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.

Showing 1-7 of 7 results.

A277614 a(n) is the coefficient of x^n/n! in exp(x + n*x^2/2).

Original entry on oeis.org

1, 1, 3, 10, 73, 426, 4951, 41308, 658785, 7149628, 144963451, 1937124696, 47660873833, 756536698360, 21888570052623, 402400189738576, 13384439813823361, 279666289640774928, 10512823691028429235, 246061359639756047008, 10314843348672697017801, 267328220273408530004896, 12363686002049118477390343, 351473836594567725961268160, 17776996370247936310502612833, 550002942283550733215994429376
Offset: 0

Views

Author

Paul D. Hanna, Nov 10 2016

Keywords

Comments

From Peter Luschny, Jan 17 2023: (Start)
a(n) is the number of connection patterns in a telephone system with n possibilities of connection and n subscribers.
The number of matchings of a complete multigraph K(n, n).
The main diagonal of A359762. (End)
Let k be a positive integer. It appears that reducing this sequence modulo k produces an eventually periodic sequence. For example, modulo 10 the sequence becomes [1, 3, 0, 3, 6, 1, 8, 5, 8, 1, 6, 3, 0, 3, 6, 1, 8, 5, 8, 1, 6, 3, 0, 3, 6, ...], with an apparent period [1, 8, 5, 8, 1, 6, 3, 0, 3, 6] of length 10 starting at a(5). - Peter Bala, Apr 16 2023

Examples

			E.g.f.: A(x) = 1 + x + 3*x^2/2! + 10*x^3/3! + 73*x^4/4! + 426*x^5/5! + 4951*x^6/6! + 41308*x^7/7! + 658785*x^8/8! + 7149628*x^9/9! + 144963451*x^10/10! + ...
The table of coefficients of x^k/k! in exp(x + n*x^2/2) begins:
n=0: 1, 1,  1,  1,   1,    1,     1,      1,       1, ...;
n=1: 1, 1,  2,  4,  10,   26,    76,    232,     764, ...;
n=2: 1, 1,  3,  7,  25,   81,   331,   1303,    5937, ...;
n=3: 1, 1,  4, 10,  46,  166,   856,   3844,   21820, ...;
n=4: 1, 1,  5, 13,  73,  281,  1741,   8485,   57233, ...;
n=5: 1, 1,  6, 16, 106,  426,  3076,  15856,  123516, ...;
n=6: 1, 1,  7, 19, 145,  601,  4951,  26587,  234529, ...;
n=7: 1, 1,  8, 22, 190,  806,  7456,  41308,  406652, ...;
n=8: 1, 1,  9, 25, 241, 1041, 10681,  60649,  658785, ...;
n=9: 1, 1, 10, 28, 298, 1306, 14716,  85240, 1012348, ...;
n=10:1, 1, 11, 31, 361, 1601, 19651, 115711, 1491281, ...; ...
in which the main diagonal forms this sequence.
In the above table, the e.g.f. of the m-th diagonal equals the e.g.f. of this sequence multiplied by ( LambertW(-x^2)/(-x^2) )^(m/2).
Example,
A(x)*sqrt(-LambertW(-x^2))/x = 1 + x + 4*x^2/2! + 13*x^3/3! + 106*x^4/4! + 601*x^5/5! + 7456*x^6/6! + 60649*x^7/7! + 1012348*x^8/8! + ...
equals the e.g.f. of the next lower diagonal in the table.
RELATED SERIES.
-LambertW(-x^2) = x^2 + 2*x^4/2! + 3^2*x^6/3! + 4^3*x^8/4! + 5^4*x^10/5! + 6^5*x^12/6! + ... + n^(n-1)*x^(2*n)/n! + ...
sqrt(-LambertW(-x^2)) = x + 3^0*x^3/(1!*2) + 5*x^5/(2!*2^2) + 7^2*x^7/(3!*2^3) + 9^3*x^9/(4!*2^4) + ... + (2*n+1)^(n-1)*x^(2*n+1)/(n!*2^n) + ...
		

Crossrefs

Programs

  • Maple
    a := n -> add(binomial(n, j) * doublefactorial(j-1) * n^(j/2), j = 0..n, 2):
    seq(a(n), n = 0..25); # Peter Luschny, Jan 17 2023
  • PARI
    {a(n) = n!*polcoeff( exp(x + n*x^2/2 + x*O(x^n)),n)}
    for(n=0,30,print1(a(n),", "))
    
  • Python
    from math import factorial, comb
    def oddfactorial(n: int) -> int:
        return factorial(2 * n) // (2**n * factorial(n))
    def a(n: int) -> int:
        return sum(comb(n, 2*j) * oddfactorial(j) * n**j for j in range(n+1))
    print([a(n) for n in range(26)]) # Peter Luschny, Jan 17 2023

Formula

E.g.f.: exp( sqrt(-LambertW(-x^2)) ) / (1 + LambertW(-x^2)).
a(n) ~ (exp(1) + (-1)^n*exp(-1)) * n^n / (sqrt(2) * exp(n/2)). - Vaclav Kotesovec, Nov 11 2016
a(n) = Sum_{j=0..n, j even} binomial(n, j) * (j - 1)!! * n^(j/2). - Peter Luschny, Jan 17 2023

A362043 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where T(n,k) = n! * Sum_{j=0..floor(n/3)} (k/6)^j * binomial(n-2*j,j)/(n-2*j)!.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 5, 1, 1, 1, 1, 4, 9, 11, 1, 1, 1, 1, 5, 13, 21, 31, 1, 1, 1, 1, 6, 17, 31, 81, 106, 1, 1, 1, 1, 7, 21, 41, 151, 351, 337, 1, 1, 1, 1, 8, 25, 51, 241, 736, 1233, 1205, 1, 1, 1, 1, 9, 29, 61, 351, 1261, 2689, 5769, 5021, 1
Offset: 0

Views

Author

Seiichi Manyama, Apr 15 2023

Keywords

Examples

			Square array begins:
  1,  1,  1,   1,   1,   1,   1, ...
  1,  1,  1,   1,   1,   1,   1, ...
  1,  1,  1,   1,   1,   1,   1, ...
  1,  2,  3,   4,   5,   6,   7, ...
  1,  5,  9,  13,  17,  21,  25, ...
  1, 11, 21,  31,  41,  51,  61, ...
  1, 31, 81, 151, 241, 351, 481, ...
		

Crossrefs

Columns k=0..2 give A000012, A190865, A001470.
Main diagonal gives A362173.
T(n,2*n) gives A362300.
T(n,6*n) gives A362301.

Programs

  • PARI
    T(n, k) = n!*sum(j=0, n\3, (k/6)^j/(j!*(n-3*j)!));

Formula

E.g.f. of column k: exp(x + k*x^3/6).
T(n,k) = T(n-1,k) + k * binomial(n-1,2) * T(n-3,k) for n > 2.
T(n,k) = n! * Sum_{j=0..floor(n/3)} (k/6)^j / (j! * (n-3*j)!).

A362319 a(n) = n! * Sum_{k=0..floor(n/5)} (n/5)^k / (k! * (n-5*k)!).

Original entry on oeis.org

1, 1, 1, 1, 1, 121, 865, 3529, 10753, 27217, 7318081, 96720625, 689990401, 3508289929, 14239793569, 5933573525881, 114415115802625, 1165402803391009, 8298505279241857, 46355961619888993, 26167218073714552321, 663290722580370585625
Offset: 0

Views

Author

Seiichi Manyama, Apr 16 2023

Keywords

Crossrefs

Programs

  • PARI
    my(N=30, x='x+O('x^N)); Vec(serlaplace(exp((-lambertw(-x^5))^(1/5))/(1+lambertw(-x^5))))

Formula

a(n) = n! * [x^n] exp(x + n*x^5/5).
E.g.f.: exp( ( -LambertW(-x^5) )^(1/5) ) / (1 + LambertW(-x^5)).

A362173 a(n) = n! * Sum_{k=0..floor(n/3)} (n/6)^k * binomial(n-2*k,k)/(n-2*k)!.

Original entry on oeis.org

1, 1, 1, 4, 17, 51, 481, 3676, 18369, 272917, 3011201, 21058236, 427112401, 6160655359, 55380250017, 1423658493076, 25361574327041, 278603741558601, 8673295084155649, 183914415577719892, 2387417408385462801, 87273239189497636171, 2146479566819857007201
Offset: 0

Views

Author

Seiichi Manyama, Apr 14 2023

Keywords

Crossrefs

Main diagonal of A362043.

Programs

  • PARI
    my(N=30, x='x+O('x^N)); Vec(serlaplace(exp((-2*lambertw(-x^3/2))^(1/3))/(1+lambertw(-x^3/2))))

Formula

a(n) = n! * [x^n] exp(x + n*x^3/6).
E.g.f.: exp( ( -2*LambertW(-x^3/2) )^(1/3) ) / (1 + LambertW(-x^3/2)).

A362314 a(n) = n! * Sum_{k=0..floor(n/4)} (n/4)^k /(k! * (n-4*k)!).

Original entry on oeis.org

1, 1, 1, 1, 25, 151, 541, 1471, 84001, 925345, 5682601, 25177681, 2245355641, 35901100951, 312222474565, 1917363070351, 232479594721921, 4873115730725761, 54830346428307601, 430468886732009185, 65997947903313461401, 1711564302775814535511
Offset: 0

Views

Author

Seiichi Manyama, Apr 15 2023

Keywords

Crossrefs

Programs

  • PARI
    my(N=30, x='x+O('x^N)); Vec(serlaplace(exp((-lambertw(-x^4))^(1/4))/(1+lambertw(-x^4))))

Formula

a(n) = n! * [x^n] exp(x + n*x^4/4).
E.g.f.: exp( ( -LambertW(-x^4) )^(1/4) ) / (1 + LambertW(-x^4)).
From Vaclav Kotesovec, Apr 18 2023: (Start)
a(n) ~ c * n^n / exp(3*n/4), where
c = cosh(1) + cos(1) if mod(n,4)=0,
c = sinh(1) + sin(1) if mod(n,4)=1,
c = cosh(1) - cos(1) if mod(n,4)=2,
c = sinh(1) - sin(1) if mod(n,4)=3. (End)

A362304 a(n) = n! * Sum_{k=0..floor(n/3)} (-n/3)^k * binomial(n-2*k,k)/(n-2*k)!.

Original entry on oeis.org

1, 1, 1, -5, -31, -99, 1201, 13231, 70785, -1362311, -21562399, -161746749, 4263108961, 87979472725, 849097038609, -28416142768649, -723086288422399, -8532476619366159, 346207723221680065, 10474480743776327179, 146105160034616914401
Offset: 0

Views

Author

Seiichi Manyama, Apr 15 2023

Keywords

Crossrefs

Programs

  • PARI
    my(N=30, x='x+O('x^N)); Vec(serlaplace(exp(lambertw(x^3)^(1/3))/(1+lambertw(x^3))))

Formula

a(n) = A362302(n,2*n).
a(n) = n! * [x^n] exp(x - n*x^3/3).
E.g.f.: exp( ( LambertW(x^3) )^(1/3) ) / (1 + LambertW(x^3)).

A362293 Expansion of e.g.f. exp( (-LambertW(-x^3))^(1/3) ).

Original entry on oeis.org

1, 1, 1, 1, 9, 41, 121, 2241, 18481, 91729, 2577681, 30833441, 215554681, 8126363961, 127462383049, 1150296157921, 54416525377761, 1056352067669921, 11684649751431841, 665061201610232769, 15390714465319910761, 201615391902487799881
Offset: 0

Views

Author

Seiichi Manyama, Apr 14 2023

Keywords

Comments

Let k be a positive integer. It appears that reducing this sequence modulo k produces an eventually periodic sequence with period a multiple of k. For example, modulo 3 the sequence becomes [1, 1, 1, 0, 2, 1, 0, 1, 1, 0, 2, 1, 0, 1, 1, 0, 2, 1, 0, ...], with an apparent period [1, 1, 0, 2, 1, 0] of length 6 starting at a(1). - Peter Bala, Apr 16 2023

Crossrefs

Programs

  • PARI
    my(N=30, x='x+O('x^N)); Vec(serlaplace(exp((-lambertw(-x^3))^(1/3))))

Formula

a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/3)} A362292(k) * binomial(n-1,3*k) * a(n-3*k-1).
a(n) ~ (1 + 2*cos(2*Pi*mod(n-1,3)/3 - sqrt(3)/2)/exp(3/2)) * n^(n-1) / (sqrt(3) * exp(2*n/3 - 1)). - Vaclav Kotesovec, Apr 18 2023
Showing 1-7 of 7 results.