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

Author

Jaroslav Krizek, Jul 30 2010, Jul 31 2010

Keywords

• nonn

Comments

Corresponding values of numbers h-1 see A179875.

Numbers h such that A175505(h) = A175505(h-1).

Numbers h such that A175506(h) = A175506(h-1).

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).

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)

Crossrefs

Cf. A179871-A179880, A179882-A179887, A179890, A179891.

Programs

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

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 *)