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.

Previous Showing 11-15 of 15 results.

A212448 Floor(4n + log(4n)).

Original entry on oeis.org

5, 10, 14, 18, 22, 27, 31, 35, 39, 43, 47, 51, 55, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100, 104, 108, 112, 116, 120, 124, 128, 132, 136, 140, 144, 148, 152, 157, 161, 165, 169, 173, 177, 181, 185, 189, 193, 197, 201, 205, 209, 213, 217, 221, 225, 229
Offset: 1

Views

Author

Mohammad K. Azarian, May 17 2012

Keywords

Crossrefs

Programs

A212449 Floor(5n + log(5n)).

Original entry on oeis.org

6, 12, 17, 22, 28, 33, 38, 43, 48, 53, 59, 64, 69, 74, 79, 84, 89, 94, 99, 104, 109, 114, 119, 124, 129, 134, 139, 144, 149, 155, 160, 165, 170, 175, 180, 185, 190, 195, 200, 205, 210, 215, 220, 225, 230, 235, 240, 245, 250, 255, 260, 265, 270, 275, 280, 285
Offset: 1

Views

Author

Mohammad K. Azarian, May 17 2012

Keywords

Crossrefs

Programs

A212450 a(n) = ceiling(n + log(n)).

Original entry on oeis.org

1, 3, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71
Offset: 1

Views

Author

Mohammad K. Azarian, May 17 2012

Keywords

Crossrefs

Programs

  • Magma
    [Ceiling(n + Log(n)): n in [1..80]]; // Vincenzo Librandi, Feb 14 2013
  • Mathematica
    Table[Ceiling[n + Log[n]], {n, 100}] (* T. D. Noe, May 21 2012 *)

A374056 a(n) = max_{i=0..n} S_4(i) + S_4(n-i) where S_4(x) = A053737(x) is the base-4 digit sum of x.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 4, 5, 6, 7, 5, 6, 7, 8, 6, 7, 8, 9, 7, 8, 9, 10, 8, 9, 10, 11, 9, 10, 11, 12, 7, 8, 9, 10, 8, 9, 10, 11, 9, 10, 11, 12, 10, 11, 12, 13, 8, 9, 10, 11, 9, 10, 11, 12, 10, 11, 12, 13, 11, 12, 13, 14, 9, 10, 11, 12, 10, 11, 12, 13, 11, 12, 13, 14
Offset: 0

Views

Author

Hermann Gruber, Jun 26 2024

Keywords

Comments

As shown in the proof of [Gruber and Holzer, lemma 9], the maximum is attained by choosing i as the largest number not exceeding n whose ternary representation is (33...3)_4. Also by lemma 6, for this choice of i we have A053737(i) = 3*floor(log_4(n+1)) and A053737(n-i) = A053737(n+1)-1, giving the formula below.

Examples

			For n=74, the maximum is attained by 63 + 11 = (333)_4 + (23)_4. Using 75=(1023)_4, comparing with the formula above, A053737(63) = 3*floor(log_4(n+1)) = 9 and A053737(11) = A053737(74+1)-1 = 5. Notice that other pairs attain the maximum as well. Namely, 43 + 31 = (223)_4 + (133)_4, as well as 47 + 27 = (233)_4 + (123)_4, and 59 + 15 = (323)_4 + (33)_4.
		

References

  • Hermann Gruber and Markus Holzer, Optimal Regular Expressions for Palindromes of Given Length. Extended journal version, in preparation, 2024.

Crossrefs

Programs

  • Mathematica
    Table[3*Floor[Log[4, k]] + DigitSum[k, 4] - 1, {k, 100}] (* Paolo Xausa, Aug 01 2024 *)
  • PARI
    a(n) = 3*logint(n+1, 4) + sumdigits(n+1, 4) - 1;

Formula

a(n) = 3*floor(log_4(n+1)) + A053737(n+1) - 1 [Gruber and Holzer, lemma 9].

A363827 Highest power of 2 dividing n which is <= sqrt(n).

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 4, 1, 2, 1, 4, 1, 2, 1, 4, 1, 2, 1, 4, 1, 2, 1, 4, 1, 2, 1, 4, 1, 2, 1, 4, 1, 2, 1, 4, 1, 2, 1, 4, 1, 2, 1, 4, 1, 2, 1, 4, 1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4
Offset: 1

Views

Author

Ilya Gutkovskiy, Oct 19 2023

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Last[Select[Divisors[n], # <= Sqrt[n] && IntegerQ[Log[2, #]] &]], {n, 100}]
    a[n_] := 2^Min[IntegerExponent[n, 2], Floor[Log2[n]/2]]; Array[a, 100] (* Amiram Eldar, Oct 19 2023 *)
  • PARI
    a(n) = if (n==1, 1, vecmax(select(x->((x^2 <= n) && (2^logint(x,2)==x)), divisors(n)))); \\ Michel Marcus, Oct 19 2023

Formula

a(n) = 2^min(A007814(n), A102572(n)). - Kevin Ryde, Oct 20 2023
Previous Showing 11-15 of 15 results.