OFFSET
1,2
COMMENTS
Also, infinitary harmonic numbers k whose harmonic mean of the infinitary divisors of k is an infinitary divisor of k.
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..135
EXAMPLE
6 is a term since the arithmetic mean of its infinitary divisors, {1, 2, 3, 6}, is 3, and 3 is also an infinitary divisor of 6.
60 is a term since the arithmetic mean of its infinitary divisors, {1, 3, 4, 5, 12, 15, 20, 60}, is 15, and 15 is also an infinitary divisor of 60.
MATHEMATICA
idivs[1] = {1}; idivs[n_] := Sort@ Flatten@ Outer[Times, Sequence @@ (FactorInteger[n] /. {p_, e_Integer} :> p^Select[Range[0, e], BitOr[e, #] == e &])]; Select[Range[10^5], IntegerQ[(r = Mean[(i = idivs[#])])] && MemberQ[i, r] &]
PROG
(PARI) isidiv(d, f) = {if (d==1, return (1)); for (k=1, #f~, bne = binary(f[k, 2]); bde = binary(valuation(d, f[k, 1])); if (#bde < #bne, bde = concat(vector(#bne-#bde), bde)); for (j=1, #bne, if (! bne[j] && bde[j], return (0)); ); ); return (1); } \\ Michel Marcus at A077609
is(n) = {my(f = factor(n), b, r); r = prod(i=1, #f~, b = binary(f[i, 2]); prod(k=1, #b, if(b[k], (f[i, 1]^(2^(#b-k))+1)/2, 1))); denominator(r) == 1 && n%r==0 && isidiv(r, f); }
CROSSREFS
KEYWORD
nonn
AUTHOR
Amiram Eldar, Mar 10 2023
STATUS
approved