OFFSET
1,1
COMMENTS
LINKS
M. F. Hasler, Table of n, a(n) for n = 1..51
EXAMPLE
Concatenation of the first A(1) = 2 terms of A = A231206, i.e., A(1..2), yields the first prime a(1) = 23. Concatenation of the next A(2) = 3 terms of A231206, i.e., A(3..6), yields a(2) = 149, etc.
The terms A(S(n) := Sum_{1 <= k <= n} A(k)) of A = A231206 in which end the primes a(n), are usually roughly twice as large as their neighbors A(S(n)-1) and A(S(n)+1) which can be chosen as smallest available numbers without the additional constraint of having to yield a prime: A231206 = (..., 4, 9, 5, ... 9, 19, 10, ... 25, 59, 26, ... 27, 63, 28, ... 33, 71, 34, ... 60, 109, 61, ... 70, 198, 72, ... 81, 179, 82, ...), where:
- 9 = A231206(S(2)=5) is twice as large as 4 and 5;
- 5 = A231206(S(3)=6) is exceptionally small: ignore this coincidence;
- 19 = A231206(S(4)=10) is twice as large as 9 and 10,
- 139 = A231206(S(5)=19) is exceptionally large: see below,
- 59 = A231206(S(6)=24) is about twice as large as 25 and 26,
- 63 = A231206(S(7)=30) is about twice as large as 27 and 28,
- 71 = A231206(S(8)=37) is about twice as large as 33 and 34,
- 109 = A231206(S(9)=37) is about twice as large as 60 and 61, etc.
To make the 5th prime, the term A231206(S(5)=19) = 139 had to be chosen much larger than its neighbors (17 and 18), which in turn leads to the 417-digit prime a(19), the concatenation of A231206(173..311). To get this prime, one has to take A231206(311) = 581 again roughly twice as large as its neighbors A231206(310) = 309 and A231206(312) = 310. This leads later again to a prime roughly twice as long as its neighbors in this sequence.
PROG
(PARI) {a=[prefix=2]; remain=pointer=1; for(n=1, 499, my(used=Set(a)); if( !remain, remain=a[pointer++]; print1(prefix", "); prefix=""); for(i=1, 1e9, setsearch(used, i) && next; remain>1 || ispseudoprime( eval( Str( prefix, i ))) || next; prefix=Str(prefix, i); a=concat(a, i); remain--; break ))}
CROSSREFS
KEYWORD
nonn,base
AUTHOR
M. F. Hasler, Nov 05 2013
STATUS
approved