A352423 Numbers that are the sum of some number of consecutive prime cubes.

8, 27, 35, 125, 152, 160, 343, 468, 495, 503, 1331, 1674, 1799, 1826, 1834, 2197, 3528, 3871, 3996, 4023, 4031, 4913, 6859, 7110, 8441, 8784, 8909, 8936, 8944, 11772, 12167, 13969, 15300, 15643, 15768, 15795, 15803, 19026, 23939, 24389, 26136, 27467, 27810, 27935, 27962, 27970, 29791

Offset: 1

Author

Michel Marcus, Apr 26 2022

Keywords

• nonn

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..10000
• Cathal O'Sullivan, Jonathan P. Sorenson, and Aryn Stahl, An Algorithm to Find Sums of Consecutive Powers of Primes, arXiv:2204.10930 [math.NT], 2022-2023. See S3 p. 10.

Crossrefs

Cf. A030078, A034707, A340771.

Programs

PARI

lista(nn) = {my(list = List(), ip = primepi(nn), vp = primes(ip)); for(i=1, ip, my(s=vp[i]^3); listput(list, s); for (j=i+1, ip, s += vp[j]^3; if (s >vp[ip]^3, break); listput(list, s); ); ); Vec(vecsort(list, , 8)); }

Python

import heapq
from sympy import prime
from itertools import islice
def agen(): # generator of terms
    p = prime(1)**3; h = [(p, 1, 1)]; nextcount = 2
    while True:
        (v, s, l) = heapq.heappop(h)
        yield v
        if v >= p:
            p += prime(nextcount)**3
            heapq.heappush(h, (p, 1, nextcount))
            nextcount += 1
        v -= prime(s)**3; s += 1; l += 1; v += prime(l)**3
        heapq.heappush(h, (v, s, l))
print(list(islice(agen(), 47))) # Michael S. Branicky, Apr 26 2022