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(n concatenated n times) - n^n.
2

%I #12 Aug 23 2014 14:06:41

%S 0,18,306,4188,52430,620010,6954234,72111672,612579510,

%T 10101010091010101010,1111111110825799440500,121212121203205111672956,

%U 13131313131010256206539060,1414141414130302134588583398

%N (n concatenated n times) - n^n.

%C Amarnath Murthy conjectured that all terms are positive for n>1. This is true. Consider the number of digits. There are n*ceiling(log_10(n+1)) digits in the concatenation, but only log_10(n^n)=n*log_10(n) in n^n. Therefore the terms are never negative. - Hauke Worpel (hw1(AT)email.com), Jun 03 2003

%H Harvey P. Dale, <a href="/A083451/b083451.txt">Table of n, a(n) for n = 1..300</a>

%F a(n) = A000461(n)-A000312(n). - _David Wasserman_, Nov 10 2004

%e a(4) = 4444 - 4^4 = 4188.

%t Table[FromDigits[Flatten[IntegerDigits/@PadRight[{},n,n]]]-n^n,{n,15}] (* _Harvey P. Dale_, Mar 25 2012 *)

%Y Cf. A083452.

%K base,nonn

%O 1,2

%A _Amarnath Murthy_ and Meenakshi Srikanth (menakan_s(AT)yahoo.com), May 01 2003

%E More terms from Hauke Worpel (hw1(AT)email.com), Jun 03 2003

%E More terms from _David Wasserman_, Nov 10 2004