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A232127 Maximal number of digits that can be appended to prime(n) preserving primality at each step. 4
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

Table of n, a(n) for n=1..90.

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}

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 May 13 12:16 EDT 2021. Contains 343839 sequences. (Running on oeis4.)