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a(1)=1; a(m+1) = Sum_{k=1..m} gcd(k, a(m+1-k)).
1

%I #16 Oct 07 2017 17:59:27

%S 1,1,2,3,5,5,9,7,9,17,15,13,17,15,27,33,21,27,29,29,35,33,29,45,55,51,

%T 73,41,59,61,57,39,63,57,59,85,49,87,65,65,77,93,83,107,117,73,91,95,

%U 137,97,117,125,115,131

%N a(1)=1; a(m+1) = Sum_{k=1..m} gcd(k, a(m+1-k)).

%H Ivan Neretin, <a href="/A053079/b053079.txt">Table of n, a(n) for n = 1..10000</a>

%e a(7) = gcd(1, a(6)) + gcd(2, a(5)) + gcd(3, a(4)) + gcd(4, a(3)) + gcd(5, a(2)) + gcd(6, a(1)) = 1 + 1 + 3 + 2 + 1 + 1 = 9.

%t Fold[Append[#1, Total@Table[GCD[#1[[#2 - k]], k], {k, #2 - 1}]] &, {1}, Range[2, 54]] (* _Ivan Neretin_, Oct 04 2017 *)

%o (PARI) lista(nn) = my(va = vector(nn)); va[1] = 1; for (m=1, nn-1, va[m+1] = sum(k=1, m, gcd(k, va[m+1-k]));); va; \\ _Michel Marcus_, Oct 05 2017

%K nonn

%O 1,3

%A _Leroy Quet_, Feb 25 2000