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.

1, 2, 4, 5, 8, 41, 60, 70, 410, 412, 416, 418, 452, 454, 458, 470, 472, 476, 478, 812, 814, 818, 830, 832, 836, 838, 872, 874, 878, 2101, 2210, 2300, 2302, 3002, 3003, 4011, 5110, 6101, 6410, 6500, 7002, 9020, 9200, 20020, 30020, 30021, 40110

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

Table of n, a(n) for n=1..47.

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

Cf. A158652, A152604, A152605, A152606, A152607, A152608, A152609.

Sequence in context: A171411 A291607 A092752 * A159260 A013103 A012973

Adjacent sequences: A152600 A152601 A152602 * A152604 A152605 A152606

Keyword

nonn,base

Author

N. J. A. Sloane, Sep 23 2009

Status

approved