A130849 a(n) is half the sum of the terms in the n-th antidiagonal of the table A130836.
0, 1, 1, 4, 2, 7, 4, 9, 8, 15, 6, 19, 13, 16, 13, 28, 15, 32, 17, 28, 27, 40, 16, 41, 34, 39, 30, 55, 28, 59, 34, 53, 50, 59, 32, 75, 57, 64, 41, 84, 47, 88, 55, 66, 72, 97, 42, 97, 71, 90, 70, 113, 65, 104, 67, 104, 97, 128, 56, 133, 103, 102, 82, 129, 89, 150, 99, 130, 100
Offset: 1
Examples
d(3, 1) = 1 d(2, 2) = 0 d(1, 3) = 1 So a(3) = 1/2 * (1 + 0 + 1) = 1
Crossrefs
Equals half the antidiagonal sums of A130836.
Programs
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Mathematica
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}]
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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
Formula
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.
Extensions
Program and corrections by Charles R Greathouse IV, Sep 02 2009
Edited by Michel Marcus, Sep 20 2018
Comments