1,4

a(n) = 1/2 * Sum_{i=0..n-1} d(n-i, i+1) where, if m = Sum_{k} p_k^e^k, and n = Sum_{k} p_k^f^k, then d(m, n) = Sum_{i=1..k} abs(e_i - f_i), the multiplicative distance between m and n.

From Ridouane Oudra, Oct 26 2025: (Start)

a(n) = Sum_{k=1..n+1} bigomega(lcm(n+1,k)/(n+1)).

a(n) = Sum_{k=1..n+1} bigomega(k/gcd(n+1,k)).

a(n) = (Sum_{k=1..n+1} bigomega(lcm(n+1,k))) - (n+1)*bigomega(n+1).

a(n) = bigomega((n+1)!) - Sum_{d|(n+1)} bigomega(d)*phi((n+1)/d).

a(n) = A022559(n+1) - A095112(n+1). (End)

d(3, 1) = 1

d(2, 2) = 0

d(1, 3) = 1

So a(3) = 1/2 * (1 + 0 + 1) = 1

MultDistance[m_, n_] := Module[{ mfac = FactorInteger[m], nfac = FactorInteger[ n]}, Plus @@ Map[(If[Length[ # ] == 1, #[[1, 2]], Abs[ #[[1, 2]] - #[[2, 2]]]]) &, Split[ Sort[Flatten[{mfac, nfac}, 1]], (#1[[1]] == #2[[1]]) &]]] DiagSum[n_] := 1/2 Sum[MultDistance[n - i, i + 1], {i, 0, n - 1}] Table[DiagSum[j], {j, 1, 1000}]

(PARI) multDist(m, n) = {if (m==n, 0, my(f=vecsort(concat(factor(m)[, 1], factor(n)[, 1]), , 8)); sum(i=1, #f, abs(valuation(m, f[i])-valuation(n, f[i]))))};

a(n)={sum(i=0, (n-1)/2, multDist(n-i, i+1))}; \\ edited by Michel Marcus, Sep 20 2018

Equals half the antidiagonal sums of A130836.

Cf. A001222, A000010, A022559, A095112.

Sequence in context: A197026 A334417 A195779 * A138754 A021963 A131914

Adjacent sequences: A130846 A130847 A130848 * A130850 A130851 A130852

nonn,easy

Jacob Woolcutt (woolcutt(AT)gmail.com), Jul 21 2007

Program and corrections by Charles R Greathouse IV, Sep 02 2009

Edited by Michel Marcus, Sep 20 2018

approved