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A038566
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Numerators in canonical bijection from positive integers to positive rationals <= 1: arrange fractions by increasing denominator then by increasing numerator.
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87
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1, 1, 1, 2, 1, 3, 1, 2, 3, 4, 1, 5, 1, 2, 3, 4, 5, 6, 1, 3, 5, 7, 1, 2, 4, 5, 7, 8, 1, 3, 7, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 5, 7, 11, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 3, 5, 9, 11, 13, 1, 2, 4, 7, 8, 11, 13, 14, 1, 3, 5, 7, 9, 11, 13, 15, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16
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OFFSET
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1,4
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COMMENTS
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Also numerators in canonical bijection from positive integers to all positive rational numbers: arrange fractions in triangle in which in the n-th row the phi(n) numbers are the fractions i/j with gcd(i,j) = 1, i+j=n, i=1..n-1, j=n-1..1. n>=2. Denominators (A020653) are obtained by reversing each row.
Also triangle in which n-th row gives phi(n) numbers between 1 and n that are relatively prime to n.
This irregular triangle gives in row n the smallest positive reduced residue system modulo n, for n >= 1. If one takes 0 for n = 1 it becomes the smallest nonnegative residue system modulo n. - Wolfdieter Lang, Feb 29 2020
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REFERENCES
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Richard Courant and Herbert Robbins. What Is Mathematics?, Oxford, 1941, pp. 79-80.
H. Lauwerier, Fractals, Princeton Univ. Press, p. 23.
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LINKS
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FORMULA
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The n-th "clump" consists of the phi(n) integers <= n and prime to n.
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EXAMPLE
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The beginning of the list of positive rationals <= 1: 1/1, 1/2, 1/3, 2/3, 1/4, 3/4, 1/5, 2/5, 3/5, .... This is A038566/A038567.
The beginning of the triangle giving all positive rationals: 1/1; 1/2, 2/1; 1/3, 3/1; 1/4, 2/3, 3/2, 4/1; 1/5, 5/1; 1/6, 2/5, 3/4, 4/3, 5/2, 6/1; .... This is A020652/A020653, with A020652(n) = A038566(n+1). [Corrected by M. F. Hasler, Mar 06 2020]
The beginning of the triangle in which n-th row gives numbers between 1 and n that are relatively prime to n:
n\k 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
1: 1
2: 1
3: 1 2
4: 1 3
5: 1 2 3 4
6: 1 5
7: 1 2 3 4 5 6
8: 1 3 5 7
9: 1 2 4 5 7 8
10: 1 3 7 9
11: 1 2 3 4 5 6 7 8 9 10
12: 1 5 7 11
13: 1 2 3 4 5 6 7 8 9 10 11 12
14: 1 3 5 9 11 13
15: 1 2 4 7 8 11 13 14
16: 1 3 5 7 9 11 13 15
17: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
18: 1 5 7 11 13 17
19: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
20: 1 3 7 9 11 13 17 19
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MAPLE
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s := proc(n) local i, j, k, ans; i := 0; ans := [ ]; for j while i<n do for k to j do if gcd(j, k) = 1 then ans := [ op(ans), k ]; i := i+1 fi od od; RETURN(ans); end; s(100);
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MATHEMATICA
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Flatten[Table[Flatten[Position[GCD[Table[Mod[j, w], {j, 1, w-1}], w], 1]], {w, 1, 100}], 2]
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PROG
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(Haskell)
a038566 n k = a038566_tabf !! (n-1) !! (k-1)
a038566_row n = a038566_tabf !! (n-1)
a038566_tabf=
zipWith (\v ws -> filter ((== 1) . (gcd v)) ws) [1..] a002260_tabl
a038566_list = concat a038566_tabf
(PARI) first(n)=my(v=List(), i, j); while(i<n, for(k=1, j, if(gcd(j, k)==1, listput(v, k); i++)); j++); Vec(v) \\ Charles R Greathouse IV, Feb 07 2013
(PARI) row(n) = select(x->gcd(n, x)==1, [1..n]); \\ Michel Marcus, May 05 2020
(SageMath)
def aRow(n):
if n == 1: return 1
return [k for k in ZZ(n).coprime_integers(n+1)]
print(flatten([aRow(n) for n in range(1, 18)])) # Peter Luschny, Aug 17 2020
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CROSSREFS
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Cf. A020652, A020653, A038566, A038567, A038568, A038569, A000010 (row lengths), A002088, A060837, A071970, A002260.
A054424 gives mapping to Stern-Brocot tree.
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KEYWORD
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nonn,frac,core,nice,tabf
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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