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A293858
Let n be even; m = n/2 and p a prime such that p<=m with n-p nonprime. The sequence contains the successive positive maxima of values n with L = primepi(m-1)-primepi(p+1)> 0.
3
16, 44, 92, 148, 368, 400, 530, 688, 992, 1052, 2228, 3562, 4952, 7102, 10262, 20684, 37052, 52394, 61456, 62828, 80144, 224648, 236476, 251806, 360524, 362534, 742856, 1655152, 1872236, 2108282, 2319728, 2707118, 8561518, 12727966, 18115354, 18245438, 21572990, 54144704
OFFSET
1,1
COMMENTS
Assuming the validity of Goldbach's Conjecture, there exists an integer L and a finite decreasing sequence of prime numbers P(i); i in {1,2,...,L}, such that P(L) < ... < P(2) < P(1) < m with n-P(i) not prime and n-P(L-1) prime, for P(L-1) prime.
The point {P(L-1), n-P(L-1)} is called the "minimal Goldbach point". The connotation of the word "minimal" is that this point lies on the line y = (-x + n) and sustains the shortest perpendicular distance to the line y = x, among all points {p,q} satisfying y=(-x+n) with prime p, 2 <= p <= m, such that n-p is prime.
Let L be the length of the set {P(1),P(2),..., P(L)}.
Notice that if m is prime then L=0. Also; if n-P(1) is prime then L=0.
LINKS
Gilmar Rodriguez Pierluissi, Table of n, a(n) for n = 1..44
J.-M. Deshouillers, A. Granville, W. Narkiewicz and C. Pomerance, An upper bound in Goldbach's problem, Math. Comp. 61 (1993), 209-213.
EXAMPLE
For n=16, previous prime of m is 7; (n-7) is not prime; previous prime of 7 is 5; n-5 is prime; L=Length({7})=1.
For n=44, previous prime of m is 19; (n-19)is not prime; previous prime of 19 is 17; n-17 is not prime; previous prime of 17 is 13; (n-13) is prime; L=Length({19, 17})= 2.
MATHEMATICA
PreviousPrime[n_]:=NextPrime[n, -1]
L[n_?EvenQ]:=Module[{m=n/2}, If[PrimeQ[m], l=0, l=Length[Drop[Most@NestWhileList[PreviousPrime, m, !PrimeQ[n-#]&], 1]]]; l]
f[n_]:=For[m=n/2, True, m--, Return[L[n]]]; For[n=16; max=-1, True, n+=2, If[f[n]>max, Print[n]; max=f[n]]]
PROG
(PARI) f(n) = {len = 0; m = n/2; if (isprime(m), return (0)); p = precprime(m-1); while (1, if (isprime(n-p), return (len)); p = precprime(p-1); len ++; ); }
lista(nn) = {lmax = 0; forstep (n=2, nn, 2, newl = f(n); if (newl > lmax, print1(n, ", "); lmax = newl); ); } \\ Michel Marcus, Oct 22 2017
CROSSREFS
Cf. A065978.
Sequence in context: A238255 A210375 A173560 * A258547 A211573 A211582
KEYWORD
nonn
AUTHOR
STATUS
approved