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)).

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, 46, 48, 52, 49, 54, 53, 51, 47, 50, 45, 46, 48, 52, 49, 54, 53, 51, 47, 50, 45, 46, 48, 52, 49, 54, 53, 51, 47, 50, 45, 46, 48, 52, 49, 54, 53, 51, 47, 50, 45

Offset

1,2

Comments

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.

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.

Links

Table of n, a(n) for n=1..71.

Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,-1,1).

Formula

From Chai Wah Wu, Jan 20 2022: (Start)

a(n) = a(n-1) - a(n-5) + a(n-6) for n > 26.

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)

Example

If starting with 12, the next number will be 12 + ((-1)*1 + (1)*2) = 13.
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.
At this point it will loop.
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.

Mathematica

NestList[#1 + Total[#2[[1 ;; -1 ;; 2]]] - Total[#2[[2 ;; -1 ;; 2]]] & @@ {#, Reverse@ IntegerDigits[#]} &, 1, 67] (* Michael De Vlieger, Dec 31 2021 *)

Prog

(Python)
def get_next_num(num):
    res = [int(x) for x in str(num)[::-1]]
    sum=0
    for i in range(len(res)):
        if(i%2)==0:
            sum+=res[i]*1
        else:
            sum+=res[i]*-1
    return num+sum
termlen=[]
for i in range(10000):
    starting_num=i
    series=[]
    current_num=starting_num
    c=0
    while(current_num not in series):
        series.append(current_num)
        current_num=get_next_num(current_num)
        c+=1
        if (c>10000000):
            print(i, "doesn't terminate")
            break
    termlen.append(len(series))
import matplotlib.pyplot as plt
import numpy as np
plt.plot(np.array(termlen))
plt.show()
import pandas as pd
s = pd.Series(termlen)
s.describe()

Crossrefs

Sequence in context: A326312 A134162 A244484 * A045776 A102252 A174838

Adjacent sequences: A350411 A350412 A350413 * A350415 A350416 A350417

Keyword

nonn,base

Author

Divyanand Valsan, Dec 30 2021

Status

approved