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A152603 a(1) = 1; thereafter, a(n) is always the smallest integer > a(n-1) not leading to a contradiction, such that any three consecutive digits in the sequence sum up to a prime.

%I A152603 #11 Feb 04 2018 03:21:55
%S A152603 1,2,4,5,8,41,60,70,410,412,416,418,452,454,458,470,472,476,478,812,
%T A152603 814,818,830,832,836,838,872,874,878,2101,2210,2300,2302,3002,3003,
%U A152603 4011,5110,6101,6410,6500,7002,9020,9200,20020,30020,30021,40110
%N A152603 a(1) = 1; thereafter, a(n) is always the smallest integer > a(n-1) not leading to a contradiction, such that any three consecutive digits in the sequence sum up to a prime.
%C A152603 Computed by Jean-Marc Falcoz.
%C A152603 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
%H A152603 Eric Angelini, <a href="http://www.cetteadressecomportecinquantesignes.com/ConsecDig.htm">Chiffres consecutifs dans quelques suites</a>
%H A152603 E. Angelini, <a href="/A152136/a152136.pdf">Chiffres consecutifs dans quelques suites</a> [Cached copy, with permission]
%F A152603 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
%o A152603 (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
%Y A152603 Cf. A158652, A152604, A152605, A152606, A152607, A152608, A152609.
%K A152603 nonn,base
%O A152603 1,2
%A A152603 _N. J. A. Sloane_, Sep 23 2009