OFFSET
1,3
COMMENTS
p always divides A316269(k, p-Kronecker(k^2-4, p)), so it's interesting to see when p^2 also divides A316269(k, p-Kronecker(k^2-4, p)).
In the following comments, let p = prime(n). Note that A316269(0, m) and A316269(1, m) is not defined, so here k must be understood as a remainder modulo p^2. because A316269(k+s*p^2, m) == A316269(k, m) (mod p^2).
Let p = prime(n). Every row contains 0, 1 and p^2 - 1. For n >= 3, the n-th row contains p - 2 numbers, whose remainders modulo p form a permutation of {0, 1, 3, 4, ..., p - 3, p - 1}.
Every row is antisymmetric, that is, k is a member iff p^2 - k is, k > 0. As a result, the sum of the n-th row is prime(n)^2*(prime(n) - 3)/2.
Equivalently, for n >= 2, row n lists 0 <= k < p^2 such that p^2 divides A316269(k, (p-Kronecker(k^2-4, p))/2), p = prime(n).
LINKS
Jianing Song, Table of n, a(n) for n = 1..29083 (primes below 600)
Jianing Song, Table of n, a(n) for n = 1..75796 (primes below 1000)
Wikipedia, Wall-Sun-Sun prime
EXAMPLE
Table starts
p = 2: 0, 1, 3,
p = 3: 0, 1, 8,
p = 5: 0, 1, 24,
p = 7: 0, 1, 10, 39, 48,
p = 11: 0, 1, 27, 36, 37, 84, 85, 94, 120,
p = 13: 0, 1, 6, 29, 34, 61, 108, 135, 140, 163, 168,
p = 17: 0, 1, 25, 45, 56, 75, 82, 132, 157, 207, 214, 233, 244, 264, 288,
p = 19: 0, 1, 42, 43, 73, 88, 106, 120, 161, 200, 241, 255, 273, 288, 318, 319, 360,
p = 23: 0, 1, 12, 15, 60, 86, 105, 141, 142, 156, 223, 306, 373, 387, 388, 424, 443, 469, 514, 517, 528,
p = 29: 0, 1, 42, 46, 80, 101, 107, 120, 226, 227, 327, 330, 358, 409, 432, 483, 511, 514, 614, 615, 721, 734, 740, 761, 795, 799, 840,
...
PROG
(PARI)
B(k, p) = (([k, -1; 1, 0]^(p-kronecker(k^2-4, p)))[1, 2])%(p^2)
forprime(p=2, 50, for(k=0, p^2-1, if(!B(k, p), print1(k, ", "))); print)
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Jianing Song, Oct 06 2018
STATUS
approved