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 11-16 of 16 results.

A354013 Expansion of e.g.f. 1/(1 + log(1-x) * (1 - log(1-x))).

Original entry on oeis.org

1, 1, 5, 32, 278, 3014, 39226, 595608, 10335888, 201785688, 4377151464, 104444584848, 2718748442208, 76668029954736, 2328328726108368, 75759574181169792, 2629417097250852480, 96963968323279825920, 3786037089608099128320, 156041617540423798782720
Offset: 0

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Author

Seiichi Manyama, May 14 2022

Keywords

Crossrefs

Programs

  • PARI
    my(N=20, x='x+O('x^N)); Vec(serlaplace(1/(1+log(1-x)*(1-log(1-x)))))
    
  • PARI
    a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=sum(j=1, i, (j-1)!*(1+2*sum(k=1, j-1, 1/k))*binomial(i, j)*v[i-j+1])); v;
    
  • PARI
    a(n) = sum(k=0, n, k!*fibonacci(k+1)*abs(stirling(n, k, 1)));

Formula

a(0) = 1; a(n) = Sum_{k=1..n} A000776(k-1) * binomial(n,k) * a(n-k).
a(n) = Sum_{k=0..n} k! * Fibonacci(k+1) * |Stirling1(n,k)|.
a(n) ~ n! / (sqrt(5) * exp((sqrt(5)-1)/2) * (1 - exp((1-sqrt(5))/2))^(n+1)). - Vaclav Kotesovec, May 15 2022

A263968 a(n) = Li_{-n}(phi) + Li_{-n}(1-phi), where Li_n(x) is the polylogarithm, phi=(1+sqrt(5))/2 is the golden ratio.

Original entry on oeis.org

-3, 4, -18, 112, -930, 9664, -120498, 1752832, -29140290, 545004544, -11325668178, 258892951552, -6456024679650, 174410345857024, -5074158021135858, 158168121299894272, -5258993667674555010, 185786981314092335104, -6949466928081909755538
Offset: 0

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Author

Vladimir Reshetnikov, Oct 30 2015

Keywords

Comments

2*Li_{-n}(phi) = a(n) - (-1)^n*A000557(n)*sqrt(5), so a(n) represents integer terms in 2*Li_{-n}(phi), and A000557(n) (with alternating signs) represents terms proportional to sqrt(5).

Examples

			For n = 4, Li_{-4}(phi) = -930 - 416*sqrt(5), so a(4) = -930 and A000557(4) = 416.
		

Crossrefs

Programs

  • Maple
    a := n -> polylog(-n,(1+sqrt(5))/2)+polylog(-n,(1-sqrt(5))/2):
    seq(round(evalf(a(n),32)), n=0..18); # Peter Luschny, Nov 01 2015
  • Mathematica
    Round@Table[PolyLog[-n, GoldenRatio] + PolyLog[-n, 1-GoldenRatio], {n, 0, 20}]
    Table[(-1)^(n+1) Sum[k! LucasL[k+2] StirlingS2[n, k], {k, 0, n}], {n, 0, 20}]
  • PARI
    vector(100, n, n--; (-1)^(n+1)*sum(k=0, n, k!*stirling(n, k, 2)*(2*fibonacci(k+1) + fibonacci(k+2)))) \\ Altug Alkan, Oct 31 2015

Formula

a(n) = (-1)^(n+1)*Sum_{k=0..n} k!*Lucas(k+2)*Stirling2(n,k), where Lucas(n) = A000032(n) and A048993(n,k) = Stirling2(n,k).
a(n) = (-1)^(n+1)*(2*A000556(n) + A000557(n)).
E.g.f.: -(1+2*exp(x))/(1+2*sinh(x)).
a(n) ~ (-1)^(n+1) * n! / log((1+sqrt(5))/2)^(n+1). - Vaclav Kotesovec, Oct 31 2015

A337555 a(0) = 1; a(n) = (1/2) * Sum_{k=1..n} binomial(n,k) * (3^k-1) * a(n-k).

Original entry on oeis.org

1, 1, 6, 43, 408, 4861, 69516, 1159663, 22108848, 474192601, 11300589876, 296237533483, 8471642214888, 262456441714741, 8756520140416236, 313017838828154503, 11935355244756882528, 483537933291091103281, 20741938090482567562596, 939180816648348685174723
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 31 2020

Keywords

Crossrefs

Programs

  • Mathematica
    a[0] = 1; a[n_] := a[n] = (1/2) Sum[Binomial[n, k] (3^k - 1) a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 19}]
    nmax = 19; CoefficientList[Series[2/(2 + Exp[x] - Exp[3 x]), {x, 0, nmax}], x] Range[0, nmax]!
  • PARI
    seq(n)={Vec(serlaplace(2 / (2 + exp(x + O(x*x^n)) - exp(3*x + O(x*x^n)))))} \\ Andrew Howroyd, Aug 31 2020

Formula

E.g.f.: 2 / (2 + exp(x) - exp(3*x)).
a(n) ~ n! / ((r+3) * log(r)^(n+1)), where r = 1.52137970680456756960408... is the real root of the equation r^3 - r = 2. - Vaclav Kotesovec, Aug 31 2020

A376111 a(0) = 1; a(n) = Sum_{k=1..n} (2^k-1) * a(k-1) * a(n-k).

Original entry on oeis.org

1, 1, 4, 35, 600, 19942, 1299768, 167796051, 43131308656, 22127283690338, 22680691426392504, 46472849736334410494, 190399379929624643874384, 1559942353285454499773312748, 25559656412925984160985399396784, 837564388804449970974724247002202883
Offset: 0

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Author

Ilya Gutkovskiy, Sep 10 2024

Keywords

Crossrefs

Programs

  • Mathematica
    a[0] = 1; a[n_] := a[n] = Sum[(2^k - 1) a[k - 1] a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 15}]
    nmax = 15; A[] = 0; Do[A[x] = 1/(1 + x A[x] - 2 x A[2 x]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

Formula

G.f. A(x) satisfies: A(x) = 1 / (1 + x * A(x) - 2 * x * A(2*x)).

A305923 Expansion of e.g.f. exp(x)/(1 - log(1 + x) - log(1 + x)^2).

Original entry on oeis.org

1, 2, 6, 21, 105, 580, 4332, 33173, 333057, 3249334, 41175698, 485901669, 7470988137, 102962077608, 1870375878472, 29342124588357, 617978798588225, 10818920340476010, 260570216908845406, 5009431835664474101, 136578252867673635369, 2844357524328057280332, 87134882338620095240484
Offset: 0

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Author

Ilya Gutkovskiy, Jun 14 2018

Keywords

Comments

Binomial transform of A005444.
Sequence is signed: first negative term is a(61).

Examples

			E.g.f.: A(x) = 1 + 2*x/1! + 6*x^2/2! + 21*x^3/3! + 105*x^4/4! + 580*x^5/5! + 4332*x^6/6! + ...
		

Crossrefs

Programs

  • Maple
    a:=series(exp(x)/(1-log(1+x)-log(1+x)^2),x=0,23): seq(n!*coeff(a,x,n),n=0..22); # Paolo P. Lava, Mar 26 2019
  • Mathematica
    nmax = 22; CoefficientList[Series[Exp[x]/(1 - Log[1 + x] - Log[1 + x]^2), {x, 0, nmax}], x] Range[0, nmax]!
    Table[Sum[Sum[Binomial[n, k] StirlingS1[k, j] j! Fibonacci[j + 1], {j, 0, k}], {k, 0, n}], {n, 0, 22}]

Formula

a(n) = Sum_{k=0..n} Sum_{j=0..k} binomial(n,k)*Stirling1(k,j)*j!*Fibonacci(j+1).
a(n) ~ (-1)^n * n! * exp(exp(-phi) - phi^2) / (sqrt(5) * (1 - exp(-phi))^(n+1)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Mar 26 2019

A364822 Expansion of e.g.f. cosh(x) / (1 - 2*sinh(x)).

Original entry on oeis.org

1, 2, 9, 56, 465, 4832, 60249, 876416, 14570145, 272502272, 5662834089, 129446475776, 3228012339825, 87205172928512, 2537079010567929, 79084060649947136, 2629496833837277505, 92893490657046167552, 3474733464040954877769, 137195165161622584426496, 5702069567580948171751185
Offset: 0

Views

Author

Mélika Tebni, Nov 07 2023

Keywords

Comments

Conjectures: For p prime (p > 2), a(p) == 2 (mod p).
For n = 2^m (m natural number), a(n) == 1 (mod n).

Crossrefs

Programs

  • Maple
    a := n -> add(binomial(n,2*k)*add(j!*combinat[fibonacci](j+2)*Stirling2(n-2*k,j), j=0..n-2*k), k=0..floor(n/2)):
    seq(a(n), n = 0 .. 20);
    # second program:
    b := proc(n) option remember; `if`(n = 0, 1, 2+2*add(binomial(n,2*k-1)*b(n-2*k+1), k=1..floor((n+1)/2))) end:
    a := proc(n) `if`(n = 0, 1, b(n)/2) end: seq(a(n), n = 0 .. 20);
    # third program:
    (1/2)*((exp(-x) + exp(x))/(1 + exp(-x) - exp(x))): series(%, x, 21):
    seq(n!*coeff(%, x, n), n = 0..20);  # Peter Luschny, Nov 07 2023
  • Mathematica
    a[n_]:=n!*SeriesCoefficient[Cosh[x]/(1 - 2*Sinh[x]),{x,0,n}]; Array[a,21,0] (* Stefano Spezia, Nov 07 2023 *)
  • PARI
    my(x='x+O('x^30)); Vec(serlaplace(cosh(x) / (1 - 2*sinh(x)))) \\ Michel Marcus, Nov 07 2023

Formula

a(n) = A000556(n) + A332257(n), for n > 0.
a(n) = (-1)^n*Sum_{k=0..floor(n/2)} A341724(n,2*k).
a(n) = (A000556(n) + A005923(n)) / 2.
a(n) ~ n! / (2*log((1 + sqrt(5))/2)^(n+1)).
Previous Showing 11-16 of 16 results.