
FORMULA

From R. J. Mathar, Mar 22 2009: (Start)
The sequence is a hybrid of two sequences at the even and odd indices with linear recurrences individually, therefore a linear recurrence in total.
For even n the Gardner reference gives the formula a(n)=n(2n^25)/3+2, which is
4,38,136,330,652,1134,1808,2706,3860,5302, n=2,4,6,8,...
with recurrence a(n)= 4 a(n1) 6 a(n2) +4 a(n3)  a(n4) and therefore with g.f. 2*(211*x4*x^2+x^3)/(x1)^4 (offset 0) (see A152110).
For n odd the Gardner reference gives a(n)= n(2n^25)/3+1, which is
0,14,76,218,472,870,1444,2226,3248,4542,6140,8074,10376,13078, n=1,3,5,7,...
with the same recurrence and with g.f. 2*x*(710*x+x^2)/(x1)^4 (offset 0).
Since the first zero does not match the sequence and should be 1, we add 1 to the g.f.:
1,14,76,218,472,870,1444,2226,3248,4542,6140,8074,10376,13078,... (see A152100),
g.f.: 12*x*(710*x+x^2)/(x1)^4.
We "aerate" both sequences by insertion of zeros at each second position,
which implies x>x^2 in the generating functions,
4,0,38,0,136,0,330,0,652,0,1134,0,1808,0,2706,0,3860,0,5302
g.f. 2*(211*x^24*x^4+x^6)/(x^21)^4 (offset 0).
1,0,14,0,76,0,218,0,472,0,870,0,1444,0,2226,0,3248,0,4542,0,6140,...
g.f. 12*x^2*(710*x^2+x^4)/(x^21)^4.
The first of these is multiplied by x to shift it right by one place:
0,4,0,38,0,136,0,330,0,652,0,1134,0,1808,0,2706,0,3860,0,5302
g.f. 2*x*(211*x^24*x^4+x^6)/(x^21)^4.
The sum of these two is
12*x^2*(710*x^2+x^4)/(x^21)^4 2*x*(211*x^24*x^4+x^6)/(x^21)^4 =
(x^55x^4+6x^3+4x^2+x+1)/((x1)^4/(x+1)).
This is exactly the Plouffe g.f. if the offset were 0.
In summary: a(n)= 3 a(n1) 2 a(n2) 2 a(n3) +3 a(n4)  a(n5), n > 6.
a(2n)= 2+2*n*(8n^25)/3, n>=1. a(2n+1)= 2n(1+8n^2+12n)/3, n>=1.
G.f.: x*(x^55x^4+6x^3+4x^2+x+1)/((x1)^4/(x+1)). (End)
