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%I #6 Mar 31 2012 10:26:03
%S 5,7,9,11,13,15,16,18,20,21,24,25,26,29,30,32,33,35,37,38,41,41,44,45,
%T 47,48,50,52,54,55,56,59,60,62,63,65,67,68,70,71,74,74,77,78,80,81,83,
%U 85,86,88,90,91,93,95,96,98,99,101,103,104,106,108,109,111,112,114,116
%N Number of partition numbers with n digits.
%C a(n)=A072212(n)-A072212(n-1) for n>=2. - _Emeric Deutsch_, Mar 19 2005
%e a(2)=7 because the number of partitions of 5,6,...,13 are 7,11,15,22,30,42,56, 77,101 and thus exactly 7 partition numbers have 2 digits.
%Y Cf. A000041, A072212.
%K nonn,base
%O 1,1
%A _Lekraj Beedassy_, Sep 07 2004
%E Corrected and extended by _Emeric Deutsch_, Mar 19 2005
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