

A308335


Palindromic primes such that sum of digits = number of digits.


1



11, 10301, 1201021, 3001003, 10000900001, 10002520001, 10013131001, 10111311101, 10301110301, 11012121011, 11020302011, 11030103011, 11100500111, 11120102111, 12000500021, 12110101121, 13100100131, 30000500003, 30011111003, 1000027200001, 1000051500001
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OFFSET

1,1


COMMENTS

Every palindrome with an even number of digits is divisible by 11, so 11 is the only term of the sequence with an even number of digits.
Every palindrome with a number of digits which is a multiple of 3 also has a sum of digits which is divisible by 3, so there is no term with 3*k digits.
So, except 11 with 2 digits, the terms of this sequence must have a number of digits that belongs to A007310.
For n > 1, the middle digit of a(n) is odd.  Chai Wah Wu, Jun 30 2019


LINKS



EXAMPLE

3001003 is a term because it is a palindromic prime that has 7 digits and its sum of its digits is 7.


MATHEMATICA

f[n_] := If[n==2, {11}, If[Mod[(n1) (n5), 6]>0, {}, Block[{h = (n  1)/2, L={}, p}, Do[p = Select[ Flatten[ Permutations /@ IntegerPartitions[ (n  c)/2, {h}, Range[0, 9]], 1], MemberQ[{1, 3, 7, 9}, Last[#]] &]; L = Join[L, Select[ FromDigits /@ (Flatten[{Reverse[#], c, #}] & /@ p), PrimeQ]], {c, 1, n2, 2}]; Sort[L]]]]; Join @@ (f /@ Range[13]) (* Giovanni Resta, Jun 06 2019 *)


PROG

(PARI) isok(p) = isprime(p) && (d=digits(p)) && (Vecrev(d) == d) && (#d == vecsum(d)); \\ Michel Marcus, Jun 29 2019


CROSSREFS



KEYWORD

nonn,base


AUTHOR



EXTENSIONS



STATUS

approved



