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Palindromes p=A002113(n) whose index n is also a palindrome and in addition a substring of p (strings in base 10).
1

%I #18 Jul 24 2024 18:11:02

%S 11,1111,12221,23332,34443,45554,56665,67776,78887,89998,111111,

%T 1222221,2333332,3444443,4555554,5666665,6777776,7888887,8999998,

%U 9101019,11111111,102020201,112121211,122222221,132323231,142424241,152525251,162626261,172727271,182828281

%N Palindromes p=A002113(n) whose index n is also a palindrome and in addition a substring of p (strings in base 10).

%C This is a proper subsequence of A248753 (where the index does not need to be palindromic).

%e 11 is a term because the eleventh palindrome is 11.

%e 1111 is a member because it is the 111th palindrome.

%e 12221 is in the sequence because the 222nd palindrome is 12221.

%t (* first load 'nthPalindrome' from A002113 and then *) nPal[n_] := nthPalindrome[n - 1]; fQ[n_] := StringPosition[ ToString[ nPal[ nPal[ k]]], ToString[ nPal[ n]]] != {}; k = 2; lst = {}; While[k < 501, If[ fQ[k], AppendTo[lst, nPal[ nPal[ k]] ]]; k++]; lst

%o (Python)

%o from itertools import count, islice

%o def A248754_gen(): # generator of terms

%o def f(n):

%o y = 10*(x:=10**(len(str(n>>1))-1))

%o return (c:=n-x)*x+int(str(c)[-2::-1] or 0) if n<x+y else (c:=n-y)*y+int(str(c)[::-1] or 0)

%o for n in count(2):

%o if str(m:=f(n)) in str(k:=f(m)):

%o yield k

%o A248754_list = list(islice(A248754_gen(),30)) # _Chai Wah Wu_, Jul 24 2024

%Y Cf. A002113, A248753.

%K nonn,base,easy

%O 1,1

%A _Robert G. Wilson v_, Oct 13 2014