A182972 Numerators of positive rationals < 1 arranged by increasing sum of numerator and denominator then by increasing numerator.

1, 1, 1, 2, 1, 1, 2, 3, 1, 3, 1, 2, 4, 1, 3, 1, 2, 3, 4, 5, 1, 5, 1, 2, 3, 4, 5, 6, 1, 3, 5, 1, 2, 4, 7, 1, 3, 5, 7, 1, 2, 3, 4, 5, 6, 7, 8, 1, 5, 7, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 3, 7, 9, 1, 2, 4, 5, 8, 10, 1, 3, 5, 7, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 1, 5, 7, 11, 1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 1, 3, 5, 7, 9, 11

Offset: 1

Author

William Rex Marshall, Dec 16 2010

Keywords

• nonn
• easy
• frac
• nice

Comments

A023022(n) and A245677(n) give number and numerator of sum of fractions a(k)/A182973(k) such that a(k) + A182973(k) = n. - Reinhard Zumkeller, Jul 30 2014

Examples

			Positive fractions < 1 listed by increasing sum of numerator and denominator, and by increasing numerator for equal sums:
1/2
1/3
1/4 2/3
1/5
1/6 2/5 3/4
1/7 3/5
1/8 2/7 4/5
1/9 3/7
1/10 2/9 3/8 4/7 5/6
1/11 5/7
1/12 2/11 3/10 4/9 5/8 6/7
1/13 3/11 5/9
1/14 2/13 4/11 7/8
1/15 3/13 5/11 7/9
1/16 2/15 3/14 4/13 5/12 6/11 7/10 8/9
1/17 5/13 7/11
1/18 2/17 3/16 4/15 5/14 6/13 7/12 8/11 9/10
1/19 3/17 7/13 9/11
(this is A182972/A182973).
		

References

• S. Cook, Problem 511: An Enumeration Problem, Journal of Recreational Mathematics, Vol. 9:2 (1976-77), 137. Solution by the Problem Editor, JRM, Vol. 10:2 (1977-78), 122-123.
• R. K. Guy, Unsolved Problems in Number Theory (UPINT), Section D11.

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
• Paul Yiu, Recreational Mathematics, 24.3.1 Appendix: Two enumerations of the rational numbers in (0,1), page 633.
• Index entries for sequences related to enumerating the rationals

Crossrefs

Cf. A182973 (denominators), A366191 (interleaved).

Cf. A020652, A038567, A182974, A182975, A182976.

Cf. A023022, A245677, A245678, A245718.

Essentially the same as A333856.

Programs

Haskell

a182972 n = a182972_list !! (n-1)
a182972_list = map fst $ concatMap q [3..] where
   q x = [(num, den) | num <- [1 .. div x 2],
                       let den = x - num, gcd num den == 1]
-- Reinhard Zumkeller, Jul 29 2014

Maple

t1:=[];
for n from 2 to 40 do
t1:=[op(t1),1/(n-1)];
for i from 2 to floor((n-1)/2) do
   if gcd(i,n-i)=1 then t1:=[op(t1),i/(n-i)]; fi; od:
od:
t1;

Mathematica

t1={}; For[n=2, n <= 40, n++, AppendTo[t1, 1/(n-1)]; For[i=2, i <= Floor[(n-1)/2], i++, If[GCD[i, n-i] == 1, AppendTo[t1, i/(n-i)]]]]; t1 // Numerator // Rest (* Jean-François Alcover, Jan 20 2015, translated from Maple *)

Pascal

program a182972;
var
  num,den,n: longint;
function gcd(i,j: longint):longint;
begin
  repeat
    if i>j then i:=i mod j else j:=j mod i;
  until (i=0) or (j=0);
  if i=0 then gcd:=j else gcd:=i;
end;
begin
  num:=1; den:=1; n:=0;
  repeat
    repeat
      inc(num); dec(den);
      if num>=den then
      begin
        inc(den,num); num:=1;
      end;
    until gcd(num,den)=1;
    inc(n); writeln(n,' ',num);
  until n=100000;
end.

Python

from itertools import count, islice
from math import gcd
def A182972_gen(): # generator of terms
    return (i for n in count(2) for i in range(1,1+(n-1>>1)) if gcd(i,n-i)==1)
A182972_list = list(islice(A182972_gen(),10)) # Chai Wah Wu, Aug 28 2023

Extensions

Corrected by William Rex Marshall, Aug 12 2013