A346215 - OEIS

%I #39 Sep 04 2021 08:26:33

%S 2,53,7,17,347,373,3181,431,1511,61,2213,14551,32779,12841,23911,

%T 31121,1764349,90001,3037297,5031121,5934899,17385223,63052261,

%U 23673281,80690807,1191279451,7594193,850110593,300482639,23554787897,8785664509,20835779213,165645916039

%N a(n) is the start of the first maximal string of n consecutive primes such that the sum of squares of pairs of consecutive primes in the string is always divisible by 10.

%e a(4) = 17 because the four consecutive primes 17, 19, 23, 29 have 17^2 + 19^2, 19^2 + 23^2, 23^2 + 29^2 are all divisible by 10, and this is maximal because 13^2 + 17^2 and 29^2 + 31^2 are not divisible by 10.

%p P:= select(isprime,[2,seq(i,i=3..10^7,2)]):

%p N:= nops(P):

%p R:= select(t -> P[t]^2 + P[t+1]^2 mod 10 = 0, [$1..N-1]):

%p nR:=nops(R):

%p V:= Vector(100): V[1]:= 2:

%p state:= 1: p:= P[R[1]];

%p for i from 2 to nR do

%p   if R[i] = R[i-1]+1 then state:= state+1

%p   else if V[state+1] = 0 then V[state+1]:= p fi;

%p        state:= 1;

%p        p:= P[R[i]];

%p   fi

%p od:

%p V:= convert(V,list):

%p member(0,V,'m'):

%p V[1..m-1];

%K nonn

%O 1,1

%A _J. M. Bergot_ and _Robert Israel_, Aug 22 2021

%E a(22)-a(33) from _Martin Ehrenstein_, Sep 01 2021