A361787 Bi-unitary arithmetic numbers k whose mean bi-unitary divisor is a bi-unitary divisor of k.

1, 6, 60, 270, 420, 630, 672, 2970, 5460, 8190, 10080, 22848, 30240, 99792, 136500, 172900, 204750, 208656, 245700, 249480, 312480, 332640, 342720, 385560, 491400, 695520, 708288, 791700, 819000, 861840, 1028160, 1037400, 1187550, 1228500, 1421280, 1528800, 1571328

Offset: 1

Author

Amiram Eldar, Mar 24 2023

Keywords

• nonn

Comments

Also, bi-unitary harmonic numbers k whose harmonic mean of the bi-unitary divisors of k is a bi-unitary divisor of k.

Examples

			6 is a term since the arithmetic mean of its bi-unitary divisors, {1, 2, 3, 6}, is 3, and 3 is also a bi-unitary divisor of 6.
60 is a term since the arithmetic mean of its bi-unitary divisors, {1, 3, 4, 5, 12, 15, 20, 60}, is 15, and 15 is also a bi-unitary divisor of 60.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..236

Crossrefs

Subsequence of A286325 and A361786.

Similar sequence: A007340, A353039, A361387.

Cf. A188999, A222266, A286324.

Programs

Mathematica

biudivQ[f_, d_] := AllTrue[f, OddQ[Last[#]] || IntegerExponent[d, First[#]] != Last[#]/2 &]; biuDivs[n_] := Module[{d = Divisors[n], f = FactorInteger[n]}, Select[d, biudivQ[f, #] &]]; Select[Range[10^5], IntegerQ[(r = Mean[(i = biuDivs[#])])] && MemberQ[i, r] &]

PARI

isbdiv(f, d) = {for (i=1, #f~, if(f[i, 2]%2 == 0 && valuation(d, f[i, 1]) == f[i, 2]/2, return(0))); 1;}
is(n) = {my(f = factor(n), r, p, e); r = prod(i=1, #f~, p = f[i, 1]; e = f[i, 2]; if(e%2, (p^(e+1)-1)/((e + 1)*(p-1)), ((p^(e+1)-1)/(p-1)-p^(e/2))/e)); denominator(r) == 1 && n%r==0 && isbdiv(f, r); }