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A083833
Palindromes p such that 5p + 1 is also a palindrome.
2
1, 2, 22, 121, 131, 222, 2222, 12021, 12121, 13031, 13131, 22222, 222222, 1202021, 1203021, 1212121, 1213121, 1302031, 1303031, 1312131, 1313131, 2222222, 22222222, 120202021, 120212021, 120303021, 120313021, 121202121, 121212121
OFFSET
1,2
COMMENTS
From Michael S. Branicky, Jun 13 2021: (Start)
All terms start and end with the digit 1, except those consisting of d 2's, which lead to palindromes of d+1 1's.
If a(n) > 1 starts with 1, then its second and second-to-last digits are either both 2 or both 3; and if it has 5 or more digits, its third and third-to-last digits are either both 0 or both 1; thus, terms in the latter case only start with 120, 121, 130, 131, and 222.
Using ^ to denote repeated concatenation, contains terms of the forms 2^d, leading to palindromes of the form 1^(d+1); (12)^d 1, leading to (60)^d 6; and (13)^d 1, leading to (65)^d 6; among other patterns.
(Conjectures)
All terms contain only the digits {0, 1, 2, 3}.
For d even, the only term with d digits is 2^d; equivalently, all terms starting with 1 have odd length. (End)
LINKS
Michael S. Branicky, Table of n, a(n) for n = 1..8217 (all terms where p has <= 26 digits)
PROG
(Python)
from itertools import product
def ispal(n): s = str(n); return s == s[::-1]
def pals(d, base=10): # all positive d-digit palindromes
digits = "".join(str(i) for i in range(base))
for p in product(digits, repeat=d//2):
if d > 1 and p[0] == "0": continue
left = "".join(p); right = left[::-1]
for mid in [[""], digits][d%2]:
t = int(left + mid + right)
if t > 0: yield t
def ok(pal): return ispal(5*pal+1)
print([p for d in range(1, 10) for p in pals(d) if ok(p)]) # Michael S. Branicky, Jun 11 2021
CROSSREFS
KEYWORD
base,nonn
AUTHOR
Amarnath Murthy and Meenakshi Srikanth (menakan_s(AT)yahoo.com), May 09 2003
EXTENSIONS
Corrected and extended by Ray Chandler, May 21 2003
STATUS
approved