A119368 Palindromes whose smallest palindromic proper multiple is the number concatenated with itself.

5, 6, 7, 8, 9, 656, 757, 767, 828, 838, 989, 57275, 58085, 65056, 65356, 65456, 65556, 68086, 69296, 73537, 74447, 75057, 75157, 75557, 76067, 76567, 76667, 79197, 82228, 82328, 82428, 83238, 85058, 86068, 86168, 86568, 87278, 87778, 89098

Offset

1,1

Comments

Palindromes where A083146 differs from A083145.

First term with an even number of digits is a(389) = 66800866. - Chai Wah Wu, Apr 08 2016

Links

Chai Wah Wu, Table of n, a(n) for n = 1..389

Example

The smallest palindromic proper multiple of 4 is 8, not the concatenation of 4 with itself, so 4 is not a term.
The smallest palindromic proper multiple of 5 is 55, the concatenation of 5 with itself, so 5 is a term.

Prog

(Python)
from itertools import count, islice
def A119368_gen(): # generator of terms
    for n in count(2):
        p = (c:=n-x)*x+int(str(c)[-2::-1] or 0) if n<(x:=10**(len(str(n>>1))-1))+(y:=10*x) else (c:=n-y)*y+int(str(c)[::-1] or 0)
        for i in count(n+1):
            if not (q:=(c:=i-x)*x+int(str(c)[-2::-1] or 0) if i<(x:=10**(len(str(i>>1))-1))+(y:=10*x) else (c:=i-y)*y+int(str(c)[::-1] or 0))%p:
                if (s:=str(q)) == (t:=str(p))*(len(s)//len(t)):
                    yield p
                break
A119368_list = list(islice(A119368_gen(), 20)) # Chai Wah Wu, Jul 11 2024

Crossrefs

Cf. A083145, A083146, A002113.

Sequence in context: A051052 A377265 A387546 * A088721 A325435 A288857

Adjacent sequences: A119365 A119366 A119367 * A119369 A119370 A119371

Keyword

base,nonn

Author

Franklin T. Adams-Watters, May 16 2006

Status

approved