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A351743
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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.
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3
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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, 115, 2, 9, 43, 28, 11, 113, 12, 19, 73, 20, 37, 41, 38, 21, 11, 148, 3, 593, 4, 297, 1, 2674, 1, 5349, 2, 2675, 1, 13376, 1, 26753, 2, 13377, 5, 13378, 3, 991, 82, 31
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OFFSET
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1,3
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COMMENTS
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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).
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.
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.
Conjecture: a(n)=1 occurs infinitely many times, and therefore so does a(n)=2.
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LINKS
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Michael De Vlieger, Annotated log-log scatterplot of a(n), n = 1..150, showing records in red, highlighting primes in green, composite prime powers in magenta, and fixed points in gold.
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FORMULA
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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.
- a(2*(n+21)-1) = a(2*n-1) for n >= 70,
- a(2*(n+21)) = 3645*a(2*n) + b(n) for n >= 70 (where b is 21-periodic).
(End)
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EXAMPLE
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a(5)=5, a prime, and s(5)=10=2*5 so a(6)=10/5=2.
a(19)=13, a prime and s(19)=169=13^2, so a(20)=1.
a(24)=2, prime, and s(24)=684=2^2*171, so a(25)=171.
a(49)=297=3^3*11, and s(49)=2673=3^5*11 so a(50)=1.
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.
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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