A342632 Number of ordered pairs (x, y) with gcd(x, y) = 1 and 1 <= {x, y} <= 2^n.
1, 3, 11, 43, 159, 647, 2519, 10043, 39895, 159703, 637927, 2551171, 10200039, 40803219, 163198675, 652774767, 2611029851, 10444211447, 41776529287, 167106121619, 668423198491, 2673693100831, 10694768891659, 42779072149475, 171116268699455, 684465093334979, 2737860308070095
Offset: 0
Keywords
Examples
Only fractions with gcd(numerator, denominator) = 1 are counted. E.g., 1/2 counts, but 2/4, 3/6, 4/8 ... do not, because they reduce to 1/2; 1/1 counts, but 2/2, 3/3, 4/4 ... do not, because they reduce to 1/1. . For n=0, the size of the grid is 1 X 1: . | 1 --+-- 1 | o Sum: 1 . For n=1, the size of the grid is 2 X 2: . | 1 2 --+---- 1 | o o 2 2 | o . 1 -- Sum: 3 . For n=2, the size of the grid is 4 X 4: . | 1 2 3 4 --+-------- 1 | o o o o 4 2 | o . o . 2 3 | o o . o 3 4 | o . o . 2 -- Sum: 11 . For n=3, the size of the grid is 8 X 8: . | 1 2 3 4 5 6 7 8 --+---------------- 1 | o o o o o o o o 8 2 | o . o . o . o . 4 3 | o o . o o . o o 6 4 | o . o . o . o . 4 5 | o o o o . o o o 7 6 | o . . . o . o . 3 7 | o o o o o o . o 7 8 | o . o . o . o . 4 -- Sum: 43
Links
- Chai Wah Wu, Table of n, a(n) for n = 0..45
- Joachim von zur Gathen and Jürgen Gerhard, Extract from "3.4. (Non-)Uniqueness of the gcd" chapter, Modern Computer Algebra, Cambridge University Press, Second Edition 2003, pp. 53-54.
Programs
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PARI
for(n=0,24,my(j=2^n);print1(2*sum(k=1,j,eulerphi(k))-1,", ")) \\ Hugo Pfoertner, Mar 17 2021
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Python
import math for n in range (0, 21): counter = 0 for x in range (1, pow(2, n)+1): for y in range(1, pow(2, n)+1): if math.gcd(y, x) == 1: counter += 1 print(n, counter)
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Python
from sympy import sieve def A342632(n): return 2*sum(t for t in sieve.totientrange(1,2**n+1)) - 1 # Chai Wah Wu, Mar 23 2021
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Python
from functools import lru_cache @lru_cache(maxsize=None) def A018805(n): if n == 1: return 1 return n*n - sum(A018805(n//j) for j in range(2, n//2+1)) - (n+1)//2 print([A018805(2**n) for n in range(25)]) # Michael S. Branicky, Mar 23 2021
Formula
Lim_{n->infinity} a(n)/2^(2*n) = 6/Pi^2 = 1/zeta(2).
Extensions
Edited by N. J. A. Sloane, Jun 13 2021