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For n>1, a(n) is the least base b>2 such that the digits of n in base b contain the digit b-1; a(1)=1.
5

%I #32 Jan 01 2021 10:31:35

%S 1,3,4,5,3,3,3,3,5,11,3,4,4,3,3,3,3,3,3,3,3,3,3,3,3,3,4,4,3,4,4,3,3,3,

%T 3,37,19,3,4,41,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,

%U 3,3,3,3,3,3,3,3,3,3,3,3,3,41,83,3,5,43

%N For n>1, a(n) is the least base b>2 such that the digits of n in base b contain the digit b-1; a(1)=1.

%C The choice b>2 in the name of this sequence comes from the fact that base 2 has the desired property for all n>1.

%H Michel Marcus, <a href="/A337535/b337535.txt">Table of n, a(n) for n = 1..10000</a>

%e a(4) = 5 since 4 = 4_5, and 5 is the only base b > 2 with digit b-1.

%e a(7) = 3 since 7 = 21_3 so containing the digit 2.

%o (PARI) isok(n, b) = vecmax(digits(n, b)) == b-1;

%o a(n) = if (n==1, return (1)); my(b=3); while(!isok(n, b), b++); b;

%Y Cf. A337496, A337536.

%K nonn,base

%O 1,2

%A _Michel Marcus_, Aug 31 2020