A372298 Primitive infinitary abundant numbers (definition 1): infinitary abundant numbers (A129656) whose all proper infinitary divisors are infinitary deficient numbers.
40, 56, 70, 72, 88, 104, 756, 924, 945, 1092, 1188, 1344, 1386, 1428, 1430, 1596, 1638, 1760, 1870, 2002, 2016, 2080, 2090, 2142, 2176, 2210, 2394, 2432, 2470, 2530, 2584, 2720, 2750, 2944, 2990, 3040, 3128, 3190, 3200, 3230, 3250, 3400, 3410, 3496, 3712, 3770
Offset: 1
Examples
40 is a term since it is an infinitary abundant number and all its proper infinitary divisors, {1, 2, 4, 5, 8, 10, 20}, are infinitary deficient numbers.
24 and 30, which are infinitary abundant numbers, are not primitive, because they are divisible by 6 which is an infinitary perfect number.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
-
Mathematica
f[p_, e_] := Module[{b = IntegerDigits[e, 2]}, m = Length[b]; Product[If[b[[j]] > 0, 1 + p^(2^(m - j)), 1], {j, 1, m}]]; isigma[1] = 1; isigma[n_] := Times @@ f @@@ FactorInteger[n]; idefQ[n_] := isigma[n] < 2*n; idivs[1] = {1}; idivs[n_] := Sort@ Flatten@ Outer[Times, Sequence @@ (FactorInteger[n] /. {p_, e_Integer} :> p^Select[Range[0, e], BitOr[e, #] == e &])]; q[n_] := Module[{d = idivs[n]}, Total[d] > 2*n && AllTrue[Most[d], idefQ]]; Select[Range[4000], q] -
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); } idivs(n) = {my(f = factor(n), d = divisors(f), idiv = []); for (k=1, #d, if (isidiv(d[k], f), idiv = concat(idiv, d[k])); ); idiv; } \\ Michel Marcus at A077609 isigma(n) = {my(f = factor(n), b); prod(i=1, #f~, b = binary(f[i, 2]); prod(k=1, #b, if(b[k], 1+f[i, 1]^(2^(#b-k)), 1)))} ; is(n) = isigma(n) > 2*n && select(x -> x < n && isigma(x) >= 2*x, idivs(n)) == [];