A147848 -id:A147848 - cp's OEIS frontend

A306653 a(n) = Sum_{m=1..n} Sum_{k=1..n} [k divides n][n/k divides m]A008683(n/k)n/k[k divides m + 2^p]A008683(k)k, where p can be a positive integer: 1,2,3,4,5,...

1, -2, -2, 2, -2, 4, -2, 0, 0, 4, -2, -4, -2, 4, 4, 0, -2, 0, -2, -4, 4, 4, -2, 0, 0, 4, 0, -4, -2, -8, -2, 0, 4, 4, 4, 0, -2, 4, 4, 0, -2, -8, -2, -4, 0, 4, -2, 0, 0, 0, 4, -4, -2, 0, 4, 0, 4, 4, -2, 8, -2, 4, 0, 0, 4, -8, -2, -4, 4, -8, -2, 0, -2, 4, 0, -4, 4, -8, -2, 0, 0, 4, -2, 8, 4

Offset: 1

Mats Granvik, Mar 03 2019

sign

It appears that for p=1 and p=2, a(n) is identically the same for all n except for n equal to multiples of 16. See A306652 for comparison.

Antti Karttunen, Table of n, a(n) for n = 1..20000
Mats Granvik, Arithmetic properties of a sum related to the first Hardy-Littlewood conjecture

Cf. A008683, A147848, A306652, A298825, A008683.

(* Dirichlet Convolution. *)
p=1;
a[n_] := 1/n*Sum[Sum[If[Mod[n, k] == 0, 1, 0]*If[Mod[m, n/k] == 0, 1, 0]*If[Mod[m + 2^p, k] == 0, 1, 0]*MoebiusMu[n/k]*n/k*MoebiusMu[k]*k, {k, 1, n}], {m, 1, n}]; a /@ Range[85]
(* conjectured faster program *)
nn = 85;
b = 4*Select[Range[1, nn, 2], SquareFreeQ];
bb = Table[DivisorSigma[0, n]*(MoebiusMu[n] + Sum[If[b[[j]] == n, LiouvilleLambda[n]*2/3, 0], {j, 1, Length[b]}]), {n, 1, nn}]

A306653(n) = (1/n)*sum(m=1, n, sumdiv(n, k, if( !(m%(n/k)) && !((m+(2^1))%k), n*moebius(n/k)*moebius(k),0))); \\ Antti Karttunen, Mar 15 2019

a(n) = 1/n * Sum_{m=1..n} Sum_{k=1..n} [k divides n]*[n/k divides m]*[k divides m + 2^p]*A008683(n/k)*n/k*A008683(k)*k, where p can be any positive integer: 1,2,3,4,5,...
a(n) = A000005(n)*(A008683(n) + Sum_{j=1..n} [4*A056911(j)=n]*A008836(n)*2/3) (conjectured formula verified for n=1..2500).
Dirichlet generating function, after Daniel Suteu in A298826 and Álvar Ibeas in A076479, appears to be: Sum_{n>=1} a(n)/n^s = (Product_{j>=1} (1 - 2*prime(j)^(-s)))*(1 + Sum_{n>=2} (1/2/2^(n*(s - 1)))). - Mats Granvik, Apr 07 2019
The conjectured Dirichlet generating function simplifies to: Sum_{n>=1} a(n)/n^s = (Product_{j>=1} (1 - 2*prime(j)^(-s)))*(1 + 2^(1 - s)/(2^s - 2)). - Steven Foster Clark, Sep 23 2022

A235384 Number of involutions in the group Aff(Z/nZ).

Original entry on oeis.org

2, 4, 6, 6, 8, 8, 16, 10, 12, 12, 24, 14, 16, 24, 28, 18, 20, 20, 36, 32, 24, 24, 64, 26, 28, 28, 48, 30, 48, 32, 52, 48, 36, 48, 60, 38, 40, 56, 96, 42, 64, 44, 72, 60, 48, 48, 112, 50, 52, 72, 84, 54, 56, 72, 128, 80, 60, 60, 144, 62, 64, 80, 100, 84

Offset: 2

Tom Edgar, Jan 08 2014

Aff(Z/nZ) is the group of functions on Z/nZ of the form x->ax+b where a and b are elements of Z/nZ and gcd(a,n)=1.
Since Aff(Z/nZ) is isomorphic to the automorphism group of the dihedral group with 2n elements (when n>=3), this is the number of automorphisms of the dihedral group with 2n elements that have order 1 or 2.
The sequence is multiplicative: a(k*m) = a(k)*a(m) if m and k are coprime.
When n=26, this is the number of affine ciphers where encryption and decryption use the same function.

			Since 18 = 2*3^2, we get a(18) = 2*(3^2+1) = 20. Since 120 = 2^3*3*5, we have a(120) = (4+2^2+2^3)*(3+1)*(5+1) = 384.

Alois P. Heinz, Table of n, a(n) for n = 2..10000
K. K. A. Cunningham, Tom Edgar, A. G. Helminck, B. F. Jones, H. Oh, R. Schwell and J. F. Vasquez, On the Structure of Involutions and Symmetric Spaces of Dihedral Groups, Note di Mat., Volume 34, No. 2, 2014.

Cf. A002618, A228179, A147848, A060594, A283796, A335005.

a:= n-> add(`if`(irem(k^2, n)=1, igcd(n, k+1), 0), k=1..n-1):
seq(a(n), n=2..100);  # Alois P. Heinz, Jan 20 2014

a[n_] := Sum[If[Mod[k^2, n] == 1, GCD[n, k+1], 0], {k, 1, n-1}]; Table[a[n], {n, 2, 100}] (* Jean-François Alcover, Mar 24 2014, after Alois P. Heinz *)
f[p_, e_] := p^e + 1; f[2, 1] = 2; f[2, 2] = 6; f[2, e_] := 3*2^(e - 1) + 4; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100, 2]  (* Amiram Eldar, Dec 05 2022 *)

A034448(n,f=factor(n))=factorback(vector(#f~,i,f[i,1]^f[i,2]+1))
a(n)=my(m=valuation(n,2)); if(m==0,1,m==1,2,m==2,6,4+3<<(m-1))*A034448(n>>m) \\ Charles R Greathouse IV, Jul 29 2016

def a(n):
    L=list(factor(n))
    if L[0][0]==2:
        m=L[0][1]
        L.pop(0)
    else:
        m=0
    order=prod([x[0]^x[1]+1 for x in L])
    if m==1:
        order=2*order
    elif m==2:
        order=6*order
    elif m>=3:
        order=(4+2^(m-1)+2^m)*order
    return order
[a(i) for i in [2..100]]

def b(n):
    sum = 0
    for a in [x for x in range(n) if ((x^2) % n == 1)]:
        sum += gcd(a+1,n)
    return sum
[b(i) for i in [2..100]]

Suppose n = 2^m*p_1^(r_1)*p_2^(r_2)*...*p_k^(r_k) where each p_i>2 is prime, r_i>0 for all i, and m>=0 is the prime factorization of n, then:
...a(n) = Product_{1<=i<=k} (p_i^(r_i)+1) if m=0,
...a(n) = 2*Product_{1<=i<=k} (p_i^(r_i)+1) if m=1,
...a(n) = 6*Product_{1<=i<=k} (p_i^(r_i)+1) if m=2,
...a(n) = (4+2^(m-1)+2^m)*Product_{1<=i<=k} (p_i^(r_i)+1) if m>=3.
a(n) = Sum_{a in row(n) of A228179} gcd(a+1,n).
Sum_{k=1..n} a(k) ~ c * n^2, where c = zeta(2)/(2*zeta(3)) = 0.684216... (A335005). - Amiram Eldar, Dec 05 2022

A306652 a(n) = Sum_{m=1..n} Sum_{k=1..n} [k divides n][n/k divides m][k divides m + 4].

Original entry on oeis.org

1, 2, 2, 4, 2, 4, 2, 6, 2, 4, 2, 8, 2, 4, 4, 10, 2, 4, 2, 8, 4, 4, 2, 12, 2, 4, 2, 8, 2, 8, 2, 14, 4, 4, 4, 8, 2, 4, 4, 12, 2, 8, 2, 8, 4, 4, 2, 20, 2, 4, 4, 8, 2, 4, 4, 12, 4, 4, 2, 16, 2, 4, 4, 14, 4, 8, 2, 8, 4, 8, 2, 12, 2, 4, 4, 8, 4, 8, 2, 20, 2, 4, 2, 16, 4

Offset: 1

Mats Granvik, Mar 03 2019

nonn

According to the first Hardy-Littlewood conjecture, the cousin primes have the same asymptotic density as the twin primes. In an analogy of Dirichlet convolutions A147848 corresponds to the twin primes while this sequence corresponds to the cousin primes, and it appears that this sequence differs from A147848 at multiples of 16. See A306653 for comparison.

Antti Karttunen, Table of n, a(n) for n = 1..20000
Mats Granvik, Arithmetic properties of a sum related to the first Hardy-Littlewood conjecture

Cf. A147848, A306652, A298825, A008683.

a[n_] := Sum[Sum[If[Mod[n, k] == 0, If[Mod[m, n/k] == 0, 1, 0], 0]*If[Mod[m + 4, k/1] == 0, 1, 0], {k, 1, n}], {m, 1, n}]; a /@Range[85] (* Dirichlet Convolution. *)
a[n_] := Sum[If[Mod[n, k] == 0, Sum[If[Mod[4, j] == 0, If[GCD[k, n/k] == j, j, 0], 0], {j, 1, n}], 0], {k, 1, n}]; a /@Range[85] (* GCD sum. *)
a[n_] := Sum[If[Mod[4, j] == 0, j*Count[Divisors[n], d_ /; GCD[d, n/d] == j], 0], {j, 1, n}]; a /@Range[85] (* After Jean-François Alcover in A034444. *)

A306652(n) = sum(m=1, n, sumdiv(n, k, !(m%(n/k)) && !((m+4)%k))); \\ Antti Karttunen, Mar 13 2019

a(n) = Sum_{m=1..n} Sum_{k=1..n} [k divides n]*[n/k divides m]*[k divides m + 4].

a(n) = Sum_{k=1..n}[k divides n]*Sum_{j=1..n}[j divides 4]*[GCD[k, n/k] = j]*j.

cp's OEIS Frontend

A306653 a(n) = Sum_{m=1..n} Sum_{k=1..n} [k divides n][n/k divides m]A008683(n/k)n/k[k divides m + 2^p]A008683(k)k, where p can be a positive integer: 1,2,3,4,5,...

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A235384 Number of involutions in the group Aff(Z/nZ).

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A306652 a(n) = Sum_{m=1..n} Sum_{k=1..n} [k divides n][n/k divides m][k divides m + 4].

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cp's OEIS Frontend

A306653 a(n) = Sum_{m=1..n} Sum_{k=1..n} [k divides n]*[n/k divides m]*A008683(n/k)*n/k*[k divides m + 2^p]*A008683(k)*k, where p can be a positive integer: 1,2,3,4,5,...

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Formula

A235384 Number of involutions in the group Aff(Z/nZ).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Sage

Sage

Formula

A306652 a(n) = Sum_{m=1..n} Sum_{k=1..n} [k divides n]*[n/k divides m]*[k divides m + 4].

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Formula

A306653 a(n) = Sum_{m=1..n} Sum_{k=1..n} [k divides n][n/k divides m]A008683(n/k)n/k[k divides m + 2^p]A008683(k)k, where p can be a positive integer: 1,2,3,4,5,...

A306652 a(n) = Sum_{m=1..n} Sum_{k=1..n} [k divides n][n/k divides m][k divides m + 4].