%I #12 Apr 03 2023 10:36:13

%S 2,3,8,3947,43968,61681

%N Numbers n such that the number of partitions of n(n+1)/2 (=A000041(A000217(n))) is prime.

%C Because asymptotically A000041(n*(n+1)/2) ~ exp(Pi*sqrt(2/3*(n*(n+1)/2))) / (4*sqrt(3)*(n*(n+1)/2)), the sum of the prime probabilities ~1/log(A000041(n*(n+1)/2)) is diverging and there are no obvious restrictions on primality; therefore, this sequence may be conjectured to be infinite.

%H Chris K. Caldwell, <a href="https://t5k.org/top20/page.php?id=54">Top twenty prime partition numbers</a>, The Prime Pages.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PartitionFunctionP.html">Partition Function P</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/IntegerSequencePrimes.html">Integer Sequence Primes</a>

%e a(3) = 8 is in the sequence because A000041(8*9/2) = 17977 is a prime.

%o (PARI) for(n=1,2000,if(ispseudoprime(numbpart(n*(n+1)/2)),print1(n,", ")))

%Y Cf. A000041, A046063, A072213, A284594, A285086, A285087.

%K nonn,hard,more

%O 1,1

%A _Serge Batalov_, Apr 09 2017