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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?

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I'm a little confused - are the general terms $a_{2r}=a_{2r-1}+a_{2r-3}$ and $a_{2r+1}=a_{2r}+a_{2r-1}$? If so then you can solve the general $a_{2r+1}=2a_{2r-1}+a_{2r-3}$, which gives the odd numbers in the sequence - and then feed that into $a_{2r}=a_{2r-1}+a_{2r-3}$ to give the even numbers. –  Mark Bennet Oct 28 '12 at 4:13
    
The bit about fractions seems to be a distraction. Linearity means you can multiply everything in your first sequence by 4, and it still satisfies the recurrence. You are then looking at an integer sequence mod 4, which will be ultimately periodic. If it hits 0 mod 4, your fractions give you an integer. –  Mark Bennet Oct 28 '12 at 4:16
    
Dear Mark. Do you imply the sequence above has infinite terms which are integers? If that is the case then a multiple sqrt2 is equal to a rational number. The k_20=12875/4 and 12875/4≈2276*sqrt2. –  Vassilis Parassidis Oct 28 '12 at 5:00
    
You start with two rational numbers $a_1$ and $a_2$ - the formulae you have determine the rest, so they are all rational. Solving the recurrence expresses $a_n$ in the form $A_i\alpha^n+B_i\beta^n$ - where $i$ represents the parity of $n$. This formula involves multiples of $\sqrt 2$, but these cancel to give the correct rational answer. –  Mark Bennet Oct 28 '12 at 7:56
    
Because all the terms are determined by $a_1$ and $a_2$ the highest number appearing in any denominator is determined by the least common multiple of the initial denominators (the least common denominator) "d". Multiply the $a_n$ by $d$ and you have a sequence of integers which you are investigating $\mod d$. This is necessarily periodic, and the question is whether it ever hits the residue class equivalent to zero. If it does so once (once the period has started), it will do so infinitely often. –  Mark Bennet Oct 28 '12 at 8:00
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1 Answer 1

One set of general solutions we get for $ a_1=\frac{1}{2n}$ and $ a_2=\frac{1}{n}$ for any positive $n$.

The integer solutions are periodic denominators of the convergents of $\sqrt{2 n^2}$.

The period of the continued fraction of $\sqrt{2 n^2}$ determines the period of the denominators which are your integer solutions.

For example for $n=6$ we get using Wolframalpha:

Denominator[Convergents[Sqrt[2 n^2], 20]]

{1, 2, 33, 68, 1121, 2310, 38081, 78472, 1293633, 2665738, 43945441, 90556620, 1492851361, 3076259342, 50713000833, 104502261008, 1722749176961, 3550000614930, 58522759015841, 120595518646612}

ContinuedFraction[Sqrt[2 n^2], 20]

{8, 2, 16, 2, 16, 2, 16, 2, 16, 2, 16, 2, 16, 2, 16, 2, 16, 2, 16, 2}

Some Pari code generating your sequences, printing the integer solutions.

g(a1,a2,n)={local(v=vector(n)); print1("g[", a1,",", a2,"] = "); v[1]=a1;v[2]=a2; for(i=3,n,v[i]=v[i+i%2-3]+v[i-1]; if(denominator(v[i])==1, print1(",",v[i]); ) );print("");v}

g(1/12,1/6,100);

Output:

g[1/12,1/6] = 2, 68, 2310, 78472, 2665738, 90556620, 3076259342, 104502261008, 3550000614930, 120595518646612, 4096697633369878, 139167124015929240,...

Goodbye!

Maybe this helps a little bit.

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