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A325904 Generator sequence for A100982.

%I A325904 #33 Oct 17 2019 13:57:55
%S A325904 1,0,-3,-8,15,-91,-54,2531,-17021,43035,-66258,1958757,-24572453,
%T A325904 146991979,-287482322,-3148566077,35506973089,-198639977241,
%U A325904 1006345648929,-8250266425561,76832268802555,-517564939540551,1890772860334557,3323588929061820,-104547561696315008,907385094824827328,-6313246535826877248
%N A325904 Generator sequence for A100982.
%C A325904 The name of this sequence is derived from its main purpose as a formula for A100982 (see link). Both formulas below stem from Mike Winkler's 2017 paper on the 3x+1 problem (see below), in which a recursive definition of A100982 and A076227 is created in 2-D space. These formulas redefine the sequences in terms of this 1-D recursive sequence.
%H A325904 Mike Winkler, <a href="https://arxiv.org/abs/1709.03385">The algorithmic structure of the finite stopping time behavior of the 3x + 1 function</a>, arXiv:1709.03385 [math.GM], 2017.
%F A325904 a(0)=1, a(1)=0, a(n) = -Sum_{k=0..n-1} a(k)*binomial(A325913(n)+n-k-2, A325913(n)-2) for n>1.
%o A325904 (Python)
%o A325904 import math
%o A325904 numberOfTerms = 20
%o A325904 L6 = [1,0]
%o A325904 def c(n):
%o A325904     return math.floor(n/(math.log2(3)-1))
%o A325904 def p(a,b):
%o A325904     return math.factorial(a)/(math.factorial(a-b)*math.factorial(b))
%o A325904 def anotherTerm(newTermCount):
%o A325904     global L6
%o A325904     for a in range(newTermCount+1-len(L6)):
%o A325904         y = len(L6)
%o A325904         newElement = 0
%o A325904         for k in range(y):
%o A325904             newElement -= int(L6[k]*p(c(y)+y-k-2, c(y)-2))
%o A325904         L6.append(newElement)
%o A325904 anotherTerm(numberOfTerms)
%o A325904 print("A325904")
%o A325904 for a in range(numberOfTerms+1):
%o A325904     print(a, "|", L6[a])
%o A325904 (SageMath)
%o A325904 @cached_function
%o A325904 def a(n):
%o A325904     if n < 2: return 0^n
%o A325904     A = floor(n/(log(3, 2) - 1)) - 2
%o A325904     return -sum(a(k)*binomial(A + n - k, A) for k in (0..n-1))
%o A325904 [a(n) for n in range(100)] # _Peter Luschny_, Sep 10 2019
%Y A325904 Cf. A020914, A076227, A100982.
%K A325904 sign
%O A325904 0,3
%A A325904 _Benjamin Lombardo_, Sep 08 2019