A279369 A mapping of rationals a/b (lowest form) to prime rationals p/q such that a/b = (p+1)/(q+1), where n (the sequence index) selects the rationals a/b from the triangle array A226314(n)/A054531(n) and a(n) selects the prime rationals p/q from the same array.

1, 12, 18, 58, 13, 74, 57, 19, 5, 72, 174, 178, 182, 429, 217, 138, 8, 225, 247, 272, 162, 825, 81, 83, 85, 849, 89, 999, 255, 1047, 23, 110, 484, 103, 1122, 288, 1383, 139, 114, 143, 1407, 32, 149, 1425, 1518, 408, 711, 176, 1677, 165, 727, 184, 1701, 188, 450, 906, 910, 914

Offset: 1

Author

Frank M Jackson and Michael B Rees, Dec 10 2016

Keywords

• nonn

Comments

Rationals a/b (lowest form) can be mapped 1-to-1 to a positive integer n where a/b is the n-th term of the triangular array A226314(n)/A054531(n). Consider two function of x, f_1 = ax-1 and f_2 = bx-1. Then by Schinzel's Hypothesis H there are infinite values of x such that f_1 and f_2 are simultaneously prime allowing a/b to be expressed using two primes p and q as a/b=(p+1)/(q+1).

By choosing the least x for generating p=f_1 and q=f_2 (see A278635) it is possible to find a unique prime rational p/q that maps to rational a/b. If n is the sequence index that selects the rational a/b from the triangular array A226314(n)/A054531(n), then a(n) selects the prime rationals p/q from the same array.

Examples

			a(7)=57 because A226314(7)/A054531(7)=1/4 and with least x=3 we have p=f_1=x-1=2 and q=f_2=4x-1=11. Therefore (p+1)/(q+1)=3/12=1/4. Also A226314(57)/A054531(57)=p/q=2/11.
		

Links

• Lance Fortnow, Counting the Rationals Quickly, Computational Complexity Weblog, Monday, March 01, 2004.
• A. Schinzel and W. Sierpiński, Sur certaines hypothèses concernant les nombres premiers, Acta Arithmetica IV., 1957.
• Wikipedia Schinzel's hypothesis H.

Crossrefs

Cf. A054531, A226314, A278635.

Programs

Mathematica

func[{i_, j_}] := {j(j-1)/2+i->{j+(i-j)/GCD[i, j], j/GCD[i, j]}}; rfunc[{i_, j_}] := {{j+(i-j)/GCD[i, j], j/GCD[i, j]}->j(j-1)/2+i}; getx[{a_, b_}] := Module[{f1, f2, x}, If[a==b, {1, 1}, (f1=a*x-1; f2=b*x-1; x=1; While[(!PrimeQ[f1]||!PrimeQ[f2])&&x<10^5, x++]; If[x==10^5, Abort[], {f1, f2}])]]; assoc=Association@Flatten[Table[func[{a, b}], {b, 1000}, {a, b}], 1]; rassoc=Association@Flatten[Table[rfunc[{a, b}], {b, 1000}, {a, b}], 1]; Table[rassoc[getx[assoc[n]]], {n, 1, 100}]