Leet Sum

Let's see what $a_{n+3}$ simplifies to in this sequence. $a_{n+3}=\frac{1337+n+3}{a_{n+2}}=$ $\frac{1337+n+3}{\frac{1337+n+2}{a_{n+1}}}=$ $\frac{(1340+n)\cdot{a_{n+1}}}{1339+n}=$ $\frac{(1340+n)\cdot{\frac{1337+n+1}{a_n}}}{1339+n}=$ $\frac{(1340+n)\cdot(1338+n)}{(1339+n)\cdot{a_n}}$ .

Multiplying by $a_n$ gives $a_n\cdot{a_{n+3}}=\frac{(1340+n)\cdot(1338+n)}{(1339+n)}=$ $\frac{(1339+n+1)\cdot(1339+n-1)}{1339+n}=$ $\frac{(1339+n)^2-1}{1339+n}=$ $1339+n-\frac{1}{1339+n}$ .

Now all we want is $\displaystyle\sum_{n=10}^{109}(1339+n-\frac{1}{1339+n})=$ $\displaystyle\sum_{n=10}^{109}(1339+n)-\sum_{n=10}^{109}\frac{1}{1339+n}$ . The first sum is clearly $1339\cdot100+\frac{100\cdot119}{2}=$ $133900+5950=$ $139850$ (only simple arithmetic sequence summation tricks were used here). The second sum is almost irrelevant, because $\displaystyle\sum_{n=10}^{109}\frac{1}{1339+n}<$ $\displaystyle\sum_{n=10}^{109}\frac{1}{1339}=$ $100\cdot\frac{1}{1339}=$ $\frac{100}{1339}<1$ . Therefore, we're subtracting a number less that $1$ from $139850$ , which tells us that $S=\lfloor{x}\rfloor$ where $x\in(139849,139850)$ , and so $S=139849$ . When $S$ is divided by $1000$ , the remainder is $849$ , which is our answer.

Let $P = a_{10} a_{13} + a_{11} a_{14} + a_{12} a_{15} + \ldots + a_{109} a_{112} = \sum_{m=10}^{109} a_m a_{m+3}$ .

The problem states that : $a_n = \frac {1337+n}{a_{n-1}}$ , which is equivalent to: $a_{n-1}a_n = 1337+n$

Plugging in n = m+1, n = m+2, and n = m+3 yields: $a_m a_{m+1} = 1337 + m + 1 = 1338 + m$ $a_{m+1} a_{m+2} = 1337 + m + 2 = 1339 + m$ $a_{m+2} a_{m+3} = 1337 + m + 3 = 1340 + m$

Thus, $\frac {(a_m a_{m+1})(a_{m+2}a_{m+3})} {a_{m+1} a_{m+2}} = a_m a_{m+3} = \frac{(1338+m)(1340+m)}{1339+m}$

$a_m a_{m+3} = \frac{(1339+m)^2 - 1}{1339+m}$

$a_m a_{m+3} = 1339 + m - \frac{1}{1339+m}$

Thus, $P = \sum_{m=10}^{109} (1339 + m) - \sum_{m=10}^{109} \frac{1}{1339 + m}$ $= 1349 + 1350 + \ldots + 1448 - (\frac{1}{1349} + \frac{1}{1350} + \ldots + \frac{1}{1448})$ $= 50 \times 2797 - (\frac{1}{1349} + \frac{1}{1350} + \ldots + \frac{1}{1448})$ $= 139850 - (\frac{1}{1349} + \frac{1}{1350} + \ldots + \frac{1}{1448})$

Note that

$0 < \frac{1}{1349} < \frac{1}{100}$

$0 < \frac{1}{1350} < \frac{1}{100}$

$\ldots$

$0 < \frac{1}{1448} < \frac{1}{100}$

Adding the above inequalities gives us: $0 < \frac{1}{1349} + \frac{1}{1350} + \ldots + \frac{1}{1448} < 1$

Consequently, $139849 < 139850 - (\frac{1}{1349} + \frac{1}{1350} + \ldots + \frac{1}{1448}) < 139850$ $139849 < P < 139850$ . $S = \lfloor a_{10} a_{13} + a_{11} a_{14} + a_{12} a_{15} + \ldots + a_{109} a_{112} \rfloor = \lfloor P \rfloor$ $S = 139849 \equiv \boxed{849} \pmod{1000}$

$a_n = \frac{1337+n}{a_{n-1}}$ $\rightarrow a_{n-1}a_n=1337+n$

$a_na_{n+3} = \frac{a_na_{n+1}a_{n+2}a_{n+3}}{a_{n+1}a_{n+2}}$ $=\frac{(1337+(n+1))(1337+(n+3))}{1337+(n+2)}$ $=\frac{((1339+n)-1)((1339+n)+1)}{1339+n}$ $=\frac{(1339+n)^2-1}{1339+n}$ $=1339+n-\frac{1}{1339+n}$

Let A = $a_{10}a_{13}+a_{11}a_{14}+\ldots+a_{109}a_{102}$

Then A = $\displaystyle \sum_{n=10}^{109} a_na_{n+3}$ $=\displaystyle \sum_{n=10}^{109} (1339+n-\frac{1}{1339+n})$ $=100(1339)+\frac{100(109+10)}{2} - \displaystyle \sum_{n=10}^{109} \frac{1}{1339+n}$ $=139850 - \displaystyle \sum_{n=10}^{109} \frac{1}{1339+n}$

Clearly, A < 139850 and A > $139850 - \frac{100}{1339+10} > 139849$

Hence S = $\lfloor A \rfloor = 139849$ and the remainder when S is divided by 1000 is 849.

Given $a_n$ we can find the next 3 numbers in the sequence as follows: $a_{n+1}=\frac {1337+n+1}{a_{n}}=\frac {1338+n}{a_n}$ $a_{n+2}=\frac {1337+n+2}{a_{n+1}}=\frac {1339+n}{ \frac {1338+n}{a_n}}=\frac {(1339+n) \times a_n}{1338+n}$ $a_{n+3}=\frac {1337+n+3}{a_{n+2}}=\frac {1340+n}{ \frac {(1339+n) \times a_n}{1338+n}}=\frac {(1340+n) \times (1338+n)}{(1339+n) \times a_n}$

To find the answer we have to calculate: $a_n \times a_{n+3}=\frac {(1339+n+1) \times (1339+n-1)}{1339+n}=\frac {(1339+n)^2}{1339+n}-\frac {1}{1339+n}$ $a_n \times a_{n+3}=1339+n- \varepsilon_n$ with $0 \leq \varepsilon_n \leq \frac {1}{1000}$ for all positive integer n.

To get S we need to calculate $\displaystyle \sum_{i=10}^{109} a_n \times a_{n+3}=\displaystyle \sum_{i=10}^{109} 1339+n-\varepsilon_n$ Which is equivalent to $133900+\displaystyle \sum_{i=10}^{109} n-\displaystyle \sum_{i=10}^{109} \varepsilon_n=139850-\gamma$ $\varepsilon_n \leq \frac {1}{1000}$ so $\gamma \leq 100/1000 \leq 1$

With this we see that $S=\left\lfloor 139850 -\gamma \right\rfloor=139849$

The remainder of S divided by 1000 are the last three digits being 849.

First we have that a(n) a(n-1) = 1337 + n a(n-1) a(n-2) = 1336 + n a(n-2)*a(n-3) = 1335 + n

Multiplying the first equation and the third we get to

a(n) a(n-1) a(n-2) a(n-3) = (1337 + n) (1335 +n)

Subistituing the second equation we get that

a(n) a(n-3) = (1337+n) (1335+n)/(1336+n)

But

1335*1337 = 1336^2 - 1

So

a(n)*a(n-3) = 1336+n - 1/(1336+n)

applying that to "n+3"

a(n)*a(n+3) = 1339+n - 1/(1339+n)

Now we can calculate the value of S

1339 + 10 + 1339 + 11 + ... + 1339 + 109 - 1/1349 - 1/1350 - ... - 1/1448

Now note that there are 99 terms ranging from -1/1339 to -1/1448, we can assure that this sum will get to a number between -1 and 0

Now for the other terms

1339 + 10 + 1339 + 11 + ... + 1339 + 109 = 100 1339 + 110 109/2 - 10*9/2 = 133900 + 5950 = 139850

So we have that

S = floor(139850 + x) where x is a number between -1 and 0

S = 139849 = 1000*139 + 849

849 is the remaining

a(1) = 1 a(2) = 1339 a(3) = 1340/1339 a(4) = 1341 1339/1340 a(5) = 1342 1340/(1341(1339) .. From the above sequence, we can find that a(n)*a(n + 3) = (1340 + n)(1338 + n)/(1339 + n) = {(1339 + n)^2 - 1}/(1339 + n) = (1339 + n) - 1/(1339 + n)

So, Summation will be given by:- [1349 + 1350 + ... + 1448 - (1/1349 + 1/1350 + .. + 1/1448)]

Since, (1/1349 + 1/1350 + .. + 1/1448) < 1, we can say that Summation will be given by:- 1349 + 1350 + ... + 1448 - 1 = 2797*50 - 1

So, remainder by 1000 will be 849

Writing out the first few terms of the sequence we get:

• $a(1) = 1$
• $a(2) = 1339$
• $a(3) = 1340/1339$
• $a(4) = (1339\cdot1341) / 1340$
• $a(5) = (1340\cdot1342) / (1339\cdot1341)$
• $a(6) = (1339\cdot1341\cdot1343) / (1340\cdot1342)$
• $a(7) = (1340\cdot1342\cdot1344) / (1339\cdot1341\cdot1343)$ ...
• $a(10) = (1339\cdot1341\cdot1343\cdot1345\cdot1347) / (1340\cdot1342\cdot1344\cdot1346)$ ...
• $a(13) = (1340\cdot1342\cdot1344\cdot1346\cdot1348\cdot1350) / (1339\cdot1341\cdot1343\cdot1345\cdot1347\cdot1349)$

Therefore

• $a(10) \cdot a(13) = (1348\cdot1350) / 1349$
• $a(11) \cdot a(14) = (1349\cdot1351) / 1350$
• $f(k) = a(k) \cdot a(k + 3) = (1338 + k)\cdot(1340 + k) / (1339 + k)$

We now need to find

$\displaystyle \sum_{k=10}^{109} f(k) = \sum_{k=0}^{99} f(k - 10) = \sum_{k=0}^{99} \frac{(1348 + k)\cdot(1350 + k)}{1349 + k}$

Let's generalize this slightly to

$\sum_{k=0}^{q} \frac{(a+k)\cdot(a+2+k)}{a+1+k}$

$\frac{(a+n)\cdot(a+2+n)}{(a+1+n)} = \frac{a^2 + 2a\cdot(n+1) + 2n + n^2}{a+1+n}$

$= \frac{a^2 + a\cdot(n+1) + a\cdot(n+1) + 2n + n^2 + 1 - 1}{a+1+n}$

$= \frac{a\cdot(a+1+n) + (n+1)\cdot(a+1+n) - 1}{a+1+n}$

$= a + (n + 1) - \frac{1}{a+1+n}$

These terms can be summed individually to give a final sum of

$(q+1)\cdot a + \frac{(q+1)\cdot(q+2)}{2} - H(a + q) + H(a)$

where $H(n)$ is the nth harmonic number.

Substituting q = 99 and a = 1348:

$(99+1)\cdot 1348 + \frac{(99+1)\cdot(99+2)}{2} - H(1348 + 99) + H(1348)$

$= 134800 + 5050 - H(1447) + H(1348)$

$H(1447) \approx 7.85480889423$

$H(1348) \approx 7.78396383039$

$134800 + 5050 - H(1447) + H(1348) = 139849.929155...$

Taking the integer part mod 1000 gives the final answer of 849.

by observing the pattern we see that

a(n) a(n + 3) = (1340 + n)(1338 + n)/(1339 + n) where n is from 10 to 109 as (109 is the last term) now we see that the fraction (1338 + n)/(1339 + n) is nearly equal to 1 but less than 1. hence for any number n ⌊n *(1338 + n)/(1339 + n)⌋=n-1 as (1338 + n)/(1339 + n)<1 hence n (1338 + n)/(1339 + n)<n

hence , it is equal to (1340 + n) - 1 = 1339 + n

now we take the ∑ of all the possible values of a

i.e ( 10 + 11 .......... + 109) = 5950

hence the sum in total will be ( required summation ) = (1338)*(100) + 5950 - 101 = 139849 ( -100 as ⌊n *(1338 + n)/(1339 + n)⌋=n-1) and -1 as the value is a fraction in the form 139849.xyz and ⌊139849.xyz⌋=139849 so the last three digits are 849 which is the required solution.

After looking at the first few terms in the sequence, I noticed that they were jumping back and forth between a number close to one, and a number close to 1339, and that number was slowly getting bigger. Ten terms in it was just under 1343, and multiplying that by the thirteenth term gave something just under 1349. The eleventh and fourteenth gave something just under 1350. So I figured I could just consider the new series:

S=1349+1350+....+1448

Which gave S = 139850, but because of the floor function, I just subtracted one. Which gave 139849, and of course dividing by 1000 just means the last three digits would be the remainder. Thus, 849.