$S_i$
be a sequence of integers obeying ${S}_{({S_i}+5)}=i1,$
for $i=0,1,2,\ldots$
and $S_0=0.$
Give a nonrecursive expression for (a) one such sequence $S_i,$
(b) all possible such sequences $S_i.$
Different expressions for different subsets of $i=0,1,\ldots$
are allowed, e.g. “$i$
is odd” or “$i$
is even”.

 Let
$a_{i+1}=5i1,$
for$i=0,1,2,\ldots.$
Find$a_5.$
 Let

Why not compute a few terms with the rule given and see if you can spot any patterns?

Which indices of the sequence could you obtain with this method? Is there a pattern to these?

Conjecture a closed form solution for just these indices (part (a)). Could you prove your conjecture?

Have you considered proving your conjecture by induction?

Instead of the usual
$P_n \implies P_{n+1},$
how about trying$P_n \implies Q$
and$Q \implies P_{n+1},$
where both implications are obtained from the rule for$S_i?$

What about the other undefined terms? Can they be defined arbitrarily?

What restrictions are there on the choice of values for
$S_{4k+2}$
and$S_{4k+3}$
terms? 
What happens if
$S_{4k+3} = a = S_{4j+2}?$

What values can’t they take? Why?

Take
$S_{4k+2},$
there are two possible forms, but only one is valid. Could you determine which one it is? 
Determine
$S_{4k+3}$
from$S_{4k+2}.$

Denote the condition
${S}_{({S_i}+5)}=i1$
with$(*).$
We want to define a sequence
$S_i$
such that$(*)$
holds. Let’s look at which terms are actually determined: starting from$S_0 = 0$
we get that$S_{S_0 + 5} = S_5 = 1.$
Then$S_{S_5 + 5} = S_4 = 4$
and$S_{S_4 + 5} = S_9 = 3.$
We can continue this process to get$S_8=8,$
$S_{13}=7,$
$S_{12}=12$
and so on.$S_0$
$S_1$
$S_2$
$S_3$
$S_4$
$S_5$
$S_6$
$S_7$
$S_8$
$S_9$
… $S_{12}$
$S_{13}$
… $0$
$?$
$?$
$?$
$4$
$1$
$?$
$?$
$8$
$3$
… $12$
$7$
… We notice that this method only gives us values for
$S_{4k}$
and$S_{4k+1},$
with the exception of$S_1.$
We could conjecture a possible solution for part (a) that$S_{4k}=4k$
and$S_{4k+1}=(4k+1)6=4k5.$
By our conjecture,$S_1=5$
from which follows that$S_0=0$
as given. Hence, we try to prove our conjecture in Proof 1 below.In order to come up with a generalised solution for part (b), we need to specify values for
$S_{4k+2}$
and$S_{4k+3}$
to define the full sequence$S_n.$
To obtain these, we can try some arbitrary values, but we may discover many cause contradictions with earlier results. We must realise that the subsequences$S_{4k+2}$
and$S_{4k+3}$
must not overlap, and that$S_{4k+2}$
and$S_{4k+3}$
cannot be of the form$4j$
or$4j+3$
(see Proof 2 below). We can choose$S_{4k+2}$
to be any$4j +1 = 4k + 4q + 1$
or any$4j + 2 = 4k + 4q + 2,$
for a fixed$q \in \mathbb{Z}.$
If$S_{4k+2} = 4k + 4q + 1$
then:$$ \begin{aligned} 4k+ 1 \overset{(\ast)}{=} S_{S_{4k+2} + 5} &= S_{4k+ 4q + 6} \\&= S_{4(k+1) + 4q + 2} \\&= 4(k+1+q) + 4q + 1 \\&= 4k + 8q + 5 \end{aligned} $$
from which we must have$1 = 8q + 5,$
which has no integer solutions.So we have to pick
$S_{4k+2} = 4k+ 4q+ 2.$
Then, applying$(\ast)$
we determine$S_{4k+3}$
as well:$$ 4k+ 1\overset{(\ast)}{=} S_{S_{4k+2} + 5} = S_{4k+4q+7} = S_{4(k+1) + 4q + 3} $$
So to obtain$S_{4k+3},$
we substitute$kq1$
for$k$
to obtain$S_{4(kq1+1) + 4q + 3} = S_{4k+3}$
$= 4(kq1)+1.$
Hence the only valid solutions are:$$ \begin{aligned} S_{4k+2} &= 4k+ 4q+ 2\\ S_{4k+3} &= 4k4q  3 \end{aligned} $$
where$q \in \mathbb{Z}$
is arbitrary. This gives us the general solution:$$ S_i = \begin{cases} i &\text{ if } i \equiv 0 \mod 4 \\ i  6 &\text{ if } i \equiv 1 \mod 4 \\ i + 4q &\text{ if } i \equiv 2 \mod 4 \\ i  6  4q &\text{ if } i \equiv 3 \mod 4 \end{cases} $$
Proof 1
We can show this by first assuming
$S_{4k+1}=4k5,$
from which we deduce that${S}_{({S_{4k+1}}+5)}={S}_{(4k5+5)}=S_{4k}.$
And then from$(*)$
we know${S}_{({S_{4k+1}}+5)}=4k,$
which shows that$S_{4k+1}=4k5 \implies S_{4k}=4k.$
Next, we assume
$S_{4k}=4k,$
from which we can deduce$S_{(S_{4k}+5)}=S_{4k+5}=S_{4(k+1)+1}.$
And from$(*)$
we know that$S_{(S_{4k}+5)}=4k1=4(k+1)5,$
which shows that$S_{4k}=4k \implies S_{4(k+1)+1}=4(k+1)5.$
By (slightly convoluted) induction, we have proven our conjecture.Proof 2
First of all, we must choose the sets
$\{t : t = S_{4k+2} \text{ for } k \in \mathbb{Z} \}$
and$\{t : t = S_{4k+3} \text{ for } k \in \mathbb{Z} \}$
to be disjoint. If$S_{4k+3} = a = S_{4j + 2}$
then$S_{S_{4k+3} + 5} = 4k+2$
and$S_{S_{4j+2} + 5} = 4j+1.$
By$(\ast)$
both of these values must be equal to$S_{a + 5},$
which is a contradiction.Secondly, for every
$k$
we must have that$S_{4k+2}, S_{4k+3} \notin \{t : t \equiv 0 \mod 4 \} \cup \{ t : t \equiv 3$
$\mod 4\}.$
Take the$S_{4k+2}$
case: if$S_{4k+2} = 4j$
then$4k+1 \overset{(\ast)}{=} S_{S_{4k+2} + 5}$
$= S_{4j+5} = S_{4(j+1) + 1}.$
By the above, this final term is equal to$4\big((j+1) 1\big)  1 = 4j 1$
, which cannot equal$4k+1.$
Similarly if$S_{4k+2} = 4j+3$
then$4k+1 \overset{(\ast)}{=} S_{S_{4k+2} + 5} = S_{4(j+1)}$
, which by the above is equal to$4(j+1).$
Once again, we obtain a contradiction. The$S_{4k+3}$
case is similar.