[seqfan] Bivariate recurrence leading to univariate one with sqrt()
Georgi Guninski
guninski at guninski.com
Tue Mar 1 12:46:13 CET 2011
Define the efficiently computable bivariate recurrence:
T(i,j)=j if i<=1
T(i,j)=i if j<=1
T(i,j)=2*T(i/2,j) if i mod 2 = 0
T(i,j)=2*T(i,floor(j/2)) otherwise
and a(n)=T(n,n)
Empirically a(n) satisfies the univariate recurrence with offset 0:
if n != 2^k and n != 2^k+1
a(n)=-1/2*sqrt(36*(a(n - 3) - 2*a(n - 2))*a(n - 1) + 28*a(n - 3)^2 - 108*a(n - 3)*a(n - 2) + 105*a(n - 2)^2 + 12*a(n - 1)^2) - 3*a(n - 3) + 9/2*a(n - 2)
else
a(n)=1/2*sqrt(36*(a(n - 3) - 2*a(n - 2))*a(n - 1) + 28*a(n - 3)^2 - 108*a(n - 3)*a(n - 2) + 105*a(n - 2)^2 + 12*a(n - 1)^2) - 3*a(n - 3) + 9/2*a(n - 2)
For the first 510 terms a(n+1)=2 A097053(n) for n>2
superseeker finds generating function with somewhat large coefficients.
fricas finds strangely looking ADE.
The sequence starts:
1, 1, 4, 6, 16, 20, 24, 28, 64, 72, 80, 88, 96, 104, 112, 120, 256, 272, 288
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