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A143548
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Irregular triangle of numbers k < p^2 such that p^2 divides k^(p-1)-1, with p=prime(n).
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11
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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
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1,3
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COMMENTS
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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 (p-1)*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..p-2. - 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^((q-1)/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
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REFERENCES
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R. Crandall and C. Pomerance, Prime Numbers: A Computational Perspective, Springer, NY, 2005
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LINKS
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EXAMPLE
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(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,
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MAPLE
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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..p-2)]))
end proc:
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MATHEMATICA
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Flatten[Table[p=Prime[n]; Select[Range[p^2], PowerMod[ #, p-1, 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, p-2]], {n, 50}]] (* Faster version from T. D. Noe, Aug 26 2008 *)
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CROSSREFS
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Cf. A039678, A056020, A056021, A056022, A056024, A056025, A056027, A056028, A056031, A056034, A056035, A096082, A138416.
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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