|
%I #10 May 26 2019 14:59:15
%S 9,8,15,14,21,20,21,10,9,8,25,24,27,26,28,9,28,27,26,9,28,27,26,25,28,
%T 27,26,25,8,9,28,27,26,25,8,21,10,9,33,32,33,8,21,10,33,32,25,8,21,10,
%U 33,32,25,28,9,8,21,10,33,32,25,28,27,26,9,8,21,10
%N Irregular array T(n, k) read by rows, where row n lists the members of n-th Fermat pseudoprime tuple. Rows are arranged first by size of largest term, then by increasing length of row, then in lexicographic order.
%C Let c_1, c_2, c_3, ..., c_u be a set C of distinct composites and let m_1, m_2, m_3, ..., m_u be a set V of variables. Then C is a Fermat pseudoprime u-tuple if there exists a mapping from the elements of C to the elements of V such that each of the following congruences is satisfied: m_1^(m_2-1) == 1 (mod m_2), m_2^(m_3-1) == 1 (mod m_3), ..., m_u^(m_1-1) == 1 (mod m_1).
%e Irregular array starts as follows:
%e 9, 8
%e 15, 14
%e 21, 20
%e 21, 10, 9, 8
%e 25, 24
%e 27, 26
%e 28, 9
%e 28, 27, 26, 9
%e 28, 27, 26, 25
%e 28, 27, 26, 25, 8, 9
%e 28, 27, 26, 25, 8, 21, 10, 9
%e 33, 32
%e 33, 8, 21, 10
%e 33, 32, 25, 8, 21, 10
%e 33, 32, 25, 28, 9, 8, 21, 10
%e 33, 32, 25, 28, 27, 26, 9, 8, 21, 10
%e 35, 34
%e 35, 34, 33, 32, 25, 6
%e 35, 9, 28, 27, 26, 25, 6
%e 35, 34, 21, 10, 33, 32, 25, 6
%e 35, 9, 8, 21, 10, 33, 32, 25, 6
%e 35, 34, 21, 10, 9, 28, 27, 26, 25, 6
%e 35, 34, 33, 8, 9, 28, 27, 26, 25, 6
%e 35, 34, 21, 10, 33, 8, 9, 28, 27, 26, 25, 6
%e 35, 34, 33, 8, 21, 10, 9, 28, 27, 26, 25, 6
%e 39, 38
%e The composites 21, 10, 9 and 8 satisfy the congruences 21^(10-1) == 1 (mod 10), 10^(9-1) == 1 (mod 9), 9^(8-1) == 1 (mod 8) and 8^(21-1) == 1 (mod 21), so 21, 10, 9, 8 is a row of the array.
%o (PARI) addtovec(vec) = my(w=[], vmax=0); for(t=1, #vec, if(vecmax(vec[t]) > vmax, vmax=vecmax(vec[t]))); for(k=1, #vec, forcomposite(c=1, vmax, if(Mod(vec[k][#vec[k]], c)^(c-1)==1, w=concat(w, [0]); w[#w]=concat(vec[k], [c])))); w
%o removefromvec(vec) = my(w=[]); for(k=1, #vec, if(vecsort(vec[k])==vecsort(vec[k], , 8), w=concat(w, [0]); w[#w]=vec[k])); w
%o printfromvec(vec) = for(k=1, #vec, if(vec[k][1]==vec[k][#vec[k]], for(t=1, #vec[k]-1, print1(vec[k][t], ", ")); print("")))
%o forcomposite(c=1, 40, my(v=[[c]]); while(#v > 0, v=addtovec(v); printfromvec(v); v=removefromvec(v)))
%Y Cf. A317721.
%K nonn,tabf
%O 1,1
%A _Felix Fröhlich_, Apr 19 2019
|
