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%I #17 Jul 07 2021 09:28:45
%S 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,
%T 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,
%U 0,4,5,2,4,5,1,2,0,0,3,10,0,3,0,2,4,0,3,0,0,6
%N Maximal number of digits that can be appended to prime(n) preserving primality at each step.
%C 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).
%H Michael S. Branicky, <a href="/A232127/b232127.txt">Table of n, a(n) for n = 1..10000</a>
%H Archimedes' Lab, <a href="http://www.archimedes-lab.org/numbers/Num24_69.html">What's Special About This Number?</a>, section about 43.
%F a(n)=A232128(A000040(n)).
%F 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))
%F 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) }
%e 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.
%o (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}
%o (Python)
%o from sympy import isprime, prime
%o def a(n):
%o pn = prime(n); ftr = {pn}; ext = 0
%o while len(ftr) > 0:
%o r1 = set(filter(isprime, (int(str(e)+d) for d in "1379" for e in ftr)))
%o ext, ftr = ext+1, r1
%o return ext - 1
%o print([a(n) for n in range(1, 91)]) # _Michael S. Branicky_, Jul 07 2021
%o (Python) # faster version for initial segment of sequence
%o from sympy import isprime, prime, primerange
%o def aupton(terms):
%o alst = []
%o for p in primerange(1, prime(terms)+1):
%o r = {p}; e = 0
%o while len(r) > 0:
%o r1 = set(filter(isprime, (int(str(e)+d) for d in "1379" for e in r)))
%o e, r = e+1, r1
%o alst.append(e - 1)
%o return alst
%o print(aupton(90)) # _Michael S. Branicky_, Jul 07 2021
%Y Cf. A232125.
%K nonn,base
%O 1,1
%A _M. F. Hasler_ and _Michel Marcus_, Nov 19 2013
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