A linear recurrence problem -- sum of rational and not integer terms of sequence. (Try 2)
What is the numerator of the sum of the rational and not integer terms generated by the linear recurrence, including the initialization terms, reduced to co-prime numerator and denominator?
a 0 = 3 7 5 , a 1 = − 7 1 , and a 2 = 3 1 , and a n = 7 7 7 a n − 1 − 7 7 7 a n − 3 where n ≥ 3 .
Assistance: a 3 = 1 5 4 and therefore would not be included in the answer as it is an integer.
Clarification: By rational and not integer terms, I meant rational numbers with denominators = 1 . I understand rational numbers as being written in the form of fractions with an integer numerator and a positive integer denominator.
Apology: In the deleted form of this problem, I had reversed the two arguments of the linear recurrence . This resulted in the answer being different from what I intended.
The answer is 253.
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2 solutions
Exactly! And that is the answer.
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Aha - this was much harder yesterday, when there were infinitely many non-integer terms, and there was no obvious solution to the recurrence relation! Led to some interesting ideas, though, I may see if I can work out a tractable problem in that form.
I regret the bad problem. I also learned a few things from my errors.
The only term which are rational and also not integer are the initialization terms. After the first six terms, all terms contain factors of 3, 7, and 37 and therefore all following terms will be integers. Here is a table of the first eight terms:
⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎛ ( 5 3 7 1 − 1 ) ( − 1 7 1 − 1 ) ( 3 − 1 ) ⎝ ⎛ 2 7 1 1 1 1 1 ⎠ ⎞ ⎝ ⎜ ⎜ ⎛ 3 1 3 3 7 8 3 1 1 1 1 ⎠ ⎟ ⎟ ⎞ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎛ 2 7 2 3 3 7 7 3 1 0 7 1 1 1 1 1 1 ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎞ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎛ 2 3 5 7 1 9 3 7 6 9 9 7 2 1 2 2 1 1 1 ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎞ ⎝ ⎜ ⎜ ⎛ 3 7 3 7 6 5 1 4 1 9 6 2 1 2 1 2 1 ⎠ ⎟ ⎟ ⎞ 3 7 5 − 7 1 3 1 1 5 4 1 1 9 7 6 9 9 3 0 6 0 2 5 4 7 2 3 0 7 6 9 7 7 0 0 5 6 1 8 2 9 8 8 0 5 2 3 8 7 ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎞
The only terms of the sequence that are not integers are a 0 , a 1 , a 2 . To see this, just note that each of their denominators is a factor of 7 7 7 ; it immediately follows that a 3 , a 4 , … onwards are integers.
We have a 0 + a 1 + a 2 = 7 7 7 2 5 3 in lowest terms, with numerator 2 5 3 .