cp's OEIS Frontend

A238530 Position of first occurrence of n in A238529 (Recursive depth of n modulo sopfr(n)).

%I A238530 #36 May 10 2021 06:01:48
%S A238530 2,8,22,166,778,4962,29922,179682,688078,7060198,42361338,674524645
%N A238530 Position of first occurrence of n in A238529 (Recursive depth of n modulo sopfr(n)).
%e A238530 The depth of 22 is 3 because 22->9->3, that is, 22 mod (11 + 2) = 9, 9 mod (3 + 3) = 3, and 3 mod (3) = 0, and 22 is the smallest number to have a depth of 3.
%e A238530 a(10) = 7060198, because 7060198->3530097->392185->78417->8665->1713->565->93->25->5
%o A238530 (Python)
%o A238530 def primfacs(n):
%o A238530     i = 2
%o A238530     primfacs = []
%o A238530     while i * i <= n:
%o A238530         while n % i == 0:
%o A238530             primfacs.append(i)
%o A238530             n = n / i
%o A238530         i = i + 1
%o A238530     if n > 1:
%o A238530         primfacs.append(n)
%o A238530     return primfacs
%o A238530 def sopfr(n):
%o A238530     plist = list(primfacs(n))
%o A238530     l = len(plist)
%o A238530     s = 0
%o A238530     while l > 0:
%o A238530         s += plist[l - 1]
%o A238530         l -= 1
%o A238530     return s
%o A238530 def sd(n):
%o A238530     d = 1
%o A238530     s = n % sopfr(n)
%o A238530     if s > 1:
%o A238530         d += sd(s)
%o A238530     return d
%o A238530 n=2
%o A238530 max=1000
%o A238530 rec = 0
%o A238530 lst = []
%o A238530 while n <= max:
%o A238530     r = sd(n)
%o A238530     if r > rec:
%o A238530         lst.append(n)
%o A238530         rec = r
%o A238530     n += 1
%o A238530 print(lst)
%K A238530 nonn,more
%O A238530 1,1
%A A238530 _J. Stauduhar_, Feb 28 2014
%E A238530 a(12) from _Michel Marcus_, Mar 26 2014