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a(n) is the number of ways the n-th prime number prime(n) can be represented as sum of two smaller odd prime numbers p1, p2 with prime(n) > p1 > (p2 minus the maximum odd prime factor of (p1-p2)).
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%I #20 Jan 09 2021 02:39:23

%S 1,1,2,3,2,2,4,3,3,5,3,6,4,4,4,7,4,4,5,6,6,6,9,7,8,8,7,7,6,11,4,11,9,

%T 7,8,9,7,13,12,6,10,15,10,9,7,13,13,11,13,10,15,10,13,14,11,13,13,12,

%U 14,17,13,13,19,9,14,19,12,8,14,22,17,14,13,16,9,15

%N a(n) is the number of ways the n-th prime number prime(n) can be represented as sum of two smaller odd prime numbers p1, p2 with prime(n) > p1 > (p2 minus the maximum odd prime factor of (p1-p2)).

%C This sequence counts the cases such that prime(n) = p1 + p2 - MaxOddPrimeFactor(p1-p2), where MaxOddPrimeFactor(m) is defined as the maximum odd prime factor of the positive integer m. If there is no odd prime factor of m, MaxOddPrimeFactor(m) is defined as 1.

%C Conjecture: a(n) > 0 when n >= 4.

%C Some nonprime odd numbers, like 27, cannot be partitioned into the form of p1 + p2 - MaxOddPrimeFactor(p1-p2).

%e When n=4, prime(4)=7, MaxOddPrimeFactor(5-3)=1, 7=5+3-1. This is the only case, so a(4)=1.

%e When n=5, prime(5)=11, MaxOddPrimeFactor(7-5)=1, 11=7+5-1. This is the only case, so a(5)=1.

%e When n=6, prime(6)=13, MaxOddPrimeFactor(11-3)=1, 13=11+3-1; and MaxOddPrimeFactor(11-5)=3, 13=11+5-3. Two cases found, so a(6)=2.

%t MaxOddPrimeFactor[m_] :=

%t Module[{factors, l, res}, factors = FactorInteger[m];

%t l = Length[factors]; res = factors[[l, 1]]; If[res == 2, res = 1];

%t res]

%t Table[p = Prime[n]; p1 = NextPrime[p/2, -1]; ct = 0;

%t While[p1 = NextPrime[p1]; p1 < p, p2 = NextPrime[p - p1, -1];

%t While[p2 = NextPrime[p2]; p2 < p1,

%t If[p == (p1 + p2 - MaxOddPrimeFactor[p1 - p2]), ct++]]]; ct, {n, 4,

%t 79}]

%K nonn

%O 4,3

%A _Lei Zhou_, Aug 17 2020