A332414 Positive integers r such that A(1,r) = A(2,r - 1) = ... = A(r,1) = 0, where A denotes the function mapping every pair of positive integers (m,n) into 1 if m * 2^(n + 2) + 1 is a prime number dividing F(n + 2) - 2, where F(n) denotes the n-th Fermat number (i.e., F(n) = A000215(n)); and into 0 otherwise.
1, 3, 4, 5, 8, 11, 12, 16, 19, 20, 21, 22, 23, 26, 28, 29, 32, 33, 34, 35, 36, 37, 38, 39, 44, 46, 47, 51, 52, 53, 55, 56, 57, 58, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 72, 74, 75, 76, 78, 80, 82, 84, 85, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 99, 100, 101
Offset: 1
Keywords
Examples
3 is a term of this sequence, because A(1,3) = A(2,2) = A(3,1) = 0.
Links
- Lorenzo Sauras Altuzarra, Some arithmetical problems that are obtained by analyzing proofs and infinite graphs, arXiv:2002.03075 [math.NT], 2020.
Programs
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Maple
A332414:=proc(n) local c, i, k, q, r, v: c:=0: i:=0: r:=1: while c < n do for k from 0 to r-1 do q:=(k+1)*2^(r-k+2)+1: if not isprime(q) or (2^(2^(r-k+2)) - 1) mod q != 0 then i:=i+1: fi: od: if i = r then v:=r: c:=c+1: fi: i:=0: r:=r+1: od: return v: end proc:
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Mathematica
Select[Range@ 29, NoneTrue[Transpose@ {#, Reverse@ #} &@ Range@ #, And[PrimeQ[#4], Mod[((#3 - 1)^#1 - 1)/(#3 - 2), #4] != 0] & @@ {#1, #2, 2^(2^(#2 + 2)) + 1, #1*2^(#2 + 2) + 1} & @@ # &] &] (* Michael De Vlieger, Feb 14 2020 *) -
PARI
isA(m, t) = ispseudoprime(q=4*m*2^t+1) && Mod(2, q)^(4*2^t)==1; isok(r) = sum(i=1, r, isA(i, r-i+1)) == 0; \\ Jinyuan Wang, Feb 18 2020
Extensions
a(17)-a(67) from Jinyuan Wang, Feb 18 2020
Comments