This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A056789 #72 Jul 25 2024 14:33:14
%S A056789 1,3,10,19,51,48,148,147,253,253,606,352,1015,738,960,1171,2313,1263,
%T A056789 3250,1869,2803,3028,5820,2784,6301,5073,6814,5458,11775,4798,14416,
%U A056789 9363,11505,11563,14898,9343,24643,16248,19276,14797,33621,14013,38830
%N A056789 a(n) = Sum_{k=1..n} lcm(k,n)/gcd(k,n).
%C A056789 For prime p, a(p) = 1 + p^2*(p-1)/2.
%C A056789 We note lcm(k,n) = k*n iff gcd(k,n) = 1 (and in general lcm(k,n) equals k*n/gcd(k,n)), and so for these values LCM/GCD = k*n. From A023896, we have that Sum_{k=1..n-1: gcd(k,n)=1} k = n*phi(n)/2, and so Sum_{k=1..n-1: gcd(k,n)=1} k*n = n * Sum_{k=1..n-1: gcd(k,n)=1} k = n^2*phi(n)/2. As this is true, certainly Sum_{k=1..n} lcm(k,n)/gcd(k,n) > n^2*phi(n)/2. - _Jon Perry_, Nov 09 2014 [Edited by _Petros Hadjicostas_, May 27 2020]
%C A056789 Conjecture: for prime p, a(p^n) = 1 + (1/2)*(p - 1)*p^2*(p^(3*n) - 1)/(p^3 - 1) for n = 1,2,3,.... Cf. A339384. - _Peter Bala_, Dec 04 2020
%C A056789 The conjecture can be proven by splitting up the sum like this: a(p^n) = 1 + Sum_{1 <= r < p^n if gcd(p,r) = 1} lcm(p^n,r)/gcd(p^n,r) + Sum_{1 <= r < p^(n-1) if gcd(p,r) = 1} lcm(p^n,p*r)/gcd(p^n,p*r) + … + Sum_{1 <= r < p if gcd(p,r) = 1} lcm(p^n,p^(n-1)*r)/gcd(p^n,p^(n-1)*r) = 1 + Sum_{1 <= r < p^n if gcd(p,r) = 1} p^n*r + Sum_{1 <= r < p^(n-1) if gcd(p,r) = 1} p^(n-1)*r + … + Sum_{1 <= r < p if gcd(p,r) = 1} p*r = 1 + p^n*(1/2)*p^n*phi(p^n) + p^(n-1)*(1/2)*p^(n-1)*phi(p^(n-1)) + … + p*(1/2)*p*phi(p) = 1 + (1/2)*(p-1)*Sum_{k=1..n} p^(3k-1) = 1 + (1/2)*(p-1)*p^2*(p^(3*n)-1)/(p^3-1). - _Sebastian Karlsson_, Dec 07 2020
%H A056789 T. D. Noe, <a href="/A056789/b056789.txt">Table of n, a(n) for n = 1..1000</a>
%F A056789 a(n) > n^2*phi(n)/2. - _Thomas Ordowski_, Nov 08 2014
%F A056789 a(n) = Sum_{k=1..n} k*n/gcd(k,n)^2. - _Thomas Ordowski_, Nov 08 2014
%F A056789 a(n) = (1/2)*Sum_{d|n} d^2*(d+1) Sum_{j|n/d} mu(j)*j^2. - _Felix A. Pahl_, Nov 23 2019
%F A056789 a(n) = 1 + Sum_{d|n, d > 1} phi(d^3)/2. - _Daniel Suteu_, Dec 10 2020
%F A056789 From _Amiram Eldar_, Oct 05 2023: (Start)
%F A056789 a(n) = (A068963(n)+1)/2.
%F A056789 Sum_{k=1..n} a(k) ~ (Pi^2/120) * n^4. (End)
%F A056789 a(n) < n^3 / 2, n > 1. - _Bill McEachen_, Jul 18 2024
%F A056789 Hence n^3/log log n << a(n) << n^3. - _Charles R Greathouse IV_, Jul 25 2024
%e A056789 a(6) = 6/1 + 6/2 + 6/3 + 12/2 + 30/1 + 6/6 = 48.
%t A056789 Table[ Sum[ LCM[k, n] / GCD[k, n], {k, 1, n}], {n, 1, 50}]
%t A056789 f[p_, e_] := p^2*(p-1)*(p^(3*e)-1)/(p^3-1)+1; a[1] = 1; a[n_] := (1 + Times @@ f @@@ FactorInteger[n])/2; Array[a, 40] (* _Amiram Eldar_, Oct 05 2023 *)
%o A056789 (Haskell)
%o A056789 a056789 = sum . a051537_row -- _Reinhard Zumkeller_, Jul 07 2013
%o A056789 (PARI) vector(50, n, sum(k=1, n, lcm(k,n)/gcd(k,n))) \\ _Michel Marcus_, Nov 08 2014
%o A056789 (PARI) a(n) = sumdiv(n, d, if(d>1, d^2*eulerphi(d)/2, 1)); \\ _Daniel Suteu_, Dec 10 2020
%Y A056789 Row sums of triangle in A051537.
%Y A056789 Cf. A023896, A068963, A339384.
%K A056789 nonn,easy,nice
%O A056789 1,2
%A A056789 _Leroy Quet_, Aug 20 2000
%E A056789 Additional comments from _Amarnath Murthy_, May 09 2002