A355458 Numbers k that set a new record m where m is the largest left-truncatable prime up to the final k (stop on reaching the final k).

1, 7, 111, 3367, 7787, 8517, 9071, 54079, 54451, 138657, 262157, 759461, 857817, 4662317, 21754021, 25400729, 41171387, 50304331, 368119693, 799245603, 938577991

Offset: 1

Author

Eder Vanzei, Jul 02 2022

Keywords

• nonn
• base
• more

Comments

If instead of comparing the values of m, we compare the number of digits concatenated to k, then there are only 3 known terms: 1, 7 and 50304331 with 19, 23 and 24 digits respectively.

Examples

			a(1) = 1 because 1 sets a record m = 89726156799336363541 and 89726156799336363541, 9726156799336363541, 726156799336363541, 26156799336363541, 6156799336363541, 156799336363541, 56799336363541, 6799336363541, 799336363541, 99336363541, 9336363541, 336363541, 36363541, 6363541, 363541, 63541, 3541, 541, 41 are all primes (the truncation stops when the final k is reached).
a(2) = 7 because for k = 2..6 no m exceeds 89726156799336363541, but for k = 7, m = 357686312646216567629137.
		

Crossrefs

Cf. A024785.

Programs

Python

from sympy import isprime
def findNewCandidates(candidates):
  newCandidates = []
  for candidate in candidates:
    for k in range(1,10):
      p = int(str(k) + str(candidate))
      if (isprime(p)):
        newCandidates.append(p)
  return newCandidates
record = 0
for k in range(1, 10**6):
  if (k % 2 == 0 or k % 5 == 0):
    continue
  toCheck = [k]
  while len(toCheck) > 0:
    lastToCheck = toCheck
    toCheck = findNewCandidates(toCheck)
  result = lastToCheck[-1]
  if (result > record):
    record = result
    print(str(k))

Extensions

a(15)-a(18) from Michael S. Branicky, Jul 02 2022

a(19)-a(21) from Michael S. Branicky, Jul 04 2022