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Based on my research I have my own hypothesis about primes.

The arithmetic mean of each subset of primes that fall between consecutive terms in the Fibonacci sequence, divided by the upper bound of the subset that is the greater Fibonacci number, gives approximately half the golden number Phi.

I have created two applications to find this relationship. One work in graphical mode and shows distribution of primes on Fibonacci spiral, and second works only in text mode.

Screenshots from graphic mode app.

Primes on Fibonacci Spiral 1

Primes on Fibonacci Spiral 2

Results from text mode app.

Some results from my console application: (Half of Phi was doubled for better comparision)


THAT APP IS CALCULATING THE PRIME NUMBERS SUBSETS BETWEEN FIBONACCI AND FINDING APPROXIMATION____ TO THE GOLDEN NUMBER_PHI._______________________ Author: Sylwester B aka Sylvi91_______________


...


Fib(13) = 233 - Fib(14) = 377 Primes list: 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 Primes: Sum = 6989 Qty = 23 Avg = 303 Approximation to the The Golden Number Phi = 1.607427


Fib(14) = 377 - Fib(15) = 610 Primes list: 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 Primes: Sum = 18165 Qty = 37 Avg = 490 Approximation to the The Golden Number Phi = 1.606557


Fib(15) = 610 - Fib(16) = 987 Primes: Sum = 43635 Qty = 55 Avg = 793 Approximation to the The Golden Number Phi = 1.606890

Fib(16) = 987 - Fib(17) = 1597 Primes: Sum = 109567 Qty = 85 Avg = 1289 Approximation to the The Golden Number Phi = 1.614277

Fib(19) = 4181 - Fib(20) = 6765 Primes: Sum = 1624217 Qty = 297 Avg = 5468 Approximation to the The Golden Number Phi = 1.616556

Fib(27) = 196418 - Fib(28) = 31781 Primes: Sum = 2499980948 Qty = 9738 Avg = 256724 Approximation to the The Golden Number Phi = 1.615577

and without listing of primes.

Fib(30) = 832040 - Fib(31) = 1346269 SUM = 40227359343 QTY = 36981 AVG = 1087784 Approx. to Phi = 1.6159979915 Fib(31) = 1346269 - Fib(32) = 2178309 SUM = 101931466309 QTY = 57909 AVG = 1760200 Approx. to Phi = 1.6161159872 Fib(32) = 2178309 - Fib(33) = 3524578 SUM = 257870996074 QTY = 90550 AVG = 2847829 Approx. to Phi = 1.6159829631 Fib(33) = 3524578 - Fib(34) = 5702887 SUM = 654441400687 QTY = 142033 AVG = 4607671 Approx. to Phi = 1.6159082233 Fib(34) = 5702887 - Fib(35) = 9227465 SUM = 1661677489343 QTY = 222855 AVG = 7456316 Approx. to Phi = 1.6161136347 Fib(35) = 9227465 - Fib(36) = 14930352 SUM = 4221069024488 QTY = 349862 AVG = 12064954 Approx. to Phi = 1.6161647093 Fib(36) = 14930352 - Fib(37) = 24157817 SUM = 10735109882717 QTY = 549903 AVG = 19521824 Approx. to Phi = 1.6161910656 Fib(37) = 24157817 - Fib(38) = 39088169 SUM = 27324559743219 QTY = 865019 AVG = 31588392 Approx. to Phi = 1.6162635809 Fib(38) = 39088169 - Fib(39) = 63245986 SUM = 69592739215201 QTY = 1361581 AVG = 51111714 Approx. to Phi = 1.6162832531 Fib(39) = 63245986 - Fib(40) = 102334155 SUM = 177416424882449 QTY = 2145191 AVG = 82704255 Approx. to Phi = 1.6163568263 Fib(40) = 102334155 - Fib(41) = 165580141 SUM = 452491851513992 QTY = 3381318 AVG = 133821146 Approx. to Phi = 1.6163912555 Fib(41) = 165580141 - Fib(42) = 267914296 SUM = 1155101203883449 QTY = 5334509 AVG = 216533743 Approx. to Phi = 1.6164403784 Fib(42) = 267914296 - Fib(43) = 433494437 SUM = 2949939781199270 QTY = 8419528 AVG = 350368783 Approx. to Phi = 1.6164857174 Fib(43) = 433494437 - Fib(44) = 701408733 SUM = 7539299522749130 QTY = 13298630 AVG = 566923023 Approx. to Phi = 1.6165268447 Fib(44) = 701408733 - Fib(45) = 1134903170 SUM = 19277443065477372 QTY = 21014892 AVG = 917322966 Approx. to Phi = 1.6165660477 Fib(45) = 1134903170 - Fib(46) = 1836311903 SUM = 49319944945730044 QTY = 33227992 AVG = 1484289058 Approx. to Phi = 1.6165979816 Fib(46) = 1836311903 - Fib(47) = 2971215073 SUM = 126244882484469729 QTY = 52565409 AVG = 2401672219 Approx. to Phi = 1.6166263027 Fib(47) = 2971215073 - Fib(48) = 4807526976 SUM = 323315442948007805 QTY = 83198799 AVG = 3886059015 Approx. to Phi = 1.6166561454 Fib(48) = 4807526976 - Fib(49) = 7778742049 SUM = 828390738449096336 QTY = 131744274 AVG = 6287869015 Approx. to Phi = 1.6166801715 Fib(49) = 7778742049 - Fib(50) = 12586269025 SUM = 2123542722014291259 QTY = 208718785 AVG = 10174181121 Approx. to Phi = 1.6167112114 Fib(50) = 12586269025 - Fib(51) = 20365011074 SUM = 5445740631692717113 QTY = 330797447 AVG = 16462462697 Approx. to Phi = 1.6167398718 Fib(51) = 20365011074 - Fib(52) = 32951280099 SUM = 13971588279376661576 QTY = 524513152 AVG = 26637250612 Approx. to Phi = 1.6167657543 Fib(52) = 32951280099 - Fib(53) = 53316291173 SUM = 35859516892843327174 QTY = 831993816 AVG = 43100701235 Approx. to Phi = 1.6167929271 Fib(53) = 53316291173 - Fib(54) = 86267571272 SUM = 92072211900316816991 QTY = 1320232935 AVG = 69739369060 Approx. to Phi = 1.6168154043 Your calculations took 918.00 seconds to run. 64.5 % from 8GB memory usage.

In both apps I have implemented Sieve of Eratosthenes to find primes and iterative function to calculate Fibonacci numbers. Using this method let me calculate subset of primes up to the value Fib(54) = 86 267 571 272. But now I facing two problems. First I have reached limits in memory of my home PC [8GB], because Eratosthenes Sieve is memory consumable. Second I have reached limits of 64 bits variables in standard C language. When the variable limit is possible to get around by using, a external library for large numbers, the memory limit does not allow you to move forward too much in the calculations.

I have two questions now.

1.) Is any other way to continue proving my findings? I'd like to check this for primes in between really large numbers, for example Fib (250) = 7896325826131730509282738943634332893686268675876375 and Fib(251) = 12776523572924732586037033894655031898659556447352249 . How to do this?

2.) Is this correct and easy to understand in form of LaTex?

$$ \displaystyle{ \frac12 \Phi \approx \frac{ \frac{p_{1} + p_{2} + ... + p_{x}}{x}}{ F{n} }} $$

Where:

$$ \displaystyle{ F{n-1} < (p_{1}, p_{2}, ... , p_{x}) \le F{n}} $$

$p_{1},p_{2},p_{x}$ - Prime numbers

$(p_{1}+p_{2}+p_{x})$ - sum of Prime numbers in a subset

$Fn-1$ - lesser value of Fibonacci sequence (lower border of primes subset)

$Fn$ - greater value of Fibonacci sequence (higher border of primes subset)

$x$ - qty of prime numbers in a subset

$\Phi$- Golden number Phi = 1.618033988749895...

Thanks in advance. Sorry for mistakes, but english is my second language.

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Hint: The average of the primes between two consecutive Fibonacci numbers $\approx$ the average of the two Fibonacci numbers:

$(987+610)/2/987\cdot 2=(987+610)/987.0= 1.61803444782$

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  • $\begingroup$ Actually not. The inaccuracy is quite significant and increases with increasing numbers. For Fib (30) and Fib (31) the mean is 1089154.5 but the average of the prime numbers in the subset is 1087784, and for Fib (53) and Fib (54) 69791931222.5 compare to 69739369060. It is hard to say that these values ​​are approximate. Other thing. The sum of two consecutive numbers of the Fibonacci sequence divided by the greater value will always give an approximation to the golden number Phi because it's golden ratio. $\endgroup$ Mar 10 at 16:41
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    $\begingroup$ $1089154/1087784.0=1.00125944121259...$ $69791931222.5/69739369060.0=1.00075369426492485...$ The other thing is exactly my point. The result of your operation will generate the golden ratio Phi because your calculation is getting closer to my calculation for greater value. $\endgroup$
    – pietfermat
    Mar 10 at 18:35
  • $\begingroup$ I've got your point of view. But I'm not sure about this because I haven't counted such large primes to prove it. I can only suppose that every next subset of primes has more numbers and larger sum and larger mean. If so, my hypothesis may be true. Thanks for your hint. $\endgroup$ Mar 10 at 19:35

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