A137243 Number of coprime pairs (a,b) with -n <= a,b <= n.
8, 16, 32, 48, 80, 96, 144, 176, 224, 256, 336, 368, 464, 512, 576, 640, 768, 816, 960, 1024, 1120, 1200, 1376, 1440, 1600, 1696, 1840, 1936, 2160, 2224, 2464, 2592, 2752, 2880, 3072, 3168, 3456, 3600, 3792, 3920, 4240, 4336, 4672, 4832, 5024, 5200, 5568
Offset: 1
Keywords
Examples
a(1) = 8 because there are eight coprime pairs (1,1), (1,0), (1,-1), (0,1), (0,-1), (-1,1), (-1,0), (-1,-1) with integral entries of absolute value <= 1. In the same way a(2) = 16 as there are sixteen coprime pairs (2,1), (2,-1), (1,2), (1,1), (1,0), (1,-1), (1,-2), (0,1), (0,-1), (-1,2), (-1,1), (-1,0), (-1,-1), (-1,-2), (-2,1), (-2,-1) of integral entries of absolute value <= 2.
Links
- Paolo Xausa, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Seiichi Manyama).
- Tom M. Apostol, Introduction to Analytic Number Theory, Springer, New York, NY, 1976, pp. 62-64.
Programs
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Mathematica
A137243[nmax_]:=8Accumulate[EulerPhi[Range[nmax]]];A137243[100] (* Paolo Xausa, Mar 25 2023 *)
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PARI
a(n)=4*sum(k=1, n, moebius(k)*(n\k)^2)+4 \\ Charles R Greathouse IV, Aug 06 2012
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Python
from functools import lru_cache @lru_cache(maxsize=None) def A137243(n): if n == 0: return 0 c, j = 0, 2 k1 = n//j while k1 > 1: j2 = n//k1 + 1 c += (j2-j)*(A137243(k1)//4-1) j, k1 = j2, n//j2 return 4*(n*(n-1)-c+j) # Chai Wah Wu, Mar 29 2021
Formula
a(n) = 4*A018805(n) + 4. - Charles R Greathouse IV, Aug 06 2012
a(n) = a(n-1) + 8*EulerPhi(n), with a(1)=8. - Juan M. Marquez, Apr 24 2015
a(n) ~ (24/Pi^2)*n^2 = 4*A059956*n^2. - Paolo Xausa, Mar 25 2023
Comments