

A116578


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


0



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, 115, 128, 0, 97, 97, 181, 181, 236, 236, 255, 88, 88, 256, 256, 392, 392, 481, 481, 512, 0, 316, 316, 601, 601, 828, 828, 973, 973, 1024
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OFFSET

0,1


COMMENTS

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 (t1)^n polynomials are solved as: (t1)^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.


LINKS



FORMULA

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}]


EXAMPLE

Triangular form of the sequence:
{2}
{0, 4}
{4, 4, 8}
{0, 11, 11, 16}
{9, 9, 25, 25, 32}
{0, 31, 31, 55, 55, 64}


MATHEMATICA

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]


CROSSREFS



KEYWORD

nonn,uned,obsc


AUTHOR



STATUS

approved



