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%I #43 Mar 30 2024 03:01:24
%S 1,3,6,8,11,17,16,20,27,31,26,44,31,45,60,48,41,75,46,80,87,73,56,108,
%T 85,87,108,116,71,165,76,112,141,115,158,192,91,129,168,196,101,239,
%U 106,188,261,157,116,256,175,235,222,224,131,297,256,284,249,199
%N a(n) = Sum_{i=1..n} gcd(2*n-i, i).
%C Sum of the GCD's of the smallest and largest parts in the partitions of 2n into exactly two parts.
%H Seiichi Manyama, <a href="/A234307/b234307.txt">Table of n, a(n) for n = 1..10000</a>
%H <a href="/index/Par#part">Index entries for sequences related to partitions</a>.
%F a(n) = Sum_{i=1..n} gcd(2*n-i, i).
%F a(n) = (A018804(2*n)-n)/2. - _Sebastian Karlsson_, Oct 03 2021
%F Conjecture: a(n) = (1/4)*Sum_{k = 1..4*n} (-1)^k *gcd(k, 8*n). - _Peter Bala_, Jan 01 2024
%F Sum_{k=1..n} a(k) ~ (Pi^2/4)*n^2 * (log(n) + 2*gamma - 1/2 + log(2)/6 - Pi^2/16 - zeta'(2)/zeta(2)), where gamma is Euler's constant (A001620). - _Amiram Eldar_, Mar 30 2024
%e a(6) = 17; the partitions of 2(6) = 12 into two parts are: (11,1),(10,2),(9,3),(8,4),(7,5),(6,6). Then a(6) = gcd(11,1) + gcd(10,2) + gcd(9,3) + gcd(8,4) + gcd(7,5) + gcd(6,6). = 1 + 2 + 3 + 4 + 1 + 6 = 17.
%p A234307:=n->add( gcd(2*n-i, i), i=1..n); seq(A234307(n), n=1..100);
%t Table[Sum[GCD[2n - i, i], {i, n}], {n, 100}]
%t f[p_, e_] := (e*(p - 1)/p + 1)*p^e; a[n_] := (Times @@ f @@@ FactorInteger[2*n] - n)/2; Array[a, 100] (* _Amiram Eldar_, Apr 28 2023 *)
%o (PARI) a(n) = sum(i=1, n, gcd(i, 2*n-i)); \\ _Michel Marcus_, Dec 23 2013
%o (PARI) a(n) = {my(f = factor(2*n)); (prod(i = 1, #f~, p = f[i, 1]; e = f[i, 2]; p^(e-1)*(p+e*(p-1))) - n)/2;} \\ _Amiram Eldar_, Mar 30 2024
%Y Cf. A001105 (sum of parts), A002378 (differences of parts).
%Y Cf. A001620, A018804, A306016.
%K nonn,easy
%O 1,2
%A _Wesley Ivan Hurt_, Dec 22 2013
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