A179876 Numbers h such that h and h-1 have same antiharmonic mean of the numbers k < h such that gcd(k, h) = 1.
2, 7, 11, 23, 47, 59, 66, 70, 78, 83, 107, 130, 167, 179, 186, 195, 211, 222, 227, 238, 255, 263, 266, 310, 322, 331, 347, 359, 366, 383, 399, 418, 438, 455, 463, 467, 470, 474, 479, 483, 494, 498, 503
Offset: 1
Keywords
Examples
For n=3: a(3) = 11; B(11) = A175505(11) / A175506(11) = 7, B(10) = A175505(10) / A175506(10) = 7.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..2047 from R. J. Mathar)
Programs
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Maple
antiHMeanGcd := proc(h) option remember; local a023896,a053818,k ; a023896 := 0 ; a053818 := 0 ; for k from 1 to h do if igcd(k,h) = 1 then a023896 := a023896+k ; a053818 := a053818+k^2 ; end if; end do: a053818/a023896 ; end proc: n := 1: for h from 2 do if antiHMeanGcd(h) = antiHMeanGcd(h-1) then printf("%d %d\n",n,h) ; n := n+1 ; end if; end do: # R. J. Mathar, Sep 26 2013
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Mathematica
hmax = 1000; antiHMeanGcd[h_] := antiHMeanGcd[h] = Module[{num = 0, den = 0, k}, For[k = 1, k <= h, k++, If[GCD[k, h] == 1, den += k; num += k^2]]; num/den]; Reap[n = 1; For[h = 2, h <= hmax, h++, If[antiHMeanGcd[h] == antiHMeanGcd[h - 1], Sow[h]; n++]]][[2, 1]] (* Jean-François Alcover, Mar 23 2020, after R. J. Mathar *)
Comments