, 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\$.$