OFFSET
1,2
COMMENTS
The previous definition was: Let n=q+q' define any state with q quarks and q' antiquarks, q,q'>0 and q==q' (mod 3). Then a(n) = sum_{q,q'} 6^q*6^q' counts all states allowing q and q' to be any of the 6 quarks or 6 antiquarks. - Colin Barker, May 14 2016
In the following q and a represent any of the 6 quarks or antiquarks.
For n = 1, we have no combination.
For combinations of 2 we have: qa, [mesons and antimesons]; the number of all possible combinations will be 6^2 = 36.
For combinations of n= 7 we have: qqqqqaa, qqaaaaa; the number of all possible combinations will be 6^5*6^2 + 6^2*6^5 =559872.
For combinations of n=8 we have: qqqqaaaa, qqqqqqqa, qaaaaaaa; the number of all possible combinations will be 6^4*6^4 + 6^7*6^1 + 6^1*6^7 = 5038848
For combinations of n=9 we have: qqqqqqaaa, qqqaaaaaa; the number of all possible combinations will be 6^6*6^3 + 6^3*6^6 = 2*6^9 = 20155392.
For combinations of n=10 we have: qqqqqqqqaa, qqqqqaaaaa, qqaaaaaaaa; the number of all possible combinations will be 3*6^10 = 181398528.
If n is even, n=2k, then its pairs are: (k+3p,k-3p), where p is an integer such that both k+3p > 0 and k-3p > 0.
If n is odd, n=2k+1, then its pairs are(k+3p+2,k-3p-1), where p is an integer such that both k+3p+2 > 0 and k-3p-1 > 0.
LINKS
Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (6,0,216,-1296).
FORMULA
a(n) = sum_{q,q'>0 , q+q'=n, q==q' (mod 3} 6^(q+q').
G.f.: 36*x^2*(1+36*x^2-6*x) / ( (36*x^2+6*x+1)*(1-6*x)^2 ). - Joerg Arndt, Mar 16 2013
From Colin Barker, May 14 2016: (Start)
a(n) = (-2^n*3^(1+n)+(-3-i*sqrt(3))*(-3-3*i*sqrt(3))^n-3*(-3+3*i*sqrt(3))^n+i*sqrt(3)*(-3+3*i*sqrt(3))^n+2^n*3^(1+n)*n)/9 where i is the imaginary unit. - Colin Barker, May 14 2016
a(n) = 6*a(n-1)+216*a(n-3)-1296*a(n-4) for n>4.
(End)
E.g.f.: 1 + ((18*x - 3)*exp(9*x) - 4*sqrt(3)*cos(Pi/6-3*sqrt(3)*x))*exp(-3*x)/9. - Ilya Gutkovskiy, May 14 2016
a(n) = 6^n*A008611(n-2). - R. J. Mathar, May 14 2016
MAPLE
A181635 := proc(n)
res := 0 ;
for q from 1 to n-1 do
a := n-q ;
if modp(a, 3) = modp(q, 3) then
res := res+6^n;
end if;
end do:
res;
end proc:
seq(A181635(n), n=1..40) ; # R. J. Mathar, May 13 2016
MATHEMATICA
LinearRecurrence[{6, 0, 216, -1296}, {0, 36, 0, 1296}, 40] (* Harvey P. Dale, Jul 18 2024 *)
PROG
(PARI) a(n) = round((-2^n*3^(1+n)+(-3-I*sqrt(3))*(-3-3*I*sqrt(3))^n-3*(-3+3*I*sqrt(3))^n+I*sqrt(3)*(-3+3*I*sqrt(3))^n+2^n*3^(1+n)*n)/9) \\ Colin Barker, May 14 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Florentin Smarandache (smarand(AT)unm.edu), Nov 03 2010
EXTENSIONS
Edited by R. J. Mathar, May 13 2016
Name changed by Colin Barker, May 14 2016
STATUS
approved