A349135 Sum of Kimberling's paraphrases (A003602) and its Dirichlet inverse.

2, 0, 0, 1, 0, 4, 0, 1, 4, 6, 0, 2, 0, 8, 12, 1, 0, 6, 0, 3, 16, 12, 0, 2, 9, 14, 12, 4, 0, 4, 0, 1, 24, 18, 24, 5, 0, 20, 28, 3, 0, 6, 0, 6, 26, 24, 0, 2, 16, 17, 36, 7, 0, 16, 36, 4, 40, 30, 0, 8, 0, 32, 36, 1, 42, 10, 0, 9, 48, 12, 0, 5, 0, 38, 46, 10, 48, 12, 0, 3, 37, 42, 0, 11, 54, 44, 60, 6, 0, 20, 56, 12

Offset: 1

Author

Antti Karttunen, Nov 13 2021

Keywords

• sign

Comments

Question: Are all terms nonnegative?

The answer to the above question is no, because A323894 (which is a prime-shifted version of this sequence) also contains negative values. For example, for n=72747675, 88062975, 130945815, 111035925 we get here a(n) = -14126242, -17546656, -14460312, -22677277. The indices are obtained by prime-shifting with A003961 the four indices mentioned in the Apr 20 2022 comment of A323894. - Antti Karttunen, Nov 30 2024

Links

• Antti Karttunen, Table of n, a(n) for n = 1..16384
• Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537

Crossrefs

Cf. A003602 (also quadrisection of this sequence), A349134, A323894 [= a(A003961(n))].

Cf. also A323882, A349126.

Programs

Mathematica

k[n_] := (n/2^IntegerExponent[n, 2] + 1)/2; d[1] = 1; d[n_] := d[n] = -DivisorSum[n, d[#]*k[n/#] &, # < n &]; a[n_] := k[n] + d[n]; Array[a, 100] (* Amiram Eldar, Nov 13 2021 *)

PARI

up_to = 16384;
DirInverseCorrect(v) = { my(u=vector(#v)); u[1] = (1/v[1]); for(n=2, #v, u[n] = (-u[1]*sumdiv(n, d, if(dA003602(n) = (1+(n>>valuation(n,2)))/2;
v349134 = DirInverseCorrect(vector(up_to,n,A003602(n)));
A349134(n) = v349134[n];
A349135(n) = (A003602(n)+A349134(n));

PARI

A349135(n) = if(1==n,2,-sumdiv(n, d, if(1==d||n==d,0,A003602(d)*A349134(n/d)))); \\ (Demonstrates the "cut convolution" formula) - Antti Karttunen, Nov 13 2021

PARI

A003602(n) = (1+(n>>valuation(n,2)))/2;
memoA349134 = Map();
A349134(n) = if(1==n,1,my(v); if(mapisdefined(memoA349134,n,&v), v, v = -sumdiv(n,d,if(dA003602(n/d)*A349134(d),0)); mapput(memoA349134,n,v); (v)));
A349135(n) = (A003602(n)+A349134(n)); \\ Antti Karttunen, Nov 30 2024

Formula

a(n) = A003602(n) + A349134(n).

a(1) = 2, and for n > 1, a(n) = -Sum_{d|n, 1A003602(d) * A349134(n/d).

For all n >= 1, a(4*n) = A003602(n). - Antti Karttunen, Dec 07 2021