cp's OEIS Frontend

A350414 a(1)=1; for n > 1, a(n) = a(n-1) + (sum of odd-indexed digits of a(n-1)) - (sum of even-indexed digits of a(n-1)).

%I A350414 #40 Jan 21 2022 14:35:38
%S A350414 1,2,4,8,16,21,20,18,25,28,34,35,37,41,38,43,42,40,36,39,45,46,48,52,
%T A350414 49,54,53,51,47,50,45,46,48,52,49,54,53,51,47,50,45,46,48,52,49,54,53,
%U A350414 51,47,50,45,46,48,52,49,54,53,51,47,50,45,46,48,52,49,54,53,51,47,50,45
%N A350414 a(1)=1; for n > 1, a(n) = a(n-1) + (sum of odd-indexed digits of a(n-1)) - (sum of even-indexed digits of a(n-1)).
%C A350414 For n > 1, a(n) is obtained from a(n-1) by adding to a(n-1) its odd-indexed digits (ones, hundreds, etc.) and subtracting its even-indexed digits (tens, thousands, etc.). After the first 21 terms, the sequence repeats with period 10.
%C A350414 It has been confirmed that if a(1) were changed to any other number <= 10^6, the resulting sequence would similarly enter a loop. If we let x=a(1) and define f(x) as the number of terms before the repeating part begins, then f(x) is symmetric.
%H A350414 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,-1,1).
%F A350414 From _Chai Wah Wu_, Jan 20 2022: (Start)
%F A350414 a(n) = a(n-1) - a(n-5) + a(n-6) for n > 26.
%F A350414 G.f.: x*(-11*x^25 - 11*x^20 - 11*x^15 - 11*x^13 - 11*x^10 - 11*x^9 - 11*x^8 - 6*x^5 - 8*x^4 - 4*x^3 - 2*x^2 - x - 1)/(x^6 - x^5 + x - 1). (End)
%e A350414 If starting with 12, the next number will be 12 + ((-1)*1 + (1)*2) = 13.
%e A350414 Then the next will be 13 + ((-1)*1 + (1)*3) = 15. This will continue as 19, 27, 32, 31, 29, 36, 39, 45, 46, 48, 52, 49, 54, 53, 51, 47, 50, 45.
%e A350414 At this point it will loop.
%e A350414 When the sequence starts with 1, it will be 1, 2, 4, 8, 16, 21, 20, 18, 25, 28, 34, 35, 37, 41, 38, 43, 42, 40, 36, 39, 45, 46, 48, 52, 49, 54, 53, 51, 47, 50, 45, ... Then this will loop. Hence I showed till 45 If you start with some other case, say a 2 digit number 26. Then, it will be 26, 30, 27, 32, 31, 29, 36, 39, 45, ... Since 45 is part of the previous loop, this will also loop. Picking any 3 digit number, say 155, it will be 155, 156, 158, 162, 159, 164, 163, 161, 157, 160, 155, ... Then it will loop. I tried for numbers up to a million and could see all loop. When plotting, then gave interesting plots with respect to the number of terms they loop after.
%t A350414 NestList[#1 + Total[#2[[1 ;; -1 ;; 2]]] - Total[#2[[2 ;; -1 ;; 2]]] & @@ {#, Reverse@ IntegerDigits[#]} &, 1, 67] (* _Michael De Vlieger_, Dec 31 2021 *)
%o A350414 (Python)
%o A350414 def get_next_num(num):
%o A350414     res = [int(x) for x in str(num)[::-1]]
%o A350414     sum=0
%o A350414     for i in range(len(res)):
%o A350414         if(i%2)==0:
%o A350414             sum+=res[i]*1
%o A350414         else:
%o A350414             sum+=res[i]*-1
%o A350414     return num+sum
%o A350414 termlen=[]
%o A350414 for i in range(10000):
%o A350414     starting_num=i
%o A350414     series=[]
%o A350414     current_num=starting_num
%o A350414     c=0
%o A350414     while(current_num not in series):
%o A350414         series.append(current_num)
%o A350414         current_num=get_next_num(current_num)
%o A350414         c+=1
%o A350414         if (c>10000000):
%o A350414             print(i,"doesn't terminate")
%o A350414             break
%o A350414     termlen.append(len(series))
%o A350414 import matplotlib.pyplot as plt
%o A350414 import numpy as np
%o A350414 plt.plot(np.array(termlen))
%o A350414 plt.show()
%o A350414 import pandas as pd
%o A350414 s = pd.Series(termlen)
%o A350414 s.describe()
%K A350414 nonn,base
%O A350414 1,2
%A A350414 _Divyanand Valsan_, Dec 30 2021