OFFSET
1,2
COMMENTS
Computed by Jean-Marc Falcoz.
From a(34)=3002 on, there starts a pattern [ 3(002){n}, ..., 2(002){n+1} ] of length 52 which then repeats forever. This allows us to write an explicit formula for any term a(n) of the sequence. - M. F. Hasler, Oct 16 2009
LINKS
Eric Angelini, Chiffres consecutifs dans quelques suites
E. Angelini, Chiffres consecutifs dans quelques suites [Cached copy, with permission]
FORMULA
a(n) = b(n)*10^[3n/52] = c(n)*10^(3n/52) with (except for smaller initial terms) 20 < b(n) < 611 and c(52k+23) = 9.89... < c(n) < c(52k) = 91.1... for all integers k > 0. - M. F. Hasler, Oct 16 2009
PROG
(PARI) A152603(n, show_all=0)={ my(a); for(i=1, n, if(i<4, a=2^i/2, my( l2d=a%100+if(i<7, 10*[1, 2, 4, 5][i-2])); while(a++, my(t=a+l2d*10^#Str(a)); forstep(d=#Str(a)-1, 0, -1, isprime(z=t\10^d%10+t\10^(d+1)%10+t\10^(d+2)%10) & next; a+=10^d-a%10^d-1; next(2)); break)); show_all&print1(a", ")); a} \\ M. F. Hasler, Oct 16 2009
CROSSREFS
KEYWORD
nonn,base
AUTHOR
N. J. A. Sloane, Sep 23 2009
STATUS
approved