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Lexicographically earliest increasing sequence such that the a(n)th term of the sequence has n divisors.
2

%I #22 Aug 05 2016 20:37:34

%S 1,2,4,9,10,11,12,13,14,16,18,64,66,100,101,112,113,1024,1025,1026,

%T 1027,1028,1029,1030,1031,1032,1033,1034,1035,1036,1037,1038,1039,

%U 1040,1041,1042,1043,1044,1045,1046,1047,1048,1049,1050,1051,1052,1053,1054,1055

%N Lexicographically earliest increasing sequence such that the a(n)th term of the sequence has n divisors.

%H Jaroslav Krizek, <a href="/A275658/b275658.txt">Table of n, a(n) for n = 1..1000</a>

%F A000005(a(a(n))) = tau(a(a(n))) = n.

%e a(1)=1 because tau(1)=1; a(2)=2 because tau(2)=2; a(3) cannot be 3 because tau(3)=2, a(3)=4 (4 is the smallest number x>3); if a(3)=4, a(4) must be the smallest number x>a(3) with 3 divisors, a(4)=9; a(9) must be number with 4 divisors and must keep increase of the sequence, a(9)=14; a(5)=10 because 10>a(4); a(6)=11; a(7)=12; a(8)=13; etc...

%Y Cf. A000005.

%K nonn

%O 1,2

%A _Jaroslav Krizek_, Aug 04 2016