A073545 - OEIS

%I #19 Feb 16 2021 09:07:35

%S 1,2,6,25,54,243,1204,3549,19544,81829,104663,663490,743764,7925355,

%T 15376922,39462786,201432540,1187707803,3034296474,8657654859,

%U 48511905236,154669032693,123533546264

%N 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).

%e a(2)=6 because 1/tau(6)+1/tau(7)+1/tau(8) = 1/4+1/2+1/4 = 1.

%t a[n_] := For[k=1, True, k++, If[Sum[1/DivisorSigma[0, k+i], {i, 0, n}]==1, Return[k]]]

%t k = 1; Table[While[Sum[1/DivisorSigma[0, k + i], {i, 0, n}] != 1, k++]; k, {n, 0, 12}] (* _Jayanta Basu_, Jul 01 2013 *)

%Y Cf. A000005.

%K nonn,more

%O 0,2

%A _Benoit Cloitre_, Aug 27 2002

%E Edited by _Dean Hickerson_, Sep 03 2002

%E 2 more terms from _Ryan Propper_, Sep 04 2005

%E a(14)-a(22) from _Donovan Johnson_, Jun 23 2010