Since I didn’t get any answers, I will restrict my question to one type of sequence and to prime numbers. The terms of this sequence are formulated as follows:
$a_1+a_2=a_3, a_1+a_3=a_4, a_3+a_4=a_5, a_3+a_5=a_6, …….$ and so on. Let’s have the positive fractions $a_1=1/f$ and $a_2=1/t$ where $f=2$ and $t$ any prime number greater than $2$. If $t=3$ we obtain the following terms $1/2, 1/3, 5/6, 11/12, 21/12, 31/12, 52/12,……$. No term of this sequence is divisible by $12$. The same happens when $t=5$, but when $t=7$ we obtain terms which are divisible by $14$.These terms are $a_8, a_{20}, a_{32}….$. If $t=11$ then the terms $a_{14}, a_{38}, a_{62},…$ are integers. If $t=13, 19, 29$, no term is an integer but if $t=23, 31,…$ we obtain terms which are integers. If $t=17$ then the term $a_{10}$ is divisible by $2t$ and the term $a_{42}=17340002$ is almost divisible, as are the other periodic terms. I applied congruent arithmetic to find which primes will have integer terms with no results, and also modified forms of the binomial which did not help either. The only thing which is constant is the period of the terms which are integer numbers after the first integer term appears. Any direction as to how to approach this problem will be greatly appreciated.
INFORMATIVE ADDENDUM
Pythagoras used rational numbers to place 6 ratios between 1 and 2. These ratios are formulated as follows. The arithmetic mean of 1 and 2 is 3/2 and the harmonic is 4/3. By dividing these two ratios we obtain $9/8$. More ratios are obtained from the powers of $(9/8)^n$, such as $(9/8)^2, (9/8)^3,(9/8)^2*(4/3), (9/8)^3*(4/3)$.
So we have put 6 ratios between 1 and 2: $1, (9/8), (9/8)^2, (4/3), (3/2), (9/8)^2*(4/3), (9/8)^3*(4/3), 2$.
Now we will put 12 ratios between 1 and 2. Let’s take the following sequence which is obtained from the method of Theon $2, 3, 5, 7, 12, 17, 29, 41, 70, 99, 169, 239, 408, 577, 985, 1393, 2378, 3363,…..$.
The first term of this sequence which is divisible by 3 is the term $a_{18}=3363$. The term $a_{17}=2378$ is also divisible by 2. From these we obtain $(3363/2378)≈2^{1/2}$ and $(2378/3363)≈1/2^{1/2}$.Now if we divide the numerators by 2 and the denominators by 3 we obtain $(2378/2)/(3363/3)≈(1/2)/(2^{(1/2)})/3$ or $1189/1121≈(9/8)^{1/2}$.
We can obtain more powers of $1189/1121$ by multiplying and dividing the terms $(a_{17}*a_{25})/(a_{18}*a_{26})$ as follows: $(2378*80782)/(2^2)/[(3363*114243)/(3^2)]=1.12499995$.
The first ratio is formulated by taking the harmonic mean of $1$ and $1.12499995$, which is $d^1=1.058823507$. By taking powers of $d^n$ we obtain the 12 ratios. So we have $1, d^1, d^2, d^3, d^4, d^5, d^6, d^7, d^8, d^9, d^{10}, d^{11}, d^{12}, 2$. If we want to put 24 ratios between 1 and 2 we then take the harmonic mean of $1$ and $1.058823507$ and we repeat the above process. This way we can put 12, 24, 48, 96, etc. ratios between 1 and 2.
Let’s formulate the terms of the sequence $k_1,k_2,\ldots,k_m$ where $m=n+3$ as follows:
$a_1+a_2=a_3$, $a_1+a_3=a_4$, $a_3+a_4=a_5$, $a_3+a_5=a_6$, $a_5+a_6=a_7$, $a_5+a_7=a_8$, . . . . . $a_n+a_{n+1}=a_{n+2}$, $a_n+a_{n+2}=a_{n+3}$.
And let’s have the positive fractions $a_1=1/f$ and $a_2=1/t$ where $f,t$ integers not both square numbers. From these two fractions we can formulate infinite terms of the above sequence. For an infinite number of pairs of fractions the above sequence has terms which are integers and for other pairs not. These integers appear as follows. If the third term is an integer then $2\cdot3+2$ terms is an integer. E.g. if $a_1=1/2$ and $a_2=1/4$ we have the terms $3/4, 5/4, 8/4, 11/4, 19/4, 27/4, 46/4, 65/4, 111/4, 157/4, 268/4$, . . . . .
The terms $k_3=8/4$, $k_{11}=268/4$, $k_{19}=9104/4$ are integers. The same thing happens if we have the pair of fractions $1/2$ and $1/10$ and so on.
My questions are: How can we tell which pairs of fractions will produce integer terms and which not? Also, will these terms which are integers appear indefinitely in these sequences or are there only a finite number of integers?