Specifically, I am interested only in partitions with at least the amount of $2$ in each bucket, but intuitively it doesn't matter?

I have no idea how to work with those complex asymptotic growth formulas. Looking for a simple approximation such as: looks exponential, or looks logarithmic.

  • $\begingroup$ How precise do you need this? It decays faster than $e^{ -n}$. $\endgroup$ – quid Oct 7 '16 at 0:01
  • $\begingroup$ @quid, the sequence I need an approximation or bound of is: $1, 1, 1, 1, 2, 4, 4, 6, 6, \dots$ are you saying that it eventually decays? $\endgroup$ – Shine On You Crazy Diamond Oct 7 '16 at 0:06
  • $\begingroup$ @quid, it's to count these babies: math.stackexchange.com/questions/1957311/… $\endgroup$ – Shine On You Crazy Diamond Oct 7 '16 at 0:07
  • $\begingroup$ I have no idea what your sequence should be. Anyway $3$ has $3$ partitions, namely $1+1+1$, $1+2$, $3$ while $3!=6$. So the third term of the sequence in OP is $1/2$. $\endgroup$ – quid Oct 7 '16 at 0:09
  • $\begingroup$ @ quid, no 1: {3}, because there must be at least 2 in each bucket. $\endgroup$ – Shine On You Crazy Diamond Oct 7 '16 at 0:10

With each part at least 2, is equivalent to "without any part =1", and is given in http://oeis.org/A002865. GF: Product_{m>1} 1/(1-x^m) recursion :a(n)= p(n)-p(n-1) and a(n) ~ Pi * exp(sqrt(2*n/3)*Pi) / (12*sqrt(2)*n^(3/2))


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