cp's OEIS Frontend

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

%I A073880 #15 Oct 07 2024 20:28:40
%S A073880 1,2,3,5,11,22,33,44,101,111,121,131,202,212,222,1001,1111,1221,1331,
%T A073880 2002,2112,2222,10001,10101,10201,10301,11011,11111,12021,13031,22222,
%U A073880 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
%N A073880 a(1) = 1, a(n) = smallest palindrome not included earlier such that a(1)+...+a(n) is a palindrome.
%C A073880 a(57) is likely to be the last term. - _Scott R. Shannon_, Oct 07 2024
%C A073880 From _Robert Israel_, Oct 07 2024: (Start)
%C A073880 Proof that a(57) is the last term.
%C A073880 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.
%C A073880 It has been checked that r must have more digits than 91899819 does.
%C A073880 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.
%C A073880 But since r + 91899819 > r, that can only happen if r + 91899819 has one more digit than r.
%C A073880 Since r has more digits than 91899819, this implies that r + 91899819 starts with 1 and r starts with 9.
%C A073880 But that's impossible because 9 + 9 ends in 8. (End)
%Y A073880 Cf. A073879.
%K A073880 nonn,base,fini,full
%O A073880 1,2
%A A073880 _Amarnath Murthy_, Aug 16 2002
%E A073880 More terms from _Giovanni Resta_, Feb 08 2006
%E A073880 a(42)-a(57) from _Scott R. Shannon_, Oct 07 2024