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Easy mate, the value of 2^(2^28) is 682106198 and I confirm your value for 2^(2^27).
Actually, I discovered the Pisano serie (Fibonacci series modulo M have a cycle) and in particular the Fibonacci series modulo 10^9+7 has a cycle of "only" 2000000016 = 2*10^9+16. So the values have also the same cycle.
It improved my submitted code by a factor 2 on last test case =).
I am now able to give you "any" gold nugget, the difficulty is now to compute the number you give modulo 2000000016.
For example the 1234567891011121314151617181967^ 1234567891011121314151617182009th (> 2^(2^106)) modulo 10^9+7 is 782429593 (I used 2 big primes numbers).
In the references we can find a pseudo code to calculate the period :
def π(m):
if m==1:
return 1
if m is prime:
p=m
if p==2: return 3
if p==5: return 20
if p≡±1 (mod 10):
D = the set of even divisors of p−1
else:
D = the set of divisors of 2p+2 that are not divisors of p+1
find the least d in D such that Ud≡I(modp) (of course, use modular exponentiation)
if Ud≡I(modp2):
print p, exit program, publish a paper - you found a Wall-Sun-Sun prime!
return d
else:
factor m as m=pe11pe22⋯penn
return LCM[pe1−11π(p1),pe2−12π(p2),…,pen−1nπ(pn)]
So the period is either a divisor of (p-1) or 2*(p+1) (except for p = 2 or 5) which can be smaller than p.
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Project Euler #137: Fibonacci golden nuggets
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Apologies Mate !!! I did some silly mistakes while mapping.
The value of 2^(2^24)th golden nugget mod 10^9+7 is 885511592.
But the value of 2^(2^25)th golden nugget mod 10^9+7 is 771000771, kindly recheck.
And the value of 2^(2^26)th golden nugget mod 10^9+7 is 268044267.
Lets Move further,I have found the value of 2^(2^27)th golden nugget mod 10^9+7 , and its value is 793677690.
Can you tell me the value of 2^(2^28)th golden nugget mod 10^9+7 ?
Easy mate, the value of 2^(2^28) is 682106198 and I confirm your value for 2^(2^27).
Actually, I discovered the Pisano serie (Fibonacci series modulo M have a cycle) and in particular the Fibonacci series modulo 10^9+7 has a cycle of "only" 2000000016 = 2*10^9+16. So the values have also the same cycle. It improved my submitted code by a factor 2 on last test case =).
I am now able to give you "any" gold nugget, the difficulty is now to compute the number you give modulo 2000000016. For example the 1234567891011121314151617181967^ 1234567891011121314151617182009th (> 2^(2^106)) modulo 10^9+7 is 782429593 (I used 2 big primes numbers).
End of the story I think =).
prime p = 10^9+7 , is not that special, so its pisano period is 2(p+1).
But there are many primes p, whose pisano period is less than p.
can you Explain?
I am not really familliar with Pisano period and it seems not so obvious. I found a calculator and some explanation here : http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibmaths.html#section6
In the references we can find a pseudo code to calculate the period :
def π(m): if m==1: return 1 if m is prime: p=m if p==2: return 3 if p==5: return 20 if p≡±1 (mod 10): D = the set of even divisors of p−1 else: D = the set of divisors of 2p+2 that are not divisors of p+1 find the least d in D such that Ud≡I(modp) (of course, use modular exponentiation) if Ud≡I(modp2): print p, exit program, publish a paper - you found a Wall-Sun-Sun prime! return d else: factor m as m=pe11pe22⋯penn return LCM[pe1−11π(p1),pe2−12π(p2),…,pen−1nπ(pn)]So the period is either a divisor of (p-1) or 2*(p+1) (except for p = 2 or 5) which can be smaller than p.