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A063987
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Irregular triangle in which n-th row gives quadratic residues modulo the n-th prime.
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23
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1, 1, 1, 4, 1, 2, 4, 1, 3, 4, 5, 9, 1, 3, 4, 9, 10, 12, 1, 2, 4, 8, 9, 13, 15, 16, 1, 4, 5, 6, 7, 9, 11, 16, 17, 1, 2, 3, 4, 6, 8, 9, 12, 13, 16, 18, 1, 4, 5, 6, 7, 9, 13, 16, 20, 22, 23, 24, 25, 28, 1, 2, 4, 5, 7, 8, 9, 10, 14, 16, 18, 19, 20, 25, 28, 1, 3, 4, 7, 9, 10, 11, 12, 16, 21, 25
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OFFSET
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1,4
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COMMENTS
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For n >= 2, row lengths are (prime(n)-1)/2. For example, since 17 is the 7th prime number, the length of row 7 is (17 - 1)/2 = 8. - Geoffrey Critzer, Apr 04 2015
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LINKS
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EXAMPLE
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Modulo the 5th prime, 11, the (11-1)/2 = 5 quadratic residues are 1,3,4,5,9 and the 5 non-residues are 2,6,7,8,10.
The irregular triangle T(n,k) begins (p is prime(n)):
n p \k 1 2 3 4 5 6 7 8 9 10 11 12 13 14
1, 2: 1
2, 3: 1
3, 5: 1 4
4, 7: 1 2 4
5, 11: 1 3 4 5 9
6: 13: 1 3 4 9 10 12
7, 17: 1 2 4 8 9 13 15 16
8, 19: 1 4 5 6 7 9 11 16 17
9, 23: 1 2 3 4 6 8 9 12 13 16 18
10, 29: 1 4 5 6 7 9 13 16 20 22 23 24 25 28
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MAPLE
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with(numtheory): for n from 1 to 20 do for j from 1 to ithprime(n)-1 do if legendre(j, ithprime(n)) = 1 then printf(`%d, `, j) fi; od: od:
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MATHEMATICA
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row[n_] := (p = Prime[n]; Select[ Range[p - 1], JacobiSymbol[#, p] == 1 &]); Flatten[ Table[ row[n], {n, 1, 12}]] (* Jean-François Alcover, Dec 21 2011 *)
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PROG
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(PARI) residue(n, m)=local(r); r=0; for(i=0, floor(m/2), if(i^2%m==n, r=1)); r
(PARI) row(n) = my(p=prime(n)); select(x->issquare(Mod(x, p)), [1..p-1]); \\ Michel Marcus, Nov 07 2020
(Python)
from sympy import jacobi_symbol as J, prime
def a(n):
p = prime(n)
return [1] if n==1 else [i for i in range(1, p) if J(i, p)==1]
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CROSSREFS
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KEYWORD
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nonn,tabf,nice,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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