%I #21 Mar 06 2021 06:24:25
%S 2,7,11,23,47,59,66,70,78,83,107,130,167,179,186,195,211,222,227,238,
%T 255,263,266,310,322,331,347,359,366,383,399,418,438,455,463,467,470,
%U 474,479,483,494,498,503
%N Numbers h such that h and h-1 have same antiharmonic mean of the numbers k < h such that gcd(k, h) = 1.
%C Corresponding values of numbers h-1 see A179875.
%C Numbers h such that A175505(h) = A175505(h-1).
%C Numbers h such that A175506(h) = A175506(h-1).
%C Antiharmonic mean B(h) of numbers k such that gcd(k, h) = 1 for numbers h >= 1 and k < h = A053818(n) / A023896(n) = A175505(h) / A175506(h).
%H Amiram Eldar, <a href="/A179876/b179876.txt">Table of n, a(n) for n = 1..10000</a> (terms 1..2047 from R. J. Mathar)
%e For n=3: a(3) = 11; B(11) = A175505(11) / A175506(11) = 7, B(10) = A175505(10) / A175506(10) = 7.
%p antiHMeanGcd := proc(h)
%p option remember;
%p local a023896,a053818,k ;
%p a023896 := 0 ;
%p a053818 := 0 ;
%p for k from 1 to h do
%p if igcd(k,h) = 1 then
%p a023896 := a023896+k ;
%p a053818 := a053818+k^2 ;
%p end if;
%p end do:
%p a053818/a023896 ;
%p end proc:
%p n := 1:
%p for h from 2 do
%p if antiHMeanGcd(h) = antiHMeanGcd(h-1) then
%p printf("%d %d\n",n,h) ;
%p n := n+1 ;
%p end if;
%p end do: # _R. J. Mathar_, Sep 26 2013
%t hmax = 1000;
%t 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];
%t 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_ *)
%Y Cf. A179871-A179880, A179882-A179887, A179890, A179891.
%K nonn
%O 1,1
%A _Jaroslav Krizek_, Jul 30 2010, Jul 31 2010