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%I #31 Sep 08 2022 08:46:14
%S 1,2,3,4,6,8,11,12,16,22,24,32,43,44,48,64,86,88,96,128,171,172,176,
%T 192,256,342,344,352,384,512,683,684,688,704,768,1024,1366,1368,1376,
%U 1408,1536,2048,2731,2732,2736,2752,2816,3072,4096,5462,5464,5472,5504
%N Positive values of m, arranged in order, such that binomial(4*m - 1, m) is odd.
%C 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.
%C Compare with A002450, which equals the values of m such that binomial(4*m + 1,m) is odd, A020988 which equals the values of m such that binomial(4*m + 2,m) is odd, A263133, which gives the values of m such that binomial(4*m + 3,m) is odd and A080674, which equals the values of m such that binomial(4*m + 4,m) is odd.
%C Compare with A118113, which appears to be the values of m such that binomial(3*m - 2,m) is odd. Cf. A003714.
%C 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.
%C 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.
%F a(n) = A263133(n) + 1.
%F 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
%e 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.
%e ........................................................
%e . Table 1: 4^n * ceiling(2^k/3) for n >= 0, k >= 1 .
%e ........................................................
%e n\k| 1 2 3 4 5 6 7 8 9
%e ---+----------------------------------------------------
%e 0 | 1 2 3 6 11 22 43 86 171 ...
%e 1 | 4 8 12 24 44 88 172 ...
%e 2 | 16 32 48 96 176 ...
%e 3 | 64 128 192 ...
%e 4 | 256 ...
%e ...
%e 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.
%e ....................................................
%e . Table 2: 2^n * (2^(2*k + 1) + 1)/3, n,k >= 0 .
%e ....................................................
%e n\k| 0 1 2 3 4 5
%e ---+----------------------------------------------
%e 0 | 1 3 11 43 171 683 ...
%e 1 | 2 6 22 86 342 1366 ...
%e 2 | 4 12 44 172 684 2732 ...
%e 3 | 8 24 88 344 1368 5464 ...
%e 4 | 16 48 176 688 2736 10928 ...
%e 5 | 32 96 352 1376 5472 21856 ...
%e 6 | 64 192 704 2752 10944 43712 ...
%e 7 | 128 384 1408 5504 21888 87424 ...
%e 8 | 256 ...
%p for n from 1 to 5000 do if mod(binomial(4*n-1, n), 2) = 1 then print(n) end if end do;
%t Select[Range[6000],OddQ[Binomial[4#-1,#]]&] (* _Harvey P. Dale_, Dec 26 2015 *)
%o (PARI) for(n=1, 1e4, if (binomial(4*n-1, n) % 2 == 1, print1(n", "))) \\ _Altug Alkan_, Oct 11 2015
%o (Magma) [n: n in [1..6000] | Binomial(4*n-1, n) mod 2 eq 1]; // _Vincenzo Librandi_, Oct 12 2015
%o (Python)
%o A263132_list = [m for m in range(1,10**6) if not ~(4*m-1) & m] # _Chai Wah Wu_, Feb 07 2016
%Y Cf. A000079, A000302, A002450, A003714, A005578, A007583, A020988, A048716, A118113, A263133.
%K nonn,easy
%O 1,2
%A _Peter Bala_, Oct 10 2015
%E More terms from _Vincenzo Librandi_, Oct 12 2015
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