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 1-9 of 9 results.

A152550 Coefficients in a q-analog of the function [LambertW(-2x)/(-2x)]^(1/2), as a triangle read by rows.

Original entry on oeis.org

1, 1, 3, 2, 12, 16, 16, 5, 55, 110, 170, 180, 130, 70, 14, 273, 728, 1443, 2145, 2640, 2614, 2200, 1485, 783, 288, 42, 1428, 4760, 11312, 20657, 32032, 42833, 50477, 52934, 49441, 41069, 29876, 19019, 10010, 4158, 1155, 132, 7752, 31008, 85272
Offset: 0

Views

Author

Paul D. Hanna, Dec 07 2008

Keywords

Comments

LambertW satisfies: [LambertW(-2x)/(-2x)]^(1/2) = exp(x*LambertW(-2x)/(-2x)).

Examples

			Triangle begins:
  1;
  1;
  3,2;
  12,16,16,5;
  55,110,170,180,130,70,14;
  273,728,1443,2145,2640,2614,2200,1485,783,288,42;
  1428,4760,11312,20657,32032,42833,50477,52934,49441,41069,29876,19019,10010,4158,1155,132;
  7752,31008,85272,181356,328440,521152,745416,969000,1159060,1278996,1307556,1238368,1085488,877240,650052,437164,262964,138320,60424,20592,4576,429;...
where row sums = (2*n+1)^(n-1) (A052750).
Row sums at q=-1 = (2*n+1)^[(n-1)/2] (A152551).
The generating function starts:
A(x,q) = 1 + x + (3 + 2*q)*x^2/faq(2,q) + (12 + 16*q + 16*q^2 + 5*q^3)*x^3/faq(3,q) + (55 + 110*q + 170*q^2 + 180*q^3 + 130*q^4 + 70*q^5 + 14*q^6)*x^4/faq(4,q) + ...
G.f. satisfies: A(x,q) = e_q( x*A(x,q)^2, q) where q-exponential series: e_q(x,q) = 1 + x + x^2/faq(2,q) + x^3/faq(3,q) +...+ x^n/faq(n,q) +...
The q-factorial of n is faq(n,q) = Product_{k=1..n} (q^k-1)/(q-1): faq(0,q)=1, faq(1,q)=1, faq(2,q)=(1+q), faq(3,q)=(1+q)*(1+q+q^2), faq(4,q)=(1+q)*(1+q+q^2)*(1+q+q^2+q^3),...
Special cases.
q=0: A(x,0) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 +... (A001764)
q=1: A(x,1) = 1 + x + 5/2*x^2 + 49/6*x^3 + 729/24*x^4 + 14641/120*x^5 +...
q=2: A(x,2) = 1 + x + 7/3*x^2 + 148/21*x^3 + 7611/315*x^4 + 872341/9765*x^5 +...
q=3: A(x,3) = 1 + x + 9/4*x^2 + 339/52*x^3 + 44521/2080*x^4 + 19059921/251680*x^5 +...

Links

Crossrefs

Cf. A052750 (row sums), A001764 (column 0), A000108 (right border), A152554.
Cf. A152551 (q=-1), A152552 (q=2), A152553 (q=3).
Cf. variants: A152290, A152555.

Programs

  • PARI
    {T(n,k)=local(e_q=1+sum(j=1,n,x^j/prod(i=1,j,(q^i-1)/(q-1))), LW2_q=sqrt(serreverse(x/(e_q+x*O(x^n))^2)/x)); polcoeff(polcoeff(LW2_q+x*O(x^n),n,x)*prod(i=1,n,(q^i-1)/(q-1))+q*O(q^k),k,q)}

Formula

G.f.: A(x,q) = Sum_{n>=0} Sum_{k=0..n*(n-1)/2} T(n,k)*q^k*x^n/faq(n,q), where faq(n,q) is the q-factorial of n.
G.f.: A(x,q) = [(1/x)*Series_Reversion( x/e_q(x,q)^2 )]^(1/2) where e_q(x,q) = Sum_{n>=0} x^n/faq(n,q) is the q-exponential function.
G.f. satisfies: A(x,q) = e_q( x*A(x,q)^2, q) and A( x/e_q(x,q)^2, q) = e_q(x,q).
G.f. at q=1: A(x,1) = (LambertW(-2*x)/(-2*x))^(1/2).
Row sums at q=+1: Sum_{k=0..n*(n-1)/2} T(n,k) = (2*n+1)^(n-1).
Row sums at q=-1: Sum_{k=0..n*(n-1)/2} T(n,k)*(-1)^k = (2*n+1)^[(n-1)/2].
Sum_{k=0..n*(n-1)/2} T(n,k)*exp(2*Pi*I*k/n) = 1 for n>=1; i.e., the n-th row sum at q = exp(2*Pi*I/n), the n-th root of unity, equals 1 for n>=1. - Vladeta Jovovic
Sum_{k=0..binomial(n,2)} T(n,k)*q^k = Sum_{pi} (2*n)!/(2*n-k+1)!*faq(n,q)/Product_{i=1..n} e(i)!*faq(i,q)^e(i), where pi runs through all nonnegative integer solutions of e(1)+2*e(2)+...+n*e(n) = n and k = e(1)+e(2)+...+e(n). - Vladeta Jovovic, Dec 04 2008

A152555 Coefficients in a q-analog of the function LambertW(-2*x)/(-2*x), as a triangle read by rows.

Original entry on oeis.org

1, 2, 7, 5, 30, 42, 42, 14, 143, 297, 462, 495, 363, 198, 42, 728, 2002, 4004, 6006, 7436, 7436, 6292, 4290, 2288, 858, 132, 3876, 13260, 31824, 58604, 91364, 122876, 145535, 153361, 143936, 120185, 87971, 56329, 29939, 12584, 3575, 429, 21318, 87210
Offset: 0

Views

Author

Paul D. Hanna, Dec 07 2008

Keywords

Examples

			Triangle begins:
  1;
  2;
  7,5;
  30,42,42,14;
  143,297,462,495,363,198,42;
  728,2002,4004,6006,7436,7436,6292,4290,2288,858,132;
  3876,13260,31824,58604,91364,122876,145535,153361,143936,120185,87971,56329,29939,12584,3575,429;
  21318,87210,242250,519384,945744,1508070,2165664,2826420,3392520,3756626,3853322,3662106,3221330,2613240,1944324,1313760,794614,420784,185640,64090,14586,1430;...
where row sums = 2*(2*n+2)^(n-1) (A097629).
Row sums at q=-1 = 2*(2*n+2)^[(n-1)/2] (A152556).
The generating function starts:
A(x,q) = 1 + 2*x + (7 + 5*q)*x^2/faq(2,q) + (30 + 42*q + 42*q^2 + 14*q^3)*x^3/faq(3,q) + (143 + 297*q + 462*q^2 + 495*q^3 + 363*q^4 + 198*q^5 + 42*q^6)*x^4/faq(4,q) + ...
The q-factorial of n is faq(n,q) = Product_{k=1..n} (q^k-1)/(q-1): faq(0,q)=1, faq(1,q)=1, faq(2,q)=(1+q), faq(3,q)=(1+q)*(1+q+q^2), faq(4,q)=(1+q)*(1+q+q^2)*(1+q+q^2+q^3), ...
Special cases.
q=0: A(x,0) = 1 + 2*x + 7*x^2 + 30*x^3 + 143*x^4 + 728*x^5 +... (A006013)
q=1: A(x,1) = 1 + 2*x + 12/2*x^2 + 128/6*x^3 + 2000/24*x^4 + 41472/120*x^5 +...
q=2: A(x,2) = 1 + 2*x + 17/3*x^2 + 394/21*x^3 + 21377/315*x^4 + 2537724/9765*x^5 +...
q=3: A(x,3) = 1 + 2*x + 22/4*x^2 + 912/52*x^3 + 126692/2080*x^4 + 56277344/251680*x^5 +...

Links

Crossrefs

Cf. A097629 (row sums), A006013 (column 0), A000108 (right border), A152559.
Cf. A152556 (q=-1), A152557 (q=2), A152558 (q=3).
Cf. variants: A152290, A152550.

Programs

  • PARI
    {T(n,k)=local(e_q=1+sum(j=1,n,x^j/prod(i=1,j,(q^i-1)/(q-1))), LW2_q=serreverse(x/(e_q+x*O(x^n))^2)/x); polcoeff(polcoeff(LW2_q+x*O(x^n),n,x)*prod(i=1,n,(q^i-1)/(q-1))+q*O(q^k),k,q)}

Formula

G.f.: A(x,q) = Sum_{n>=0} Sum_{k=0..n*(n-1)/2} T(n,k)*q^k*x^n/faq(n,q), where faq(n,q) is the q-factorial of n.
G.f.: A(x,q) = (1/x)*Series_Reversion( x/e_q(x,q)^2 ) where e_q(x,q) = Sum_{n>=0} x^n/faq(n,q) is the q-exponential function.
G.f. satisfies: A(x,q) = e_q( x*A(x,q), q)^2 and A( x/e_q(x,q)^2, q) = e_q(x,q)^2.
G.f. at q=1: A(x,1) = LambertW(-2*x)/(-2*x).
Row sums at q=+1: Sum_{k=0..n*(n-1)/2} T(n,k) = 2*(2*n+2)^(n-1).
Row sums at q=-1: Sum_{k=0..n*(n-1)/2} T(n,k)*(-1)^k = 2*(2*n+2)^[(n-1)/2].
Sum_{k=0..n*(n-1)/2} T(n,k)*exp(2*Pi*I*k/n) = 2 for n>=1; i.e., the n-th row sum at q = exp(2*Pi*I/n), the n-th root of unity, equals 2 for n>=1. - Vladeta Jovovic
Sum_{k=0..binomial(n,2)} T(n,k)*q^k = Sum_{pi} 2*(2*n+1)!/(2*n-k+2)!*faq(n,q)/Product_{i=1..n} e(i)!*faq(i,q)^e(i), where pi runs over all nonnegative integer solutions to e(1)+2*e(2)+...+n*e(n) = n and k = e(1)+e(2)+...+e(n). - Vladeta Jovovic, Dec 07 2008

A152800 Irregular triangle read by rows: the q-analog of the Euler numbers; expansion of the arithmetic inverse of the q-cosine of x.

Original entry on oeis.org

1, 1, 0, 1, 2, 1, 1, 0, 0, 1, 3, 5, 8, 10, 10, 9, 7, 5, 2, 1, 0, 0, 0, 1, 4, 10, 21, 36, 55, 78, 101, 122, 138, 145, 143, 134, 117, 95, 72, 50, 32, 18, 9, 3, 1, 0, 0, 0, 0, 1, 5, 16, 41, 87, 164, 283, 452, 679, 967, 1311, 1700, 2118, 2540, 2937, 3282, 3546, 3706, 3751, 3676, 3487
Offset: 0

Views

Author

Paul D. Hanna, Dec 26 2008

Keywords

Comments

The q-cosine is cos_q(x,q) = Sum_{n>=0} (-1)^n*x^(2n)/faq(2n,q) and faq(n,q) = Product_{k=1..n} (q^k-1)/(q-1) is the q-factorial of n.

Examples

			Nonzero coefficients in row n range from x^(n-1) to x^(2n(n-1)) for n>0.
Triangle begins:
  1;
  1;
  0,1,2,1,1;
  0,0,1,3,5,8,10,10,9,7,5,2,1;
  0,0,0,1,4,10,21,36,55,78,101,122,138,145,143,134,117,95,72,50,32,18,9,3,1;
  0,0,0,0,1,5,16,41,87,164,283,452,679,967,1311,1700,2118,2540,2937,3282,3546,3706,3751,3676,3487,3202,2842,2436,2014,1602,1223,894,622,409,253,145,76,35,14,4,1;
  ...
Explicit expansion of g.f.:
1/cos_q(x,q) = 1 + x^2/faq(2,q) + x^4*(q + 2*q^2 + q^3 + q^4)/faq(4,q) +
x^6*(q^2 + 3*q^3 + 5*q^4 + 8*q^5 + 10*q^6 + 10*q^7 + 9*q^8 + 7*q^9 + 5*q^10 + 2*q^11 + q^12)/faq(6,q) +
x^8*(q^3 + 4*q^4 + 10*q^5 + 21*q^6 + 36*q^7 + 55*q^8 + 78*q^9 + 101*q^10 + 122*q^11 + 138*q^12 + 145*q^13 + 143*q^14 + 134*q^15 + 117*q^16 + 95*q^17 + 72*q^18 + 50*q^19 + 32*q^20 + 18*q^21 + 9*q^22 + 3*q^23 + q^24)/faq(8,q) +...

Links

Crossrefs

Cf. A000364 (row sums=Euler numbers); A152801, A152802, A152803, A152804.
Cf. A152290, A152550.

Programs

  • PARI
    {T(n,k)=polcoeff(polcoeff(1/sum(m=0,n,(-1)^m*x^(2*m)/prod(j=1,2*m,(q^j-1)/(q-1))+x*O(x^(2*n+1))),2*n,x)*prod(j=1,2*n,(q^j-1)/(q-1)),k,q)}
for(n=0,8,for(k=0,2*n*(n-1),print1(T(n,k),", "));print(""))

Formula

G.f.: 1/cos_q(x,q) = Sum_{n>=0} Sum_{k=0..2n(n-1)} T(n,k)*q^k*x^(2n)/faq(2n,q).
G.f.: 1/cos(x) = Sum_{n>=1} Sum_{k=0..2n(n-1)} T(n,k)*x^(2n)/(2n)!.
Sum_{k=0..2n(n-1)} T(n,k) = A000364(n).
Sum_{k=0..2n(n-1)} T(n,k)*(-1)^k = 1 for n>=0.
Sum_{k=0..2n(n-1)} T(n,k)*I^k = (-1)^[n/2] for n>=0 where I^2=-1.
Sum_{k=0..2n(n-1)} T(n,k)*exp(2*Pi*I*k/n) = 1 for n>0.

A139755 Table of q-derangement numbers of type A, by rows.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 3, 5, 7, 8, 8, 6, 4, 2, 1, 4, 9, 16, 24, 32, 37, 38, 35, 28, 20, 12, 6, 2, 1, 1, 5, 14, 30, 54, 86, 123, 160, 191, 210, 214, 202, 176, 141, 104, 69, 41, 21, 9, 3, 1, 6, 20, 50, 104, 190, 313, 473, 663, 868, 1068, 1240, 1362, 1417, 1398, 1307, 1157, 968
Offset: 1

Views

Author

Jonathan Vos Post, Jun 13 2008

Keywords

Comments

This sequence is from Table 1.1 of Chen and Wang, p. 2. Abstract: We show that the distribution of the coefficients of the q-derangement numbers is asymptotically normal. We also show that this property holds for the q-derangement numbers of type B.
Number of terms in row n appears to be A084265(n+2). - N. J. A. Sloane, Jul 20 2008
T(n,k) is the number of derangements in the set S(n) of permutations of {1,2,...,n} having major index equal to k. Example: T(4,3)=2 because we have 4312 (descent positions 1 and 2) and 2341 (descent position 3). - Emeric Deutsch, May 04 2009

Examples

			The table begins:
==============================================================================
k=...|.1.|.2.|.3.|..4.|..5.|..6.|..7.|..8.|..9.|.10.|.11.|.12.|.13.|.14.|.15.|
==============================================================================
n=2..|.1.|
n=3..|.1.|.1.|
n=4..|.1.|.2.|.2.|..2.|..1.|..1.|
n=5..|.1.|.3.|.5.|..7.|..8.|..8.|..6.|..4.|..2.|
n=6..|.1.|.4.|.9.|.16.|.24.|.32.|.37.|.38.|.35.|.28.|.20.|.12.|..6.|..2.|..1.|
===============================================================================
Number of terms in rows 2..22: [1,2,6,9,15,20,28,35,45,54,66,77,91,104,120,135,153,170,190,209,231].
From _Paul D. Hanna_, Jun 20 2009: (Start)
For row n=4, the sum over powers of I, a 4th root of unity, is:
1*I + 2*I^2 + 2*I^3 + 2*I^4 + 1*I^5 + 1*I^6 = -1. (End)

Links

Crossrefs

Cf. A000166, A152290.
Cf. diagonals: A141753, A141754.

Programs

  • Mathematica
    T[n_, k_] := SeriesCoefficient[QFactorial[n, q] Sum[(-1)^m q^(m(m-1)/2)/ QFactorial[m, q], {m, 0, n}], {q, 0, k}];
Table[T[n, k], {n, 2, 8}, {k, 1, n(n-1)/2 - Mod[n, 2]}] // Flatten (* Jean-François Alcover, Jul 26 2018 *)
  • PARI
    T(n,k)=if(k<1 || k>n*(n-1)/2-(n%2),0,polcoeff( prod(j=1,n,(1-q^j)/(1-q))*sum(k=0,n,(-1)^k*q^(k*(k-1)/2)/if(k==0,1,prod(j =1,k,(1-q^j)/(1-q)))),k,q)) \\ Paul D. Hanna, Jul 07 2008

Formula

T(n,k) = [q^k] { [n]q! * Sum{m=0..n} (-1)^m*q^(m(m-1)/2) / [m]q! } for n>=2 and 1<k<M(n), where M(n) = number of terms in row n = n*(n-1)/2 - (n mod 2); here, the q-factorial of n is denoted [n]_q! = Product{j=1..n} (1-q^j)/(1-q). - Paul D. Hanna, Jul 07 2008
From Paul D. Hanna, Jun 20 2009: (Start)
For row n>1, the sum over powers of the n-th root of unity = -1:
-1 = Sum_{k=1..n*(n-1)/2} T(n,k)*exp(2*Pi*I*k/n), where I^2=-1.
(End)

Extensions

More terms from Paul D. Hanna, Jul 07 2008

A152291 a(n) = (n+1)^floor((n-1)/2).

Original entry on oeis.org

1, 1, 1, 4, 5, 36, 49, 512, 729, 10000, 14641, 248832, 371293, 7529536, 11390625, 268435456, 410338673, 11019960576, 16983563041, 512000000000, 794280046581, 26559922791424, 41426511213649, 1521681143169024, 2384185791015625
Offset: 0

Views

Author

Paul D. Hanna, Dec 02 2008

Keywords

Crossrefs

Cf. A152290, A000272.
This is for Coxeter type A what A078707 is for Coxeter type B.

Programs

  • Magma
    [(n+1)^((n-1) div 2): n in [0..30]]; // Vincenzo Librandi, May 31 2015
  • PARI
    a(n)=(n+1)^floor((n-1)/2)
  • PARI
    vector(30, n, n--; (n+1)^((n-1)\2)) \\ Michel Marcus, Jun 01 2015

Formula

Row sums of A152290 at q=-1: a(n) = Sum_{k=0..n(n-1)/2} A152290(n,k)*(-1)^k.
a(n) = denominator((-3+(-1)^n)*((1-sqrt(1+n+1/(1+n)))^n-(1+sqrt(1+n+1/(1+n)))^n)/(8*sqrt(1+n+1/(1+n)))). - Gerry Martens, May 31 2015

A121774 Number of n-bead necklaces with n+1 colors, divided by (n+1), for n>0, with a(0)=1.

Original entry on oeis.org

1, 1, 2, 6, 33, 260, 2812, 37450, 597965, 11111134, 235796238, 5628851294, 149346730841, 4361070182716, 139013934267864, 4803839602537336, 178901440745010273, 7143501829211426576, 304465936544543927890, 13797052631578947368422, 662424832016591020302673, 33591880889828764020700500
Offset: 0

Views

Author

Paul D. Hanna, Aug 20 2006

Keywords

Links

Crossrefs

Cf. A000010, A121773, A056665, A152290.

Programs

  • Mathematica
    a[n_] := DivisorSum[n, EulerPhi[n/#] * (n+1)^(#-1) &] / n; a[0] = 1; Array[a, 20, 0] (* Amiram Eldar, Aug 15 2023 *)
  • PARI
    a(n)=if(n==0,1,(1/n)*sumdiv(n,d,eulerphi(n/d)*(n+1)^(d-1)))
  • PARI
    /* a(n) = Sum_{k=0..[n/2]} A152290(n, n*k):  */
{A152290(n,k)=local(e_q=1+sum(j=1,n,x^j/prod(i=1,j,(q^i-1)/(q-1))),LW_q=serreverse(x/e_q+x^2*O(x^n))/x); polcoeff(polcoeff(LW_q+x*O(x^n),n,x)*prod(i=1,n,(q^i-1)/(q-1))+q*O(q^k),k,q)}
{a(n)=sum(k=0,n\2,A152290(n, n*k))}
for(n=0,20,print1(a(n),", ")) \\ Paul D. Hanna, Jul 18 2013

Formula

a(n) = (1/n)*Sum_{d|n} phi(n/d)*(n+1)^(d-1), for n>0, with a(0)=1.
a(n) = Sum_{k=0..[n/2]} A152290(n, n*k), where A152290 is a triangle of coefficients in a q-analog of the LambertW function. - Paul D. Hanna, Jul 18 2013
a(n) = A121773(n)/(n+1). - Amiram Eldar, Aug 15 2023

A152282 Coefficients in a q-analog of the LambertW function at q=2: A(x) = Sum_{n>=0} a(n)*x^n/faq(n,2) where faq(n,q) = q-factorial of n.

Original entry on oeis.org

1, 1, 4, 43, 1076, 58746, 6772360, 1619251271, 794625904404, 795206398610710, 1615965837952912216, 6649024230536100958062, 55277445682961080929146824, 927088288759058912165347148404, 31329256772332779793923906186541200
Offset: 0

Views

Author

Paul D. Hanna, Dec 02 2008

Keywords

Examples

			G.f.: A(x) = 1 + x + 4/3*x^2 + 43/21*x^3 + 1076/315*x^4 + 58746/9765*x^5 +...
G.f. satisfies: A(x) = e_q( x*A(x), 2) where the q-exponential series is:
e_q(x,q) = 1 + x + x^2/faq(2,q) + x^3/faq(3,q) +...+ x^n/faq(n,q) +...
The q-factorial of n is faq(n,q) = Product_{k=1..n} (q^k-1)/(q-1):
faq(0,q)=1, faq(1,q)=1, faq(2,q)=(1+q), faq(3,q)=(1+q)*(1+q+q^2), faq(4,q)=(1+q)*(1+q+q^2)*(1+q+q^2+q^3), ...
Also, the logarithm of the g.f. begins:
log(A(x)) =  A(x)*x/(2-1) - A(x)^2*x^2/(2*(2^2-1)) + A(x)^3*x^3/(3*(2^3-1)) - A(x)^4*x^4/(4*(2^4-1)) + A(x)^5*x^5/(5*(2^5-1)) +...

Crossrefs

Cf. A152290, A152283 (q=3).

Programs

  • PARI
    {a(n,q=2)=local(e_q=1+sum(j=1,n,x^j/prod(i=1,j,(q^i-1)/(q-1))),LW_q=serreverse(x/e_q+x^2*O(x^n))/x); polcoeff(LW_q+x*O(x^n),n,x)*prod(i=1,n,(q^i-1)/(q-1))}
  • PARI
    {a(n)=local(A=1+x);for(i=1,n,A=exp(sum(m=1,n,-(A+x*O(x^n))^m*(-x)^m/(m*(2^m-1)))));prod(k=1,n,2^k-1)*polcoeff(A,n)}

Formula

G.f. satisfies: A(x) = e_q( x*A(x), 2) and A( x/e_q(x,2) ) = e_q(x,2) where e_q(x,q) = Sum_{n>=0} x^n/faq(n,q) is the q-exponential function.
G.f.: A(x) = (1/x)*Series_Reversion( x/e_q(x,2) ).
G.f. satisfies: A(x) = exp( Sum_{n>=1} -A(x)^n*(-x)^n/(n*(2^n-1)) ). [Paul D. Hanna, Oct 24 2011]
a(n) = Sum_{k=0..n(n-1)/2} A152290(n,k)*2^k.
a(n) = faq(n,2)*Sum_{pi} n!/((n-k+1)!*Product_{i=1..n} (e(i)!*faq(i,2)^e(i))), where pi runs through all nonnegative integer solutions of e(1)+2*e(2)+...+n*e(n)=n and k=e(1)+e(2)+...+e(n). [Vladeta Jovovic, Dec 03 2008]

A152283 Coefficients in a q-analog of the LambertW function at q=3: A(x) = Sum_{n>=0} a(n)*x^n/faq(n,3) where faq(n,q) = q-factorial of n.

Original entry on oeis.org

1, 1, 5, 92, 5621, 1093236, 663362421, 1242109529088, 7129029760138649, 124860091946887218320, 6652206059042029394600021, 1075572123264132205051327968256, 526826946994724781414669857330392909
Offset: 0

Views

Author

Paul D. Hanna, Dec 02 2008

Keywords

Examples

			G.f.: A(x) = 1 + x + 5/4*x^2 + 92/52*x^3 + 5621/2080*x^4 + 1093236/251680*x^5 +...
G.f. satisfies: A(x) = e_q( x*A(x), 3) where the q-exponential series is:
e_q(x,q) = 1 + x + x^2/faq(2,q) + x^3/faq(3,q) +...+ x^n/faq(n,q) +...
e_q(x,3) = 1 + x + x^2/4 + x^3/52 + x^4/2080 + x^5/251680 +...
The q-factorial of n is faq(n,q) = Product_{k=1..n} (q^k-1)/(q-1):
faq(0,q)=1, faq(1,q)=1, faq(2,q)=(1+q), faq(3,q)=(1+q)*(1+q+q^2), faq(4,q)=(1+q)*(1+q+q^2)*(1+q+q^2+q^3), ...

Crossrefs

Cf. A152290, A152282 (q=2).

Programs

  • PARI
    {a(n,q=3)=local(e_q=1+sum(j=1,n,x^j/prod(i=1,j,(q^i-1)/(q-1))),LW_q=serreverse(x/e_q+x^2*O(x^n))/x); polcoeff(LW_q+x*O(x^n),n,x)*prod(i=1,n,(q^i-1)/(q-1))}

Formula

G.f. satisfies: A(x) = e_q( x*A(x), 3) and A( x/e_q(x,3) ) = e_q(x,3) where e_q(x,q) = Sum_{n>=0} x^n/faq(n,q) is the q-exponential function.
G.f.: A(x) = (1/x)*Series_Reversion( x/e_q(x,3) ).
a(n) = Sum_{k=0..n(n-1)/2} A152290(n,k)*3^k.
a(n) = faq(n,3)*Sum_{pi} n!/((n-k+1)!*Product_{i=1..n} (e(i)!*faq(i,3)^e(i))), where pi runs through all nonnegative integer solutions of e(1)+2*e(2)+...+n*e(n)=n and k=e(1)+e(2)+...+e(n). [From Vladeta Jovovic, Dec 03 2008]

A386011 Total number of inversions in all parking functions of length n.

Original entry on oeis.org

0, 1, 18, 300, 5400, 108045, 2408448, 59521392, 1620000000, 48230748225, 1560833556480, 54591962772204, 2053129541019648, 82648417236328125, 3546584706554265600, 161642713497024891840, 7799116552647941947392, 397183826482614347896737
Offset: 1

Views

Author

Kyle Celano, Jul 14 2025

Keywords

Examples

			a(2)=1 because in the 3 parking functions of length 2 (11, 12, 21), there is 1 inversion: (1,2).

Links

Crossrefs

Cf. A001809, A000272, A240796.

Programs

  • Mathematica
    Table[Binomial[n,2] * n*(n+1)^(n-2)/2, {n, 0, 18}]

Formula

a(n) = binomial(n,2) * n*(n+1)^(n-2)/2.
a(n) = Sum_{k=0..binomial(n,2)} A152290(n,k)*k.
a(n) = binomial(n,2)*A055865(n)/2.
Showing 1-9 of 9 results.

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