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.

A306709 For n > 1, a(n) = gcd(A001414(n), A167344(n)) where A001414(n) is the sum of primes p dividing n (with repetition) and A167344(n) = b(n) is the totally multiplicative sequence with b(p) = (p-1)*(p+1) = p^2 - 1; a(1) = 0.

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 1, 3, 2, 1, 1, 1, 1, 9, 8, 1, 1, 8, 1, 9, 2, 1, 1, 9, 2, 3, 1, 1, 1, 2, 1, 1, 2, 1, 12, 2, 1, 3, 16, 1, 1, 12, 1, 15, 1, 1, 1, 1, 2, 12, 4, 1, 1, 1, 16, 1, 2, 1, 1, 12, 1, 3, 1, 3, 18, 16, 1, 3, 2, 2, 1, 12, 1, 3, 1, 1, 18, 18, 1, 1, 4, 1, 1, 2, 2, 9, 32, 1, 1, 1, 4, 27
Offset: 1

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Author

Juri-Stepan Gerasimov, May 20 2019

Keywords

Comments

Positions of records: 0, 1, 8, 14, 35, 39, 65, 87, ...

Crossrefs

Programs

Formula

a(n) = gcd(sopfr(n), A003958(n)*A003959(n)) for n > 1; a(1) = 0.
a(p) = 1 for all primes p. - Antti Karttunen, Jan 03 2021

A335915 Fully multiplicative with a(2) = 1, and a(p) = A000265(p-1)*A000265(p+1) = A000265(p^2 - 1), for odd primes p.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 3, 1, 1, 3, 15, 1, 21, 3, 3, 1, 9, 1, 45, 3, 3, 15, 33, 1, 9, 21, 1, 3, 105, 3, 15, 1, 15, 9, 9, 1, 171, 45, 21, 3, 105, 3, 231, 15, 3, 33, 69, 1, 9, 9, 9, 21, 351, 1, 45, 3, 45, 105, 435, 3, 465, 15, 3, 1, 63, 15, 561, 9, 33, 9, 315, 1, 333, 171, 9, 45, 45, 21, 195, 3, 1, 105, 861, 3, 27, 231, 105, 15, 495, 3, 63, 33, 15, 69
Offset: 1

Views

Author

Antti Karttunen, Jul 09 2020

Keywords

Comments

For all i, j: A324400(i) = A324400(j) => a(i) = a(j) => A336118(i) = A336118(j).

Crossrefs

Programs

  • PARI
    A000265(n) = (n>>valuation(n,2));
    A335915(n) = { my(f=factor(n)); prod(k=1,#f~,if(2==f[k,1],1,(A000265(f[k,1]-1)*A000265(f[k,1]+1))^f[k,2])); };

Formula

Completely multiplicative with a(2) = 1, and for odd primes p, a(p) = A000265(p-1)*A000265(p+1).
For all n >= 1, A335904(a(n)) = A336118(n).
For all n >= 0, a(2^n) = a(3^n) = 1, a(5^n) = a(7^n) = 3^n.
a(n) = A336466(n) * A336467(n). - Antti Karttunen, Jan 31 2021

A340323 Multiplicative with a(p^e) = (p + 1) * (p - 1)^(e - 1).

Original entry on oeis.org

1, 3, 4, 3, 6, 12, 8, 3, 8, 18, 12, 12, 14, 24, 24, 3, 18, 24, 20, 18, 32, 36, 24, 12, 24, 42, 16, 24, 30, 72, 32, 3, 48, 54, 48, 24, 38, 60, 56, 18, 42, 96, 44, 36, 48, 72, 48, 12, 48, 72, 72, 42, 54, 48, 72, 24, 80, 90, 60, 72, 62, 96, 64, 3, 84, 144, 68, 54
Offset: 1

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Author

Keywords

Comments

Starting with any integer and repeatedly applying the map x -> a(x) reaches the fixed point 12 or the loop {3, 4}.

Examples

			a(2^s) = 3 for all s>0.
		

Crossrefs

Programs

  • Maple
    f:= proc(n) local  t;
      mul((t[1]+1)*(t[1]-1)^(t[2]-1),t=ifactors(n)[2])
    end proc:
    map(f, [$1..100]); # Robert Israel, Jan 07 2021
  • Mathematica
    fa[n_]:=fa[n]=FactorInteger[n];
    phi[1]=1; phi[p_, s_]:= (p + 1)*( p - 1)^(s - 1)
    phi[n_]:=Product[phi[fa[n][[i, 1]], fa[n][[i, 2]]], {i,Length[fa[n]]}];
    Array[phi, 245]
  • PARI
    A340323(n) = if(1==n,n,my(f=factor(n)); prod(i=1,#f~,(f[i,1]+1)*((f[i,1]-1)^(f[i,2]-1)))); \\ Antti Karttunen, Jan 06 2021

Formula

a(n) = A167344(n) / A340368(n) = A048250(n) * A326297(n). - Antti Karttunen, Jan 06 2021
Sum_{k=1..n} a(k) ~ c * n^2, where c = (zeta(6)/(2*zeta(2)*zeta(3))) * Product_{p prime} (1 + 2/p^2) = 0.56361239505... . - Amiram Eldar, Nov 12 2022

A340368 Multiplicative with a(p^e) = (p - 1) * (p + 1)^(e-1).

Original entry on oeis.org

1, 1, 2, 3, 4, 2, 6, 9, 8, 4, 10, 6, 12, 6, 8, 27, 16, 8, 18, 12, 12, 10, 22, 18, 24, 12, 32, 18, 28, 8, 30, 81, 20, 16, 24, 24, 36, 18, 24, 36, 40, 12, 42, 30, 32, 22, 46, 54, 48, 24, 32, 36, 52, 32, 40, 54, 36, 28, 58, 24, 60, 30, 48, 243, 48, 20, 66, 48, 44, 24, 70, 72, 72, 36, 48, 54, 60, 24, 78, 108, 128, 40
Offset: 1

Views

Author

Antti Karttunen, Jan 06 2021

Keywords

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := (p - 1)*(p + 1)^(e - 1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Nov 12 2022 *)
  • PARI
    A340368(n) = if(1==n,n,my(f=factor(n)); prod(i=1,#f~,(f[i,1]-1)*((f[i,1]+1)^(f[i,2]-1))));

Formula

a(n) = A167344(n) / A340323(n).
a(n) = A173557(n) * A327564(n).
Sum_{k=1..n} a(k) ~ c * n^2, where c = (1/2) * Product_{p prime} (1 - 3/p^2 + 2/p^4) / Product_{p prime} (1 - 2/p^2 - 1/p^3) = 0.4313799748... . - Amiram Eldar, Nov 12 2022
Showing 1-4 of 4 results.