A350916 Positive integers k such that (k+1)^4 has a divisor congruent to -1 modulo k.

1, 2, 3, 5, 9, 11, 14, 17, 29, 35, 41, 43, 59, 65, 69, 125, 134, 139, 174, 194, 339, 386, 449, 461, 681, 901, 937, 1169, 1322, 1325, 1715, 1971, 2211, 3054, 6395, 7989, 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

Offset: 1

Author

Max Alekseyev, Jan 21 2022

Keywords

• nonn

Comments

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.

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).

Only terms 1, 2, 5, 9, 11, 14, 29 are shared between two or more sequences, all others come from exactly one sequence.

Links

• Max Alekseyev, Table of n, a(n) for n = 1..5000

Crossrefs

Union of A350917, A350919, A350920, A350921, A350922, A350923, A103974, A350924, A350925, A350926.

Cf. A001570.

Programs

PARI

{ for(k=1,10^6, fordiv((k+1)^4,d, if(Mod(d,k)==-1, print1(k,", "); break)) ); }