%I #45 Nov 08 2024 05:51:09
%S 1,2,4,8,6,32,64,12,256,512,24,2048,36,30,16384,32768,96,72,262144,
%T 192,1048576,2097152,60,8388608,216,768,67108864,288,1536,536870912,
%U 1073741824,120,576,8589934592,6144,34359738368,68719476736,180,864
%N Smallest k such that 1/k can be written as a sum of exactly 2 unit fractions in n ways.
%C From _Jianing Song_, Aug 30 2021: (Start)
%C a(n) is the smallest number whose square has exactly 2n-1 divisors.
%C a(n) is the earliest occurrence of 2n-1 in A048691. (End)
%H David W. Wilson, <a href="/A016017/b016017.txt">Table of n, a(n) for n = 1..1000</a> (terms 122, 365 and 608 corrected by _Amiram Eldar_, Nov 08 2024)
%F a(n+1) <= 2^n.
%F From _Labos Elemer_, May 22 2001: (Start)
%F a(n) = sqrt(A061283(n)).
%F a(n) = sqrt(Min{k| A000005(k)=2n-1}).
%F a((p+1)/2) = 2^((p-1)/2) = 2^A005097(i) if p is the i-th odd prime. [Corrected by _Jianing Song_, Aug 30 2021] (End)
%F a(n) is the least k such that (tau(k^2) + 1)/2 = n. - _Vladeta Jovovic_, Aug 01 2001
%e a(1)=1 and a(2)=2 because 1/2 = 1/3 + 1/6 = 1/4 + 1/4.
%e a(3)=4 because 1/4 = 1/5 + 1/20 = 1/6 + 1/12 = 1/8 + 1/8.
%e a(4)=8 because 1/8 = 1/9 + 1/72 = 1/10 + 1/40 = 1/12 + 1/24 = 1/16 + 1/16.
%e a(5)=6 because 1/6 = 1/7 + 1/42 = 1/8 + 1/24 = 1/9 + 1/18 = 1/10 + 1/15 = 1/12 + 1/12.
%t f[j_, n_] := (Times @@ (j(Last /@ FactorInteger[n]) + 1) + j - 1)/j; t = Table[0, {50}]; Do[a = f[2, n]; If[a < 51 && t[[a]] == 0, t[[a]] = n; Print[{a, n}]], {n, 2^30}] (* _Robert G. Wilson v_, Aug 03 2005 *)
%o (PARI) a(n) = {k = 1; while (numdiv(k^2) != (2*n-1), k++); return (k); }; \\ _Amiram Eldar_, Jan 07 2019 after _Michel Marcus_ at A071571
%Y Identical to A071571 shifted right.
%Y Cf. A000005, A000290, A005408, A005179, A003680, A037992, A055079, A048691.
%K nonn,changed
%O 1,2
%A _Robert G. Wilson v_
%E Entry revised by _N. J. A. Sloane_, Aug 14 2005
%E Offset corrected by _David W. Wilson_, Dec 27 2018