|
| |
| |
|
|
|
1, 0, 0, 1, 0, 0, 0, 2, -1, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, -2, 0, 0, 0, 0, 4, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 0, 0, -2, 0, 0, 0, 0, 0, 0, 0, 0, -3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, -2
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,8
|
|
|
COMMENTS
|
Positions of nonzero entries appear to be given by A001694.
The limit lim_{k->oo} Sum_{n=1..k} a(n)/n appears to converge to some number.
Sum_{n=1..30000} a(n)/n = 1.31897... This appears to be in agreement with:
(1/30000)*Sum_{n<=30000} log(A014963(n))*log(A014963(n+2)) = 1.30351...
If the limit can be proven to converge to a number greater than 1, then it is true that Sum_{n<=X} log(A014963(n))*log(A014963(n+2)) > X, where ">" means "greater than" as in usual inequality notation.
The twin prime conjecture, according to Terence Tao on his blog, is that Sum_{n<=X} log(A014963(n))*log(A014963(n+2)) >> X where ">>" means "asymptotically greater than". There he also says that the first Hardy Littlewood conjecture states that Sum_{n<=X} log(A014963(n))*log(A014963(n+h)) = S(h)*X + o(X), where "S(h)" is the singular series.
Compare this to the prime number theorem which is lim_{X->oo} (Sum_{n<=X} log(A014963(n)))/X = 1.
(End)
|
|
|
LINKS
|
|
|
|
FORMULA
|
The Dirichlet generating function, after Daniel Suteu above and Álvar Ibeas in A076479, appears to be: Sum_{n>=1} a(n)/n^s = zeta(s)^2*(Product_{j>=1} (1 - 2*prime(j)^(-s)))*(1 + Sum_{n>=2} ((1/2)/2^(n*(s - 1)))).
Conjectured formula: Let b(n) = 4*A056911(n) and c(n) = A000005(n)*A008683(n) + Sum_{j=1..length(b(1..N))} [b(j)=n]*A008836(n)*2/3) then a(n) = Sum_{k=1..n}[k|n] c(n/k)*A000005(k). (End)
Conjecture: a(n) = (-1)^(n+1) * Sum_{d|n} mu(d)*tau(d)*tau(n/d). - Ridouane Oudra, Nov 19 2019
The conjectured Dirichlet generating function simplifies to: Sum_{n>=1} a(n)/n^s = zeta(s)^2*(Product_{j>=1} (1 - 2*prime(j)^(-s)))*(1 + 2^(1 - s)/(2^s - 2)). - Steven Foster Clark, Sep 12 2022
|
|
|
MATHEMATICA
|
nn = 90;
A = Table[Boole[Mod[n, k] == 0], {n, nn}, {k, nn}];
B = Table[If[Mod[k, n] == 0, MoebiusMu[n]*n, 0], {n, nn}, {k, nn}];
T = (A.B);
TwinMangoldt = Table[a = T[[All, kk]];
F1 = Table[If[Mod[n, k] == 0, a[[n/k]], 0], {n, nn}, {k, nn}];
b = T[[All, kk + 2]];
F2 = Table[ If[Mod[n, k] == 0, b[[n/k]], 0], {n, nn}, {k, nn}];
(F1.F2)[[All, 1]], {kk, nn - 2}];
TT = Transpose[TwinMangoldt];
Table[Sum[TT[[n, k]], {k, n}]/n, {n, nn - 2}]
nn = 108; 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}];
cc = Table[Sum[If[Mod[n, k] == 0, bb[[n/k]]*DivisorSigma[0, k], 0], {k, 1,
n}], {n, 1, nn}]
(* This faster conjectured program agrees with Antti Karttunen's precomputed \
list of numbers. *)(* (End) *)
(* Dirichlet generating function *) s=7; nn=2500; N[Zeta[s]^2*Product[(1 - 2 Prime[j]^(-s)), {j, 1, nn}]*(1 + Sum[1/2/2^(n*(s - 1)), {n, 2, nn}]), 40] (* Mats Granvik, Apr 06 2019 *)
|
|
|
PROG
|
(PARI)
up_to = 256;
DirConv(ma, h) = { my(u = matsize(ma)[1], md = matrix(u, u)); for(n=1, u-h, for(k=1, u, md[n, k] = sumdiv(k, d, ma[n, d]*ma[n+h, k/d]))); (md); };
A298826list(up_to) = { my(h=2, matA = matrix(up_to+h, up_to+h, n, k, !(n%k)), matB = matrix(up_to+h, up_to+h, n, k, (!(k%n))*moebius(n)*n), matT = matA*matB, matD = DirConv(matT, 2)); vector(up_to, i, (1/i)*sum(j=1, i, matD[j, i])); };
v298826 = A298826list(up_to);
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
sign
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
