

A320162


Irregular triangle read by rows: row n lists 0 <= k < p^2 such that p^2 divides A172236(k, pKronecker(k^2+4, p)), p = prime(n).


2



0, 0, 4, 5, 0, 7, 18, 0, 12, 20, 24, 25, 29, 37, 0, 5, 18, 19, 24, 43, 78, 97, 102, 103, 116, 0, 2, 14, 45, 70, 82, 87, 99, 124, 155, 167, 0, 24, 38, 40, 63, 83, 103, 105, 184, 186, 206, 226, 249, 251, 265, 0, 31, 37, 63, 79, 100, 137, 144, 150, 180, 181, 211, 217, 224, 261, 282, 298, 324, 330
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OFFSET

1,3


COMMENTS

p always divides A172236(k, pKronecker(k^2+4, p)), so it's interesting to see when p^2 also divides A172236(k, pKronecker(k^2+4, p)). If this holds, then p is called a kWallSunSun prime (and thus being a (k + s*p^2)WallSunSun prime for all integer s). Specially, there is no WallSunSun prime below 10^14 implies that there is no 1 in the first pi(10^14) rows.
Note that A172236(0, m) is not defined, so here k must be understood as a remainder modulo p^2. because A172236(k+s*p^2, m) == A172236(k, m) (mod p^2).
Let p = prime(n). Every row contains 0. For n >= 2, if p == 3 (mod 4), then the nth row contains p numbers, whose remainders modulo p form a permutation of {0, 1, 2, 3, ..., p  2, p  1}. If p == 1 (mod 4), then the nth row contains p  2 numbers, whose remainders modulo p form a permutation of {0, 1, 2, 3, ..., p  2, p  1} \ {+2*((p  1)/2)! mod p}.
Every row is antisymmetric, that is, k is a member iff p^2  k is, k > 0. As a result, the sum of the nth row is prime(n)^2*(prime(n)  1)/2 if prime(n) == 3 (mod 4) and prime(n)^2*(prime(n)  3)/2 if prime(n) == 1 (mod 4).
Equivalently, if p = prime(n) == 1 (mod 4), then row n lists 0 <= k < p^2 such that p^2 divides A172236(k, (pKronecker(k^2+4, p))/2).  Jianing Song, Jul 06 2019


LINKS

Jianing Song, Table of n, a(n) for n = 1..29193 (primes below 600)
Jianing Song, Table of n, a(n) for n = 1..75966 (primes below 1000)
Wikipedia, WallSunSun prime


EXAMPLE

Table starts
p = 2: 0,
p = 3: 0, 4, 5,
p = 5: 0, 7, 18,
p = 7: 0, 12, 20, 24, 25, 29, 37,
p = 11: 0, 5, 18, 19, 24, 43, 78, 97, 102, 103, 116,
p = 13: 0, 2, 14, 45, 70, 82, 87, 99, 124, 155, 167,
p = 17: 0, 24, 38, 40, 63, 83, 103, 105, 184, 186, 206, 226, 249, 251, 265,
p = 19: 0, 31, 37, 63, 79, 100, 137, 144, 150, 180, 181, 211, 217, 224, 261, 282, 298, 324, 330,
p = 23: 0, 21, 30, 38, 40, 70, 79, 89, 111, 149, 198, 248, 281, 331, 380, 418, 440, 450, 459, 489, 491, 499, 508,
p = 29: 0, 15, 40, 41, 49, 51, 56, 64, 74, 84, 126, 182, 204, 381, 460, 637, 659, 715, 757, 767, 777, 785, 790, 792, 800, 801, 826,
...


PROG

(PARI)
B(k, p) = (([k, 1; 1, 0]^(pkronecker(k^2+4, p)))[1, 2])%(p^2)
forprime(p=2, 50, for(k=0, p^21, if(!B(k, p), print1(k, ", "))); print)


CROSSREFS

Cf. A143548, A172236, A320161 (discriminant k^24).
Cf. A238736 (primes p such that 2 occurs in the corresponding row).
Sequence in context: A318740 A240160 A249860 * A200013 A178219 A322232
Adjacent sequences: A320159 A320160 A320161 * A320163 A320164 A320165


KEYWORD

nonn,tabf


AUTHOR

Jianing Song, Oct 06 2018


STATUS

approved



