A068061 Palindromic numbers j that are not of the form k + reverse(k) for any k.

1, 3, 5, 7, 9, 111, 131, 151, 171, 191, 212, 232, 252, 272, 292, 313, 333, 353, 373, 393, 414, 434, 454, 474, 494, 515, 535, 555, 575, 595, 616, 636, 656, 676, 696, 717, 737, 757, 777, 797, 818, 838, 858, 878, 898, 919, 939, 959, 979, 999, 10101, 10301, 10501

Offset: 1

Author

Klaus Brockhaus, Feb 15 2002

Keywords

• base
• easy
• nonn

Comments

Intersection of A002113 and A067031. Every palindrome with an even number of digits is of the form k + reverse(k), for example 123321 = 123000 + 000321, so the sequence has no terms with an even number of digits.

It seems that the terms follow a strict pattern: x1x', x3x', x5x', x7x', x9x', y1y', y3y', y5y', y7y', y9y' and so on. x' is reverse(x). Apart from the first 5 terms in the sequence, the surrounding terms (x and y) simply iterate over the positive integers. - Dmitry Kamenetsky, Mar 10 2017

Every palindrome with an odd number of digits is of the form k + reverse(k) if the central digit is even, for example 1234321 = 1232000 + 0002321, so no term with an odd number of digits has an even central digit. - A.H.M. Smeets, Feb 01 2019

Examples

			9 belongs to this sequence, since there is no k such that k + reverse(k) = 9 (cf. A067031).
		

Links

• Michel Marcus, Table of n, a(n) for n = 1..500

Crossrefs

Cf. A002113, A067031, A068062.

Programs

PARI

isok(n) = {if (Pol(d=digits(n)) == Polrev(d), for (k=1, n-1, if (k + fromdigits(Vecrev(digits(k))) == n, return (0));); 1;);} \\ Michel Marcus, Mar 12 2017