A181635 Expansion of 36*x^2*(1+36*x^2-6*x) / ((36*x^2+6*x+1)*(1-6*x)^2).

0, 36, 0, 1296, 15552, 46656, 559872, 5038848, 20155392, 181398528, 1451188224, 6530347008, 52242776064, 391820820480, 1880739938304, 14105549537280, 101559956668416, 507799783342080, 3656158440062976, 25593109080440832, 131621703842267136, 921351926895869952, 6317841784428822528, 33168669368251318272, 227442304239437611008

Offset: 1

Author

Florentin Smarandache (smarand(AT)unm.edu), Nov 03 2010

Keywords

• nonn
• easy

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).

Crossrefs

Cf. A181633, A181685.

Programs

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 *)

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

Formula

a(n) = Sum_{q>0, 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

Extensions

Edited by R. J. Mathar, May 13 2016

Name changed by Colin Barker, May 14 2016