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.

Showing 1-4 of 4 results.

A074939 Even numbers such that base 3 representation contains no 2.

Original entry on oeis.org

0, 4, 10, 12, 28, 30, 36, 40, 82, 84, 90, 94, 108, 112, 118, 120, 244, 246, 252, 256, 270, 274, 280, 282, 324, 328, 334, 336, 352, 354, 360, 364, 730, 732, 738, 742, 756, 760, 766, 768, 810, 814, 820, 822, 838, 840, 846, 850, 972, 976, 982, 984, 1000, 1002
Offset: 0

Views

Author

Benoit Cloitre, Oct 04 2002; Nov 15 2003

Keywords

Comments

Even numbers in A005836; n such that binomial(2n,n) == 1 (mod 3).
Sum of an even number of distinct powers of 3. - Emeric Deutsch, Dec 03 2003

Crossrefs

Intersection of A005843 and A005836.

Programs

  • Mathematica
    Select[2*Range[0,600],DigitCount[#,3,2]==0&] (* Harvey P. Dale, Dec 10 2016 *)
  • Python
    def A074939(n): return int(bin((n.bit_count()&1)+(n<<1))[2:],3) # Chai Wah Wu, Jun 26 2025

Formula

a(n) = A083094(n)/2; a(n) mod 3 = A010060(n); n such that coefficient of x^n equals 1 in Product_{k>=0} (1 - x^(3^k)).
a(n) + A074938(n) = A055246(n+1). - Philippe Deléham, Jul 10 2005

A083097 a(n) = A083096(n)/6.

Original entry on oeis.org

0, 2, 5, 6, 14, 15, 18, 20, 41, 42, 45, 47, 54, 56, 59, 60, 122, 123, 126, 128, 135, 137, 140, 141, 162, 164, 167, 168, 176, 177, 180, 182, 365, 366, 369, 371, 378, 380, 383, 384, 405, 407, 410, 411, 419, 420, 423, 425, 486, 488, 491, 492, 500
Offset: 1

Views

Author

Benoit Cloitre, Apr 22 2003

Keywords

Comments

Is this the same as A083095? - Andrew S. Plewe, May 30 2007
Apparently (2*a(n)) mod 3 = A010060(n-1), the Thue-Morse sequence.

Crossrefs

Programs

A083094 Numbers k such that Sum_{j=0..k} (binomial(k,j) mod 3) is odd.

Original entry on oeis.org

0, 8, 20, 24, 56, 60, 72, 80, 164, 168, 180, 188, 216, 224, 236, 240, 488, 492, 504, 512, 540, 548, 560, 564, 648, 656, 668, 672, 704, 708, 720, 728, 1460, 1464, 1476, 1484, 1512, 1520, 1532, 1536, 1620, 1628, 1640, 1644, 1676, 1680, 1692, 1700, 1944, 1952
Offset: 1

Views

Author

Benoit Cloitre, Apr 22 2003

Keywords

Comments

Apparently a(n)/2 (mod 3) = A010060(n), the Thue-Morse sequence.

Crossrefs

Cf. A010060, A051638, A074939, A083093, A083095 (gives a b-file of 16384 terms).

Programs

  • Mathematica
    Select[Range[0, 2000],OddQ[Sum[Mod[Binomial[#, j], 3], {j, 0, #}]] &] (* Paul F. Marrero Romero, Dec 28 2024 *)
  • PARI
    isok(n) = sum(k=0, n, binomial(n,k) % 3) % 2; \\ Michel Marcus, Dec 05 2013
    
  • Python
    def A083094(n): return int(bin(((m:=n-1).bit_count()&1)+(m<<1))[2:],3)<<1 # Chai Wah Wu, Jun 26 2025

Formula

a(n) = 4*A083095(n). - Hugo Pfoertner, Jan 12 2025
Numbers that are multiples of 4 and such that base-3 digits contain no 1's, or equivalently, numbers such that base-3 digits contains an even number of 2's and no 1's, i.e. a(n) = 2*A074939(n-1). This characterization can be derived from the formula in A051638. - Chai Wah Wu, Jun 26 2025

Extensions

More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Apr 29 2003

A122485 Values of A083097(k) such that A083097(k) = A083097(k+1) - 1.

Original entry on oeis.org

5, 14, 41, 59, 122, 140, 167, 176, 365, 383, 410, 419, 491, 500, 527, 545, 1094, 1112, 1139, 1148, 1220, 1229, 1256, 1274, 1463, 1472, 1499, 1517, 1580, 1598, 1625, 1634, 3281, 3299, 3326, 3335, 3407, 3416, 3443, 3461, 3650, 3659, 3686, 3704, 3767, 3785, 3812
Offset: 1

Views

Author

Alexander Adamchuk, Sep 15 2006

Keywords

Comments

A083097(n) = A083095(n) = A083096(n)/6 = A083094(n)/4, where A083096 are the Numbers k such that 3 divides Sum_{j=1..k} C(2*j,j) = A066796(k).
All terms are of the form 9*m + 5 and belong to A017221 with m = {0, 1, 4, 6, 13, 15, 18, 19, 40, 42, ...}.
Corresponding numbers m such that a(m) = A083097(m) are A129771 (evil odd numbers).

Examples

			A083097 begins {0, 2, 5, 6, 14, 15, 18, 20, 41, 42, 45, 47, 54, 56, 59, 60, ...}.
So a(1) = 5 because 5 = A083097(3) = A083097(3+1) - 1.
a(2) = 14 because 14 = A083097(5) = A083097(5+1) - 1.
		

Crossrefs

Formula

a(n) = A083097(A129771(n)).

Extensions

More terms from R. J. Mathar, Jan 17 2008
More terms from Jinyuan Wang, Jan 22 2022
Edited by Michel Marcus, Jan 22 2022
Showing 1-4 of 4 results.