Ok guys, I was just tinkering around with primes and found that:

$P = ((A\cdot2^n)+1)$

where P is a prime, n is a nonzero positive integer, and A is either a prime or a composite number. What i found is that this formula is feasible to find primes as long as you have the correct combination of primes to form composites for A, and suitable n.

Now, what's interesting is, with P increasingly bigger, i've found that:

Using the formula above, If: $A_1$ denotes a single prime, $A_2$ is a composite formed by two primes, $A_3$ is a composite formed by three primes, etc. Then if $A_k$ is a composite formed by $k$ primes, the number of $P$ with small $k$, would be bigger than the number of $P$ with big $k$ along the number line. (be aware that "big $k$" could mean a prime that is multiplied many times/big power of a prime)

In simple terms, If A is just a single prime, with A and n varied, then the P from that would be more numerous than the P that is formed from a composite A (and the more primes/more power of the same prime used to form A, the less that P would appear). My question is, is this behavior of primes actually hold true? and can anyone prove it?

for example, using the formula, from 2 to 97 (i've tried by hand) there are:

$P(A_1)$ (Prime from a single prime): $2,3,5,7,11,17,23,29,47,53,59,67,83,89,97, $(15 primes)
$P(A_2)$ (Prime from multiplication of two primes): $13,19,31,41,43,61,71,73,79$ (9 primes)
$P(A_3)$ (prime from multiplication of three primes): $37$ (1 prime, with $A_1 = 2\cdot3\cdot3$, and n = $1$)

constructive criticisms and corrections are welcome. Thank you.

  • 3
    Well, if $P$ is just a random odd integre, then as $P$ gets large, so will $A$, and so will the probability the $A$ as more and more prime factors ... Or, if you only test up to $P<100$, you certainly cannot find a case of $A_{26}$ – Hagen von Eitzen Apr 22 '17 at 7:30
  • @HagenvonEitzen thanks for your reply. With $A_26$, i was doing it by hand so i really don't know. But i found that (with the limit of 97), there are actually more P with small prime factors than larger ones (with large primes like 97, there is only one prime factor (3), but, the $2^n$ is large (32)), so an alternative explanation can be made that while P is getting larger, there are more small A's with bigger $2^n$, than ones with bigger prime factors (and small $2^n$ ). but if that is true, i don't know why. – Geoffry Gifari Apr 22 '17 at 7:45

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