A308335 Palindromic primes such that sum of digits = number of digits.

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

Offset: 1

Author

Bernard Schott, May 20 2019

Keywords

• nonn
• base

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

Examples

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

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..10000
• G. L. Honaker, Jr. and Chris Caldwell, Prime Curios!, 10201021.

Crossrefs

Intersection of A000040 (primes), A002113 (palindromes) and A061384 (sum of digits = number of digits).

Intersection of A002385 and A061384.

Intersection of A069710 and A002113.

Programs

Mathematica

f[n_] := If[n==2, {11}, If[Mod[(n-1) (n-5), 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, n-2, 2}]; Sort[L]]]]; Join @@ (f /@ Range[13]) (* Giovanni Resta, Jun 06 2019 *)

PARI

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

Extensions

a(6)-a(21) from Jon E. Schoenfield, May 20 2019