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%I #4 Mar 30 2012 17:31:16
%S 1,2,1,4,2,1,8,5,2,1,16,11,5,2,1,32,24,12,5,2,1,64,51,27,12,5,2,1,128,
%T 107,60,28,12,5,2,1,256,222,131,63,28,12,5,2,1,512,457,282,140,64,28,
%U 12,5,2,1,1024,935,601,307,143,64,28,12,5,2,1,2048,1904,1270,666,316,144
%N Triangle read by rows: T(n,m) = number of binary numbers n+1 digits long which have m 1's as a substring.
%e T(4,2)=11 because of the sixteen binary digits which are 5 long, {10000, 10001, 10010, 10011, 10100, 10101, 10110, 10111, 11000, 11001, 11010, 11011, 11100, 11101, 11110, 11111}, 11 have "11" as a substring.
%e Triangle begins:
%e n\m
%e 0 1 0 0 0 0 0 0 0 0 0
%e 1 2 1 0 0 0 0 0 0 0 0
%e 2 4 2 1 0 0 0 0 0 0 0
%e 3 8 5 2 1 0 0 0 0 0 0
%e 4 16 11 5 2 1 0 0 0 0 0
%e 5 32 24 12 5 2 1 0 0 0 0
%t T[n_, m_] := Length[ Select[ StringPosition[ #, ToString[(10^m - 1)/9]] & /@ Table[ ToString[ FromDigits[ IntegerDigits[i, 2]]], {i, 2^n, 2^(n + 1) - 1}], # != {} &]]; Flatten[ Table[ T[n, m], {n, 0, 11}, {m, n + 1}]]
%Y First column = A000079 = Powers of 2, the second column = A027934 = number of compositions of n with at least one even part and the last column = A045623 = number of 1's in all compositions of n+1.
%Y Cf. A000079, A027934, A045623, A109434, A109435.
%K base,nonn,tabl
%O 0,2
%A _Robert G. Wilson v_, Jun 27 2005
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