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# interesting theorems

last edited by 4 years ago
` \\ ~~~~   for  oeis edits _-_`

watch DYK 9,12,28&29

http://en.wikipedia.org/wiki/Portal:Mathematics/Did_you_know/28

xmas 2013  distinguishing FP from primes.

for consec integers, (a^2-b^2)=(a+b).    Andrica's conjecture says a-b<sqrt(a)+sqrt(b)  where a,b consec primes

for consec composites, either get 2 or 1  (2 iff skip an intervening odd (prime)).  Eliminating all evens, result is the sequence 9,15,21,25,27,33,35,39,45,49,51,55,57,...

from 17,29,31,41,43,53,67,71,79,89,97,101,103,109,113,  no hit.  These are from consec composites where (a+b) is prime.

However, primes MISSED by the procedure are A079149,    3,5,7,11,13,23,37,47,59,61,73,83,107,...

Proizvolov's identity  ref http://en.wikipedia.org/wiki/Proizvolov%27s_identity

xa − yb = 1  only soln in natl numbers is 3^2-2^3  (Catalan)

Zeckendorf's theorem  ref  http://en.wikipedia.org/wiki/Zeckendorf%27s_theorem

A conjecture by Sun:

Every odd integer n ≥ 5 can be written in the form p + x(x + 1) where p is a prime and x a positive integer.

in other words,  every odd n>5 is a prime plus product of 2 consecutive integers.   Thus, every prime n>5 = a prime plus prod of 2 consecutive integer.

Related, via Lemoine  every odd n>5 = p+2q  where p,q are primes (see A046927)

With primes, (2) useful properties (does incl False hits)

Mod(P,30) must reside in (1,7,11,13.17,19,23,29)

Fermat's Little Theorem  Mod( (2^(p-1), P) =1

I have implemented this in Pari/GP, though size limitation seems ~ 525000000  (5E8).  %correct >99.9

Misc approx

Pi/Euler~ 2*e  ie  5.443  vs  5.437

Euler*(Phi+e) ~ Pi/e + e/sqrt(2)   ie  3.0802  vs  3.0778

Erdos-Borwein ~ (sqrt(2)/Euler+Pi/e )-2   ie  1.6067  vs 1.6058

Mill's constant*2 -1 ~ Phi  ie  1.613 vs 1.618

I am producing prime sequences from concatenated sequences like Mill's prime.  Submissions due shortly, 2+

**** versions of 1  (9/2/2016) ***

It is well known that 0.999....equals 1.  It is interesting to think further.

In the same vein, 1.000...1  should also=1.  Let a=0.999....  and b=1/a=1.000....1

We know  a *1/a =1.   Of course a-a=0.  This implies 1.000...1 - 0.999... should =0

Generally alternate forms of the same number in the same form is troublesome.

However, the basic laws of our math systems doesn't require or presume such forms can't exist.

1/0.999 =1.001001001...

1/0.9999= 1.000100010001...

etc

Similarly, (1+a)-(1-a)=2a  eg (1+0.20)-(1-0.20) = 1.2-0.80=0.40=2*0.20. Note 1/0.8 != 1.20

So, with a=0.000...1, 2a=0.000...2