

A347402


Lexicographically earliest sequence of distinct terms > 0 such that the product n * a(n) forms a palindrome in base 10.


3



1, 2, 3, 11, 101, 37, 23, 29, 19, 0, 4, 21, 38, 18, 35, 17, 16, 14, 9, 0, 12, 22, 7, 88, 209, 26, 703, 31, 8, 0, 28, 66, 47, 121, 15, 77, 6, 13, 154, 0, 187, 143, 277, 48, 1129, 99, 33, 44, 239, 0, 291, 406, 132, 518, 91, 377, 303, 364, 219, 0, 442, 386, 287, 333, 777
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OFFSET

1,2


COMMENTS

When n ends with a zero, we have a(n) = 0 in the sequence.


LINKS



EXAMPLE

For n = 7 we have a(7) = 23 and 7 * 23 = 161 is a palindrome in base 10; indeed, at n=7, multiples 7 * 1 = 7 and 7 * 11 = 77 are palindromes but 1 and 11 have already appeared in the sequence. The next palindrome multiple is 7 * 23 = 161 and 23 has not yet appeared so a(7) = 23;
for n = 8 we have a(8) = 29 and 8 * 29 = 232 is a palindrome in base 10;
for n = 9 we have a(9) = 19 and 9 * 19 = 171 is a palindrome in base 10;
for n = 10 we have a(10) = 0 and 10 * 0 = 0 is a palindrome in base 10;
for n = 11 we have a(11) = 4 and 11 * 4 = 44 is a palindrome in base 10; etc.


MATHEMATICA

a[1]=1; a[n_]:=a[n]=If[Mod[n, 10]==0, 0, (k=1; While[!PalindromeQ[n*k]MemberQ[Array[a, n1], k], k++]; k)]; Array[a, 65] (* Giorgos Kalogeropoulos, May 05 2022 *)


PROG

(Python)
def ispal(n): s = str(n); return s == s[::1]
def aupton(terms):
alst, seen = [1], {1}
for n in range(2, terms+1):
if n%10 == 0: alst.append(0); continue
an = 1
while an in seen or not ispal(n * an): an += 1
alst.append(an); seen.add(an)
return alst


CROSSREFS



KEYWORD

base,nonn


AUTHOR



STATUS

approved



