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A073545
Least k such that 1/tau(k) + 1/tau(k+1) + 1/tau(k+2) + ... + 1/tau(k+n) is equal to 1 (where tau(k)=A000005(k) is the number of divisors of k).
0
1, 2, 6, 25, 54, 243, 1204, 3549, 19544, 81829, 104663, 663490, 743764, 7925355, 15376922, 39462786, 201432540, 1187707803, 3034296474, 8657654859, 48511905236, 154669032693, 123533546264
OFFSET
0,2
EXAMPLE
a(2)=6 because 1/tau(6)+1/tau(7)+1/tau(8) = 1/4+1/2+1/4 = 1.
MATHEMATICA
a[n_] := For[k=1, True, k++, If[Sum[1/DivisorSigma[0, k+i], {i, 0, n}]==1, Return[k]]]
k = 1; Table[While[Sum[1/DivisorSigma[0, k + i], {i, 0, n}] != 1, k++]; k, {n, 0, 12}] (* Jayanta Basu, Jul 01 2013 *)
CROSSREFS
Cf. A000005.
Sequence in context: A173609 A294945 A353363 * A103063 A220191 A030228
KEYWORD
nonn,more
AUTHOR
Benoit Cloitre, Aug 27 2002
EXTENSIONS
Edited by Dean Hickerson, Sep 03 2002
2 more terms from Ryan Propper, Sep 04 2005
a(14)-a(22) from Donovan Johnson, Jun 23 2010
STATUS
approved