aliquot sum formula - Page 2

VBCurtis · 2020-07-09, 01:19

Quote:

Originally Posted by drmurat

how big number is not important it gives correct valie

Prove it.

drmurat · 2020-07-09, 05:02

Quote:

Originally Posted by VBCurtis

Prove it.

I wiill try to prove . if anyone show me a sample . which is not correct . I will glad to see that sample

Happy5214 · 2020-07-09, 16:29

Quote:

Originally Posted by drmurat

my way is a bit faster
if A = 28 = 4×7 = 2^2 × 7
my formula for number A = 2 ^ n * m ( m is prime)
B = A + 2 * ( 2^ n - 1 ) - ( m - 1 )
B = 28 + 2 * ( 3) - ( 7-1)
B = 28 + 6 - 6
B= 28

A = 2^ 400.000 * 5
B= A + 2 * ( 2^400.000 - 1 ) - ( 5 - 1)

what do you think ?

Quote:

Originally Posted by VBCurtis

Prove it.

Replacing m with p (for my sanity), we get the following:

$B = \displaystyle\sum_{i=0}^n 2^i + p \displaystyle\sum_{i=0}^{n-1} 2^i = (2^{n+1} - 1) + p(2^n - 1) = 2^{n+1} - 1 + p \cdot 2^n - p = p \cdot 2^n + 2^{n+1} - p - 2 + 1 = p \cdot 2^n + 2(2^n - 1) - p + 1 = A + 2(2^n - 1) - (p - 1)$

drmurat · 2020-07-09, 17:32

Quote:

Originally Posted by Happy5214

Replacing m with p (for my sanity), we get the following:

$B = \displaystyle\sum_{i=0}^n 2^i + p \displaystyle\sum_{i=0}^{n-1} 2^i = (2^{n+1} - 1) + p(2^n - 1) = 2^{n+1} - 1 + p \cdot 2^n - p = p \cdot 2^n + 2^{n+1} - p - 2 + 1 = p \cdot 2^n + 2(2^n - 1) - p + 1 = A + 2(2^n - 1) - (p - 1)$

is this proof ? or known rule

drmurat · 2020-07-10, 11:34

any idea . is this proof ?

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
I Think I Have A "Prime Generating Formula" (without the formula)	MathDoggy	Miscellaneous Math	13	2019-03-03 17:11
P-1 formula (help wanted)	Prime95	Data	18	2012-02-12 18:07
New σ for Aliquot	JohnFullspeed	Aliquot Sequences	18	2011-08-20 21:11
serious bug in aliquot.ub	Andi47	Aliquot Sequences	3	2009-03-08 10:18
New LLT formula	hoca	Math	7	2007-03-05 17:41

2020-07-10, 11:34	#16
drmurat "murat" May 2020 turkey 1110011₂ Posts	any idea . is this proof ?