A232127 Maximal number of digits that can be appended to prime(n) preserving primality at each step.

7, 7, 7, 7, 1, 6, 3, 8, 6, 6, 3, 6, 1, 5, 3, 0, 6, 5, 5, 4, 6, 1, 1, 0, 2, 4, 9, 0, 4, 0, 5, 1, 1, 5, 3, 1, 2, 1, 0, 2, 0, 4, 2, 3, 7, 5, 2, 3, 4, 3, 5, 4, 5, 0, 4, 3, 4, 5, 3, 1, 1, 5, 1, 2, 2, 0, 6, 3, 0, 4, 5, 2, 4, 5, 1, 2, 0, 0, 3, 10, 0, 3, 0, 2, 4, 0, 3, 0, 0, 6

Offset

1,1

Comments

Consider chains (p^(0),p^(1),p^(2),...p^(L)) of primes such that p^(k-1) = floor(p^(k)/10), or otherwise said, p^(k+1) is obtained from p^(k) by appending a digit. Then a(n) is one less than the number of primes in the longest possible such chain with p^(0)=prime(n).

Links

Michael S. Branicky, Table of n, a(n) for n = 1..10000

Archimedes' Lab, What's Special About This Number?, section about 43.

Formula

a(n)=A232128(A000040(n)).

a(n) > 0 if and only if there is a prime p between 10*prime(n)+1 and 10*prime(n)+9, in which case a(n) >= 1+a(primepi(p))

a(n) = max { L in N | exists (p[0],...,p[L]) in P^(L+1) (P = the primes A000040), such that p[0] = prime(n) and for k=1,...,L : p[k-1] = floor(p[k]/10) }

Example

a(14)=5 because for prime(14)=43, one can add at most 5 digits to the right preserving primality at each step: 439 is prime, 4391 is prime, 43913 is prime, 439133 is prime, 4391339 is prime. There is no longer chain possible starting with 43.

Prog

(PARI) {howfar(p)=my(m); forstep(d=1, 9, 2, d==5&&next; isprime(p*10+d)||next; m=max(1+howfar(10*p+d), m)); m}
(Python)
from sympy import isprime, prime
def a(n):
  pn = prime(n); ftr = {pn}; ext = 0
  while len(ftr) > 0:
    r1 = set(filter(isprime, (int(str(e)+d) for d in "1379" for e in ftr)))
    ext, ftr = ext+1, r1
  return ext - 1
print([a(n) for n in range(1, 91)]) # Michael S. Branicky, Jul 07 2021
(Python) # faster version for initial segment of sequence
from sympy import isprime, prime, primerange
def aupton(terms):
  alst = []
  for p in primerange(1, prime(terms)+1):
    r = {p}; e = 0
    while len(r) > 0:
      r1 = set(filter(isprime, (int(str(e)+d) for d in "1379" for e in r)))
      e, r = e+1, r1
    alst.append(e - 1)
  return alst
print(aupton(90)) # Michael S. Branicky, Jul 07 2021

Crossrefs

Cf. A232125.

Sequence in context: A216159 A276615 A103983 * A195413 A083947 A269349

Adjacent sequences: A232124 A232125 A232126 * A232128 A232129 A232130

Keyword

nonn,base

Author

M. F. Hasler and Michel Marcus, Nov 19 2013

Status

approved