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 1-3 of 3 results.

A263132 Positive values of m such that binomial(4*m - 1, m) is odd.

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 11, 12, 16, 22, 24, 32, 43, 44, 48, 64, 86, 88, 96, 128, 171, 172, 176, 192, 256, 342, 344, 352, 384, 512, 683, 684, 688, 704, 768, 1024, 1366, 1368, 1376, 1408, 1536, 2048, 2731, 2732, 2736, 2752, 2816, 3072, 4096, 5462, 5464, 5472, 5504
Offset: 1

Views

Author

Peter Bala, Oct 10 2015

Keywords

Comments

This sequence, when viewed as a set, equals the set of numbers of the form 4^n * ceiling(2^k/3) for n >= 0, k >= 1, i.e., the product subset in Z of A000302 and A005578 regarded as sets. See the example below.
Equivalently, this sequence, when viewed as a set, equals the set of numbers of the form 2^n * (2^(2*k + 1) + 1)/3 for n,k >= 0, i.e., the product subset in Z of A000079 and A007583 regarded as sets. See the example below.
2*a(n) gives the values of m such that binomial(4*m - 2,m) is odd. 4*a(n) gives the values of m such that binomial(4*m - 3,m) is odd (other than m = 1) and also the values of m such that binomial(4*m - 4,m) is odd.

Examples

			1) Notice how this sequence can be read from Table 1 below by moving through the table in a sequence of 'knight moves' (1 down and 2 to the left) starting from the first row. For example, starting at 11 on the top row we move in a series of knights moves 11 -> 12 -> 16, then return to the top row at 22 and move 22 -> 24 -> 32, return to the top row at 43 and move 43 -> 44 -> 48 -> 64, then return to top row at 86 and so on.
........................................................
.   Table 1: 4^n * ceiling(2^k/3) for n >= 0, k >= 1   .
........................................................
n\k|   1    2    3    4     5     6    7    8     9
---+----------------------------------------------------
0  |   1    2    3    6    11    22   43   86   171 ...
1  |   4    8   12   24    44    88  172  ...
2  |  16   32   48   96   176    ...
3  |  64  128  192  ...
4  | 256  ...
...
2) Notice how this sequence can be read from Table 2 below in a sequence of 'knight moves' (2 down and 1 to the left) starting from the first two rows. For example, starting at 43 in the first row we jump 43 -> 44 -> 48 -> 64, then return to the second row at 86 and jump 86 -> 88 -> 96 -> 128, followed by 171 -> 172 -> 176 -> 192 -> 256, and so on.
....................................................
.   Table 2: 2^n * (2^(2*k + 1) + 1)/3, n,k >= 0   .
....................................................
n\k|   0    1     2     3      4      5
---+----------------------------------------------
0  |   1    3    11    43    171    683  ...
1  |   2    6    22    86    342   1366  ...
2  |   4   12    44   172    684   2732  ...
3  |   8   24    88   344   1368   5464  ...
4  |  16   48   176   688   2736  10928  ...
5  |  32   96   352  1376   5472  21856  ...
6  |  64  192   704  2752  10944  43712  ...
7  | 128  384  1408  5504  21888  87424  ...
8  | 256 ...

Crossrefs

Cf. A000079, A000267, A000302, A005578, A007583, A048716, A384688.
Other odd binomials: A002450 (4*m+1,m), A020988 (4*m+2,m), A263133 (4*m+3,m), A080674 (4*m+4,m), A118113 (3*m-2,m), A003714 (3*m,m).

Programs

  • Magma
    [n: n in [1..6000] | Binomial(4*n-1, n) mod 2 eq 1]; // Vincenzo Librandi, Oct 12 2015
  • Maple
    for n from 1 to 5000 do if mod(binomial(4*n-1, n), 2) = 1 then print(n) end if end do;
  • Mathematica
    Select[Range[6000],OddQ[Binomial[4#-1,#]]&] (* Harvey P. Dale, Dec 26 2015 *)
  • PARI
    for(n=1, 1e4, if (binomial(4*n-1, n) % 2 == 1, print1(n", "))) \\ Altug Alkan, Oct 11 2015
  • PARI
    a(n) = my(r,s=sqrtint(4*n-3,&r)); (1<Kevin Ryde, Jun 14 2025
  • Python
    A263132_list = [m for m in range(1,10**6) if not ~(4*m-1) & m] # Chai Wah Wu, Feb 07 2016

Formula

a(n) = A263133(n) + 1.
m is a term if and only if m AND NOT (4*m-1) = 0 where AND and NOT are bitwise operators. - Chai Wah Wu, Feb 07 2016
a(n) = (2^A000267(n-1) + 2^A384688(n-1)) / 3. - Kevin Ryde, Jun 14 2025

Extensions

More terms from Vincenzo Librandi, Oct 12 2015

A263133 Numbers m such that binomial(4*m + 3, m) is odd.

Original entry on oeis.org

0, 1, 2, 3, 5, 7, 10, 11, 15, 21, 23, 31, 42, 43, 47, 63, 85, 87, 95, 127, 170, 171, 175, 191, 255, 341, 343, 351, 383, 511, 682, 683, 687, 703, 767, 1023, 1365, 1367, 1375, 1407, 1535, 2047, 2730, 2731, 2735, 2751, 2815, 3071, 4095, 5461, 5463, 5471, 5503
Offset: 1

Views

Author

Peter Bala, Oct 11 2015

Keywords

Comments

The even terms in the sequence are A020988. If m is a term in the sequence then 2*m + 1 is also a term in the sequence. Repeatedly applying the transformation m -> 2*m + 1 to the terms of A020988 produces all the terms of this sequence. See the example below.
2*a(n) gives the values of m such that binomial(4*m + 6, m) is odd.

Examples

			1) This sequence can be read from Table 1 below in a sequence of 'knight moves' (2 down and 1 to the left) starting from the first two rows. For example, starting at 42 in the first row we jump 42 -> 43 -> 47 -> 63, then return to the second row at 85 and jump 85 -> 87 -> 95 -> 127, followed by 170 -> 171 -> 175 -> 191 -> 255, and so on.
...........................................................
. Table 1. 2^n*ceiling((2^(2*k + 1) - 1)/3) - 1, n,k >= 0 .
...........................................................
  n\k|   0    1    2    3    4    5
  ---+---------------------------------
   0 |   0    2   10   42  170  682 ...
   1 |   1    5   21   85  341  ...
   2 |   3   11   43  171  683  ...
   3 |   7   23   87  343  ...
   4 |  15   47  175  687  ...
   5 |  31   95  351  ...
   6 |  63  191  703  ...
   7 | 127  383  ...
   8 | 255  767  ...
   9 | 511  ...
   ...
The first row of the table is A020988. The columns of the table are obtained by repeatedly applying the transformation m -> 2*m + 1 to the entries in the first row.
2) Alternatively, this sequence can be read from Table 2 below by starting with a number on the top row and moving in a series of 'knight moves' (1 down and 2 to the left) through the table as far as you can, before returning to the next number in the top row and repeating the process. For example, starting at 10 in the first row we move 10 -> 11 -> 15, then return to the top row at 21 and move 21 -> 23 -> 31, before returning to the top row at 42 and so on.
........................................................
. Table 2. (4^n)*ceiling(2^k/3) - 1 for n >= 0, k >= 1 .
........................................................
n\k|    1    2    3    4     5     6     7     8    9   10
---+---------------------------------------------------------
  0|    0    1    2    5    10    21    42    85  170  682...
  1|    3    7   11   23    43    87   171   343  683  ...
  2|   15   31   47   95   175   351   687  1375  ...
  3|   63  127  191  383   703  1407  2751  5503  ...
  4|  255  511  767 1535  2815  5631 11007 22015  ...
  5| 1023 2047 3071 6143 11263 22527 44031 88063  ...
  6| 4095 ...
  ...
The first row of the table is A000975. The columns of the table are obtained by repeatedly applying the transformation m -> 4*m + 3 to the entries in the first row.

Crossrefs

Cf. A000975.
Other odd binomials: A263132 (4*m-1,m), A002450 (4*m+1,m), A020988 (4*m+2,m), A080674 (4*m+4,m), A118113 (3*m-2,m), A003714 (3*m,m).

Programs

  • Magma
    [n: n in [0..6000] | Binomial(4*n+3, n) mod 2 eq 1]; // Vincenzo Librandi, Oct 12 2015
  • Maple
    for n from 1 to 4096 do if mod(binomial(4*n+3, n), 2) = 1 then print(n) end if end do;
  • Mathematica
    Select[Range[0,5600],OddQ[Binomial[4#+3,#]]&] (* Harvey P. Dale, Apr 15 2019 *)
  • PARI
    for(n=0, 1e4, if (binomial(4*n+3, n) % 2 == 1, print1(n", "))) \\ Altug Alkan, Oct 11 2015
  • PARI
    a(n) = my(r,s=sqrtint(4*n-3,&r)); (1<Kevin Ryde, Jul 06 2025
  • Python
    A263133_list = [m for m in range(10**6) if not ~(4*m+3) & m] # Chai Wah Wu, Feb 07 2016

Formula

a(n) = A263132(n) - 1.
m is a term if and only if m AND NOT (4*m+3) = 0 where AND and NOT are bitwise operators. - Chai Wah Wu, Feb 07 2016
a(n) = (2^A000267(n) + 2^A384688(n))/3 - 1, for n >= 1. - Kevin Ryde, Jul 06 2025

Extensions

More terms from Vincenzo Librandi, Oct 12 2015

A173195 Values of k such that 4^x + 4^y + 4^z = k^2 with arbitrary integers x <= y <= z.

Original entry on oeis.org

3, 6, 9, 12, 18, 24, 33, 36, 48, 66, 72, 96, 129, 132, 144, 192, 258, 264, 288, 384, 513, 516, 528, 576, 768, 1026, 1032, 1056, 1152, 1536, 2049, 2052, 2064, 2112, 2304, 3072, 4098, 4104, 4128, 4224, 4608, 6144, 8193, 8196, 8208, 8256, 8448, 9216, 12288
Offset: 1

Views

Author

Michel Lagneau, Feb 12 2010

Keywords

Comments

We prove that the solutions of 4^x + 4^y + 4^z = k^2 are (x,y,2y-x-1), for any arbitrary integer x,y. We calculate z. 4^x + 4^y + 4^z is square if positive integers m and odd integer t are such as : 1 + 4^(y-x) + 4^(z-x) = (1 + t*2^m)^2, that's why : (1 + 4^(z-y)*( 4^(y-x)) = t(1 + t*2^(m+1)) t.2^(m+1), and then m = 2y - 2x - 1. If we report this value in the precedent equation, we obtain : t-1 = (2^(z-2y+x+1) + t)(2^(z-2y+x+1) - t) * 4^(y-x-1). Because t is odd, z = 2y - x - 1. Finally, this values gives the square (2^x + 2^(2y-x-1))^2 = k^2.
From Frederik P.J. Vandecasteele, Jun 06 2025: (Start)
For a given n, the exponents are x = A384688(n-1), y = A138099(n), z = A000267(n-1) so that a(n) = 2^A384688(n-1) + 2^A000267(n-1).
Terms are all and only those k whose binary expansion is two 1 bits an odd distance apart. (End)

Examples

			x = 0, y = 1 then z = 1, and k = 3.
x = 1, y = 2 then z = 2, and k = 6.
x = 0, y = 2 then z = 3, and k = 9.

References

  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976.
  • J. M. De Koninck, A. Mercier, 1001 problèmes en théorie classique des nombres. Ellipses, 2004.
  • H. N. Shapiro, Introduction to the Theory of Numbers, John Wiley & Sons, 1983.

Links

Crossrefs

Cf. A263132.
Subsequence of A018900.
Cf. A384688, A138099, A000267.

Programs

  • Maple
    for x from 0 to 1000 do :for y from x to 1000 do: n := evalf(2^x + 2^(2*y-x-1)): print (n) ; od :od :
  • Mathematica
    Take[Union[Select[Sqrt[Flatten[Table[(2^x + 2^(2*y - x - 1))^2, {x, 0, 13}, {y, 0, 13}]]], IntegerQ]],49] (* Jean-François Alcover, Sep 13 2011 *)

Formula

k = 2^x + 2^(2y-x-1), and z = 2y - x - 1.
Conjecture: a(n) = 3*A263132(n). - George Beck, May 05 2021
Showing 1-3 of 3 results.

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