A383866 The sum of divisors d of n having the property that for every prime p dividing n the p-adic valuation of d is either 0 or an infinitary divisor of the p-adic valuation of n.

1, 3, 4, 7, 6, 12, 8, 11, 13, 18, 12, 28, 14, 24, 24, 19, 18, 39, 20, 42, 32, 36, 24, 44, 31, 42, 31, 56, 30, 72, 32, 35, 48, 54, 48, 91, 38, 60, 56, 66, 42, 96, 44, 84, 78, 72, 48, 76, 57, 93, 72, 98, 54, 93, 72, 88, 80, 90, 60, 168, 62, 96, 104, 79, 84, 144

Offset: 1

Author

Amiram Eldar, May 13 2025

Keywords

• nonn
• easy
• mult

Comments

First differs from A383864 at n = 256.

The sum of divisors d of n such that each is a unitary divisor of an exponential infinitary divisor of n (see A383760).

Analogous to the sum of (1+e)-divisors (A051378) as exponential infinitary divisors (A383760, A361175) are analogous to exponential divisors (A322791, A051377).

The number of these divisors is A383865(n).

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A036537, A051377, A051378, A077609, A322791, A361175, A383760, A383864, A383865.

Programs

Mathematica

infdivs[n_] := If[n == 1, {1}, Sort@ Flatten@ Outer[Times, Sequence @@ (FactorInteger[n] /. {p_, m_Integer} :> p^Select[Range[0, m], BitOr[m, #] == m &])]];  (* Michael De Vlieger at A077609 *)
f[p_, e_] := 1 + Total[p^infdivs[e]]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

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); }
infdivs(n) = {d = divisors(n); f = factor(n); idiv = []; for (k=1, #d, if (isidiv(d[k], f), idiv = concat(idiv, d[k])); ); idiv; } \\ Michel Marcus at A077609
a(n) = {my(f = factor(n), d); prod(i = 1, #f~, d = infdivs(f[i, 2]); 1 + sum(j = 1, #d, f[i, 1]^d[j]));}

Formula

Multiplicative with a(p^e) = 1 + Sum_{d infinitary divisor of e} p^d.

a(n) <= A051378(n), with equality if and only if all the exponents in the prime factorization of n are in A036537.

Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = Product_{p prime} f(1/p) = 1.52187097260174705015..., and f(x) = (1-x) * (1 + Sum_{k>=1} (1 + Sum{d infinitary divisor of k} x^(2*k-d))).