

A117078


a(n) is the smallest k such that prime(n+1) = prime(n) + (prime(n) mod k), or 0 if no such k exists.


77



0, 0, 3, 0, 3, 9, 3, 5, 17, 3, 25, 11, 3, 13, 41, 47, 3, 11, 7, 3, 67, 5, 7, 9, 31, 3, 9, 3, 5, 33, 41, 25, 3, 43, 3, 29, 151, 53, 7, 167, 3, 19, 3, 7, 3, 17, 199, 73, 3, 5, 227, 3, 11, 7, 251, 257, 3, 53, 7, 3, 13, 31, 101, 3, 103, 101, 13, 109, 3, 5, 347, 9, 19, 367, 5, 13, 127, 131, 131, 19, 3
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OFFSET

1,3


COMMENTS

There is a unique decomposition of the primes: provided the weight a(n) is > 0, we have prime(n) = weight * level + gap, or A000040(n)=a(n)*A117563(n)+A001223(n).
a(n) is the smallest divisor of A118534(n) greater than A001223(n) (gap).
a(n) = 0 only for primes 2, 3 and 7. Conjecture: 2, 3 and 7 are the only primes for which log(A000040(n)) < sqrt(A001223(n)).
a(n) > 0 if and only if 2*prime(n+1) < 3*prime(n).  Thomas Ordowski, Nov 25 2013


LINKS



EXAMPLE

For n = 1 we have prime(n) = 2, prime(n+1) = 3; there is no k such that 3  2 = 1 = (2 mod k), hence a(1) = 0.
For n = 3 we have prime(n) = 5, prime(n+1) = 7; 3 is the smallest k such that 7  5 = 2 = (5 mod k), hence a(3) = 3.
For n = 19 we have prime(n) = 67, prime(n+1) = 71; 7 is the smallest k such that 71  67 = 4 = (67 mod k), hence a(19) = 7.


MATHEMATICA

f[n_] := Block[{a, p = Prime@n, np = Prime[n + 1]}, a = Min@ Select[ Divisors[2p  np], # > np  p &]; If[a == Infinity, 0, a]]; Array[f, 80] (* Robert G. Wilson v, May 08 2006 *)


PROG

(PARI) {m=78; for(n=1, m, p=prime(n); d=prime(n+1)p; k=0; j=1; while(k==0&&j<p, if(p%j!=d, j++, k=j)); print1(k, ", "))}


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



