A073880 a(1) = 1, a(n) = smallest palindrome not included earlier such that a(1)+...+a(n) is a palindrome.

1, 2, 3, 5, 11, 22, 33, 44, 101, 111, 121, 131, 202, 212, 222, 1001, 1111, 1221, 1331, 2002, 2112, 2222, 10001, 10101, 10201, 10301, 11011, 11111, 12021, 13031, 22222, 100001, 101101, 102201, 103301, 110011, 111111, 120021, 130031, 20202, 1000001, 1001001, 1002001, 1003001, 1010101, 1011101, 1012101, 1020201, 2042402, 10000001, 10011001, 10022001, 10033001, 10100101, 10111101, 10200201, 10300301

Offset: 1

Author

Amarnath Murthy, Aug 16 2002

Keywords

• nonn
• base
• fini
• full

Comments

a(57) is likely to be the last term. - Scott R. Shannon, Oct 07 2024

From Robert Israel, Oct 07 2024: (Start)

Proof that a(57) is the last term.

Sum_{i=1..57} a(i) = 91899819. Suppose r is a palindrome not in {a(1),...,a(56)} such that 91899819 + r is a palindrome.

It has been checked that r must have more digits than 91899819 does.

r can't end in 0 because it's a palindrome. If r ends (and therefore starts) with digit x, then r + 91899819 ends (and starts) with x-1.

But since r + 91899819 > r, that can only happen if r + 91899819 has one more digit than r.

Since r has more digits than 91899819, this implies that r + 91899819 starts with 1 and r starts with 9.

But that's impossible because 9 + 9 ends in 8. (End)

Crossrefs

Cf. A073879.

Extensions

More terms from Giovanni Resta, Feb 08 2006

a(42)-a(57) from Scott R. Shannon, Oct 07 2024