A347402 - OEIS

A347402

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

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

Chai Wah Wu, Table of n, a(n) for n = 1..10000

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, n-1], 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

print(aupton(100)) # Michael S. Branicky, Aug 30 2021