A348664 Numbers whose binary expansion is not rich.

203, 211, 300, 308, 333, 357, 395, 406, 407, 419, 422, 423, 459, 467, 556, 564, 600, 601, 604, 616, 617, 628, 653, 666, 667, 669, 690, 709, 714, 715, 723, 741, 779, 787, 790, 791, 803, 811, 812, 813, 814, 815, 820, 835, 838, 839, 844, 845, 846, 847, 851, 869

Offset

1,1

Comments

A word of length k is "rich" if it contains, as contiguous subsequences, exactly k + 1 distinct palindromes (including the empty word).

There are A225681(k)/2 terms with k binary digits.

Links

Rémy Sigrist, Table of n, a(n) for n = 1..10000

Formula

{k: A137397(k) <= A070939(k)}. - Michael S. Branicky, Oct 29 2021

Example

For n = 203:
- the binary expansion of 203 is "11001011" and has 8 binary digits,
- we have the following 8 palindromes: "", "0", "1", "00", "11", "010", "101", "1001"
- so 203 is not rich, and belongs to this sequence.
For n = 204:
- the binary expansion of 204 is "11001100" and has 8 binary digits,
- we have the following 9 palindromes: "", "0", "1", "00", "11", "0110", "1001", "001100", "110011"
- so 204 is rich, and does not belong to this sequence.

Mathematica

Select[Range@1000, Length@Select[Union[Subsequences[s=IntegerDigits[#, 2]]], PalindromeQ]<=Length@s&] (* Giorgos Kalogeropoulos, Oct 29 2021 *)

Prog

(PARI) is(n) = { my (b=binary(n), p=select(w->w==Vecrev(w), setbinop((i, j)->b[i..j], [1..#b]))); #b!=#p }
(Python)
def ispal(s): return s == s[::-1]
def ok(n):
  s = bin(n)[2:]
  return len(s) >= 1 + len(set(s[i:j] for i in range(len(s)) for j in range(i+1, len(s)+1) if ispal(s[i:j])))
print([k for k in range(870) if ok(k)]) # Michael S. Branicky, Oct 29 2021

Crossrefs

Cf. A206926, A216264, A225681, A070939, A137397.

Sequence in context: A198981 A259330 A090486 * A228320 A346899 A247921

Adjacent sequences: A348661 A348662 A348663 * A348665 A348666 A348667

Keyword

nonn,base

Author

Rémy Sigrist, Oct 28 2021

Status

approved