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.

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

Offset: 1

Author

David James Sycamore, Jun 07 2022

Keywords

• nonn

Comments

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.

Examples

			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.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000
• 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.

Crossrefs

Cf. A000005, A000040, A011655, A079978.

Programs

Mathematica

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 *)

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

Formula

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.

Conjectures from Rémy Sigrist, Jun 20 2022: (Start)

- 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)

Extensions

More terms from Michael De Vlieger, Jun 08 2022