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Old 2020-07-09, 01:19   #12
VBCurtis
 
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"Curtis"
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Quote:
Originally Posted by drmurat View Post
how big number is not important it gives correct valie
Prove it.
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Old 2020-07-09, 05:02   #13
drmurat
 
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Quote:
Originally Posted by VBCurtis View Post
Prove it.
I wiill try to prove . if anyone show me a sample . which is not correct . I will glad to see that sample
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Old 2020-07-09, 16:29   #14
Happy5214
 
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Quote:
Originally Posted by drmurat View Post
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 View Post
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)
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Old 2020-07-09, 17:32   #15
drmurat
 
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Quote:
Originally Posted by Happy5214 View Post
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
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Old 2020-07-10, 11:34   #16
drmurat
 
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any idea . is this proof ?
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