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a(n) = 10^( (10^n-1)/9 + n) + 1.
0

%I #3 Oct 26 2013 13:37:13

%S 101,10000000000001,

%T 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001

%N a(n) = 10^( (10^n-1)/9 + n) + 1.

%C Makowski observes that A230093(a(n)) >= 2 for all n >= 1.

%D Makowski, Andrzej. On Kaprekar's "junction numbers''. Math. Student 34 1966 77 (1967). MR0223292 (36 #6340)

%H <a href="/index/Coi#Colombian">Index entries for Colombian or self numbers and related sequences</a>

%Y Cf. A006064, A230093, A230094, A230100.

%K nonn,base

%O 1,1

%A _N. J. A. Sloane_, Oct 26 2013