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A080437 For n < 5, a(n) = n-th prime. For n >= 5, let m = n-th prime. If m is a k-digit prime then a(n) = smallest prime obtained by inserting at least one digit between every pair of digits of m. There are (k-1) places where digit insertion takes place and a(n) contains at least 2k-1 digits. 2

%I

%S 2,3,5,7,101,103,107,109,223,229,311,307,401,433,457,503,509,601,607,

%T 701,733,709,823,809,907,10061,10093,10007,10009,10103,10247,10301,

%U 10337,10369,10429,10501,10567,10613,10607,10723,10709,10831,11941

%N For n < 5, a(n) = n-th prime. For n >= 5, let m = n-th prime. If m is a k-digit prime then a(n) = smallest prime obtained by inserting at least one digit between every pair of digits of m. There are (k-1) places where digit insertion takes place and a(n) contains at least 2k-1 digits.

%C At least up to n = 10^5, one inserted digit per position suffices. - _Robert Israel_, Feb 12 2016

%H Matthew M. Conroy and Robert Israel, <a href="/A080437/b080437.txt">Table of n, a(n) for n = 1..10000</a> (n = 1..168 from Matthew M. Conroy)

%H Math Overflow, <a href="http://mathoverflow.net/questions/230844/can-we-always-attain-another-prime-via-inserting-digits-between-the-digits-of-a">"Can we always attain another prime via inserting digits between the digits of a fixed prime?"</a>

%p f:= proc(n) local p, Lp, q0, x, Lx, k, i, q;

%p # This function attempts to insert one digit in each position.

%p p:= ithprime(n);

%p if p < 10 then return p fi;

%p Lp:= convert(p,base,10);

%p k:= nops(Lp);

%p q0:= add(100^(i-1)*Lp[i],i=1..k);

%p for x from 0 to 10^k-1 do

%p Lx:= convert(10^k+x,base,10);

%p q:= q0 + 10*add(100^(i-1)*Lx[i],i=1..k-1);

%p if isprime(q) then return q fi

%p od:

%p error("Need more than one digit");

%p end proc:

%p map(f, [$1..100]); # _Robert Israel_, Feb 12 2016

%Y Cf. A080436.

%K base,nonn

%O 1,1

%A _Amarnath Murthy_, Feb 21 2003

%E More terms from _Matthew Conroy_ Sep 18 2007

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Last modified November 19 20:42 EST 2019. Contains 329323 sequences. (Running on oeis4.)