A152606 -id:A152606 - 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

A152605 a(1) = 1; thereafter a(n) is always the smallest integer > a(n-1) not leading to a contradiction, such that any five consecutive digits in the sequence sum up to a prime.

Original entry on oeis.org

1, 2, 3, 4, 7, 12, 30, 51, 83, 231, 232, 312, 323, 327, 413, 414, 530, 541, 701, 811, 812, 1101, 2110, 3011, 6301, 7030, 7103, 8110, 9011, 21011, 21013, 21017, 21019, 21053, 21055, 21059, 21071, 21073, 21077, 21079, 21413, 21415, 21419

Offset: 1

N. J. A. Sloane, Sep 23 2009

Computed by Jean-Marc Falcoz.

From a(116)=6100011 on, there starts a pattern of 75 terms which then repeats indefinitely (with duplication of a substring of 5 digits in the middle of each term). - 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, A152603, A152604, A152606, A152607, A152608, A152609.

A152605(n,show_all=0,s=[1, 2, 3, 4, 7, 12, 30, 51, 83, 231, 232, 312, 323, 327, 413, 414, 530, 541, 701, 811, 812, 1101])={ my(a); for(i=1,n, if(i<=#s,a=s[i], my(ld=a%10^4); while(a++,my(t=a+ld*10^#Str(a));forstep(d=#Str(a)-1,0,-1,isprime(sum(j=d,d+4,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.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

PARI

Formula