A088787 - OEIS

%I #9 Aug 16 2024 11:49:11

%S 61,97,227,449,661,797,881,1307,2797,3877,5477,5657,8297,8669,8849,

%T 8861,9887,11257,12517,15287,16487,17387,19427,20287,20897,23197,

%U 23957,25717,26227,30637,36947,39217,41687,41927,42197,42797,45127,45827,51637

%N Numbers which are primes and which remain prime for three successive applications of incrementing each digit by 2 with carries ignored.

%C Imagine that each digit of n is on a wheel on a combination lock and each wheel is being rotated two notches for each application of the function. Thus for the digits 0 to 7, the replacement digit is simply the digit+2, but for 8 the replacement digit is 0 and for 9 the replacement digit is 1. Thus 227 -> 449 -> 661 -> 883 -> 5 (initial zeros are dropped on results.)

%e a(3)=227 because (i) 227 is prime and (ii) incrementing each digit of 227 by 2 yields 449 which is prime and (iii) incrementing each digit of 449 by 2 (ignoring carries) yields 661 which is prime and (iv) incrementing each digit of 661 by 2 yields 883 which is prime

%t Select[Prime@Range@10000, AllTrue[First@Last@Reap[n = #; Do[n=Sow[FromDigits@ReplaceAll[IntegerDigits@n,k_:>Mod[2+k,10]]],3]], PrimeQ] &](* _Hans Rudolf Widmer_, Aug 11 2024 *)

%o Function Cycle222(long: p): long {i: integer; r: long; c: integer; for i=floor(log10(p)) to 0 step -1; c = floor(p/10^(i-1)) mod 10; r = r*10 + ((c+2) mod 10); next i; return r; } p=2; for i=1 to 90; n=sequence.count; do while sequence.count=n; j=cycle222(p); k=cycle222(j); if isprime(j) and isprime(k) and isprime(cycle222(k)) then sequence.add(p) else p=nextPrimeAfter(p); loop; next i;

%Y Cf. A088785, A088786, A088788.

%K base,nonn

%O 1,1

%A Chuck Seggelin (barkeep(AT)plastereddragon.com), Oct 16 2003