, 2026

Suppose the first five terms of the sequence are $x$, $y$, 20, $z$, 57.

The sum of the third term, $20$, and
the fourth term, $z$, is equal to the
fifth term, $57$.

Thus, $20+z=57$, or $z=37$.

Similarly, $y+20=37$, and so the
second term is $y=37-20=17$.

Finally, $x=20-y$, or $x=20-17=3$.

(The first five terms of this sequence are $3$, $17$, $20$, $37$, $57$.)
The first six terms of this Dunbar sequence are $a$, $b$, $a+b$, $a+2b$, $2a+3b$, $3a+5b$.

Thus, we want to find all possible ordered pairs of positive integers
$(a,b)$ for which $a<b$ and $3a+5b=104$.

Since $5b=104-3a$, then $104-3a$ must be a multiple of $5$ (since $5b$ is a multiple of $5$).

The smallest positive integer $a$ for
which $104-3a$ is a multiple of 5 is
$a=3$.

When $a=3$, $5b=104-3(3)=95$ and so $b=\dfrac{95}{5}=19$.

In this case, the first six terms of the sequence are $3$, $19$, $22$, $41$, $63$, $104$.

The second smallest positive integer $a$ for which $104-3a$ is a multiple of $5$ is $a=8$.

When $a=8$, $5b=104-3(8)=80$ and so $b=16$.

In this case, the first six terms of the sequence are $8$, $16$, $24$, $40$, $64$, $104$.

The third smallest positive integer $a$ for which $104-3a$ is a multiple of $5$ is $a=13$.

When $a=13$, $5b=104-3(13)=65$ and so $b=13$.

However, $(a,b)=(13,13)$ does not
satisfy the condition $a<b$, and
for all values of $a$ greater than
$13$, $a>b$.

Thus, there are exactly two ordered pairs $(a,b)$ satisfying the given
conditions.

These are $(3,19)$ and $(8,16)$.
Solution 1:

The first eight terms of this Dunbar sequence are $c$, $d$, $c+d$, $c+2d$, $2c+3d$, $3c+5d$, $5c+8d$, $8c+13d$.

Thus, we want to find all possible ordered pairs of positive integers
$(c,d)$ for which $c<d$ and $(5c+8d)(8c+13d)=41\,440$.

Suppose that we let the sixth, seventh and eighth terms of the sequence
be the positive integers $R$, $S$ and $T$, respectively.

Then $R<S<T$ and $S\times T=41\,440$.

Since $S<T$ and $S\times T=41\,440$, then $S^2<41\,440$ and so $S<\sqrt{41\,440}$ or $S\leq 203$ (since $S$ is a positive integer).

Also, since $R<S$ and $T=S+R$, then $T<2S$.

So, $41\,440=S $\times T<S\times(2S)=2S^2$$, and thus$41\,440<2S^2$or$$\sqrt{\dfrac{41\,440}{2}}<S\$,$ which
gives $144\leq S$.

Summarizing, we have $\$S\times$
T=41\,440$, where$$144\leq S\leq$
203$.

Since $S$ and $T$ are positive integers, then $(S,T)$ is a factor pair of $41\,440=2^5\times5\times7\times37$.

If $S$ is a factor of $2^5\times5\times7\times37$ and $144\leq S\leq 203$, then the possible
values of $S$ are $2^2\times37=148$, $2^5\times5=160$, and $5\times37=185$. Thus, the factor pairs
$(S,T)$ are $(148,280)$, $(160,259)$, and $(185,224)$.

If the seventh term is $S=148$ and
the eighth term is $T=280$, then the
sixth term is $280-148=132$, the
fifth term is $148-132=16$, the
fourth term is $132-16=116$, and the
third term is $16-116=-100$.

Each term must be a positive integer, and so $(S,T)\neq(148,280)$.

For each of the two remaining possible factor pairs $(S,T)$, we similarly work backward in an
attempt to determine $(c,d)$, the
first two terms of the sequence.

We summarize this work in the table that follows.

Factor pair $\boldsymbol{(S,T)}$
Term $\boldsymbol{8}$
Term $\boldsymbol{7}$
Term $\boldsymbol{6}$
Term $\boldsymbol{5}$
Term $\boldsymbol{4}$
Term $\boldsymbol{3}$
Term $\boldsymbol{2}$
Term $\boldsymbol{1}$

$(148,280)$
$280$
$148$
$132$
$16$
$116$
$-100$

$(160,259)$
$259$
$160$
$99$
$61$
$38$
$23$
$15$
$8$

$(185,
224)$
$224$
$185$
$39$
$147$
$-108$

Thus, $(c,d)=(8,15)$ is the only
pair for which the product of the seventh term and the eighth term is
equal to $41\,440$.

Solution 2:

We begin as we did in Solution 1 by letting the seventh term be $S=5c+8d$ and the eighth term be $T=8c+13d$, so $S\times T=41\,440$.

Next, we work backward to determine the first term, $c$, and the second term, $d$, in terms of $S$ and $T$.

With the eighth term equal to $T$ and
the seventh term equal to $S$, the
sixth term is $T-S$.

Then, the fifth term is $S-(T-S)=2S-T$, the fourth term is $(T-S)-(2S-T)=2T-3S$, and the third term is
$(2S-T)-(2T-3S)=5S-3T$.

Finally, the second term is $(2T-3S)-(5S-3T)=5T-8S=d$, and the first
term is $(5S-3T)-(5T-8S)=13S-8T=c$.

Given that $c>0$, it follows
that $13S-8T>0$ or $13S>8T$.

Since $S\times T=41\,440$, then $T=\dfrac{41\,440}{S}$. Substituting, we
get $13S>8\times\dfrac{41\,440}{S}$.

$S$ is a positive integer and so
simplifying, we get $S^2>\dfrac{8\times41\,440}{13}$ or
$S\geq 160$.

Given that $c<d$, it follows
that $13S-8T<5T-8S$ or $21S<13T$.

Substituting $T=\dfrac{41\,440}{S}$,
we get $21S<13\times\dfrac{41\,440}{S}$.

$S$ is a positive integer and so
simplifying, we get $S^2<\dfrac{13\times41\,440}{21}$ or
$S\leq 160$.

Therefore, $S=160$ and $T=\dfrac{41\,440}{160}=259$.

Substituting the values of $S$ and
$T$, the first term is $c=13S-8T=13(160)-8(259)=8$, and the second
term is $d=5T-8S=5(259)-8(160)=15$.

We can confirm that the first $8$
terms of the sequence are $8$, $15$, $23$, $38$, $61$, $99$, $160$, $259$, and that $160\times259=41\,440$.