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Numbers k such that tau(k) == 1 (mod(2*(tau(prime(k) - k + 1)))), where tau(k) = A000005(k).
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%I #17 Dec 21 2024 00:44:57

%S 1,196,225,441,484,625,1089,1156,1225,1296,1444,3025,3249,3844,4225,

%T 5929,6561,7225,7396,7569,8281,11236,12321,13225,13924,15129,16641,

%U 17689,20164,21025,24025,25281,25600,34225,34969,40401,42025,47089,50625,51076,55225

%N Numbers k such that tau(k) == 1 (mod(2*(tau(prime(k) - k + 1)))), where tau(k) = A000005(k).

%C 118 terms < 5*10^5 were found.

%C All terms are squares because their number of divisors is odd (see formula field in A000005: a(n) is odd iff n is square).

%e 1 is a term because tau(1) = 1, tau(2 -1 + 1) = 2 and 1 modulo 4 is 1.

%e 196 is a term because tau(196) = 9, tau(1193 - 196 + 1) = 4 and 9 modulo 8 is 1.

%e 200 is not a term because tau(200) = 12, tau(1223 -200 + 1) = 11 and 12 modulo 22 = 12.

%o (PARI)

%o isok(k)=my(d_1=numdiv(k),d_2=numdiv(prime(k)-k+1));d_1%(2*d_2)==1;

%o for(k=1,1000,if(isok(k),print1(k", ")))

%Y Cf. A000005, A000040, A011757.

%Y Cf. A378789, A378794.

%K nonn,easy

%O 1,2

%A _Claude H. R. Dequatre_, Dec 07 2024