A152550 Coefficients in a q-analog of the function [LambertW(-2x)/(-2x)]^(1/2), as a triangle read by rows.
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
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
- Paul D. Hanna, Rows 0 to 30 of the triangle, flattened.
- Eric Weisstein's World of Mathematics, q-Exponential Function.
- Eric Weisstein's World of Mathematics, q-Factorial.
Crossrefs
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
Comments