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A308335
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Palindromic primes such that sum of digits = number of digits.
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1
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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
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1,1
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COMMENTS
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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
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LINKS
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EXAMPLE
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3001003 is a term because it is a palindromic prime that has 7 digits and its sum of its digits is 7.
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MATHEMATICA
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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 *)
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PROG
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(PARI) isok(p) = isprime(p) && (d=digits(p)) && (Vecrev(d) == d) && (#d == vecsum(d)); \\ Michel Marcus, Jun 29 2019
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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