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.

Previous Showing 11-19 of 19 results.

A364398 G.f. satisfies A(x) = 1 + x/A(x)^3*(1 + 1/A(x)).

Original entry on oeis.org

1, 2, -14, 162, -2270, 35234, -582958, 10076354, -179802046, 3287029698, -61246957902, 1158889656930, -22207636788894, 430106644358242, -8405699952109166, 165557885912786818, -3282954949273886590, 65487784219460233602, -1313225110482709157518
Offset: 0

Views

Author

Seiichi Manyama, Jul 22 2023

Keywords

Crossrefs

Programs

  • Maple
    A364398 := proc(n)
        if n = 0 then
            1;
        else
            (-1)^(n-1)*add( binomial(n,k) * binomial(4*n+k-2,n-1),k=0..n)/n ;
        end if;
    end proc:
    seq(A364398(n),n=0..70); # R. J. Mathar, Jul 25 2023
  • Mathematica
    nmax = 18; A[] = 1; Do[A[x] = 1+x/A[x]^3*(1+1/A[x]) + O[x]^(nmax+1) // Normal, {nmax}]; CoefficientList[A[x], x] (* Jean-François Alcover, Oct 21 2023 *)
  • PARI
    a(n) = if(n==0, 1, (-1)^(n-1)*sum(k=0, n, binomial(n, k)*binomial(4*n+k-2, n-1))/n);

Formula

a(n) = (-1)^(n-1) * (1/n) * Sum_{k=0..n} binomial(n,k) * binomial(4*n+k-2,n-1) for n > 0.
D-finite with recurrence 2*n*(462919*n -714364)*(4*n-3) *(2*n-1)*(4*n-1)*a(n) +(625365036*n^5 -2723245780*n^4 +4202103460*n^3 -2471353250*n^2 +81675089*n +289227120)*a(n-1) +(-484851248*n^5 +5501638270*n^4 -25122933600*n^3 +57439557800*n^2 -65490996232*n +29691239955)*a(n-2) +(2*n-5)*(652184*n -1103659)*(4*n-13) *(n-3)*(4*n-11)*a(n-3)=0. - R. J. Mathar, Jul 25 2023
a(n) ~ c*(-1)^(n-1)*256^n*27^(-n)*2F1([1-n, 4*n], [3*n], -1)*n^(-3/2), with c = sqrt(3/(32*Pi)). - Stefano Spezia, Oct 21 2023

A365847 Expansion of (1/x) * Series_Reversion( x*(1-x)^4/(1+x)^4 ).

Original entry on oeis.org

1, 8, 96, 1368, 21440, 356968, 6197408, 110947768, 2033381760, 37963483592, 719495148768, 13806129179928, 267693334199616, 5236670783633960, 103227182363423008, 2048451544990578552, 40888361539777714944, 820400146864231266184
Offset: 0

Views

Author

Seiichi Manyama, Sep 20 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n, binomial(4*n+k+3, k)*binomial(4*(n+1), n-k))/(n+1);

Formula

a(n) = (1/(n+1)) * Sum_{k=0..n} binomial(4*n+k+3,k) * binomial(4*(n+1),n-k).
Conjecture: g.f.: B^4, where B is the g.f. of A260332.

A260331 Labelings of n diamond-shaped posets with 4 vertices per diamond where the labels follow the poset relations.

Original entry on oeis.org

1, 2, 280, 277200, 1009008000, 9777287520000, 207786914375040000, 8508874143657888000000, 611958228411875304960000000, 72094798889203029677337600000000, 13177487340968529764423766528000000000, 3577714168047637768100581459885056000000000, 1392303245637418713834022280928868392960000000000
Offset: 0

Views

Author

Manda Riehl, Jul 29 2015

Keywords

Comments

By diamond-shaped poset with 4 vertices, we mean a poset on four elements with e_1 < e_2, e_1 < e_3, e_2 < e_4, e_3 < e_4, and no order relations between e_2 and e_3. In the Hasse diagram for such a poset, we have a least element, two elements in the level above, and one element in the top level, so the diagram resembles a diamond.

Examples

			For a single diamond (n=1) poset with 4 vertices, we must label the least element 1 and the greatest element 4, and the two central elements can be labeled either 2, 3 or 3, 2 respectively. Thus the associated permutations are 1234 and 1324.
		

Crossrefs

Programs

Formula

a(n) = (4n)!/12^n.

Extensions

More terms from Michael De Vlieger, Apr 06 2016
a(4) corrected by Georg Fischer, May 08 2021

A363380 G.f. satisfies A(x) = 1 + x * A(x)^4 * (1 + A(x)^2).

Original entry on oeis.org

1, 2, 20, 284, 4712, 85392, 1638112, 32699472, 672188768, 14133399744, 302535052160, 6570819330688, 144442463464704, 3207564324825600, 71848240540852224, 1621452789508328704, 36831997860270007808, 841470878382566444032
Offset: 0

Views

Author

Seiichi Manyama, May 29 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n, binomial(n, k)*binomial(4*n+2*k+1, n)/(4*n+2*k+1));

Formula

a(n) = Sum_{k=0..n} binomial(n,k) * binomial(4*n+2*k+1,n)/(4*n+2*k+1).

A371676 G.f. satisfies A(x) = 1 + x * A(x)^2 * (1 + A(x)^(1/2))^2.

Original entry on oeis.org

1, 4, 40, 524, 7824, 126228, 2143544, 37750812, 683194912, 12628104740, 237388091208, 4524456276524, 87228274533040, 1698091537435444, 33332913873239640, 659038408936005692, 13112372856351746112, 262338658739430857796, 5274545338183090647656
Offset: 0

Views

Author

Seiichi Manyama, Apr 02 2024

Keywords

Crossrefs

Programs

  • PARI
    a(n, r=2, t=4, u=1) = r*sum(k=0, n, binomial(n, k)*binomial(t*n+u*k+r, n)/(t*n+u*k+r));

Formula

G.f. satisfies A(x) = ( 1 + x * A(x)^2 * (1 + A(x)^(1/2)) )^2.
a(n) = 2 * Sum_{k=0..n} binomial(n,k) * binomial(4*n+k+2,n)/(4*n+k+2).

A260579 Labelings of n diamond-shaped posets with 4 vertices per diamond where the labels follow the poset relations whose associated reading permutation avoids 321 in the classical sense.

Original entry on oeis.org

1, 2, 106, 5976, 387564, 27247446, 2020632046, 155622020610, 12327937844924, 998103225615208
Offset: 0

Views

Author

Manda Riehl, Jul 29 2015

Keywords

Comments

By diamond-shaped poset with 4 vertices, we mean a poset on four elements with e_1 < e_2, e_1 < e_3, e_2 < e_4, e_3 < e_4, and no order relations between e_2 and e_3. In the Hasse diagram for such a poset, we have a least element, two elements in the level above, and one element in the top level, so the diagram resembles a diamond. The associated permutation is the permutation obtained by reading the labels of each poset by levels left to right, starting with the least element.
Additional terms were provided by David Bevan.

Examples

			For a single diamond (n=1) poset with 4 vertices, we must label the least element 1 and the greatest element 4, and the two central elements can be labeled either 2, 3 or 3, 2 respectively. Thus the associated permutations are 1234 and 1324.
		

Crossrefs

A364826 G.f. satisfies A(x) = 1 - x*A(x)^4 * (1 - 3*A(x)).

Original entry on oeis.org

1, 2, 22, 338, 6038, 117570, 2420758, 51833106, 1142472150, 25749801986, 590737764118, 13748997055826, 323842714201622, 7704914865207362, 184899022770465558, 4470200057557410834, 108776308617293352534, 2662072268791363675650
Offset: 0

Views

Author

Seiichi Manyama, Aug 09 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = (-1)^n*sum(k=0, n, (-3)^k*binomial(n, k)*binomial(4*n+k+1, n)/(4*n+k+1));

Formula

a(n) = (-1)^n * Sum_{k=0..n} (-3)^k * binomial(n,k) * binomial(4*n+k+1,n) / (4*n+k+1).
a(n) = (1/n) * Sum_{k=0..n-1} 2^(n-k) * binomial(n,k) * binomial(5*n-k,n-1-k) for n > 0.
a(n) = (1/n) * Sum_{k=1..n} 2^k * 3^(n-k) * binomial(n,k) * binomial(4*n,k-1) for n > 0.

A371700 G.f. satisfies A(x) = 1 + x * A(x)^6 * (1 + A(x)).

Original entry on oeis.org

1, 2, 26, 482, 10450, 247554, 6208970, 162064322, 4356511138, 119788611458, 3353361311738, 95251219926690, 2738421518770546, 79531905952256642, 2329955712706784682, 68770993359030211458, 2043143866891345880898, 61050342965542475675906
Offset: 0

Views

Author

Seiichi Manyama, Apr 03 2024

Keywords

Crossrefs

Programs

  • PARI
    a(n, r=1, t=6, u=1) = r*sum(k=0, n, binomial(n, k)*binomial(t*n+u*k+r, n)/(t*n+u*k+r));

Formula

a(n) = Sum_{k=0..n} binomial(n,k) * binomial(6*n+k+1,n)/(6*n+k+1).
a(n) = (1/n) * Sum_{k=0..n-1} (-1)^k * 2^(n-k) * binomial(n,k) * binomial(7*n-k,n-1-k) for n > 0.
a(n) = (1/n) * Sum_{k=1..n} 2^k * binomial(n,k) * binomial(6*n,k-1) for n > 0.

A378840 G.f. A(x) satisfies A(x) = ( 1 + x * A(x)^(4/3) * (1 + A(x)^(1/3)) )^3.

Original entry on oeis.org

1, 6, 66, 902, 13794, 225990, 3878946, 68854278, 1253647938, 23283474310, 439394508162, 8401507608966, 162413310158626, 3169029168475206, 62330703549363810, 1234503404283308038, 24599422679682518658, 492824963618477891334, 9920626149798702401730
Offset: 0

Views

Author

Seiichi Manyama, Dec 09 2024

Keywords

Crossrefs

Programs

  • PARI
    a(n, r=3, t=4, u=1) = r*sum(k=0, n, binomial(n, k)*binomial(t*n+u*k+r, n)/(t*n+u*k+r));

Formula

a(n) = 3 * Sum_{k=0..n} binomial(n,k) * binomial(4*n+k+3,n)/(4*n+k+3).
Conjecture: g.f.: B(x)^3, where B(x) is the g.f. of A260332.
Previous Showing 11-19 of 19 results.