A266378 Recursive 2-parameter sequence allowing calculation of the Möbius function.

-1, -1, 0, -1, -1, -1, -1, -1, -1, 0, 1, 2, 2, 1, -1, -3, -6, -10, -14, -17, -17, -14, -9, -4, -1, -1, -3, -5, -6, -5, -2, 3, 9, 14, 16, 14, 9, 4, 1, 0, 0, -1, -4, -9, -15, -20, -22, -19, -10, 5, 24, 43, 58, 67, 70, 67, 58, 44, 28, 14, 5, 1, 0, -1, -5, -14, -29, -49, -71, -90, -100, -95, -70, -23, 44, 126, 216, 305, 382, 436, 459, 449, 411

Offset

3,12

Comments

The a(n,m) forms a table where each row has (n-2)*(n-3)/2+1 = A000124(n-3) elements.

The index of the first row is n=3 and the index of the first column is m=0.

The right diagonal a(n, A000217(n-3)) = A008683(n), Möbius numbers, for n>=3.

Links

Table of n, a(n) for n=3..85.

Gevorg Hmayakyan, Recursive Formula Related To The Mobius Function

Formula

a(n,m) = a(n, m-1)+a(n-1, m)-a(n-1, m-n+1)-a(n-1, nu(n-1))*(T(n-2, m-1)-T(n-2, m-2)), where T(n,m) are coefficients of A008302, nu(n)=(n-2)*(n-3)/2, a(n,0)=-1, a(n,m)=0 if m<0 and m>nu(n).

Möbius(n) = a(n,nu(n)).

Easy to prove:

a(n+1,1)-a(n,1) = -1 - Möbius(n), n>=3 and accordingly

a(n,1) = 3 - n - A002321(n-1).

a(n+1,2)-a(n,2) = 2 - n - A002321(n) - Möbius(n)*(n-3), n>3 and accordingly

a(n,2) = -1 - (n-1)*(n-4)/2 - (n-3)*A002321(n-1).

a(n,-1+(n-2)*(n-3)/2) = Möbius(n+1) + Möbius(n)*(n-3).

a(n,-2+(n-2)*(n-3)/2) = Möbius(n+2) + Möbius(n+1)*(n-3) + (1/2)*Möbius(n)*(n-1)*(n-4) and in general:

a(n,-k+(n-2)*(n-3)/2)=f(n,k), for n>k.

a(n,-k-n+(n-2)*(n-3)/2)=f(n,n+k) + f(n,k) + h(n, k), for n>k,

where:

f(n,k)=Sum_{m=0..k}{Möbius(n+k-m)*(T(n-1,m)-T(n-1,m-1))}

h(n,s)=Sum_{k=1..n}{Sum_{i=0..k+1)}{(-1)^(i+1)*binomial(k+1, i)*f(n+k, s-2*k+1-i)}}

Easy to see:

a(n,2)-(n-3)*a(n,1)=(n-3)*(n-4)/2

Conjecture:

Sum_{m=0..(n-2)*(n-3)/2}{a(n,m)} = -A068337(n-1).

Sum_{m=0..(n-2)*(n-3)/2}{a(n,m)*m^k*(-1)^m} = 0, n>2*k+4.

Example

The first few rows are
-1;
-1,  0;
-1, -1, -1,  -1;
-1, -1,  0,   1,   2,   2,   1;
-1, -3, -6, -10, -14, -17, -17, -14, -9, -4, -1;

Maple

T := proc(n, k) option remember; if k=0 then 1 else if (n=1 and k=1) then 0 else if (k<0 or k>binomial(n, 2)) then 0 else T(n-1, k)+T(n, k-1)-T(n-1, k-n) end if end if end if end proc:
nu := n->((n-2)*(n-3))/(2):
a := proc (n, m) option remember; if m = 0 then -1 elif m < 0 or nu(n) < m then 0 else procname(n, m-1)+procname(n-1, m)-procname(n-1, m-n+1)-procname(n-1, nu(n-1))*(T(n-2, m-1)-T(n-2, m-2)) end if end proc:
L := seq(seq(a(N, r), r = 0 .. nu(N)), N = 3 .. 20);

Crossrefs

Cf. A002321, A008302, A008683, A068337.

Sequence in context: A007799 A122888 A371830 * A092113 A331485 A342061

Adjacent sequences: A266375 A266376 A266377 * A266379 A266380 A266381

Keyword

sign,tabf

Author

Gevorg Hmayakyan, Dec 28 2015

Status

approved