cp's OEIS Frontend

A375495 a(n) = number of different ways of selecting the minimum number of operations chosen from f(x) = 3x+1 and g(x) = floor(x/2) needed to reach n when starting from 1.

%I A375495 #28 Jun 02 2025 15:27:58
%S A375495 1,1,1,2,1,4,1,1,6,1,3,1,14,1,2,5,5,1,28,1,1,4,1,9,5,9,1,48,1,1,2,3,
%T A375495 10,1,1,15,5,23,12,2,1,131,1,3,1,4,6,3,20,5,2,1,1,27,5,43,34,25,1,4,1,
%U A375495 1,332,1,5,5,2,1,10,8,12,3,37,5,5,4,10,2,1,1,39,5,63,68,67
%N A375495 a(n) = number of different ways of selecting the minimum number of operations chosen from f(x) = 3x+1 and g(x) = floor(x/2) needed to reach n when starting from 1.
%C A375495 The minimum number of operations is A375494(n) and that minimum is attained by a(n) different sequences of operations.
%o A375495 (Python)
%o A375495 from itertools import product
%o A375495 seq = [None for _ in range(200)]
%o A375495 num = [   0 for _ in range(len(seq))]
%o A375495 for L in range(0, 23):
%o A375495    for P in product((True, False), repeat=L):
%o A375495       x = 1
%o A375495       for upward in P:
%o A375495          x = 3*x+1 if upward else x//2
%o A375495       if x < len(seq):
%o A375495          if num[x] == 0 or L < seq[x]:
%o A375495             seq[x], num[x] = L, 1
%o A375495          elif L == seq[x]:
%o A375495             num[x] += 1
%o A375495 print(', '.join([str(x) for x in num]))
%Y A375495 Cf. A375494 (number of operations), A375496 (indices of 1's).
%K A375495 nonn
%O A375495 0,4
%A A375495 _Russell Y. Webb_, Aug 18 2024