A119512 Determinant of n X n matrix of first n^2 terms of A000020 number of primitive polynomials of degree n over GF(2).
2, 2, 244, -80544, 2895473496576
Offset: 1
Examples
a(2) = 2 = |2 1| |2 2|.
This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
a(2) = 2 = |2 1| |2 2|.
with(numtheory): phi(2^n-1)/n;
Table[EulerPhi[(2^n - 1)]/n, {n, 1, 50}]
a(n)=eulerphi(2^n-1)/n \\ Hauke Worpel (thebigh(AT)outgun.com), Jun 10 2008
The first few are x, x+1; x^2+x+1; x^3+x+1, x^3+x^2+1; ... Note that x is irreducible but not primitive.
Do[a = Reverse[ IntegerDigits[n, 2]]; b = {0}; l = Length[a]; k = 1; While[k < l + 1, b = Append[b, a[[k]]*x^(k - 1) ]; k++ ]; b = Apply[Plus, b]; c = Factor[b, Modulus -> 2]; If[b == c, Print[ FromDigits[ IntegerDigits[n, 2]]]], {n, 3, 250, 2} ]
seq(N, p=2, maxdeg=oo) = {
my(a = List(), k=0, d=0);
while (d++ <= maxdeg,
for (n=p^d, 2*p^d-1, my(f=Mod(Pol(digits(n,p)),p));
if(polisirreducible(f), listput(a, subst(lift(f),'x,10)); k++);
if(k >= N, break(2))));
Vec(a);
};
seq(27) \\ Gheorghe Coserea, May 28 2018
The first few are x+1; x^2+x+1; x^3+x+1, x^3+x^2+1; ... Note that x is irreducible but not primitive.
car = 2; maxDegree = 13;
okQ[{1, 1}] = True;
okQ[coefs_List] := Module[{P}, P = coefs.x^Range[Length[coefs]-1, 0, -1]; coefs[[1]] == 1 && IrreduciblePolynomialQ[P, Modulus -> car] && PrimitivePolynomialQ[P, car]];
FromDigits /@ Select[Table[IntegerDigits[k, car], {k, car+1, car^(maxDegree + 1)}], okQ] (* Jean-François Alcover, Sep 09 2019 *)
a(19) = 8 because A000019(19) = 8. a(20) = 24000 because A000020(20) = 24000.
for m from 1 do
url:= sprintf("https://oeis.org/A%06d/b%06d.txt",m,m);
S:= URL:-Get(url);
L:= StringTools[Split](S,"\n");
for t in L do
g:= sscanf(t, "%d %d");
if nops(g) = 2 and g[1] = m then
a[m]:= g[2];
break
fi;
od;
if not assigned(a[m]) then break fi;
od:
seq(a[i],i=1..m-1); # Robert Israel, May 31 2015
n=3719, sigma(n)=3720, phi(n)=3718, a(n)=p(sigma(n))=34847.
Do[g=n;a=Prime[u=DivisorSigma[1,n]]; b=Prime[w=EulerPhi[n]];s=a-b; If[Equal[s,6],Print[{n,a,b,u,w,u-w}]; ta=Append[ta,a]],{n,1,10000}] ta=Delete[ta,1]
Prime[DivisorSigma[1,#]]&/@Select[Range[5000],Prime[DivisorSigma[ 1,#]] == Prime[ EulerPhi[#]]+6&] (* Harvey P. Dale, Sep 22 2016 *)
p=2;q=3;forprime(r=5,1e6,if(r-p==6 && isprime(primepi(q)), print1(r", "));p=q;q=r) \\ Charles R Greathouse IV, May 15 2013
a(3)=20 because the third sequence not begining with a "1" is A000020.
For n = 3 = 11 in binary, the polynomial is 1+x+x^2 and the 2 shift register sequences are {00..., 01101...}.
For n = 4 = 100 in binary, the polynomial is 1+x^3 and the 4 shift register sequences are {000..., 001001..., 011011..., 111...}.
For n = 6 = 110 in binary, the polynomial is 1+x^2+x^3 and the 2 shift register sequences are {000..., 0010111001...}.
For n = 10 = 1010 in binary, the polynomial is 1+x^2+x^4 and the 4 shift register sequences are {0000..., 0001010001..., 0011110011..., 0110110...}.
For n = 11 = 1011 in binary, the polynomial in 1+x+x^2+x^4. Using a Fibonacci LSFR, if the current state of the register is 0001, the next input bit is 0+0+1=1, and the next state is 0011. If the current state is 0100, the next input bit is 0+0+0=0, and the next state is 1000. The 4 shift register sequences are {0000..., 00011010001..., 00101110010..., 1111...}.
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