A373045 Prime concatenated analog clock numbers read counterclockwise. Version 2: hours > 9 may be split into individual digits.

2, 3, 5, 7, 11, 19, 43, 101, 211, 1019, 1987, 2111, 21211, 76543, 101987, 432121, 4321211, 12111019, 6543212111, 9876543212111, 1019876543212111019, 654321211101987654321211, 4321211101987654321211101, 6543212111019876543212111, 76543212111019876543212111019

Offset: 1

Author

Eric Angelini and Michael S. Branicky, May 20 2024

Keywords

• nonn
• base

Comments

In this version, the digits of 10, 11, and 12 may be split, in contrast to A036343.

a(47) has 1499 digits.

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..46
• Eric Angelini, Philip Guston's primes
• Tiziano Mosconi, in reply to Carlos Rivera, Puzzle 19: Primes on a clock, primepuzzles.net, Aug 13 2001.

Crossrefs

Cf. A036343, A373044.

Programs

PARI

A373045_row(r)={my(d=concat([digits(i)|i<-[1..12]]), p); Set([p| s<-[1..#d], d[s]&& isprime(p=fromdigits([d[(s-i)%#d+1]| i<-[1..r]]))])}\\ r-digit-terms
A373045_upto_length(L)=concat([A373045_row(r)|r<-[1..L]]) \\ M. F. Hasler, May 21 2024

Python

import heapq
from sympy import isprime
from itertools import islice
def agen(): # generator of terms
    digits = [1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 0, 1, 1, 1, 2]
    h = [(digits[i], i) for i in range(len(digits))]
    found = set()
    while True:
        v, last = heapq.heappop(h)
        if v not in found and isprime(v):
            found.add(v)
            yield v
        nxt = (last-1)%len(digits)
        heapq.heappush(h, (v*10+digits[nxt], nxt))
print(list(islice(agen(), 25))) # Michael S. Branicky, May 20 2024