A345161 If n = Product (p_j^k_j) then a(n) = max (nextprime(p_j) - p_j), where nextprime = A151800.

1, 1, 2, 1, 2, 2, 4, 1, 2, 2, 2, 2, 4, 4, 2, 1, 2, 2, 4, 2, 4, 2, 6, 2, 2, 4, 2, 4, 2, 2, 6, 1, 2, 2, 4, 2, 4, 4, 4, 2, 2, 4, 4, 2, 2, 6, 6, 2, 4, 2, 2, 4, 6, 2, 2, 4, 4, 2, 2, 2, 6, 6, 4, 1, 4, 2, 4, 2, 6, 4, 2, 2, 6, 4, 2, 4, 4, 4, 4, 2, 2, 2, 6, 4, 2, 4, 2, 2, 8, 2

Offset: 1

Author

Ilya Gutkovskiy, Aug 26 2021

Keywords

• nonn

Examples

			a(39) = a(3 * 13) = a(prime(2) * prime(6)), prime(3) - prime(2) = 5 - 3 = 2, prime(7) - prime(6) = 17 - 13 = 4, so a(39) = max(2, 4) = 4.
		

Links

• Ilya Gutkovskiy, Plot of (1/prime(n)) * Sum_{k=1..n} a(k)
• Index entries for sequences computed from indices in prime factorization
• Index entries for primes, gaps between

Crossrefs

Cf. A000079 (positions of 1's), A000720, A001223, A006530, A061395, A063095, A151800, A327441, A347101, A347102.

Programs

Mathematica

a[n_] := Max @@ (NextPrime[#[[1]]] - #[[1]] & /@ FactorInteger[n]); Table[a[n], {n, 1, 90}]

Formula

If n = Product (p_j^k_j) then a(n) = max (prime(pi(p_j) + 1) - p_j), where pi = A000720.

a(2^j*n) = a(n).

a(n^j) = a(n), j > 0.

a(prime(n)^j) = A001223(n), j > 0.

a(n!) = A327441(n).

a(prime(n)#) = A063095(n).

2 + Sum_{k=1..n-1} a(prime(k)^j) = prime(n), j > 0.

Sum_{d|n} mu(n/d) * a(d) = 0 if n is an even number or an odd number divisible by a square > 1.