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Search: id:A136673
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| A136673 |
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Triangle of coefficients from a polynomial recursion for Galois field GF(2^n) polynomials: p(x,n)=(x+1)*p(x,n-1)-x*p(x,n-2); or f(x,n)=x^n+x+1;. |
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+0
1
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| 2, 1, 1, 2, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1
(list; table; graph; listen)
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OFFSET
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1,1
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COMMENT
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Row sums are:
{3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3}
The result is very dependent on the two initial polynomials.
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REFERENCES
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Taylor L. Booth, Sequential Machines and Automata Theory, John Wiley and Sons, Inc., 1967, Appendix I
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FORMULA
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p(x,0)=2+x;p(x,1)=1+2*x; p(x,n)=(x+1)*p(x,n-1)-x*p(x,n-2); or f(x,n)=x^n+x+1;
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EXAMPLE
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{2, 1},
{1, 2},
{1, 1, 1},
{1, 1, 0, 1},
{1, 1, 0, 0, 1},
{1, 1, 0, 0, 0, 1},
{1, 1, 0, 0, 0, 0, 1},
{1, 1, 0, 0, 0, 0, 0, 1},
{1, 1, 0, 0, 0, 0, 0, 0, 1},
{1, 1, 0, 0, 0, 0, 0, 0, 0, 1},
{1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1}
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MATHEMATICA
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p[x, 0] = 2 + x; p[x, 1] = 1 + 2*x; p[x_, n_] := p[x, n] = (x + 1)*p[x, n - 1] - x*p[x, n - 2]; Table[ExpandAll[p[x, n]], {n, 0, 10}]; a = Table[CoefficientList[p[x, n], x], {n, 0, 10}] Flatten[a]
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CROSSREFS
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Cf. A057764, A103204.
Sequence in context: A015491 A015776 A067461 this_sequence A097588 A030382 A120889
Adjacent sequences: A136670 A136671 A136672 this_sequence A136674 A136675 A136676
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KEYWORD
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nonn,uned,tabl
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AUTHOR
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Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Apr 05 2008
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