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%I #8 Dec 05 2013 19:55:21
%S 1,2,2,1,3,1,2,2,1,4,1,1,2,3,1,1,4,1,2,1,3,3,1,1,1,4,2,2,1,1,2,3,2,2,
%T 1,1,6,1,2,1,1,1,2,5,1,2,1,1,1,4,1,4,1,2,1,1,4,3,1,3,1,1,1,1,5,3,2,1,
%U 2,1,1,1,2,4,3,2,1,2,1,1,1,6,1,3,2,2,1,1,1,1,2,5,1,3,2,2,1,1,1,1,6,1,4,1,2
%N Table T(n,k) read by rows, giving number of occurrences of the remainder k when n is divided by i=1,2,3,...,n.
%C The n-th row adds to n.
%F Let a(m) be the m-th term in the sequence. Then m=f(n)+k where f(1)=1 and f(n+1)=f(n)+floor((n+1)/2). n is the number being divided by the various i's and k is the remainder under consideration. f(n) has the generating function F(x)= (x(1+2x^2-2x^3))/((1-x)^2(1+x^2)) - Bruce Corrigan (scentman(AT)myfamily.com), Oct 22 2002
%F G.f. for k-th column: Sum_{m>0} x^((k+1)*m+k)/(1-x^m). - _Vladeta Jovovic_, Dec 16 2002
%e The table begins
%e 1
%e 2
%e 2 1
%e 3 1
%e 2 2 1
%e 4 1 1
%e 2 3 1 1
%e 4 1 2 1
%Y Cf. A023645 for T(n, 2) and A072527 for T(n, 3).
%Y Cf. A230374, A230399.
%K nonn,tabl
%O 1,2
%A _Amarnath Murthy_, Aug 01 2002
%E Edited by Bruce Corrigan (scentman(AT)myfamily.com), Oct 22 2002
%E More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Apr 25 2003
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