cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A329545 After a(1) = 1, add the even terms and subtract the odd ones, the result must always be a palindrome. This is the lexicographically earliest sequence of distinct positive integers with this property.

Original entry on oeis.org

1, 2, 3, 4, 18, 11, 5, 16, 13, 7, 6, 14, 15, 26, 22, 33, 17, 28, 25, 36, 35, 9, 8, 58, 55, 44, 46, 10, 20, 30, 73, 77, 66, 24, 40, 50, 103, 79, 68, 34, 23, 81, 48, 47, 80, 83, 72, 54, 43, 85, 52, 49, 38, 37, 70, 64, 53, 87, 32, 27, 60, 57, 90, 12, 45, 59, 92, 42, 75, 61, 94, 62, 95, 63, 74, 69, 194
Offset: 1

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Author

Eric Angelini and Jean-Marc Falcoz, Nov 16 2019

Keywords

Comments

Negative palindromes are not allowed. After 50000 terms, the smallest unused integers are 964, 1020, 1029, 1031, 1038, 1041, 1047, 1051, ... and the largest used is 173410. The largest palindrome produced so far is 309903. Is the sequence a permutation of the integers > 0?
a(63411) = 964. Rémy Sigrist's comment in A329544 shows that terms in A104444 are not in the sequence. Conjecture: Sequence is a permutation of positive integers not in A104444. - Chai Wah Wu, Dec 11 2019

Examples

			The sequence starts with 1, smallest positive integer.
1 + 2 = 3 (palindrome)
1 + 2 - 3 = 0 (palindrome)
1 + 2 - 3 + 4 = 1 (palindrome)
1 + 2 - 3 + 4 + 18 = 22 (palindrome)
1 + 2 - 3 + 4 + 18 - 11 = 11 (palindrome)
1 + 2 - 3 + 4 + 18 - 11 - 5 = 6 (palindrome)
1 + 2 - 3 + 4 + 18 - 11 - 5 + 16 = 22 (palindrome), etc.
		

Crossrefs

Cf. A329544 (same idea, but where the odd integers are added and the even ones are subtracted).
Cf. A002113 (palindromes), A086862 (first differences of palindromes).
Cf. A104444.

Programs

  • PARI
    A329545_vec(N, u=1, U, a, s=2, d)={vector(N, n, a=u; while(bittest(U, a-u)|| Vecrev(d=digits(s+(-1)^a*a))!=d|| (a>s&&bittest(a, 0)), a++); s+=(-1)^a*a; U+=1<<(a-u); while(bittest(U, 0), U>>=1; u++); a)} \\ M. F. Hasler, Nov 16 2019