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%I #17 Oct 08 2023 09:05:08
%S 2,13,43,157,401,2969,7237,30697,57397,68239,576019,967019,225769,
%T 6590069,10942949,21235127,57401779,186564317,154836067,117219967,
%U 2598759227,7470538489,28594410847,59107046659,240456558467,34511350409,193861351357,249423946921,368059259143
%N First of a sequence of exactly n consecutive primes whose squares have the same last digit.
%C a(n) > 4.5*10^11 for n >= 30. - _David A. Corneth_, Oct 07 2023
%e a(3) = 43 because the 3 consecutive primes 43, 47, 53 all have squares ending in 9, while the primes 41 and 59 preceding 43 and following 53 have squares ending in 1.
%p N:= 16: # for a(1) .. a(N)
%p V:= Vector(N):
%p p:= 2: q:= 2: count:= 0: d:= 4: i:= 1:
%p while count < N do
%p p:= nextprime(p);
%p if p^2 mod 10 <> d then
%p if i <= N and V[i] = 0 then
%p V[i]:= q; count:= count+1;
%p fi;
%p q:= p; i:= 1; d:= p^2 mod 10;
%p else
%p i:= i+1;
%p fi
%p od:
%p convert(V,list);
%o (PARI)
%o upto(n) = {
%o my(res = [], ld = 4, streak = 1);
%o forprime(p = 3, n,
%o nd = p^2 % 10;
%o if(nd == ld,
%o streak++
%o ,
%o if(streak > #res,
%o res = concat(res, vector(streak - #res, i, oo))
%o );
%o if(res[streak] == oo,
%o c = p;
%o for(i = 1, streak,
%o c = precprime(c-1);
%o );
%o res[streak] = c;
%o );
%o streak = 1;
%o
%o );
%o ld = nd
%o ); res
%o } \\ _David A. Corneth_, Oct 07 2023
%Y Cf. A054681.
%K nonn,base
%O 1,1
%A _Robert Israel_ and the late _J. M. Bergot_, Oct 06 2023
%E More terms from _David A. Corneth_, Oct 07 2023
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