A343813 Number of partitions of prime(n) containing at least one prime.

1, 2, 5, 12, 48, 88, 269, 450, 1176, 4355, 6558, 20958, 43412, 61733, 122194, 324532, 820827, 1107647, 2652517, 4655220, 6133664, 13751210, 23192039, 49730098, 132657130, 213646624, 270244858, 429702432, 540212859, 848899870, 3905568236, 5952945182, 11078643138

Offset: 1

Author

Paolo Xausa, Apr 30 2021

Keywords

• nonn

Examples

			a(4) = 12 because there are 12 partitions of prime(4) = 7 that contain at least one prime. These partitions are [7], [5,2], [5,1,1], [4,3], [4,2,1], [3,3,1], [3,2,2], [3,2,1,1], [3,1,1,1,1], [2,2,2,1], [2,2,1,1,1], [2,1,1,1,1,1].
		

Links

• David A. Corneth, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000040, A058698, A235945.

Programs

Mathematica

nterms=20;Table[Total[Map[If[Count[#, _?PrimeQ]>0,1,0] &,IntegerPartitions[Prime[n]]]],{n,1,nterms}]

PARI

forprime(p=2,59,my(m=0); forpart(X=p, for(k=1,#X, if(isprime(X[k]),m++;break))); print1(m,", ")) \\ Hugo Pfoertner, Apr 30 2021

PARI

seq(n)={my(p=primes(n), m=p[#p]); vecextract(Vec(1/eta(x+O(x*x^m)) - 1/prod(k=1, m, 1-if(!isprime(k), x^k) + O(x*x^m)), -m), p)} \\ Andrew Howroyd, Apr 30 2021

Python

from sympy.utilities.iterables import partitions
from sympy import sieve, prime
def A343813(n):
    p = prime(n)
    pset = set(sieve.primerange(2,p+1))
    return sum(1 for d in partitions(p) if len(set(d)&pset) > 0) # Chai Wah Wu, May 01 2021

Formula

a(n) = A235945(A000040(n)).