A179767 - OEIS

%I #15 Mar 19 2021 10:18:52

%S 3,19,7,179,1447,32059,55171,17231,32611,644823367,8870650619,

%T 10808693851,26813406071

%N a(n) is the smallest prime of the form 4k + 3 such that the first n iterations of the map p -> 4p + 3 are prime with the next iteration being composite.

%C This sequence is finite and complete. Proof:

%C Suppose a(9) = p exists. Then, we obtain the sequence of 10 primes:

%C   E = {4p + 3 -> 16p + 15 -> 64p + 63 ->...-> (2^20)p + (2^22 - 1)}.

%C The prime divisors or 2^(2*q) - 1 for q = 1,2,...,11 are

%C     (2^2 - 1) -> {3};

%C     (2^4 - 1) -> {3, 5};

%C     (2^6 - 1) -> {3, 7};

%C     (2^8 - 1) -> {3, 5, 17};

%C     (2^10 - 1) -> {3, 11, 31};

%C     (2^12 - 1) -> {3, 5, 7, 13};

%C     (2^14 - 1) -> {3, 43, 127};

%C     (2^16 - 1) -> {3, 5, 17, 257};

%C     (2^18 - 1) -> {3, 7, 19, 73};

%C     (2^20 - 1) -> {3, 5, 11, 31, 41};

%C     (2^22 - 1) -> {3, 23, 89, 683}.

%C But p == r (mod 32) where r is element of the set {3, 7, 11, 15, 19, 23, 27, 31}, and one of the ten numbers of E is divisible by r. For example, 27 | (2^18)p + 2^18 - 1 if p == 27 (mod 32).

%C Remark: the map p -> 4p + 1 is not interesting because the corresponding sequence contains only two numbers: a(0) = 5 and a(1) = 13 if we consider only 2 iterations {4p + 1 -> 16p + 5 -> 64p + 21}: if p==0 (mod 3) => 64p + 21 is composite, if p==1 (mod 3) => 16p + 5 is composite and if  p==2 (mod 3) => 4p + 1 is composite.

%C From _Michael S. Branicky_, Mar 19 2021: (Start)

%C Proof of finiteness is incorrect.  Flaw is last sentence: "For example, ...". Specifically, 27 does not divide quantity unless 27 | k where p = 32*k + 27.

%C No further terms < 10^11. (End)

%e a(0) = 3 because 4*3 + 3 = 15 is composite => 0 iteration;

%e a(1) = 19 because 4*19 + 3 = 79 is prime => 1 iteration;

%e a(2) = 7 -> 31 -> 127 are primes => 2 iterations;

%e a(3) = 179 -> 719 -> 2879 -> 11519 are primes => 3 iterations;

%e a(8) = 32611 -> 130447 -> 521791 -> 2087167 -> 8348671 -> 33394687 -> 133578751 -> 534315007 -> 2137260031 are primes => 8 iterations.

%p with(numtheory):for m from 0 to 8 do: ii:=0:for i from 1 to 50000 do : n:=ithprime(i):if

%p   irem(n,4) = 3 then nn:=n: id:=0:k:=0:for it from 1 to 8 do: p:=4*nn+3: if type

%p   (p,prime)=true and id=0 then k:=k+1:nn:=p:else id:=1:fi:od:if k=m and ii=0 then

%p   ii:=1:print(n):else fi:else fi:od:od:

%o (Python)

%o from sympy import isprime

%o def iters(p):

%o   c = 0

%o   while isprime(4*p + 3): p, c = 4*p + 3, c + 1

%o   return c

%o def a(n):

%o   k = 0

%o   while True:

%o     p, k = 4*k + 3, k + 1

%o     if isprime(p) and iters(p) == n: return p

%o print([a(n) for n in range(9)]) # _Michael S. Branicky_, Mar 19 2021

%K nonn,more,hard

%O 0,1

%A _Michel Lagneau_, Jan 10 2011

%E a(9)-a(12) from _Michael S. Branicky_, Mar 19 2021