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

1,1

Comments

Question: Are all terms nonnegative?

Links

Antti Karttunen, Table of n, a(n) for n = 1..16384

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

Formula

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

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

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

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 *)

Prog

(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(d<n, v[n/d]*u[d], 0)))); (u) }; \\ Compute the Dirichlet inverse of the sequence given in input vector v.
A003602(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

Crossrefs

Cf. A003602 (also quadrisection of this sequence), A349134.

Cf. also A323882, A349126.

Sequence in context: A349913 A346236 A323365 * A353336 A349126 A340188

Adjacent sequences: A349132 A349133 A349134 * A349136 A349137 A349138

Keyword

nonn

Author

Antti Karttunen, Nov 13 2021

Status

approved