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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A354253 Expansion of e.g.f. 1/sqrt(9 - 8 * exp(x)).

Original entry on oeis.org

1, 4, 52, 1108, 32980, 1261204, 58928212, 3253363348, 207225008980, 14958174725524, 1206698072485972, 107589343503498388, 10505997552329149780, 1115087729794287434644, 127819745001180490920532, 15736779719362919373550228, 2071062794354825889656471380
Offset: 0

Views

Author

Seiichi Manyama, May 21 2022

Keywords

Comments

Conjecture: Let k be a positive integer. The sequence obtained by reducing a(n) modulo k is eventually periodic with the period dividing phi(k) = A000010(k). For example, modulo 7 we obtain the sequence [1, 4, 3, 2, 3, 0, 0, 4, 3, 2, 3, 0, 0, 4, 3, 2, 3, 0, 0, ...], with a preperiod of length 1 and an apparent period thereafter of 6 = phi(7). Cf. A354242. - Peter Bala, Jul 08 2022

Crossrefs

Programs

  • PARI
    my(N=20, x='x+O('x^N)); Vec(serlaplace(1/sqrt(9-8*exp(x))))
    
  • PARI
    my(N=20, x='x+O('x^N)); Vec(serlaplace(sum(k=0, N, binomial(2*k, k)*(2*(exp(x)-1))^k)))
    
  • PARI
    a(n) = sum(k=0, n, 2^k*(2*k)!*stirling(n, k, 2)/k!);

Formula

E.g.f.: Sum_{k>=0} binomial(2*k,k) * (2 * (exp(x) - 1))^k.
a(n) = Sum_{k=0..n} 2^k * (2*k)! * Stirling2(n,k)/k!.
a(n) ~ sqrt(2) * n^n / (3 * exp(n) * log(9/8)^(n + 1/2)). - Vaclav Kotesovec, Jun 04 2022
Conjectural o.g.f. as a continued fraction of Stieltjes type: 1/(1 - 4*x/(1 - 9*x/(1 - 12*x/(1 - 18*x/(1 - 20*x/(1 - 27*x/(1 - ... - (8*n-4)*x/(1 - 9*n*x/(1 - ...))))))))). - Peter Bala, Jul 08 2022
a(0) = 1; a(n) = Sum_{k=1..n} (8 - 4*k/n) * binomial(n,k) * a(n-k). - Seiichi Manyama, Sep 09 2023
a(0) = 1; a(n) = 4*a(n-1) - 9*Sum_{k=1..n-1} (-1)^k * binomial(n-1,k) * a(n-k). - Seiichi Manyama, Nov 17 2023

A292783 Square array A(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of e.g.f. 1/sqrt(1 - 2*k*x).

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 3, 0, 1, 3, 12, 15, 0, 1, 4, 27, 120, 105, 0, 1, 5, 48, 405, 1680, 945, 0, 1, 6, 75, 960, 8505, 30240, 10395, 0, 1, 7, 108, 1875, 26880, 229635, 665280, 135135, 0, 1, 8, 147, 3240, 65625, 967680, 7577955, 17297280, 2027025, 0, 1, 9, 192, 5145, 136080, 2953125, 42577920, 295540245, 518918400, 34459425, 0
Offset: 0

Views

Author

Ilya Gutkovskiy, Sep 23 2017

Keywords

Examples

			E.g.f. of column k: A_k(x) = 1 + k*x/1! + 3*k^2*x^2/2! + 15*k^3*x^3/3! + 105*k^4*x^4/4! + 945*k^5*x^5/5! + 10395*k^6*x^6/6! +
Square array begins:
1,    1,      1,       1,       1,        1,  ...
0,    1,      2,       3,       4,        5,  ...
0,    3,     12,      27,      48,       75,  ...
0,   15,    120,     405,     960,     1875,  ...
0,  105,   1680,    8505,   26880,    65625,  ...
0,  945,  30240,  229635,  967680,  2953125,  ...
		

Crossrefs

Columns k=0..4 give A000007, A001147, A001813, A011781, A144828.
Rows n=0.2 give A000012, A001477, A033428.
Main diagonal gives A292784.
Cf. A131182.

Programs

  • Mathematica
    Table[Function[k, n! SeriesCoefficient[1/Sqrt[1 - 2 k x], {x, 0, n}]][j - n], {j, 0, 10}, {n, 0, j}] // Flatten
    Table[Function[k, SeriesCoefficient[1/(1 + ContinuedFractionK[-i k x, 1, {i, 1, n}]), {x, 0, n}]][j - n], {j, 0, 10}, {n, 0, j}] // Flatten

Formula

O.g.f. of column k: 1/(1 - k*x/(1 - 2*k*x/(1 - 3*k*x/(1 - 4*k*x/(1 - 5*k*x/(1 - ...)))))), a continued fraction.
E.g.f. of column k: 1/sqrt(1 - 2*k*x).
A(n,k) = k^n*A001147(n).

A345103 a(n) = 1 + 4 * Sum_{k=0..n-1} binomial(n,k) * a(k) * a(n-k-1).

Original entry on oeis.org

1, 5, 61, 1277, 37741, 1437725, 67013101, 3693540317, 234974905261, 16945434018845, 1366008048556141, 121721015465713757, 11880107754103150381, 1260413749895624939165, 144427420001275864755181, 17776090894283922227621597, 2338833689096321086977341101, 327585830473259220341296486685
Offset: 0

Views

Author

Ilya Gutkovskiy, Jun 08 2021

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := a[n] = 1 + 4 Sum[Binomial[n, k] a[k] a[n - k - 1], {k, 0, n - 1}]; Table[a[n], {n, 0, 17}]
    nmax = 17; CoefficientList[Series[Exp[x]/Sqrt[9 - 8 Exp[x]], {x, 0, nmax}], x] Range[0, nmax]!
    Table[Sum[Sum[Binomial[n, k] StirlingS2[k, j] 4^j (2 j - 1)!!, {j, 0, k}], {k, 0, n}], {n, 0, 17}]
  • PARI
    N=20; x='x+O('x^N); Vec(serlaplace(exp(x)/sqrt(9-8*exp(x)))) \\ Seiichi Manyama, Oct 20 2021

Formula

E.g.f.: exp(x) / sqrt(9 - 8 * exp(x)).

A156654 Triangle T(n, k) = coefficients of p(x,n), where p(x,n) = ((1-x)^(2*n+1)/x^n) * Sum_{j >= n} ( (2*j+1)^n * binomial(j, n) * x^j ), read by rows.

Original entry on oeis.org

1, 3, 1, 25, 22, 1, 343, 515, 101, 1, 6561, 14156, 5766, 396, 1, 161051, 456197, 299342, 49642, 1447, 1, 4826809, 16985858, 15796159, 4592764, 371239, 5090, 1, 170859375, 719818759, 878976219, 383355555, 58474285, 2550165, 17481, 1, 6975757441, 34264190872, 52246537948, 31191262504, 7488334150, 660394024, 16574428, 59032, 1
Offset: 0

Views

Author

Roger L. Bagula, Feb 12 2009

Keywords

Examples

			Triangle begins as:
          1;
          3,         1;
         25,        22,         1;
        343,       515,       101,         1;
       6561,     14156,      5766,       396,        1;
     161051,    456197,    299342,     49642,     1447,       1;
    4826809,  16985858,  15796159,   4592764,   371239,    5090,     1;
  170859375, 719818759, 878976219, 383355555, 58474285, 2550165, 17481, 1;
		

Crossrefs

Programs

  • Magma
    m:= 40; R:=PowerSeriesRing(Rationals(), m);
    T:= func< n | Coefficients(R!( ((1-x)^(2*n+1)/x^n)*(&+[ (2*j+1)^n*Binomial(j, n)*x^j: j in [n..m]] ) )) >;
    [T(n): n in [0..12]]; // G. C. Greubel, Apr 02 2021
    
  • Mathematica
    p[x_, n_]:= ((1-x)^(2*n+1)/x^n)*Sum[(2*j+1)^n*Binomial[j, n]*x^j, {j,n,2*n}];
    Table[CoefficientList[Series[p[x,n], {x,0,n}], x], {n,0,12}]//Flatten (* modified by G. C. Greubel, Apr 02 2021 *)
  • Sage
    def p(n, x): return ((1-x)^(2*n+1)/x^n)*sum( (2*j+1)^n*binomial(j, n)*x^j for j in (n..2*n) )
    flatten([[( p(n, x) ).series(x, n+1).list()[k] for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Apr 02 2021

Formula

Define p(x,n) = ((1-x)^(2*n+1)/x^n) * Sum_{j >= n} ( (2*j+1)^n * binomial(j, n) * x^j ) then the triangle is defined by T(n, k) = coefficients of p(x,n) for row n and column k.
Sum_{k=0..n} T(n,k) = 2^(n-1) * n! * Catalan(n-1) = A144828(n) = A052714(n+1). - G. C. Greubel, Apr 02 2021

Extensions

Edited by G. C. Greubel, Apr 02 2021
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