|
| |
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
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 (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.
|
|
|
LINKS
|
Table of n, a(n) for n=0..54.
|
|
|
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
|
Sequence in context: A200291 A175457 A049797 * A078050 A134271 A094403
Adjacent sequences: A116575 A116576 A116577 * A116579 A116580 A116581
|
|
|
KEYWORD
|
nonn,uned,obsc
|
|
|
AUTHOR
|
Roger L. Bagula, Mar 21 2006
|
|
|
STATUS
|
approved
|
| |
|
|
