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%I #15 Jun 29 2017 13:45:37
%S 3,7,5,7,5,7,31,29,23,19,5,7,31,47,29,23,19,5,23,19,31,359,397,5,7,5,
%T 7,113,109,31,29,139,137,23,47,19,31,47,29,139,137,7,5,509,199,463,5,
%U 7,31,29,139,331,47,47,29,23,19,5,139,773,109,5,7,113,947,241
%N a(n) = least prime p such that the triple of consecutive primes (p, q, r) has the property that p + prime(n+2) = q + prime(n+1) = r + prime(n).
%C The sequence is generalizable to M (rather than 3) conditions. For instance, with four equalities we obtain the following sequence: least prime p such that the 4-tuple of consecutive primes (p, q, r, t) has the property that p + prime(n+3) = q + prime(n+2) = r + prime(n+1) = t + prime(n). The corresponding sequence is 5, 7, 5, 7, 31, 29, 23, 19, 17, 7, ... for n = 3, 4, ...
%e a(2) = 3 because the triple of consecutive primes (3, 5, 7) has the property that 3 + prime(2+2) = 5 + prime(2+1) = 7 + prime(2) = 10. For n = 2, p = 3 is unique.
%e a(3) = 7 because the triple of consecutive primes (7, 11, 13) has the property that 7 + prime(3+2) = 11 + prime(3+1) = 13 + prime(3) = 18. For n = 3, the other values of p are 13, 37, 67, 97, 103, 193, ... where p == 3 or 7 (mod 10).
%p nn:=100:
%p for i from 2 to nn do:
%p p1:=ithprime(i):p2:=ithprime(i+1):p3:=ithprime(i+2):
%p ii:=0:
%p for j from 2 to nn while(ii=0)do:
%p q1:=ithprime(j):q2:=ithprime(j+1):q3:=ithprime(j+2):
%p if p1+q3=p2+q2 and p1+q3=p3+q1 and p2+q2=p3+q1
%p then
%p ii:=1: printf(`%d, `,q1):
%p else
%p fi:
%p od:
%p od:
%Y Cf. A000040.
%K nonn,easy
%O 2,1
%A _Michel Lagneau_, May 29 2017
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