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2020-07-09, 01:19   #12
VBCurtis

"Curtis"
Feb 2005
Riverside, CA

52·173 Posts

Quote:
 Originally Posted by drmurat how big number is not important it gives correct valie
Prove it.

2020-07-09, 05:02   #13
drmurat

"murat"
May 2020
turkey

3516 Posts

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

2020-07-09, 16:29   #14
Happy5214

"Alexander"
Nov 2008
The Alamo City

17816 Posts

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:

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

2020-07-09, 17:32   #15
drmurat

"murat"
May 2020
turkey

1101012 Posts

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

 2020-07-10, 11:34 #16 drmurat   "murat" May 2020 turkey 3516 Posts any idea . is this proof ?

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