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SOLVED: This is false

Beginning with 3, add the next consecutive prime (2) and then take that sum (5) and add that to next consecutive prime (3) to get (8), and so on...

$$ 3 + 2 = 5 $$ $$ 5 + 3 = 8 $$ $$ 8 + 5 = 13 $$ $$ 13 + 7 = 20 $$ $$ 20 + 11 = 31 $$ $$ 31 + 13 = 44 $$ $$ 44 + 17 = 61 $$ $$ 61 + 19 = 80 $$ $$ 80 + 23 = 103 $$

Could you prove that every odd number in this sequence is prime AND that every even number (except 2) minus 1 is prime as well? i.e (8-1=7)(20-1=19)

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6  
There is an error on the third line. It should be $5+5=10$. This may influence your conjectures. –  ant11 Aug 11 at 21:27
2  
It should contain $2+3+5=10$, which is already a contradiction, since $10-1$ is not prime. $8$ is not in this sequence. –  Thomas Andrews Aug 11 at 21:33

2 Answers 2

up vote 4 down vote accepted

I will assume the suggested sequence is:

Starting with $5$, add primes starting from $3$; equivalently, starting from $3$, add primes starting from $2$, the first one.

However, the $21$st partial sum of primes is $712$, and $712+3=715$ which is obviously divisible by $5$.

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1  
Ah, you worked with the sequence he wrote down instead of the sequence he said. I like your answer, especially as we cover both reasonable interpretations of his question. +1 to you –  mixedmath Aug 11 at 21:38
    
This is the correct intended sequence. Thanks! I should of checked it more myself :\ Will keep working at it! –  quanta Aug 11 at 23:06

I love elementary conjectures, and I always try to encourage students in my elementary number theory classes to conjecture away, right or wrong. Here, it happens that both of your conjectures are wrong. The first several primes are

$$2,3,5,7,11,13,17,19,23,$$

with partial sums

$$2,5,10,17,28,41,58,\color{#aa0000}{77}, 100.$$

I've highlighted $77$ in $\color{#aa0000}{\text{red}}$ because it's not prime. Also, every even number except for $2$ on this list fails your second conjecture, as $9, 27, 57,$ and $99$ are all composite.

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