|
|
A248753
|
|
Palindromes p = A002113(n) whose index n is a substring of p.
|
|
1
|
|
|
11, 1111, 12221, 23332, 34443, 45554, 56665, 67776, 78887, 89998, 101101, 111111, 121121, 131131, 141141, 151151, 161161, 171171, 181181, 191191, 1020201, 1121211, 1222221, 1323231, 1424241, 1525251, 1626261, 1727271, 1828281, 1929291, 2030302, 2131312
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
|
|
LINKS
|
|
|
EXAMPLE
|
11 is a term because the eleventh palindrome is 11.
12221 is in the sequence because the 222nd palindrome is 12221.
101101 is a member because it is the 1101st palindrome.
|
|
MATHEMATICA
|
(* first load 'nthPalindrome' from A002113 and then *) nPal[n_] := nthPalindrome[n - 1]; fQ[n_] := StringPosition[ ToString[ nPal[ n]], ToString[ n]] != {}; k = 1; lst = {}; While[k < 3001, If[fQ[k], AppendTo[lst, nPal[ k]]]; k++]; lst
|
|
PROG
|
(Python)
from itertools import count, islice
def A248753_gen(startvalue=2): # generator of terms >= startvalue
def f(n):
y = 10*(x:=10**(len(str(n>>1))-1))
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)
for n in count(max(2, startvalue)):
if str(n) in str(k:=f(n)):
yield k
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,base,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|