
2^n*a(n+1)/(n+1)! = A082590(n).
2^n*a(n+1)/(n+1) = A076729(n). (End)
a(n) ~ 2^(n+1/2) * n^n / exp(n). - Vaclav Kotesovec, Dec 20 2013
a(n) = (2*n)! * [z^(2*n)] 2*u*exp(u)*hypergeom([1/2], [3/2], u), where u = (z/2)^2. - Peter Luschny, Mar 14 2023

A112336 A number triangle related to the central binomial coefficients.

Original entry on oeis.org

1, 2, 2, 6, 4, 4, 20, 12, 8, 8, 70, 40, 24, 16, 16, 252, 140, 80, 48, 32, 32, 924, 504, 280, 160, 96, 64, 64, 3432, 1848, 1008, 560, 320, 192, 128, 128, 12870, 6864, 3696, 2016, 1120, 640, 384, 256, 256, 48620, 25740, 13728, 7392, 4032, 2240, 1280, 768, 512, 512
Offset: 0

Views

Author

Paul Barry, Sep 04 2005

Keywords

Comments

Row sums are A082590. Diagonal sums are A106189.

Examples

			Triangle begins
1;
2,2;
6,4,4;
20,12,8,8;
70,40,24,16,16;
		

Formula

Number triangle T(n, k)=2^k*binomial(2(n-k), n-k); Column k has g.f. (2x)^k/sqrt(1-4x).

A187887 Riordan matrix (1/((1-x)*sqrt(1-4*x)),x/(1-x)).

Original entry on oeis.org

1, 3, 1, 9, 4, 1, 29, 13, 5, 1, 99, 42, 18, 6, 1, 351, 141, 60, 24, 7, 1, 1275, 492, 201, 84, 31, 8, 1, 4707, 1767, 693, 285, 115, 39, 9, 1, 17577, 6474, 2460, 978, 400, 154, 48, 10, 1, 66197, 24051, 8934, 3438, 1378, 554, 202, 58, 11, 1, 250953, 90248, 32985, 12372, 4816, 1932, 756, 260, 69, 12, 1
Offset: 0

Views

Author

Emanuele Munarini, Mar 15 2011

Keywords

Comments

Row sums are A082590.
First column is A006134.

Examples

			Triangle begins:
1,
3,1,
9,4,1,
29,13,5,1,
99,42,18,6,1,
351,141,60,24,7,1,
1275,492,201,84,31,8,1,
		

Programs

  • Mathematica
    Select[Flatten[Table[Sum[Binomial[n-i,k]Binomial[2i,i],{i,0,n}],{n,0,10},{k,0,10}]],#!=0&] (* Harvey P. Dale, Jul 05 2012 *)
  • Maxima
    create_list(sum(binomial(n-i,k)*binomial(2*i,i), i,0,n),n,0,8,k,0,n);

Formula

a(n,k) = [x^n] 1/((1-x)*sqrt(1-4*x))*(x/(1-x))^k.
Recurrence: a(n+1,k+1) = a(n,k+1) + a(n,k).
a(n,k) = sum(binomial(n-i,k)*binomial(2*i,i),i=0..n).
G.f.: 1/(sqrt(1-4*x)*(1-x-x*y)).

Extensions

Mathematica program corrected by Harvey P. Dale, Jul 05 2012
Comment added and comment corrected by Michel Marcus, Jun 23 2013

A228197 Number of n-edge ordered trees with bicolored boundary edges.

Original entry on oeis.org

1, 2, 8, 36, 160, 692, 2928, 12200, 50304, 205940, 838928, 3405496, 13788736, 55723592, 224863712, 906365136, 3649978880, 14687731572, 59067989072, 237424661016, 953914608320, 3831159414552, 15381896102432, 61739966366256, 247750559632640, 993955865320392, 3986890331450528
Offset: 0

Views

Author

Louis Shapiro, Aug 20 2013

Keywords

Examples

			When n=3 edges there are A000108(3)= 5 ordered trees. Four of these consist of three boundary edges each contributing 2^3 trees to the count. The last, UDUDUD, has two boundary edges giving the last 2^2 trees for a total of 36.
		

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(1-2*x-2*x*Sqrt[1-4*x])/((4*x-1)*(2*x-1)), {x, 0, 20}], x] (* Vaclav Kotesovec, Aug 23 2013 *)
    Table[2^(2n)-2^n*JacobiP[n-1,1/2,-n,3],{n,0,20}] (* Benedict W. J. Irwin, Sep 16 2016 *)
  • PARI
    x = 'x + O('x^66);
    C = serreverse( x/( 1/(1-x) ) ) / x; \\ Catalan A000108
    B = (1-4*x)^(-1/2); \\ central binomial coefficients
    gf = (1+4*x^2*B^2*C)/(1-2*x);
    Vec(gf) \\ Joerg Arndt, Aug 21 2013

Formula

G.f.: (1+4*x^2*B^2*C)/(1-2*x), C is the Catalan g.f. (see A000108) and B =(1-4*x)^(-1/2) is the g.f. for the central binomial coefficients (A000984).
a(n) ~ 4^n * (1-1/(sqrt(Pi*n))). - Vaclav Kotesovec, Aug 23 2013
Conjecture: (-n+1)*a(n) +2*(5*n-8)*a(n-1) +4*(-8*n+17)*a(n-2) +16*(2*n-5)*a(n-3)=0. - R. J. Mathar, Aug 25 2013
a(n) = 2^(2*n)-2^n*JacobiP(n-1,1/2,-n,3) = 2^(2*n)-2*A082590(n-1), which satisfies the above conjecture. - Benedict W. J. Irwin, Sep 16 2016
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