

A143548


Irregular triangle of numbers k < p^2 such that p^2 divides k^(p1)1, with p=prime(n).


11



1, 1, 8, 1, 7, 18, 24, 1, 18, 19, 30, 31, 48, 1, 3, 9, 27, 40, 81, 94, 112, 118, 120, 1, 19, 22, 23, 70, 80, 89, 99, 146, 147, 150, 168, 1, 38, 40, 65, 75, 110, 131, 134, 155, 158, 179, 214, 224, 249, 251, 288, 1, 28, 54, 62, 68, 69, 99, 116, 127, 234, 245, 262, 292, 293, 299, 307, 333, 360
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OFFSET

1,3


COMMENTS

Row n begins with 1 and has prime(n)1 terms. The first differences of each row are symmetric. For k > p^2, the solutions are just shifted by m*p^2 for m > 0. An open question is whether every integer appears in this sequence. For instance, 2 does not appear until the prime 1093 and 5 does not appear until the prime 20771.
For row n > 1, the sum of the terms in row n is (p1)*p^2*(p+1)/2, which is A138416.  T. D. Noe, Aug 24 2008, corrected by Robert Israel, Sep 27 2016
Max Alekseyev points out that there is a much faster method of computing these numbers. Let p=prime(n) and let r be a primitive root of p (see A001918 and A060749). Then the terms in row n are r^(k*p) (mod p^2) for k=0..p2.  T. D. Noe, Aug 26 2008
The numbers in this sequence are the bases to Euler pseudoprimes q, which are squares of prime numbers, such that n^((q1)/2) == +1 mod q. An exception is the first number 9 = 3*3, which is, following the strict definition in Crandall and Pomerance, no Fermat pseudoprime and hence no Euler pseudoprime.  Karsten Meyer, Jan 08 2011
For row n > 1, the sum is zero modulo p^2 (rows are antisymmetric due to Binomial Theorem).  Peter A. Lawrence, Sep 11 2016


REFERENCES

R. Crandall and C. Pomerance, Prime Numbers: A Computational Perspective, Springer, NY, 2005


LINKS



EXAMPLE

(2) 1,
(3) 1, 8,
(5) 1, 7, 18, 24,
(7) 1, 18, 19, 30, 31, 48,
(11) 1, 3, 9, 27, 40, 81, 94, 112, 118, 120,
(13) 1, 19, 22, 23, 70, 80, 89, 99, 146, 147, 150, 168,
(17) 1, 38, 40, 65, 75, 110, 131, 134, 155, 158, 179, 214, 224, 249, 251, 288,


MAPLE

f:= proc(n) local p, j, x;
p:= ithprime(n);
x:= numtheory:primroot(p);
op(sort([seq(x^(i*p) mod p^2, i=0..p2)]))
end proc:


MATHEMATICA

Flatten[Table[p=Prime[n]; Select[Range[p^2], PowerMod[ #, p1, p^2]==1&], {n, 50}]] (* T. D. Noe, Aug 24 2008 *)
Flatten[Table[p=Prime[n]; r=PrimitiveRoot[p]; b=PowerMod[r, p, p^2]; Sort[NestList[Mod[b*#, p^2]&, 1, p2]], {n, 50}]] (* Faster version from T. D. Noe, Aug 26 2008 *)


CROSSREFS

Cf. A039678, A056020, A056021, A056022, A056024, A056025, A056027, A056028, A056031, A056034, A056035, A096082, A138416.


KEYWORD

nonn,tabf


AUTHOR



STATUS

approved



