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A155801
Nontrivial "Strobogrammatic" primes, the same "upside-down" in at least one base b with 2 <= b <= 10.
0
3, 5, 7, 11, 13, 17, 31, 37, 43, 73, 101, 107, 127, 181, 257, 313, 443, 619, 757, 1093, 1193, 1297, 1453, 1571, 1619, 1787, 1831, 1879, 2801, 4889, 5113, 5189, 5557, 5869, 5981, 6211, 6827, 7607, 7759, 7919, 8191
OFFSET
1,1
COMMENTS
I have to say "nontrivial" because every nonnegative integer is strobogrammatic in base 1. Strobogrammatic binary primes == primes in A006995 == A016041. Strobogrammatic primes in base 3 = 13, 757, 1093, 9103, ... == primes strobogrammatic in bases 2 and 3. For bases 2 < k < 8 we have that every strobogrammatic prime in base k must also be strobogrammatic in base 2 and hence palindromic in base 2. Hence we have, for example, strobogrammatic base-4 primes = A056130 = "Palindromic primes in bases 2 and 4."
Strobogrammatic primes in base 5 = 31, 19531, 394501, 472631, ... == primes strobogrammatic in base 2 and base 5. Strobogrammatic primes base 6 = 7, 37, 43, 1297, 55987, ... == primes strobogrammatic in base 2 and base 6. Note that 1101011 (base 6) = 18881 (base 10) which is strobogrammatic base 10 but not prime base 6 nor 10 (though prime base 2). Strobogrammatic primes base 7 = 2801, 134807, this last being strobogrammatic prime in bases 2, 4 and 7. Strobogrammatic primes base 8 = 73, 262657, 295433, ... Strobogrammatic primes base 9 break the above pattern, as they can have the digit 8 and are A068188 (tetradic primes). Strobogrammatic primes base 10 == A007597. Except sometimes for the first element, these (for the same range of k) must all have an odd number of digits.
FORMULA
A000040 INTERSECTION A155584[1 < k < 11, n].
EXAMPLE
5189 = 1101011 (base 6) which numeral string is the same upside-down (and backwards). 11, 101, 181 and 619 are strobogrammatic base 10, the conventional interpretation of the word.
KEYWORD
base,nonn
AUTHOR
Jonathan Vos Post, Jan 27 2009
STATUS
approved