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 A232127 Maximal number of digits that can be appended to prime(n) preserving primality at each step. 5
 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 (list; graph; refs; listen; history; text; internal format)
 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,...,p[L]) in P^(L+1) (P = the primes A000040), such that p = 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

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Last modified September 24 23:43 EDT 2022. Contains 356951 sequences. (Running on oeis4.)