A173290 Partial sums of A001615.
1, 4, 8, 14, 20, 32, 40, 52, 64, 82, 94, 118, 132, 156, 180, 204, 222, 258, 278, 314, 346, 382, 406, 454, 484, 526, 562, 610, 640, 712, 744, 792, 840, 894, 942, 1014, 1052, 1112, 1168, 1240, 1282, 1378, 1422, 1494, 1566, 1638, 1686, 1782, 1838, 1928, 2000, 2084
Offset: 1
Keywords
References
- W. Hürlimann, Dedekind's arithmetic function and primitive four squares counting functions, Journal of Algebra, Number Theory: Advances and Applications, Volume 14, Number 2, 2015, Pages 73-88; http://scientificadvances.co.in; DOI: http://dx.doi.org/10.18642/jantaa_7100121599
Links
- Enrique Pérez Herrero, Table of n, a(n) for n = 1..5000
- W. Hürlimann, Dedekind's arithmetic function and primitive four squares counting functions, Journal of Algebra, Number Theory: Advances and Applications, Volume 14, Number 2, 2015, Pages 73-88.
Crossrefs
Programs
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Magma
[(&+[MoebiusMu(k)^2*Floor(n/k)*Floor(1 + n/k): k in [1..n]])/2: n in [1..60]]; // G. C. Greubel, Nov 23 2018
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Maple
with(numtheory): a:=n->(1/2)*add(mobius(k)^2*floor(n/k)*floor(1+n/k),k=1..n); seq(a(n),n=1..55); # Muniru A Asiru, Nov 24 2018
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Mathematica
Table[Sum[DirichletConvolve[j, MoebiusMu[j]^2, j, k], {k,1,n}], {n,60}] (* G. C. Greubel, Nov 23 2018 *) psi[n_] := If[n==1, 1, n*Times@@(1 + 1/FactorInteger[n][[;;,1]])]; Accumulate[Array[psi, 50]] (* Amiram Eldar, Nov 23 2018 *)
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PARI
S(n) = sum(k=1, sqrtint(n), moebius(k)*(n\(k*k))); \\ see: A013928 a(n) = sum(k=1, sqrtint(n), k*(k+1) * (S(n\k) - S(n\(k+1))))/2 + sum(k=1, n\(1+sqrtint(n)), moebius(k)^2*(n\k)*(1+n\k))/2; \\ Daniel Suteu, Nov 23 2018
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Sage
def A173290(n) : return add(k*mul(1+1/p for p in prime_divisors(k)) for k in (1..n)) [A173290(n) for n in (1..52)] # Peter Luschny, Jun 10 2012
Formula
a(n) = Sum_{i=1..n} A001615(i) = Sum_{i=1..n} (n * Product_{p|n, p prime} (1 + 1/p)).
a(n) = 15*n^2/(2*Pi^2) + O(n*log(n)). - Enrique Pérez Herrero, Jan 14 2012
a(n) = Sum_{i=1..n} A063659(i) * floor(n/i). - Enrique Pérez Herrero, Feb 23 2013
a(n) = (1/2)*Sum_{k=1..n} mu(k)^2 * floor(n/k) * floor(1+n/k), where mu(k) is the Moebius function. - Daniel Suteu, Nov 19 2018
a(n) = (Sum_{k=1..floor(sqrt(n))} k*(k+1) * (A013928(1+floor(n/k)) - A013928(1+floor(n/(k+1)))) + Sum_{k=1..floor(n/(1+floor(sqrt(n))))} mu(k)^2 * floor(n/k) * floor(1+n/k))/2. - Daniel Suteu, Nov 23 2018
Comments