20100422, 03:46  #111 
Aug 2006
13526_{8} Posts 
I'm looking at your posts on the other forum. You write:
Code:
#include <math.h> #include <stdio.h> #include <limits.h> #pragma precision=log10l(ULLONG_MAX)/2 typedef enum { FALSE=0, TRUE=1 } BOOL; BOOL is_prime( int p ){ if( p == 2 ) return TRUE; else if( p <= 1  p % 2 == 0 ) return FALSE; else { BOOL prime = TRUE; const int to = sqrt(p); int i; for(i = 3; i <= to; i+=2) if (!(prime = p % i))break; return prime; } } BOOL is_mersenne_prime( int p ){ if( p == 2 ) return TRUE; else { const long long unsigned m_p = ( 1LLU << p )  1; long long unsigned s = 4; int i; for (i = 3; i <= p; i++){ s = (s * s  2) % m_p; } return s == 0; } } int main(int argc, char **argv){ const int upb = log2l(ULLONG_MAX)/2; int p; printf(" Mersenne primes:\n"); for( p = 2; p <= upb; p += 1 ){ if( is_prime(p) && is_mersenne_prime(p) ){ printf (" M%u",p); } } printf("\n"); } Also, I would be wary of the call to sqrt; it may not round properly for sufficiently large numbers. Just for kicks, here's how I'd code the basic trial division prime tester: Code:
int isprime(int p ) { if(!(p&1)) return p == 2; if(p%3 == 0) return p == 3; if(p < 25) return p > 1; int i, to = (int)sqrt(p + 0.5); for(i = 5; i <= to; i += 6) { if (p%i == 0) return 0; if (p%(i+2) == 0) return 0; } return 1; } Last fiddled with by CRGreathouse on 20100422 at 03:51 
20101212, 20:53  #112 
May 2004
New York City
1088_{16} Posts 
Another possibly interesting variation:
2^2 + 3^3 + 5^5 + ... + p^p = 10^{m}K What is the smallest prime p such that the sum of squares of all primes up to p is a multiple of 10 (or 100 or 1000 and so on). (I think somewhere in these series we'll get numeric pointers to empirical evidence connecting some number theoretical conjectures.) 
20101212, 21:19  #113  
Aug 2006
2·29·103 Posts 
Quote:
Why? 

20101212, 21:22  #114 
May 2004
New York City
10210_{8} Posts 
Something big? Now wouldn't that be cool to see ...

20101212, 21:43  #115 
Aug 2006
1011101010110_{2} Posts 

20101212, 21:52  #116 
May 2004
New York City
2^{3}·23^{2} Posts 
So far, this (sum p^p) sequence is:
11 751 1129 361649 Couldn't have done that without a computer. 
20101213, 01:39  #117  
"Nathan"
Jul 2008
Maryland, USA
5×223 Posts 
Methinks your calculator needs new batteries
Quote:
2^2 + 3^3 = 4 + 27 = 31 2^2 + 3^3 + 5^5 = 4 + 27 + 3125 = 3156 2^2 + 3^3 + 5^5 + 7^7 = 826699 The next term brings the sum to 285 billion and change (i.e. 285e9), and the term after that brings the sum to over 303 trillion (i.e. 303e12). Looking at the sum mod 10, we see that 2^2 = 4, 3^3 = 7, 5^5 = 5, 7^7 = 3, 11^11 = 1. This adds up to 0 (mod 10). Hence, 2^2 + 3^3 + 5^5 + 7^7 + 11^11 = 285,312,497,210 would be the first time the sum is divisible by 10. Likely, the way to attack the problem for mod 10^n, would be to ask questions about the residues of p^p (mod 10^n) in general. 

20101213, 02:13  #118 
May 2004
New York City
10210_{8} Posts 
I suppose my batteries would have been sufficient to get
that first term, but for the second I would have needed my homebuilt multiprecision (computerresident) calculator, written in C but currently in mothballs. Does the sum for 361649 really end in four zeros? Maybe those sums should be in the same table. This could get interesting. 
20101213, 02:54  #119 
Aug 2006
2×29×103 Posts 
Five, actually  it will be fourth and fifth on your list:
11, 751, 1129, 361649, 361649, 12462809, 12462809, 1273183931, 1273183931. Would someone check me on these? The repeating entries are freaking me out. Last fiddled with by CRGreathouse on 20101213 at 03:08 Reason: update list 
20101213, 03:32  #120 
May 2004
New York City
2^{3}·23^{2} Posts 
Well, so far the sequence of sum(p^p)mod(10^n)
(I think it needs a better name) is: 11 751 1129 361649 361649 12462809 12462809 1273183931 1273183931 Any bets that the tenth and eleventh terms are equal, and whether the twelfth breaks THAT pattern? 
20101213, 03:43  #121 
Aug 2006
13526_{8} Posts 
Nothing to 10^10.

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