A116578 - OEIS

%I #7 Jan 27 2013 12:41:32

%S 2,0,4,4,4,8,0,11,11,16,9,9,25,25,32,0,31,31,55,55,64,28,28,79,79,115,

%T 115,128,0,97,97,181,181,236,236,255,88,88,256,256,392,392,481,481,

%U 512,0,316,316,601,601,828,828,973,973,1024

%N Integerization of a truncated Pascal root structure with a power of two level pumping.

%C I used a backward representation of the roots so that the least comes first: the results behaves like an economics or population curve. When taken as Modulo two one can see a pattern like that of Pascal's triangle in the zeros and ones. The alternating (t-1)^n polynomials are solved as: (t-1)^n=1 and instead of the 2^n coefficients, the roots are used for sequence. It is a unique new approach to the problem of Pascal's triangle.

%F a(n) = Table[Table[Floor[2^(n - 1)*Abs[x]] /. NSolve[(x - 1)^n - 1 == 0.x][[m]], {m, n, 1, -1}], {n, 1, 10}]

%e Triangular form of the sequence:

%e {2}

%e {0, 4}

%e {4, 4, 8}

%e {0, 11, 11, 16}

%e {9, 9, 25, 25, 32}

%e {0, 31, 31, 55, 55, 64}

%t Table[Table[Floor[2^(n - 1)*Abs[x]] /. NSolve[(x - 1)^n - 1 == 0.x][[m]], {m, n, 1, -1}], {n, 1, 10}] Flatten[a]

%K nonn,uned,obsc

%O 0,1

%A _Roger L. Bagula_, Mar 21 2006