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Min(|d(k+1-i) - d(i)|, for i = 1..k, ), where d(1),..,d(k) are the divisors of prime(n) + 1.

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a(n)=A056737(A008864(n)).

Cf. A008864, A002378, A056737.

proposed

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Antti Karttunen: a(n) = Axxxxxx(A008864(n)) ? Do we have sequence Axxxxxx? (please include it on crossref & formula-lines), thanks!

allocated Min(|d(k+1-i) - d(i)|, for i = 1..k, where d(1),..,d(k) are the divisors Clarkof Kimberlingprime(n) + 1.

2, 0, 1, 2, 1, 5, 3, 1, 2, 1, 4, 17, 1, 7, 2, 3, 4, 29, 13, 1, 35, 2, 5, 1, 7, 11, 5, 3, 1, 13, 8, 1, 17, 4, 5, 11, 77, 37, 2, 23, 3, 1, 4, 95, 7, 10, 49, 2, 7, 13, 5, 1, 11, 4, 37, 10, 3, 1, 137, 41, 67, 7, 8, 11, 155, 47, 79, 13, 17, 11, 53, 2, 7, 5, 1, 8

1,1

Clark Kimberling, <a href="/A285124/b285124.txt">Table of n, a(n) for n = 1..10000</a>

prime(6) + 1 = 14 has divisors 1,2,7,14, so that k=4 and d(k+1-i) - d(i) ranges through {--13,-5,5,13}, so that a(6) = 5.

f[n_] := f[n] = Prime[n]+1;

Table[Divisors[f[n]] - Reverse[Divisors[f[n]]], {n, 1, 10}]

Table[Min[Abs[Divisors[f[n]] - Reverse[Divisors[f[n]]]]], {n, 1, 100}]

Cf. A008864, A002378.

allocated

nonn,easy

Clark Kimberling, Apr 11 2017

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allocated for Clark Kimberling

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