\\ ~~~~ for oeis edits _-_
watch DYK 9,12,28&29
http://en.wikipedia.org/wiki/Portal:Mathematics/Did_you_know/9 Millenium
http://en.wikipedia.org/wiki/Portal:Mathematics/Did_you_know/12 largest prime
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.
Ref http://Z. W. Sun, On sums of primes and triangular numbers, Journal of Combinatorics and Number Theory 1(2009), no. 1
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
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