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A351743 a(1)=1. Thereafter, a(n+1) is the greatest divisor of s(n) which is prime to a(n), where s(n) is the n-th partial sum.

%I A351743 #43 Jun 20 2022 13:18:44
%S A351743 1,1,2,1,5,2,3,5,4,3,1,28,1,57,2,29,5,6,13,1,170,1,341,2,171,5,172,3,
%T A351743 115,2,9,43,28,11,113,12,19,73,20,37,41,38,21,11,148,3,593,4,297,1,
%U A351743 2674,1,5349,2,2675,1,13376,1,26753,2,13377,5,13378,3,991,82,31
%N A351743 a(1)=1. Thereafter, a(n+1) is the greatest divisor of s(n) which is prime to a(n), where s(n) is the n-th partial sum.
%C A351743 It follows from the definition that a(n)|s(n) for all n, and that the terms in any run of three consecutive terms are pairwise coprime. Furthermore a(n), s(n) have period 3 parity cycles: a(n) has 1,1,0,... (A011655; n>=1), and s(n) has 1,0,0,... (A079978; n>=0), (0 means even and 1 means odd).
%C A351743 Equivalent definition 1. a(1)=1, a(n+1) is the product of the maximal prime powers of all primes p which divide s(n) but do not divide a(n), namely: a(n+1) = Product_{p^k|s(n), p^(k+1) does not divide s(n), p does not divide a(n), k>=1} p^k. Thus a(n)=1, the empty product, iff a(n) and s(n) have the same prime divisors (squarefree kernel). a(1)=a(2)=1 is the only occurrence of adjacent 1's.
%C A351743 Equivalent definition 2. a(1)=1, a(n+1) is the quotient of s(n) and the product of all prime powers q^r, where q|a(n), q^r divides s(n) but q^r+1 does not. A necessary (but not sufficient) condition for a(n)=1 (n>1) is that a(n-1) and s(n-1) have the same parity. If a(n)=1 with a(n+1) even then a(n+2)=1 and a(n+4)=2. If a(n)=1 with a(n+1) odd then a(n+2)=2. Thus 1's (after a(2)) occur in pairs separated by an even term, or as singletons followed by an odd term > 1. The first occurrence of the latter is a(151). In either case 2 soon follows.
%C A351743 Conjecture: a(n)=1 occurs infinitely many times, and therefore so does a(n)=2.
%H A351743 Michael De Vlieger, <a href="/A351743/b351743.txt">Table of n, a(n) for n = 1..10000</a>
%H A351743 Michael De Vlieger, <a href="/A351743/a351743.png">Annotated log-log scatterplot of a(n)</a>, n = 1..150, showing records in red, highlighting primes in green, composite prime powers in magenta, and fixed points in gold.
%F A351743 If a(n) = prime p, then a(n+1) = s(n)/p^k, where k is the greatest power of p dividing s(n), and if a(n)^k = s(n) then a(n+1)=1.
%F A351743 Conjectures from _Rémy Sigrist_, Jun 20 2022: (Start)
%F A351743 - a(2*(n+21)-1) = a(2*n-1) for n >= 70,
%F A351743 - a(2*(n+21)) = 3645*a(2*n) + b(n) for n >= 70 (where b is 21-periodic).
%F A351743 (End)
%e A351743 a(5)=5, a prime, and s(5)=10=2*5 so a(6)=10/5=2.
%e A351743 a(19)=13, a prime and s(19)=169=13^2, so a(20)=1.
%e A351743 a(24)=2, prime, and s(24)=684=2^2*171, so a(25)=171.
%e A351743 a(49)=297=3^3*11, and s(49)=2673=3^5*11 so a(50)=1.
%e A351743 a(150)=473394=2*3*257*307, s(150)=2840364=2^2*3^2*257*307. Since these two numbers have the same squarefree kernel, a(151)=1, then a(152)=2840365 (odd), a(153)=2.
%t A351743 nn = 120; a[1] = m = s = 1; Do[k = s; While[GCD[m, k] > 1, k /= GCD[m, k]]; m = a[n] = k; s += k, {n, 2, nn}]; Array[a, nn] (* _Michael De Vlieger_, Jun 20 2022, after PARI *)
%o A351743 (PARI) { s=0; for (n=1, 67, if (n==1, v=1, w=s; while (1 < g=gcd(w,v), w /= g); v=w); s+=v; print1 (v", ")) } \\ _Rémy Sigrist_, Jun 19 2022
%Y A351743 Cf. A000005, A000040, A011655, A079978.
%K A351743 nonn
%O A351743 1,3
%A A351743 _David James Sycamore_, Jun 07 2022
%E A351743 More terms from _Michael De Vlieger_, Jun 08 2022