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%I #16 Aug 14 2023 08:46:33
%S 1,2,3,5,9,11,14,17,29,35,41,43,59,65,69,125,134,139,174,194,339,386,
%T 449,461,681,901,937,1169,1322,1325,1715,1971,2211,3054,6395,7989,
%U 8857,9077,10849,11483,12545,13082,20909,21506,23861,35233,54734,62210,66923,89045,129494,143289,172899,174725,203321,332315,375129,390051,426389,493697,561513,982094
%N Positive integers k such that (k+1)^4 has a divisor congruent to -1 modulo k.
%C For (k+1)^3 similar sequence is finite {1, 2, 3, 5, 9, 11, 14}, while for (k+1)^2 it is just {1, 2, 3, 5}. Starting with power 4 (this sequence), the number of values of k is infinite. One series of values for power 6 is given by A001570.
%C Formed by the union of 10 linear recurrent sequences satisfying b(n) = q*b(n-1) - b(n-2) - 4: A350919 (q=3), A350920 (q=4), A350921 (q=6), A350922 (q=7), A350923 (q=10), A103974 (q=14), A350924 (q=16), A350925 (q=16), A350926 (q=23), A350917 (q=23). Each of them give identities (b(n)+1)^4 = (b(n)*b(n-1)-1) * (b(n)*b(n+1)-1).
%C Only terms 1, 2, 5, 9, 11, 14, 29 are shared between two or more sequences, all others come from exactly one sequence.
%H Max Alekseyev, <a href="/A350916/b350916.txt">Table of n, a(n) for n = 1..5000</a>
%o (PARI) { for(k=1,10^6, fordiv((k+1)^4,d, if(Mod(d,k)==-1, print1(k,", "); break)) ); }
%Y Union of A350917, A350919, A350920, A350921, A350922, A350923, A103974, A350924, A350925, A350926.
%Y Cf. A001570.
%K nonn
%O 1,2
%A _Max Alekseyev_, Jan 21 2022
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