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a(1) = 2, a(2) = 1; for n >= 3, a(n) = least prime not included earlier that divides the concatenation of all previous terms.
4

%I #14 Nov 30 2019 02:05:12

%S 2,1,3,71,7,10177,2100001,101770000001,4603,13,107,4013,23,

%T 3097349301044927552199565217412468305904367,1847,37,367767021959,

%U 54371,3229,17,520063,29,389,8059,732713,11,7123120001,137,294563,1656881076199062425029,313583,4817,277

%N a(1) = 2, a(2) = 1; for n >= 3, a(n) = least prime not included earlier that divides the concatenation of all previous terms.

%C Conjecture:(1) Every concatenation is squarefree. (2) This is a rearrangement of the noncomposite numbers other than 5.

%C Conjecture (1) is false. 3^2 divides the concatenation for a(22) and a(30). - _Sean A. Irvine_, Nov 25 2009

%e a(4) = 71 as 213 = 3*71.

%Y Cf. A096098.

%K base,nonn

%O 1,1

%A _Amarnath Murthy_, Jun 24 2004

%E More terms from _Sean A. Irvine_, Nov 25 2009

%E a(30)-a(33) from _Chai Wah Wu_, Nov 29 2019