A Judagrali series: problem.

[r034802n]

A Judagrali series:
Find the 4th prime number N such that N=NumberOfDigit s(Q) where N is prime AND Q is
prime AND where Q=2p-1 where P is prime.
Example the 1st prime number N is 2, because 2=NumberOfDigit s(31) because 2 is prime and
because 31=2^5-1 and because 5 is prime.
We are looking for the 2nd,3rd, 4th prime number N in the Judagrali series as well as its
corresponding Q and P values.
Hints:
1 The 4th N corresponds to a very large M. Take this into account when doing the
calculation.
2 Computational efficiency is important.
3 Object orientation and inheritance MUST be used. i.e. N is a prime, and a subset
(subclass) of Q, Q is also prime and a subset of P which is also prime.

[r035198x]

A Judagrali series:
Find the 4th prime number N such that N=NumberOfDigit s(Q) where N is prime AND Q is
prime AND where Q=2p-1 where P is prime.
Example the 1st prime number N is 2, because 2=NumberOfDigit s(31) because 2 is prime and
because 31=25-1 and because 5 is prime.
We are looking for the 2nd,3rd, 4th prime number N in the Judagrali series as well as its
corresponding Q and P values.
Hints:
1 The 4th N corresponds to a very large M. Take this into account when doing the
calculation.
2 Computational efficiency is important.
3 Object orientation and inheritance MUST be used. i.e. N is a prime, and a subset
(subclass) of Q, Q is also prime and a subset of P which is also prime.

Doesn't look right here. You've got

because 31=25-1 and because 5 is prime.

That cannot be correct.

[r034802n]

Doesn't look right here. You've got


That cannot be correct.

it is 2 to the power 5 not 25

[r035198x]

So Q = (2^p) - 1 ?

[r034802n]

So Q = (2^p) - 1 ?

yes Q = (2^p) - 1 and p is prime. M is also equal to the N I am searching for.

[JosAH]

@OP: Who/what is/was "Judagrali" ?

kind regards,

Jos

[r034802n]

[QUOTE=JosAH]@OP: Who/what is/was "Judagrali" ?

kind regards,

Jos[/QUOTE


its a name used for the series.

[JosAH]

[QUOTE=r034802n]

@OP: Who/what is/was "Judagrali" ?

kind regards,

Jos[/QUOTE


its a name used for the series.

Is it a new series? I can't find it; not even in Sloane's Mathematical Series
encyclopedia. (or maybe I'm suffering from temporary blindness again ;-)

kind regards,

Jos

[JosAH]

Found them:

p: 5 n: 2 q: 31
p: 7 n: 3 q: 127
p: 61 n: 19 q: 2305843009213693951
p: 521 n: 157 q: 6864797660130609714981900799081393217269435300143305409394463459185543183397656052122559640661454554977296311391480858037121987999716643812574028291115057151

kind regards,

Jos

[r035198x]

Are you allowed to use java.math.BigIn teger?

[r035198x]

Found them:

p: 5 n: 2 q: 31
p: 7 n: 3 q: 127
p: 61 n: 19 q: 2305843009213693951
p: 521 n: 157 q: 6864797660130609714981900799081393217269435300143305409394463459185543183397656052122559640661454554977296311391480858037121987999716643812574028291115057151

kind regards,

Jos

What! Does the J in Jos stand for Judagrali or something?
You obviously didn't count fingers here so we await your secret algorithm.

[r035198x]

Argh, the Ns are Mersenne primes ...

[JosAH]

Argh, the Ns are Mersenne primes ...

Yep, but that doesn't buy you much; you have to find a prime 2^p-1 (p must be
prime as well) *and* its decimal representation must have a prime number of digits.

kind regards,

Jos (<--- aka the Evil Prime Hunter Strikes Again! ;-)

[JosAH]

What! Does the J in Jos stand for Judagrali or something?
You obviously didn't count fingers here so we await your secret algorithm.

I used my toes as well ;-)

kind regards,

Jos

ps. hint: RPL

[r035198x]

I used my toes as well ;-)

kind regards,

Jos

ps. hint: RPL

Bah $%##%*!6