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Zerofree numbers k such that A061486(k) is prime.
3

%I #12 Feb 13 2018 00:31:08

%S 2,3,5,7,11,12,13,15,16,18,19,21,23,25,27,29,31,32,34,35,37,43,45,51,

%T 52,53,54,56,57,58,59,61,65,72,73,75,78,79,81,85,87,89,91,92,95,97,98,

%U 212,213,216,218,219,223,225,229,232,233,235,236,239,243,245,249,255,256,269,272,273,278,283

%N Zerofree numbers k such that A061486(k) is prime.

%H Robert Israel, <a href="/A137923/b137923.txt">Table of n, a(n) for n = 1..10000</a>

%F Members of the sequence are numbers n = X(1)...X(r) for which digits the following equation holds: (X(1) + ... + X(r)) + (X(1)*X(2) + ... + X(r-1)*X(r)) + ... + (X(1)*...*X(r)) = p, where p is a prime number, X(i) is the i-th digit of n, and every digit is nonzero.

%e 2-digit numbers are of the form X(1)X(2).

%e The equation is then X(1) + X(2) + X(1)*X(2) = p, where p is prime and both digits are nonzero. Power set is {();(1);(2);(1,2)}, so indices of digits in the equation are running through the power set.

%e Following numbers n are solutions of the equation:

%e 11 because 1 + 1 + 1*1 = 3;

%e 12 (and its reverse, 21) because 1 + 2 + 1*2 = 5;

%e 13 (and its reverse, 31) because 1 + 3 + 1*3 = 7;

%e 15 (and its reverse, 51) because 1 + 5 + 1*5 = 11;

%e 16 (and its reverse, 61) because 1 + 6 + 1*6 = 13;

%e 18 (and its reverse, 81) because 1 + 8 + 1*8 = 17;

%e 19 (and its reverse, 91) because 1 + 9 + 1*9 = 19;

%e 23 (and its reverse, 32) because 2 + 3 + 2*3 = 11;

%e 25 (and its reverse, 52) because 2 + 5 + 2*5 = 17;

%e ...

%p filter := proc (n) local L; L := convert(n, base, 10); not has(L, 0) and isprime(add(add(convert(L[i .. j], `*`), i = 1 .. j), j = 1 .. nops(L))) end proc:

%p select(filter, [$1..1000]); # _Robert Israel_, Feb 11 2018

%Y Cf. A052382, A061486.

%K easy,nonn,base

%O 1,1

%A _Ctibor O. Zizka_, Apr 30 2008

%E More terms from _Robert Israel_, Feb 11 2018