cp's OEIS Frontend

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.

Showing 1-4 of 4 results.

A206284 Numbers that match irreducible polynomials over the nonnegative integers.

Original entry on oeis.org

3, 6, 9, 10, 12, 18, 20, 22, 24, 27, 28, 30, 36, 40, 42, 44, 46, 48, 50, 52, 54, 56, 60, 66, 68, 70, 72, 76, 80, 81, 88, 92, 96, 98, 100, 102, 104, 108, 112, 114, 116, 118, 120, 124, 126, 130, 132, 136, 140, 144, 148, 150, 152, 154, 160, 162, 164, 168, 170
Offset: 1

Views

Author

Clark Kimberling, Feb 05 2012

Keywords

Comments

Starting with 1, which encodes 0-polynomial, each integer m encodes (or "matches") a polynomial p(m,x) with nonnegative integer coefficients determined by the prime factorization of m. Write m = prime(1)^e(1) * prime(2)^e(2) * ... * prime(k)^e(k); then p(m,x) = e(1) + e(2)x + e(3)x^2 + ... + e(k)x^k.
Identities:
p(m*n,x) = p(m,x) + p(n,x),
p(m*n,x) = p(gcd(m,n),x) + p(lcm(m,n),x),
p(m+n,x) = p(gcd(m,n),x) + p((m+n)/gcd(m,n),x), so that if A003057 is read as a square matrix, then
p(A003057,x) = p(A003989,x) + p(A106448,x).
Apart from powers of 3, all terms are even. - Charles R Greathouse IV, Feb 11 2012
Contains 2*p^m and p*2^m if p is an odd prime and m is in A052485. - Robert Israel, Oct 09 2016

Examples

			Polynomials having nonnegative integer coefficients are matched to the positive integers as follows:
   m    p(m,x)    irreducible
  ---------------------------
   1    0         no
   2    1         no
   3    x         yes
   4    2         no
   5    x^2       no
   6    1+x       yes
   7    x^3       no
   8    3         no
   9    2x        yes
  10    1+x^2     yes
		

Crossrefs

Cf. A052485, A206285 (complement), A206296.
Positions of ones in A277322.
Terms of A277318 form a proper subset of this sequence. Cf. also A277316.
Other sequences about factorization in the same polynomial ring: A206442, A284010.
Polynomial multiplication using the same encoding: A297845.

Programs

  • Maple
    P:= n -> add(f[2]*x^(numtheory:-pi(f[1])-1), f =  ifactors(n)[2]):
    select(irreduc @ P, [$1..200]); # Robert Israel, Oct 09 2016
  • Mathematica
    b[n_] := Table[x^k, {k, 0, n}];
    f[n_] := f[n] = FactorInteger[n]; z = 400;
    t[n_, m_, k_] := If[PrimeQ[f[n][[m, 1]]] && f[n][[m, 1]]
    == Prime[k], f[n][[m, 2]], 0];
    u = Table[Apply[Plus,
        Table[Table[t[n, m, k], {k, 1, PrimePi[n]}], {m, 1,
          Length[f[n]]}]], {n, 1, z}];
    p[n_, x_] := u[[n]].b[-1 + Length[u[[n]]]]
    Table[p[n, x], {n, 1, z/4}]
    v = {}; Do[n++; If[IrreduciblePolynomialQ[p[n, x]],
    AppendTo[v, n]], {n, z/2}]; v  (* A206284 *)
    Complement[Range[200], v]      (* A206285 *)
  • PARI
    is(n)=my(f=factor(n));polisirreducible(sum(i=1, #f[,1], f[i,2]*'x^primepi(f[i,1]-1))) \\ Charles R Greathouse IV, Feb 12 2012

Extensions

Introductory comments edited by Antti Karttunen, Oct 09 2016 and Peter Munn, Aug 13 2022

A284010 a(n) = least natural number with the same prime signature polynomial p(n,x) has when it is factored over Z. Polynomial p(n,x) has only nonnegative integer coefficients that are encoded in the prime factorization of n.

Original entry on oeis.org

0, 0, 2, 0, 4, 2, 8, 0, 2, 2, 16, 2, 32, 6, 6, 0, 64, 2, 128, 2, 6, 2, 256, 2, 4, 6, 2, 2, 512, 2, 1024, 0, 30, 6, 12, 2, 2048, 6, 6, 2, 4096, 2, 8192, 2, 6, 2, 16384, 2, 8, 2, 30, 2, 32768, 2, 12, 2, 30, 30, 65536, 2, 131072, 6, 6, 0, 60, 2, 262144, 2, 30, 2, 524288, 2, 1048576, 6, 6, 2, 24, 6, 2097152, 2, 2, 6, 4194304, 6, 12, 6, 6, 2, 8388608, 4, 24, 2, 210
Offset: 1

Views

Author

Antti Karttunen, Mar 20 2017

Keywords

Comments

Let p(n,x) be the completely additive polynomial-valued function such that p(prime(n),x) = x^(n-1) as defined by Clark Kimberling in A206284. To compute a(n), factor p(n,x) over Z and collect the exponents of its irreducible polynomial factors using them as exponents of primes (in Z) as 2^e1 * 3^e2 * 5^e3 * ..., with e1 >= e2 >= e3 >= ...

Examples

			For n = 7 = prime(4), the corresponding polynomial is x^3, which factorizes as (x)(x)(x), thus a(7) = 2^3 = 8.
For n = 14 = prime(4) * prime(1), the corresponding polynomial is x^3 + 1, which factorizes as (x + 1)(x^2 - x + 1), thus a(14) = 2^1 * 3^1 = 6.
For n = 90 = prime(3) * prime(2)^2 * prime(1), the corresponding polynomial is x^2 + 2x + 1, which factorizes as (x + 1)^2, thus a(90) = 2^2 = 4.
		

Crossrefs

Cf. A046523, A206284 (positions of 2's), A206442, A277322, A284011, A284012.
Cf. also A260443, A278233, A278243.

Programs

  • PARI
    \\ After Charles R Greathouse IV's code in A046523 and A277322:
    pfps(n) = { my(f=factor(n)); sum(i=1, #f~, f[i, 2] * 'x^(primepi(f[i, 1])-1)); };
    A284010(n) = { if(!bitand(n, (n-1)), 0, my(p=0, f=vecsort(factor(pfps(n))[, 2], ,4)); prod(i=1, #f, (p=nextprime(p+1))^f[i])); }

Formula

a(2^n) = 0. [By an explicit convention.]
Other identities. For all n >= 1:
A284011(n) = a(A260443(n)).

A277322 a(n) = number of irreducible polynomial factors (counted with multiplicity) in the polynomial (with nonnegative integral coefficients) constructed from the prime factorization of n.

Original entry on oeis.org

0, 0, 1, 0, 2, 1, 3, 0, 1, 1, 4, 1, 5, 2, 2, 0, 6, 1, 7, 1, 2, 1, 8, 1, 2, 2, 1, 1, 9, 1, 10, 0, 3, 2, 3, 1, 11, 2, 2, 1, 12, 1, 13, 1, 2, 1, 14, 1, 3, 1, 3, 1, 15, 1, 3, 1, 3, 3, 16, 1, 17, 2, 2, 0, 4, 1, 18, 1, 3, 1, 19, 1, 20, 2, 2, 1, 4, 2, 21, 1, 1, 2, 22, 2, 3, 2, 2, 1, 23, 2, 4, 1, 4, 2, 4, 1, 24, 1, 2, 1, 25, 1, 26, 1, 2
Offset: 1

Views

Author

Antti Karttunen, Oct 09 2016

Keywords

Comments

Let p(n,x) be the completely additive polynomial-valued function such that p(prime(n),x) = x^(n-1) as defined by Clark Kimberling in A206284. Then this sequence is the number of irreducible factors in p(n,x), counted with multiplicity.

Examples

			For n = 7 = prime(4), the corresponding polynomial is x^3, which factorizes as (x)(x)(x), thus a(7) = 3.
For n = 14 = prime(4) * prime(1), the corresponding polynomial is x^3 + 1, which factorizes as (x + 1)(x^2 - x + 1), thus a(14) = 2.
For n = 90 = prime(3) * prime(2)^2 * prime(1), the corresponding polynomial is x^2 + 2x + 1, which factorizes as (x + 1)^2, thus a(90) = 2.
pfps(660) = pfps(2^2*3*5*11) = pfps(2^2) + pfps(3) + pfps(5) + pfps(11) = 2 + x + x^2 + x^4 which is irreducible, so a(660) = 1.
For n = 30030 = Product_{i=1..6} prime(i), the corresponding polynomial is x^5 + x^4 + x^3 + x^2 + x + 1, which factorizes as (x+1)(x^2 - x + 1)(x^2 + x + 1), thus a(30030) = 3.
		

Crossrefs

Cf. A206442 (gives the number of irreducible polynomial factors without multiplicity), A206284 (positions of 1's, i.e., irreducible polynomials).

Programs

  • PARI
    allocatemem(2^29);
    A064989(n) = {my(f); f = factor(n); if((n>1 && f[1,1]==2), f[1,2] = 0); for (i=1, #f~, f[i,1] = precprime(f[i,1]-1)); factorback(f)};
    pfps(n) = if(1==n, 0, if(!(n%2), 1 + pfps(n/2), 'x*pfps(A064989(n))));
    A277322 = n -> if(!bitand(n,(n-1)), 0, vecsum(factor(pfps(n))[,2]));
    for(n=1, 121121, write("b277322.txt", n, " ", A277322(n)));
    
  • PARI
    pfps(n)=my(f=factor(n)); sum(i=1,#f~, f[i,2] * 'x^(primepi(f[i,1])-1))
    A277322(n) = if(1==n, 0, vecsum(factor(pfps(n))[, 2])); \\ Charles R Greathouse IV, test for one added by Antti Karttunen, Oct 09 2016

Formula

a(2^n) = 0. [By an explicit convention.]
a(A000040(n)) = n-1.
a(A007188(n)) = n.
a(A260443(n)) = A277013(n).

A206719 Number of distinct irreducible factors of the polynomial p(n,x) defined at A206073.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 2, 2, 1, 1, 3, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 1, 2, 2, 2, 3, 1, 2, 2, 2, 1, 3, 1, 2, 3, 2, 1, 2, 2, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 3, 1, 2, 3, 1, 2, 1, 3, 1, 3, 1, 2, 3, 2, 1, 3, 1, 2, 1, 2, 1, 3, 2, 2, 1, 2, 1, 4, 1, 2, 2, 2, 2, 2, 1, 3, 2
Offset: 1

Views

Author

Clark Kimberling, Feb 11 2012

Keywords

Comments

The polynomials having coefficients in {0,1} are enumerated as in A206074 (and A206073).

Examples

			p(1,n) = 1, so a(1)=0
p(2,n) = x, so a(2)=1
p(6,n) = x(1+x), so a(6)=2
p(18,n) = x(x+1)(1-x+x^2), so a(18)=3
p(90,n) = x(1+x)(1+x^2)(1-x+x^2), so a(90)=4
		

Crossrefs

Programs

  • Mathematica
    t = Table[IntegerDigits[n, 2], {n, 1, 1000}];
    b[n_] := Reverse[Table[x^k, {k, 0, n}]]
    p[n_, x_] := p[n, x] = t[[n]].b[-1 + Length[t[[n]]]]
    TableForm[Table[{n, p[n, x],
       FactorList[p[n, x]], -1 + Length[FactorList[p[n, x]]]}, {n, 1, 9}]]
    Table[Length[FactorList[p[n, x]]], {n, 1, 120}]
  • PARI
    A206719(n) = { my(f = factor(Pol(binary(n)))); (#f~); }; \\ Antti Karttunen, Dec 16 2017
Showing 1-4 of 4 results.