

A271268


Concatenate sum of digits of previous term and product of digits of previous term, starting with 8.


1



8, 88, 1664, 17144, 17112, 1214, 88, 1664, 17144, 17112, 1214, 88, 1664, 17144, 17112, 1214, 88, 1664, 17144, 17112, 1214, 88, 1664, 17144, 17112, 1214, 88, 1664, 17144, 17112, 1214, 88, 1664, 17144, 17112, 1214, 88, 1664, 17144, 17112, 1214, 88, 1664, 17144
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OFFSET

1,1


COMMENTS

Each term is created by calculating the sum of the digits of the previous number, and the product of its digits. The results are concatenated to give the new number. Starting with 8, the second number is 88. The third number is generated as follows: 8+8=16, 8x8=64, which gives 1664. Continuing this way, the 7th number in this sequence becomes 88, equal to the second number of the sequence. Therefore, the pattern 88, 1664, 17144, 17112, 1214, ... repeats itself indefinitely.


LINKS

Peter Kagey, Table of n, a(n) for n = 1..10000


MATHEMATICA

NestList[FromDigits@ Join[IntegerDigits@ Total@ #, IntegerDigits[Times @@ #]] &@ IntegerDigits@ # &, 8, 48] (* Michael De Vlieger, Aug 26 2016 *)
PadRight[{8}, 50, {1214, 88, 1664, 17144, 17112}] (* Harvey P. Dale, Oct 04 2017 *)


PROG

(Haskell)
a271268 = 8 : cycle [88, 1664, 17144, 17112, 1214]
 Correction by Peter Kagey, Aug 25 2016
(Python)
from functools import reduce
from operator import mul
def product(seq):
return reduce(mul, seq, 1)
def conversion(n):
n = str(n)
return str(sum(int(i) for i in n)) + \
str(product(int(i) for i in n))
def a271268(n):
if n == 1:
return 8
else:
r = 8
while n > 1:
r = conversion(r)
n = 1
return int(r)


CROSSREFS

Cf. A271220.
Sequence in context: A036917 A003497 A051605 * A006750 A082783 A137144
Adjacent sequences: A271265 A271266 A271267 * A271269 A271270 A271271


KEYWORD

nonn,base


AUTHOR

Sander Claassen, Apr 03 2016


STATUS

approved



