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 81-90 of 130 results. Next

A334786 a(n) is the total number of down steps before the first up step in all 4_2-Dyck paths of length 5*n. A 4_2-Dyck path is a lattice path with steps (1, 4), (1, -1) that starts and ends at y = 0 and stays above the line y = -2.

Original entry on oeis.org

0, 3, 16, 115, 950, 8510, 80388, 788392, 7950930, 81935425, 859005840, 9132977490, 98240702586, 1067197649840, 11691092372000, 129011823098160, 1432744619523530, 16000911127589355, 179590878292003200, 2024687100104286525, 22917687021180660940
Offset: 0

Views

Author

Sarah Selkirk, May 11 2020

Keywords

Examples

			For n = 1, there are the 4_2-Dyck paths UDDDD, DUDDD, DDUDD. Before the first up step there are a(1) = 0 + 1 + 2 = 3 down steps in total.
		

Crossrefs

Programs

  • Mathematica
    a[0] = 0; a[n_] := 4 * Binomial[5*n, n]/(n+1) - 2 * Binomial[5*n+2, n]/(n+1); Array[a, 21, 0]

Formula

a(0) = 0 and a(n) = 4 * binomial(5*n, n)/(n+1) - 2 * binomial(5*n+2, n)/(n+1) for n > 0.
a(n) ~ c*2^(-8*n)*5^(5*n)/n^(3/2), where c = (7/16)*sqrt(5/(2*Pi)). - Stefano Spezia, Oct 19 2022

A334787 a(n) is the total number of down steps before the first up step in all 4_3-Dyck paths of length 5*n. A 4_3-Dyck path is a lattice path with steps (1, 4), (1, -1) that starts and ends at y = 0 and stays above the line y = -3.

Original entry on oeis.org

0, 6, 34, 251, 2105, 19040, 181076, 1784728, 18067803, 186754590, 1962728460, 20910164730, 225308533359, 2451112021568, 26885549373440, 297008527319440, 3301615350645935, 36903975448964670, 414518195957729886, 4676429192392769805, 52965796433899543810
Offset: 0

Views

Author

Sarah Selkirk, May 11 2020

Keywords

Examples

			For n = 1, there are the 4_3-Dyck paths UDDDD, DUDDD, DDUDD, DDDUD. Before the first up step there are a(1) = 0 + 1 + 2 + 3 = 6 down steps in total.
		

Crossrefs

Programs

  • Mathematica
    a[0] = 0; a[n_] := 4 * Binomial[5*n, n]/(n+1) - Binomial[5*n+3, n]/(n+1); Array[a, 21, 0]

Formula

a(0) = 0 and a(n) = 4*binomial(5*n, n)/(n+1) - binomial(5*n+3, n)/(n+1) for n > 0.
a(n) ~ c*2^(-8*n)*5^(5*n)/n^(3/2), where c = (131/128)*sqrt(5/(2*Pi)). - Stefano Spezia, Oct 19 2022

A346579 a(n) = (1/(5*n)) * Sum_{d|n} mu(n/d) * binomial(5*d,d).

Original entry on oeis.org

1, 4, 30, 240, 2125, 19776, 192129, 1922496, 19692504, 205444500, 2175519379, 23322637440, 252631900235, 2760767859780, 30400169155500, 336977763170048, 3757141504436392, 42107201575798248, 474084628585822412, 5359833703935374000, 60823006052351537106, 692556314455384443196
Offset: 1

Views

Author

Ilya Gutkovskiy, Jul 24 2021

Keywords

Comments

Inverse Euler transform of A002294.
Moebius transform of A261498.

Crossrefs

Programs

  • Mathematica
    Table[(1/(5 n)) Sum[MoebiusMu[n/d] Binomial[5 d, d], {d, Divisors[n]}], {n, 22}]
  • PARI
    a(n) = sumdiv(n, d, moebius(n/d)*binomial(5*d,d))/(5*n); \\ Michel Marcus, Jul 24 2021

A346681 a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(5*k,k) / (4*k + 1).

Original entry on oeis.org

1, 0, 5, 30, 255, 2275, 21476, 210404, 2120041, 21830314, 228713056, 2430255074, 26128088701, 283703487059, 3106713300821, 34270543858459, 380471319687826, 4247891403168599, 47665096853113576, 537244509843680309, 6079834137116933061, 69054467456964456599
Offset: 0

Views

Author

Ilya Gutkovskiy, Jul 29 2021

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Sum[(-1)^(n - k) Binomial[5 k, k]/(4 k + 1), {k, 0, n}], {n, 0, 21}]
    nmax = 21; A[] = 0; Do[A[x] = 1/(1 + x) + x (1 + x)^4 A[x]^5 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
  • PARI
    a(n) = sum(k=0, n, (-1)^(n-k)*binomial(5*k, k)/(4*k + 1)); \\ Michel Marcus, Jul 29 2021

Formula

G.f. A(x) satisfies: A(x) = 1 / (1 + x) + x * (1 + x)^4 * A(x)^5.
a(n) ~ 5^(5*n + 11/2) / (3381 * sqrt(Pi) * n^(3/2) * 2^(8*n + 7/2)). - Vaclav Kotesovec, Jul 30 2021

A346766 a(n) = Sum_{k=0..n} Stirling2(n,k) * binomial(5*k,k) / (4*k + 1).

Original entry on oeis.org

1, 1, 6, 51, 531, 6331, 83532, 1195452, 18316582, 297727712, 5099398853, 91554269703, 1715910362408, 33457504204403, 676778172939139, 14168046060375184, 306327815585165519, 6827996259530724139, 156654003923243040925, 3694188118839057258940, 89428870506038692255920
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 02 2021

Keywords

Comments

Stirling transform of A002294.

Crossrefs

Programs

  • Mathematica
    Table[Sum[StirlingS2[n, k] Binomial[5 k, k]/(4 k + 1), {k, 0, n}], {n, 0, 20}]
    nmax = 20; CoefficientList[Series[Sum[(Binomial[5 k, k]/(4 k + 1)) x^k/Product[1 - j x, {j, 0, k}], {k, 0, nmax}], {x, 0, nmax}], x]
    nmax = 20; CoefficientList[Series[HypergeometricPFQ[{1/5, 2/5, 3/5, 4/5}, {1/2, 3/4, 1, 5/4}, 3125 (Exp[x] - 1)/256], {x, 0, nmax}], x] Range[0, nmax]!
  • PARI
    a(n) = sum(k=0, n, stirling(n, k, 2)*binomial(5*k, k)/(4*k + 1)); \\ Michel Marcus, Aug 03 2021

Formula

G.f.: Sum_{k>=0} ( binomial(5*k,k) / (4*k + 1) ) * x^k / Product_{j=0..k} (1 - j*x).

A346936 a(n) = Sum_{d|n} mu(n/d) * binomial(5*d,d) / (4*d+1).

Original entry on oeis.org

1, 4, 34, 280, 2529, 23712, 231879, 2330160, 23950320, 250540836, 2658968129, 28558319744, 309831575759, 3390416555996, 37377257156716, 414741861215840, 4628362722856424, 51912988232308104, 584909606696793884, 6617078646710069720, 75134301594081157746, 855968478539048248916
Offset: 1

Views

Author

Ilya Gutkovskiy, Aug 08 2021

Keywords

Comments

Moebius transform of A002294.

Crossrefs

Programs

  • Mathematica
    Table[Sum[MoebiusMu[n/d] Binomial[5 d, d]/(4 d + 1), {d, Divisors[n]}], {n, 22}]

A364865 G.f. satisfies A(x) = 1 + x*A(x)^4 / (1 + x*A(x)^4).

Original entry on oeis.org

1, 1, 3, 11, 43, 170, 657, 2392, 7675, 17603, -11898, -529678, -4783303, -33099464, -201744488, -1130700432, -5917753701, -28985131575, -131668554663, -540199800203, -1862208441834, -4014999475540, 10784817197302, 222255824910088, 1973412557775753
Offset: 0

Views

Author

Seiichi Manyama, Aug 11 2023

Keywords

Crossrefs

Programs

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

Formula

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

A382000 E.g.f. A(x) satisfies A(x) = 1 + x*exp(2*x)*A(x)^5.

Original entry on oeis.org

1, 1, 14, 342, 12872, 659280, 42828912, 3375009568, 312860626304, 33361836534144, 4023352486200320, 541461682626399744, 80448618080927609856, 13079749459734097573888, 2309915877337042992324608, 440332184936376095626076160, 90117169223076699520606896128
Offset: 0

Views

Author

Seiichi Manyama, Mar 12 2025

Keywords

Crossrefs

Programs

  • PARI
    a(n) = n!*sum(k=0, n, (2*k)^(n-k)*binomial(5*k+1, k)/((5*k+1)*(n-k)!));

Formula

a(n) = n! * Sum_{k=0..n} (2*k)^(n-k) * A002294(k)/(n-k)!.
a(n) ~ 2^(n-3) * n^(n-1) * sqrt(5*(1 + LambertW(512/3125))) / (exp(n) * LambertW(512/3125)^n). - Vaclav Kotesovec, Mar 22 2025

A386367 a(n) = Sum_{k=0..n-1} binomial(5*k,k) * binomial(5*n-5*k-2,n-k-1).

Original entry on oeis.org

0, 1, 13, 163, 2021, 24930, 306655, 3765448, 46182101, 565939603, 6931070490, 84845250370, 1038235255415, 12700966517968, 155336699256808, 1899439862390640, 23222289820948405, 283872591297526505, 3469680960837171415, 42404345427419774621, 518193229118757697930
Offset: 0

Views

Author

Seiichi Manyama, Jul 19 2025

Keywords

Examples

			(1/5) * log( Sum_{k>=0} binomial(5*k,k)*x^k ) = x + 13*x^2/2 + 163*x^3/3 + 2021*x^4/4 + 4986*x^5 + ...
		

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n-1, binomial(5*k, k)*binomial(5*n-5*k-2, n-k-1));
    
  • PARI
    my(N=30, x='x+O('x^N), g=x*sum(k=0, N, binomial(5*k+3, k)/(k+1)*x^k)); concat(0, Vec(g*(1-g)/(1-5*g)^2))

Formula

G.f.: g*(1-g)/(1-5*g)^2 where g*(1-g)^4 = x.
L.g.f.: Sum_{k>=1} a(k)*x^k/k = (1/5) * log( Sum_{k>=0} binomial(5*k,k)*x^k ).
G.f.: (g-1)/(5-4*g)^2 where g=1+x*g^5.
a(n) = Sum_{k=0..n-1} binomial(5*k-2+l,k) * binomial(5*n-5*k-l,n-k-1) for every real number l.
a(n) = Sum_{k=0..n-1} 4^(n-k-1) * binomial(5*n-1,k).
a(n) = Sum_{k=0..n-1} 5^(n-k-1) * binomial(4*n+k-1,k).

A386566 a(n) = Sum_{k=0..n-1} binomial(5*k-1,k) * binomial(5*n-5*k,n-k-1).

Original entry on oeis.org

0, 1, 14, 181, 2284, 28506, 353630, 4370584, 53882392, 663116347, 8150224204, 100073884670, 1227826127020, 15055154471696, 184508186225552, 2260299193652496, 27679951219660080, 338872887728053465, 4147618793911034330, 50753529798492061819, 620942367878256638264
Offset: 0

Views

Author

Seiichi Manyama, Jul 26 2025

Keywords

Examples

			(1/4) * log( Sum_{k>=0} binomial(5*k-1,k)*x^k ) = x + 7*x^2 + 181*x^3/3 + 571*x^4 + 28506*x^5/5 + ...
		

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n-1, binomial(5*k-1, k)*binomial(5*n-5*k, n-k-1));
    
  • PARI
    my(N=30, x='x+O('x^N), g=sum(k=0, N, binomial(5*k, k)/(4*k+1)*x^k)); concat(0, Vec(g*(g-1)/(5-4*g)^2))

Formula

G.f.: g*(g-1)/(5-4*g)^2 where g=1+x*g^5.
G.f.: g/(1-5*g)^2 where g*(1-g)^4 = x.
L.g.f.: Sum_{k>=1} a(k)*x^k/k = (1/4) * log( Sum_{k>=0} binomial(5*k-1,k)*x^k ).
a(n) = Sum_{k=0..n-1} binomial(5*k-1+l,k) * binomial(5*n-5*k-l,n-k-1) for every real number l.
a(n) = Sum_{k=0..n-1} 4^(n-k-1) * binomial(5*n,k).
a(n) = Sum_{k=0..n-1} 5^(n-k-1) * binomial(4*n+k,k).
Conjecture D-finite with recurrence 196608*n*(4*n-3)*(2*n-1)*(18270873280*n -32560150837) *(4*n-1)*a(n) +1280*(-1399185802400000*n^5 +1022280893000000*n^4 +17669158913120000*n^3 -48968110172924750*n^2 +49502057719349955*n -17877514345852392)*a(n-1) +125000*(-61298198200000*n^5 +1447969779032500*n^4 -7721498995066250*n^3 +17474948768595875*n^2 -18352567310653770*n +7399184154389181)*a(n-2) +48828125*(5*n-11) *(5*n-14)*(4958243695*n -6717884799) *(5*n-13)*(5*n-12)*a(n-3)=0. - R. J. Mathar, Jul 30 2025
Previous Showing 81-90 of 130 results. Next