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%I #19 Nov 16 2017 15:52:25
%S 2,3,4,5,7,8,10,14,36,37,39,42,43,44,46,47
%N Numbers n such that floor(sqrt(1)) + floor(sqrt(2)) + floor(sqrt(3)) + ... + floor(sqrt(n)) is prime.
%C The sequence is complete. Indeed, let s(n) be the sum of floor(sqrt(k)) for k from 1 to n. It is easy to verify that s(n^2+j), for 0 <= j < (n+1)^2-n^2, is equal to n(j+1) + n(4n+1)(n-1)/6, which is always divisible by n or by n/6 for n > 6. - _Giovanni Resta_, Mar 26 2014
%e 2 is a term because floor(sqrt(1))+floor(sqrt(2)) = 1+1 = 2 is prime;
%e 14 is a term because floor(sqrt(1))+ ... +floor(sqrt(14)) = 1+1+1+2+2+2+2+2+3+3+3+3+3+3 = 31 is prime.
%p A214036:=proc(q) local a,n; a:=0;
%p for n from 1 to q do a:=a+floor(sqrt(n)); if isprime(a) then print(n); fi; od; end:
%p A214036(10^10);
%p Alternative program:
%p A214036_bis:=proc(q) local a,j,n; a:=0;
%p for n from 1 to q do for j from 1 to 2*n+1 do
%p a:=a+n; if isprime(a) then print(n^2+j-1); fi;
%p od; od; end:
%p A214036_bis(10^10);
%t Position[Accumulate[Table[Floor[Sqrt[n]],{n,50}]],_?PrimeQ]//Flatten (* _Harvey P. Dale_, Apr 14 2017 *)
%o (PARI)
%o default(realprecision,66);
%o sm = 0; /* sum(n>=1, floor(sqrt(n)) */
%o for (n=1, 10^9, sm+=sqrtint(n); if (isprime(sm), print1(n,", ")));
%o /* _Joerg Arndt_, Mar 07 2013 */
%Y Cf. A220953.
%K nonn,fini,full
%O 1,1
%A _Paolo P. Lava_, Mar 06 2013
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