# [seqfan] Fibonaroot

Eric Angelini Eric.Angelini at kntv.be
Sat Jul 13 17:01:50 CEST 2013

Hello SeqFans,

Consider the set P* as the union of A000040 (the prime numbers)
and A028835 (numbers with a prime Digital Root -- the DR of the
integer N being the iterated sum of the digits of N).

The sequence S(1) herunder has a(1)=1 and a(2)=1 (the seed);
then, from a(3) on, we have for a(n) the sum of the two previous
terms of S.

With a twist.

The twist being that when the said sum does NOT belong to P*, we
replace it with its digital root (DR). Let's see how S starts:

S(1)=1,1,2,3,5,8,13,21,34,...

All terms > 9 are part of P* but now the sum 21+34 is not (=55);
we replace thus 55 by its DR, which is 1:

S(1)=1,1,2,3,5,8,13,21,34,1,...

The sum 34+1 is not part of P* (=35); we replace it by its DR (=8):
S(1)=1,1,2,3,5,8,13,21,34,1,8...

We proceed mechanically like this -- and quickly enter into a loop:

S(1)=1,1,2,3,5,8,13,21,34,1,8,9,17,8,25,6,31,37,68,6,74,8,1,9,1,1,2,...

If we call S(k) a sequence with the seed [1,k], we see that S(2) ends
also in a loop:

S(2)=1,2,3,... (as the fate of a sequence containing two consecutive
terms already seen in a previous sequence is known).

S(3)=1,3,4,7,11,9,2,11,13,6,19,25,8,6,14,20,34,9,43,52,95,147,8,155,...
S(4)=1,4,5,9,14,23,37,6,43,4,47,6,53,59,4,9,13,4,17,21,38,59,97,156,...

I guess S(3) and S(4) will end soon in a loop too -- but is there a
S(k) sequence which doesn't?

[Please forgive me if this is old hat -- and/or filled with typos, as
always]
Best,
É.