A very simple property of Mersenne numbers Tony Reix 2009, 11th of
A very simple property of Mersenne numbers. Tony Reix. 2009, 11th of April. â· Version 0.1 â. Let show that: 2q â 1 = 1 + 6(1 + 22 + 24 + 26 + ... + 22qâ1. 2. ) .
A very simple property of Mersenne numbers Tony Reix 2009, 11th of April. I Version 0.1 J q−1
Let show that: 2q − 1 = 1 + 6(1 + 22 + 24 + 26 + ... + 22 2 ) . Which is equivalent to : q−1 q−1 2 2 X X 2q − 1 = 1 + 6 (2i )2 = 1 + 2 × 3 22i (I) i=0
i=0
First : 2q − 1 = 2q − 2 + 1 = 2(2q−1 − 1) + 1 . Since q is an odd number : q − 1 = 2n . Now, show: 22n − 1 = 3
n−1 X
22i (II)
i=0
22n
With n = 1, we have: − 1 = 3 = 3 × 1 = 3(22×0 ) . With n = 2, we have: 22n − 1 = 15 = 3 × 5 = 3(1 + 22×1 ) = 3(1 + 22×(n−1) ) . Thus, property (II) is true for ranks n = 1 and 2 . Suppose that property (II) is true at rank n . Then, at rank n + 1, we have: 22(n+1) − 1 = 4 × 22n − 1 = 22n − 1 + 3 × 22n . And thus: n−1 n X X 2(n+1) 2i 2n 2×0 2×1 2×(n−1) 2×n 2 −1 = 3 2 +3×2 = 3(2 +2 +...+2 +2 =3 22i i=0
i=0
Which shows that the property (II) is also true at rank n + 1 and thus that it is true for any n > 0 . And so the property (I) is true for any odd q > 2 . As an example: 211 − 1 = 1 + 2((25 )2 − 1) = 1 + 2 × 3(1 + 22 + 42 + 82 + 162 ) which produces a nice figure: a square of side 1 + 2 identical squares of side 25 missing a square of side 1 each. And each of the nearly squares is made of 3 times a set of squares of side (2i )2 with i = 0..4.
A property dealing with the order of 3 modulo a Mersenne prime. Tony Reix ([email protected]). ZetaX (AOPS forum). 2008, 8th of March. In May of 2006, ...
order to try proving a faster Primality test of Fermat numbers. Though these properties are detailed in the ... V 2 n = 4(2U2 n + (â1)n) or V 2 n â 8U2 n = (â1)n4. 2 ...
Conjecture 1 (Mersenne numbers) is mine, based on my work on the use of the Cycles of the Digraph under x2 − 2 modulo a Mersenne prime for primality ...
1. Definition of the LLT numbers. Let's say: L(x) = x2 − 2 , L1 = L , Lm = L◦Lm−1 = L◦L◦L... ◦ L. ... C− m the sum of the negative coefficients of the polynomial Lm(x) . ... −2 (mod Fn) for: m > 1 ( LLT.5). C+ m ≡ 3 (mod 10) for: m ≥ 1 ( LLT.6). 2n−1
Fermat theorem, the first part of the conjecture has been proved. Now, how can ... All references to theorems apply to properties of Lucas Sequences as given.
and Si = 2S 2 iâ1 â 1 for i = 1,2,3,... q â 2 . The proof is based on the chapters 4 (The Lucas Functions) and 8.4 (The. Lehmer Functions) of the book âÃdouard ...
G1Ã4+0[4] mod 5 = G1Ã16+0[16] mod 17 = G1Ã256+0[256] mod 257 = 0. G2Ã4+0[4] mod 5 = G2Ã16+0[16] mod 17 = G2Ã256+0[256] mod 257 = 2. G2Ã4+1[4] ...
I'm looking for a complete proof for the following conjecture: Conjecture 1 Mq = 2q â 1 . S0 = 32 + 1/32 ... Here after, (a | b) is the Legendre symbol. All references ...
This paper provides a proof of a LLT-like test for Fermat numbers, based .... Proof: Since p is a prime, and by Fermat little theorem, we have: 2pâ1 â¡ 1 (mod p).
The proof is based on the chapters 4 (The Lucas Functions) and 8.4 (The ... Proof: Since p is a prime, and by Fermat little theorem, we have: 2pâ1 â¡ 1 (mod p).
based on iteration of the quadratic maps x â x2 and x â x2 â2 over a finite .... x does not necessary exist in Fp, but, since it is a quadratic equation, for.
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serie exhibits the same kind of properties than the Fermat numbers. an = 4n + 2n .... The proof of (FLS.7) by Saouter provides nearly all we need for proving ... Since by Fermat little theorem we have: 2pâ1 â¡ 1 (mod p), then Ï also divides pâ1
other, but also to introduce you to new concepts, that you can find in programming languages that you do not know (yet), as functional programming languages.
such as vision or touch, or are there supramodal or amodal ways of ..... geometrical and/or functional properties that are perceived over and above the spatial.
This paper provides the proof of two new primality tests for Fermat numbers, .... Since Vn = 3n + 1 as shown in page 1, and by the previous theorem, we have: Fn ...
Tricky. Standard. Simple. Easy. Very Easy. Modifier. -40%. -30%. -20%. -10%. -. +10%. +20% ..... Mark is moving but not consciously evading the marksman.
