Gevorg Hmayakyan - cp's OEIS frontend

A366900 a(n) is the number of real roots of the derivative of the cyclotomic polynomial Phi(n, 1/x).

0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 2, 1, 1, 3, 0, 1, 1, 1, 2, 3, 1, 1, 2, 1, 1, 1, 2, 1, 3, 1, 0, 3, 1, 3, 2, 1, 1, 3, 2, 1, 3, 1, 2, 3, 1, 1, 2, 1, 1, 3, 2, 1, 1, 3, 2, 3, 1, 1, 4, 1, 1, 3, 0, 3, 3, 1, 2, 3, 3, 1, 2, 1, 1, 3, 2, 3, 3, 1, 2, 1, 1, 1, 4, 3, 1, 3

Offset: 1

Gevorg Hmayakyan, Oct 26 2023

nonn

c[n_, y_] := Limit[D[Cyclotomic[n, 1/x], x], x -> y]; Table[Length[Solve[c[n, x] == 0, x, Reals]], {n, 1, 128}]

a(n)=my(v=valuation(n,2)); 2*omega(n>>v) - (v <= 1 && n > 2) \\ Andrew Howroyd, Oct 27 2023

For n = 2^m, a(n) = 0;
For odd n = p^m, a(n) = 1;
For odd n = p1^r1*p2^r2*...*pm^rm, a(n) = 2m-1;
For n = 2*p1^r1*p2^r2*...*pm^rm, a(n) = 2m-1 if p1, ..., pm are odd;
For n = 2^r*p1^r1*p2^r2*...*pm^rm, a(n) = 2m if p1, ..., pm are odd and r > 1.

A339210 a(n) = Sum_{k=0..(n+1)/2} A010815(n*k).

Original entry on oeis.org

0, 0, 1, 1, 1, 0, 2, 1, 1, 0, 2, 0, 2, 0, 0, 1, 2, 0, 2, 1, 1, 1, 2, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 0, 0, 1, 2, 0, 0, 0, 2, 0, 2, 0, 2, 1, 2, 0, 2, 1, 2, 0, 2, 0, 2, 1, 2, 0, 2, 0, 2, 0, 0, 1, 1, -1, 2, 1, 0, 2, 2, 0, 2, 1, 1, 1, 1, 0, 2, 2, 1, 0, 2, 0, 1, 0

Offset: 1

Gevorg Hmayakyan, Nov 26 2020

sign

Conjecture: a(p) = 2 for prime p > 5. Tested for p up to 100000.
From Sebastian Karlsson, Dec 12 2020: (Start)
Proof: The terms of the sum are A010815(n*k) = (-1)^m if n*k = m*(3*m+-1)/2, 0 otherwise. If n*k = m*(3*m+-1)/2 and n is a prime p, then p (>2) divides either m or 3*m+-1. If p divides m, then either m = 0 or p <= m and p < p+1 < 3*m+-1. For m = 0 we get the term (-1)^0 = 1. If m is not 0, then p*k <= p*(p+1)/2 < m*(3*m+-1)/2, so p*k cannot be expressed as m*(3*m+-1)/2.
On the contrary, if p divides 3*m+-1, let 3*m+-1 = q*p were q is an integer. We get: m*(3*m+-1)/2 = q*p*(q*p+-1)/6. If p > 5 and q >= 2 then q*p*(q*p+-1)/6 >= 2*p*(2*p+-1)/6 > p*(p+1)/2 >= p*k. Thus, p*k cannot be expressed as m*(3*m+-1)/2. If q = 1, i.e., if p = 3*m+-1, then m*(3*m+-1)/2 = p*(p+-1)/6. Thus, for k = (p+-1)/6 (which always is an integer if p >= 5) we get the only term (except for when k = 0) for which A010815(p*k) is not 0. When k = (p+-1)/6, m = (p+-1)/3 and hence A010815(p*k) = 1 (since (p+-1)/3 is even if p > 2). This shows that a(p) = 1+0+...+0+1+0+...+0 = 2. (End)

Cf. A010815.

d[ n_] := With[ {m = Sqrt[24 n + 1]}, If[ IntegerQ[m], KroneckerSymbol[ 12, m], 0]];
a[ n_] := Sum[d[n*k], {k, 0, (n + 1)/2}]

f(n) = if( issquare( 24*n + 1, &n), kronecker( 12, n)); \\ A010815
a(n) = sum(k=0, (n+1)/2, f(n*k)); \\ Michel Marcus, Nov 27 2020

A337760 Irregular triangle where T(n,k) are the coefficients of expansion 2^(n-1) Product_{k=1..n} sin(kt) = Sum_{k=1..n(n+1)/2} T(n,k)cos(kt) for even n and 2^(n-1) Product_{k=1..n} sin(kt) = Sum_{k=1..n(n+1)/2} T(n,k)sin(kt) for odd n.

Original entry on oeis.org

0, 1, 0, 1, 0, -1, 0, 0, 1, 0, 1, 0, -1, 1, 0, 0, 0, 0, 0, -1, 0, -1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, -1, 0, -1, 0, 1, 0, 1, 0, 1, 0, 0, 0, -2, 0, 0, 0, -1, 0, 0, 0, 0, 0, 1, 0, 1, 0, -1, 0, 0, 0, 0, 2, 0, 1, 0, 1, 0, 0, 0, -1, 0, -1, 0, 0, 0, -1, 0, 0

Offset: 1

Gevorg Hmayakyan, Sep 18 2020

This coefficients appear in Euler totient function exact formula.

			sin(t) = sin(t),
2*sin(t)*sin(2*t) = cos(t)-cos(3*t),
4*sin(t)*sin(2*t)*sin(3*t) = sin(2*t)+sin(4*t)-sin(6*t),
8*sin(t)*sin(2*t)*sin(3*t)*sin(4*t) = 1-cos(6*t)-cos(8*t)+cos(10*t),
...
and corresponding table is:
0, 1
0, 1, 0, -1
0, 0, 1,  0, 1, 0, -1
1, 0, 0,  0, 0, 0, -1,  0, -1, 0, 1
0, 1, 0,  1, 0, 1,  0,  0,  0, 0, 0, -1, 0, -1, 0, 1
0, 1, 0,  1, 0, 0,  0, -2,  0, 0, 0, -1, 0,  0, 0, 0, 0, 1, 0, 1, 0, -1
...

an := proc (n, r) option remember;
if n < 0 or r < 0 then
0
elif n = 1 then
if r = 1 then
1
else
0
end if;
elif r=0 and n mod 2 = 0 then
procname(n-1, n-r)
else
procname(n-1, n-r)+(-1)^n*(procname(n-1, n+r)-procname(n-1, r-n))
end if
end proc

Table[Expand[2^(n-1)*TrigReduce[Product[Sin[k*t],{k,1,n}]]],{n,1,10}]

T(1, 1) = 1,
T(n, r) = 0 if r < 0 or r > n*(n+1)/2,
T(n, 0) = T(n - 1, n) if n is even,
T(n, 0) = 0 if n is odd,
T(n, r)  = T(n - 1, n - r) + (-1)^n*(T(n - 1, n + r) - T(n - 1, r - n)).

A275966 a(n) is the real part of -ISum_{d|n}(mobius(d)I^(n/d)), I=sqrt(-1), mobius(n) is A008683.

Original entry on oeis.org

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

Offset: 1

Gevorg Hmayakyan, Mar 19 2017

It seems the nonzero coefficients are powers of 2.

This function is multiplicative with a(p^n) = Re(I^(p^n+1) - I^(p^(n-1)+1)).

			a(4) = -Re(I*(mobius(1)*I^4 + mobius(2)*I^2 + mobius(4)*I)) = Re((I^4-I^2)*I) = Re(2*I) = 0.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
Wikipedia, Dirichlet beta function
Wikipedia, Riemann zeta function

Cf. A008683, A101455.

a(n):=-Re(I*add(numtheory:-mobius(d)*I^(n/d), d = numtheory:-divisors(n))).

Table[-Re[I *  Sum[MoebiusMu[d] * (I^(n/d)), {d, Divisors[n]}]], {n, 81}] (* Indranil Ghosh, Mar 19 2017 *)

a(n)=my(f=factor(n)); prod(i=1,#f~, if(f[i,1]==2, if(f[i,2]==1,-1,0), if(f[i,1]%4==3, 2*(-1)^f[i,2], 0))) \\ Charles R Greathouse IV, Mar 22 2017

a(p^n) = (-1)^n*2, for prime p=3 mod 4.
a(p^n) = 0, for prime p=1 mod 4.
a(2) = -1, a(2^n) = 0 for n > 1.
a(n) = -Re(I*Sum_{d|n}(mobius(d)*I^(n/d))).
Dirichlet g.f.: Sum_{n >= 1} a(n)/n^s = beta(s)/zeta(s), where beta(s) and zeta(s) are Dirichlet Beta and Riemann zeta functions accordingly.
Sum_{n >= 1} a(n)/n^s = (1-2^(-s))*Product_{p=3 mod 4}(p^s-1)/(p^s+1), where p runs over prime numbers.
Sum_{n>=1} mobius(n)/(z^n-I) = Sum_{n >= 1} b(n)/z^n. a(n)=Re(b(n)).
Sum_{n>=1} a(n)/(z^n-1) = z/(z^2+1)
Sum_{d|n} a(d) = A101455(n). - Gevorg Hmayakyan, Dec 27 2017

A284059 The absolute values of A275966.

Original entry on oeis.org

1, 1, 2, 0, 0, 2, 2, 0, 2, 0, 2, 0, 0, 2, 0, 0, 0, 2, 2, 0, 4, 2, 2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 4, 0, 0, 0, 0, 2, 0, 0, 0, 4, 2, 0, 0, 2, 2, 0, 2, 0, 0, 0, 0, 2, 0, 0, 4, 0, 2, 0, 0, 2, 4, 0, 0, 4, 2, 0, 4, 0, 2, 0, 0, 0, 0, 0, 4, 0, 2, 0, 2, 0, 2, 0, 0, 2, 0

Offset: 1

Gevorg Hmayakyan, Mar 19 2017

This is multiplicative function with a(p^n) = |Re(I^(p^n+1) - I^(p^(n-1)+1))|.

			a(9) = |Re(I*(mobius(1)*I^9 + mobius(3)*I^3 + mobius(9)*I))| = |Re((I^10 - I^4))| = |-2| = 2.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Cf. A089491, A275966.

a(n):=abs(Re(I*add(numtheory:-mobius(d)*I^(n/d), d = numtheory:-divisors(n)))).

Table[Abs@ Re[I* Sum[MoebiusMu[d] * I^(n/d), {d, Divisors[n]}]], {n, 87}] (* Indranil Ghosh, Mar 19 2017 *)
f[p_, e_] := If[Mod[p, 4] == 1, 0, 2]; f[2, e_] := If[e == 1, 1, 0]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Jan 27 2024 *)

a(n)=my(f=factor(n)); prod(i=1,#f~, if(f[i,1]==2, if(f[i,2]==1,1,0), if(f[i,1]%4==3, 2, 0))) \\ Charles R Greathouse IV, Mar 22 2017

a(n) = |Re(I*Sum_{d|n}(mobius(d)*I^(n/d)))|.

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 3/Pi = 0.954929... (A089491). - Amiram Eldar, Jan 27 2024

A282938 Recursive 2-parameter sequence allowing calculation of the Möbius function (not the same as A266378).

Original entry on oeis.org

1, -1, 1, -1, -1, 2, -1, 0, 1, -2, 1, 0, -1, 2, -1, -1, 3, -2, -1, 1, -1, 1, 1, 0, -2, 1, 1, -2, 1, 0, -1, 1, -1, 3, -1, 0, -1, -2, 1, 1, 1, -1, -1, 3, -2, -1, 1, -1, 2, -2, 1, 1, 0, 0, -1, -3, 2, 2, 0, 0, -2, 1, 0, 0, 1, -3, 2, 1, -1, 1, -2, 2, -2, 2, -2, -1

Offset: 1

Gevorg Hmayakyan, Feb 25 2017

The a(n,m) forms a table where each row has (n-1)*(n-2)/2+1 = A000124(n-2) elements.
The index of the first row is n=1 and the index of the first column is m=0.
The right diagonal a(n, A000217(n-2)) = A008683(n), Möbius numbers, for n>=1.

			The first few rows starting from 1 follow:
  1
  -1
  1, -1
  -1, 2, -1, 0
  1, -2, 1, 0, -1, 2, -1
  -1, 3, -2, -1, 1, -1, 1, 1, 0, -2, 1
  1, -2, 1, 0, -1, 1, -1, 3, -1, 0, -1, -2, 1, 1, 1, -1
  -1, 3, -2, -1, 1, -1, 2, -2, 1, 1, 0, 0, -1, -3, 2, 2, 0, 0, -2, 1, 0, 0

Cf. A000124, A000217, A008683, A231599.

nu[n_]:=(n-1)*(n-2)/2
U[n_, m_] := U[n, m] = If[n > 1, U[n - 1, m - n + 1] - U[n - 1, m], 0]
U[1, m_] := U[1, m] = If[m == 0, 1, 0]
a[n_, m_] := a[n, m] = If[(m < 0) || (nu[n] < m), 0, a[n - 1, m - n + 1] - a[n - 1, m] - a[n - 1, nu[n - 1]]*U[n - 1, m - 1]]
a[1, m_] := a[1, m] = If[m == 0, 1, 0]
Table[Table[a[n, m], {m, 0, nu[n]}], {n, 1, 30}]
Table[a[n, nu[n]], {n, 1, 50}]

a(n,m) = a(n-1, m-n+1) - a(n-1, m) - a(n-1, nu(n-1))*U(n-1, m-1),
where U(n,m) are coefficients of A231599, nu(n)=(n-1)*(n-2)/2, a(1,0)=1, a(n,m)=0 if m<0 and m>nu(n).
Möbius(n) = a(n,nu(n)).

A282634 Recursive 2-parameter sequence allowing the Ramanujan's sum calculation.

Original entry on oeis.org

1, 1, -1, 2, -1, -1, 2, 0, -2, 0, 4, -1, -1, -1, -1, 2, 1, -1, -2, -1, 1, 6, -1, -1, -1, -1, -1, -1, 4, 0, 0, 0, -4, 0, 0, 0, 6, 0, 0, -3, 0, 0, -3, 0, 0, 4, 1, -1, 1, -1, -4, -1, 1, -1, 1, 10, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, 4, 0, 2, 0, -2, 0, -4, 0

Offset: 1

Gevorg Hmayakyan, Feb 20 2017

a(n,0) = phi(n), where phi(n) is Euler's totient function A000010(n).

a(n,1) = mu(n), where mu(n) is the Möbius function A008683(n).

			The few first rows follow:            c_n(t)
  t   0   1   2   3   4   5   6     |  t   1   2   3   4   5   6   7
n                                   |n
1     1;                            |1     1;
2     1, -1;                        |2    -1,  1;
3     2, -1, -1;                    |3    -1, -1,  2;
4     2,  0, -2,  0;                |4     0, -2,  0,  2;
5     4, -1, -1, -1, -1;            |5    -1, -1, -1, -1,  4;
6     2,  1, -1, -2, -1,  1;        |6     1, -1, -2, -1,  1,  2;
7     6, -1, -1, -1, -1, -1, -1;    |7    -1, -1, -1, -1, -1, -1,  6;
      ...                           |     ...
[Edited by _Seiichi Manyama_, Mar 05 2018]

Seiichi Manyama, Rows n=1..140 of triangle, flattened
Gevorg Hmayakyan, On The Moebius and Euler Totient Functions Calculation.
Charles A. Nicol, On Restricted Partitions and a Generalization Of The Euler Totient and The Moebius Function, PNAS 39(9) (1953), 963-968.

Cf. A000010 (phi(n)), A008683 (mu(n)), A054532, A054533, A054534, A054535, A231599.

b[n_, m_] := b[n, m] = If[n > 1, b[n - 1, m] - b[n - 1, m - n + 1], 0]
b[1, m_] := b[1, m] = If[m == 0, 1, 0]
nt[n_, t_] := Round[(n - 1)/2 - t/n]
a[n_, t_] := Sum[b[n, k*n + t], {k, 0, nt[n, t]}]
Flatten[Table[Table[a[n, m], {m, 0, n - 1}], {n, 1, 20}]]

a(n,t) = Sum(b(n, k*n + t), k=0..N(n, t)), where b(n,k) = A231599(n-1,k) and N(n,t) = [(n - 1)/2 - t/n].
a(n,t) = c_n(t) for t >= 1, where c_n(t) is a Ramanujan's sum A054533.
a(n,t) = a(n,-t)
From Seiichi Manyama, Mar 05 2018: (Start)
a(n,t) = c_n(n-t) = Sum_{d | gcd(n,n-t)} d*mu(n/d) for 0 <= t <= n-1.
So a(n,t) = Sum_{d | gcd(n,t)} d*mu(n/d) for 1 <= t <= n-1. (End)

A282283 Recursive 2-parameter sequence allowing calculation of the Euler Totient function.

Original entry on oeis.org

0, 1, -1, 1, 2, -4, 2, -4, 10, -6, -2, 2, 6, -16, 10, 4, -6, 8, -10, 4, -10, 28, -18, -8, 10, -10, 10, -2, 8, -10, 0, 2, 12, -34, 22, 10, -12, 12, -22, 30, -30, 6, 10, -10, 8, 0, 6, -14, 6, -18, 52, -34, -16, 18, -18, 34, -36, 20, 10, -6, -2, 4, -28, 18, 8

Offset: 0

Gevorg Hmayakyan, Feb 11 2017

The a(n,m) forms a table where each row has (n*(n-3)+4)/2 = A152947(n) elements.
The index of the first row is n=1 and the index of the first column is m=0.
The right diagonal a(n, A152947(n)) = A000010(n), Euler Totient function.

			The first few rows are:
0, 1;
-1, 1;
2, -4, 2;
-4, 10, -6, -2, 2;
6, -16, 10, 4, -6, 8, -10, 4;
-10, 28, -18, -8, 10, -10, 10, -2, 8, -10, 0, 2;
12, -34, 22, 10, -12, 12, -22, 30, -30, 6, 10, -10, 8, 0, 6, -14, 6;

Cf. A000010, A152947, A231599.

U[n_, m_] := U[n, m] = If[n > 1, U[n - 1, n*(n - 1)/2 - m]*(-1)^n - U[n - 1, m], 0]
U[1, m_] := U[1, m] = If[m == 0, 1, 0]
Q[n_, m_] := U[n, m - 2] - 2*U[n, m - 1] + U[n, m]
nu[n_]:=(n-1)*n/2+2-n
a[n_, m_] := a[n, m] = If[(m < 0) || (nu[n] < m), 0, a[n - 1, m - n + 1] - a[n - 1, m] - a[n - 1, nu[n - 1]]*Q[n - 1, m]]
a[1, m_] := a[1, m] = If[m == 1, 1, 0]
Table[Table[a[n, m], {m, 0, nu[n]}], {n, 1, 20}]
Table[a[n, nu[n]], {n, 1, 50}]

nu(n) = (n*(n-3)+4)/2
Q(n,m) = 2*A231599(n,m-1)-A231599(n,m-2)-A231599(n,m)
a(n, m) = a(n - 1, m - n + 1) - a(n - 1, m) - a(n - 1, nu(n - 1))*Q(n - 1, m) if (m < 0) or (nu(n) < m)
a(1,m)=1 if m=1 and 0 otherwise.
a(n,nu(n))= A000010(n)

A280665 Recursive 1-parameter sequence a(n) allowing calculation of the Möbius function.

Original entry on oeis.org

1, 0, 0, -1, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, -1, 2, -1, 0, 0, 0, 0, 0, 0, 0, 0, -1, 2, -1, 0, 1, -2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, -2, 0, 1, 1, -1, 1, -1, -2, 3, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 1, 1, -2, -1, 0, -1, 3, -1, 1, -1, 0, 1, -2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, -2, 0, 0, 2, 2, -3, -1, 0, 0, 1, 1, -2, 2, -1, 1, -1

Offset: 1

Gevorg Hmayakyan, Jan 06 2017

sign

This sequence is generated from A266378 by excluding the second recursion parameter.

			Möbius(2) = a(c(1)+2) and because the c(1)=2 => a(c(1)+2)= a(4). l(4)=2, K(4)=1 so l(4)-2<K(4) and l(4)*(l(4)-1)/2>=K(4) and a(4)=u(4)-v(4)-x(4)
p(4)=c(l(4)-2)=c(0)=0
u(4)=a(p(4)+K(4)+1)=a(2)=0
v(4)=a(p(4)+K(4)-l(4)+2)=a(1)=1
x(4)=a(p(4)+l(4)-1)*T(l(4)-1,l(4)*(l(4)-1)/2-K(4))=a(1)*T(1,0)=0, as T(1,0)=0.
a(4)=u(4)-v(4)-x(4)=0-1-0=-1.

Cf. A002321, A003600, A008302, A266378.

l := n->floor((1/3)*(81+81*n+3*sqrt(1104+1458*n+729*n^2))^(1/3)-5/(81+81*n+3*sqrt(1104+1458*n+729*n^2))^(1/3)):
c := n->(1/6)*n*(n^2+3*n+8):
K := n->n-1-c(l(n)-1):
A := (n, z)->z*(product(z^i-1, i = 1 .. n-1)):
T := (n, k)->coeff(eval(A(n, z)), z, k):
p := n->c(l(n)-2):
u := n->a(p(n)+K(n)+1):
v := n->a(p(n)+K(n)-l(n)+2):
x := n->a(p(n)+l(n)-1)*T(l(n)-1, (1/2)*l(n)*(l(n)-1)-K(n)):
a := proc (n) option remember; if K(n) <= l(n)-2 or (1/2)*l(n)*(l(n)-1) < K(n) then 0 else u(n)-v(n)-x(n) end if end proc:
a(2) := 0:
a(1) := 1:

l(n) = floor((1/3)*(81+81*n+3*sqrt(729*n^2+1458*n+1104))^(1/3)-5/(81+81*n+3*sqrt(729*n^2+1458*n+1104))^(1/3))
c(n) = n*(n^2+3*n+8)/6 = A003600(n)
K(n) = n - 1 - c(l(n) - 1)
T(n,m) are coefficients of A008302
p(n) = c(l(n)-2)
u(n) = a(p(n)+K(n)+1)
v(n) = a(p(n)+K(n)-l(n)+2)
x(n) = a(p(n)+l(n)-1)*T(l(n)-1,l(n)*(l(n)-1)/2-K(n))
a(1) = 1
a(2) = 0
if (l(n)-2 >= K(n) or (1/2)*l(n)*(l(n)-1) < K(n)) then a(n) = 0 else a(n) = u(n)-v(n)-x(n)
Möbius(n) = a(c(n-1)+n)
A100198(n-2) = a(c(n-1)-n), for n>3.

A279817 a(1) = -1; for n>1, Sum_{d|n} a(n-d+1) = 0.

Original entry on oeis.org

-1, 1, 1, 0, 1, 0, 1, -1, 0, 1, 1, -2, 1, 1, -1, 1, 1, -3, 1, -3, 1, 2, 1, -6, 0, 0, 0, 0, 1, -2, 1, -2, -1, 5, -1, -4, 1, 3, 0, 3, 1, -3, 1, -7, -3, 10, 1, -9, 0, -10, 2, -4, 1, -7, 2, 6, -1, 4, 1, -25, 1, 2, -2, 4, -1, -11, 1, 6, -1, 13, 1, -20, 1, -3, -4, 0

Offset: 1

Gevorg Hmayakyan, Dec 19 2016

sign

			When n is any prime p, we have Sum_{d|p} a(p-d+1) = 0, so a(p-1+1) + a(p-p+1) = 0, hence a(p)=1.
For n=4, we have a(4-1+1) + a(4-2+1) + a(4-4+1) = 0, so a(4) + a(3) + a(1) = 0, hence a(4)=0.

a := proc (n) option remember; -add(a(n-d+1), d = `minus`(numtheory:-divisors(n), {1})) end proc; a(1) := -1; seq(simplify(a(i)), i = 1 .. 1000)

For primes p and q:
a(p) = 1.
If p^2 - p + 1 is prime then a(p^2) = 0.
If p*q - p + 1 and p*q - q + 1 are primes then a(p*q) = -1.

cp's OEIS Frontend

User: Gevorg Hmayakyan

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