A152605 -id:A152605 - cp's OEIS frontend

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

N. J. A. Sloane, Sep 23 2009

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

Eric Angelini, Chiffres consecutifs dans quelques suites
E. Angelini, Chiffres consecutifs dans quelques suites [Cached copy, with permission]

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

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

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

A152606 a(1) = 1; thereafter a(n) is always the smallest integer > a(n-1) not leading to a contradiction, such that any six consecutive digits in the sequence sum up to a prime.

Original entry on oeis.org

1, 2, 3, 4, 5, 8, 9, 21, 45, 83, 89, 450, 503, 630, 701, 810, 901, 2101, 2103, 4121, 6301, 6303, 6503, 6901, 43030, 70103, 81010, 90101, 210101, 210103, 210107, 210109, 210143, 210145, 210149, 210161, 210163, 210167, 210169, 210503

Offset: 1

N. J. A. Sloane, Sep 23 2009

Computed by Jean-Marc Falcoz.

From a(269) = 1010001010 on, there starts a pattern of 104 terms, which then repeats indefinitely (with 6 digits in the middle of each term duplicated). - M. F. Hasler, Oct 16 2009

Eric Angelini, Chiffres consecutifs dans quelques suites
E. Angelini, Chiffres consecutifs dans quelques suites [Cached copy, with permission]

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

a(n, show_all=0, s=[1, 2, 3, 4, 5, 8, 9, 21, 45, 83, 89, 450, 503, 630, 701, 810, 901, 2101, 2103, 4121, 6301, 6303, 6503, 6901, 43030])={ my(a,nd=#Str(s[ #s])); for(i=1,n, if( i<=#s, a=s[i], my(ld=a%10^nd); while(a++,my(t=a+ld*10^#Str(a));forstep(d=#Str(a)-1,0,-1,isprime(sum(j=d,d+nd,t\10^j%10))&next;a+=10^d-a%10^d-1; next(2));break));show_all & print1(a", "));a} \\ M. F. Hasler, Oct 16 2009

cp's OEIS Frontend

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.

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