A373044 Prime concatenated analog clock numbers read clockwise. Version 2: hours > 9 are split in 2 digits.

2, 3, 5, 7, 11, 23, 67, 89, 101, 4567, 10111, 67891, 89101, 789101, 4567891, 23456789, 56789101, 1234567891, 45678910111, 12345678910111, 1112123456789101, 23456789101112123, 112123456789101112123, 891011121234567891011, 4567891011121234567891

Offset: 1

Author

Eric Angelini and Michael S. Branicky, May 20 2024

Keywords

• nonn
• base

Comments

In this version, the numbers 10, 11, and 12 may be split up into individual digits, in contrast to A036342.

a(59) has 1325 digits.

Examples

			101 is a term here using the digits 1 and 0 from 10 and the first 1 of 11.
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..58
• 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. A036342, A373045.

Programs

PARI

A373044_row(r)={my(d=concat([digits(i)|i<-[1..12]]), p); Set([p| s<-[1..#d], d[s]&& isprime(p=fromdigits([d[i%#d+1]| i<-[s-1..s+r-2]]))])}\\ r-digit-terms
A373044_upto_length(L)=concat([A373044_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)))