塔形素数猜想
∀ p ∈ P , ∃ k ∈ N + , k < p p s . t . ∀ n ∈ N , f n ( p , k ) = p f n − 1 ( p , k ) − k ∈ P . ( 其 中 f 0 ( p , k ) = p p − k ) \forall p \in \mathbb{P} , \exists k \in \mathbb{N}^+, k < p^p s.t. \forall n \in \mathbb{N}, f_n(p,k) = p^{f_{n-1}(p,k)}-k\in \mathbb{P}. (其中f_0(p,k)=p^p-k) ∀p∈P,∃k∈N+,k<pps.t.∀n∈N,fn(p,k)=pfn−1(p,k)−k∈P.(其中f0(p,k)=pp−k)
e g : f 0 ( 2 , 1 ) = 2 2 − 1 = 3 ∈ P eg:f_0(2,1) = 2^2-1 = 3 \in \mathbb{P} eg:f0(2,1)=22−1=3∈P
f 1 ( 2 , 1 ) = 2 f 0 ( 2 , 1 ) − 1 = 2 3 − 1 = 7 ∈ P f_1(2,1) = 2^{f_0(2,1)}-1 = 2^3-1 = 7 \in \mathbb{P} f1(2,1)=2f0(2,1)−1=23−1=7∈P
f 2 ( 2 , 1 ) = 2 f 1 ( 2 , 1 ) − 1 = 2 7 − 1 = 127 ∈ P f_2(2,1) = 2^{f_1(2,1)}-1 = 2^7-1 = 127 \in \mathbb{P} f2(2,1)=2f1(2,1)−1=27−1=127∈P
f 3 ( 2 , 1 ) = 2 f 2 ( 2 , 1 ) − 1 = 2 127 − 1 ∈ P f_3(2,1) = 2^{f_2(2,1)}-1 = 2^{127}-1 \in \mathbb{P} f3(2,1)=2f2(2,1)−1=2127−1∈P 第12个梅森素数
f 4 ( 2 , 1 ) = 2 f 3 ( 2 , 1 ) − 1 ∈ P ? f_4(2,1) = 2^{f_3(2,1)}-1 \in \mathbb{P} ? f4(2,1)=2f3(2,1)−1∈P?
f 0 ( 3 , 4 ) = 3 3 − 4 = 5 ∈ P f_0(3,4) = 3^3-4 = 5 \in \mathbb{P} f0(3,4)=33−4=5∈P
f 1 ( 3 , 4 ) = 3 f 0 ( 3 , 4 ) − 1 = 3 5 − 4 = 239 ∈ P f_1(3,4) = 3^{f_0(3,4)}-1 = 3^5-4 = 239 \in \mathbb{P} f1(3,4)=3f0(3,4)−1=35−4=239∈P
f 2 ( 3 , 4 ) = 3 f 1 ( 3 , 4 ) − 1 = 3 239 − 4 ∈ P ？ f_2(3,4) = 3^{f_1(3,4)}-1 = 3^{239}-4 \in \mathbb{P} ？ f2(3,4)=3f1(3,4)−1=3239−4∈P？