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%I #26 Sep 08 2022 08:46:14
%S 0,1,2,3,5,7,10,11,15,21,23,31,42,43,47,63,85,87,95,127,170,171,175,
%T 191,255,341,343,351,383,511,682,683,687,703,767,1023,1365,1367,1375,
%U 1407,1535,2047,2730,2731,2735,2751,2815,3071,4095,5461,5463,5471,5503
%N Numbers m such that binomial(4*m + 3, 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; and A080674, which gives the values of m such that binomial(4*m + 4, m) is odd.
%C Compare with A263132, which lists the values of m such that binomial(4*m - 1, m) is odd.
%C The sequence of even values of a(n) is [0, 2, 10, 42, 170, ...] = A020998. 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 A020998 produces all the terms of this sequence. See the example below.
%C 2*a(n) gives the values of m such that binomial(4*m + 6, m) is odd.
%F a(n) = A263132(n) - 1.
%F 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
%e 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.
%e ...........................................................
%e . Table 1. 2^n*ceiling((2^(2*k + 1) - 1)/3) - 1, n,k >= 0 .
%e ...........................................................
%e n\k| 0 1 2 3 4 5
%e ---+---------------------------------
%e 0 | 0 2 10 42 170 682 ...
%e 1 | 1 5 21 85 341 ...
%e 2 | 3 11 43 171 683 ...
%e 3 | 7 23 87 343 ...
%e 4 | 15 47 175 687 ...
%e 5 | 31 95 351 ...
%e 6 | 63 191 703 ...
%e 7 | 127 383 ...
%e 8 | 255 767 ...
%e 9 | 511 ...
%e ...
%e 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.
%e 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.
%e ........................................................
%e . Table 2. (4^n)*ceiling(2^k/3) - 1 for n >= 0, k >= 1 .
%e ........................................................
%e n\k| 1 2 3 4 5 6 7 8 9 10
%e ---+---------------------------------------------------------
%e 0| 0 1 2 5 10 21 42 85 170 682...
%e 1| 3 7 11 23 43 87 171 343 683 ...
%e 2| 15 31 47 95 175 351 687 1375 ...
%e 3| 63 127 191 383 703 1407 2751 5503 ...
%e 4| 255 511 767 1535 2815 5631 11007 22015 ...
%e 5| 1023 2047 3071 6143 11263 22527 44031 88063 ...
%e 6| 4095 ...
%e ...
%e 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.
%p for n from 1 to 4096 do if mod(binomial(4*n+3, n), 2) = 1 then print(n) end if end do;
%t Select[Range[0,5600],OddQ[Binomial[4#+3,#]]&] (* _Harvey P. Dale_, Apr 15 2019 *)
%o (PARI) for(n=0, 1e4, if (binomial(4*n+3, n) % 2 == 1, print1(n", "))) \\ _Altug Alkan_, Oct 11 2015
%o (Magma) [n: n in [0..6000] | Binomial(4*n+3, n) mod 2 eq 1]; // _Vincenzo Librandi_, Oct 12 2015
%o (Python)
%o A263133_list = [m for m in range(10**6) if not ~(4*m+3) & m] # _Chai Wah Wu_, Feb 07 2016
%Y Cf. A000975, A002450, A020988, A080674, A263133.
%K nonn,easy
%O 1,3
%A _Peter Bala_, Oct 11 2015
%E More terms from _Vincenzo Librandi_, Oct 12 2015
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