A290254 The viabin numbers of the self-conjugate integer partitions.

0, 1, 5, 6, 19, 21, 26, 28, 71, 75, 85, 89, 102, 106, 116, 120, 271, 279, 299, 307, 333, 341, 361, 369, 398, 406, 426, 434, 460, 468, 488, 496, 1055, 1071, 1111, 1127, 1179, 1195, 1235, 1251, 1309, 1325, 1365, 1381, 1433, 1449, 1489, 1505, 1566, 1582, 1622

Offset: 1

Author

Emeric Deutsch, Aug 23 2017

Keywords

• nonn

Comments

For the definition of viabin number see comment in A290253.

The binary representation of a(n) is obtained by concatenating the binary representation of (n-1) and the reversed bit-flipped binary representation of (n-1) and then dropping the last bit. This suggests that it would have been more natural to index from 0. - Peter J. Taylor, Sep 24 2021

Examples

			19 is in the sequence. Indeed, binary (19) = 10011 and so the southeast border of the Ferrers board of the corresponding integer partition is ENNEEN, where E = (1,0) and N = (0,1). This leads to the self-conjugate integer partition [3,1,1].
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..16384

Crossrefs

Cf. A059894, A290253.

Programs

Maple

a := proc (n) local i, m, r: m, r := n, 0: for i from 0 while 1 < m do r := 2*r+1-irem(m, 2, 'm') end do: r+2^i end proc: SC := {0}: for n to 3000 do if a(n) = n then SC := `union`(SC, {n}) else  end if end do: SC; # first part of the program taken from A059894.

Mathematica

nmax = 3000; (* nmax=3000 gives 64 terms *)
a[n_] := Module[{i, m = n, r = 0}, For[i = 0, 1 < m, i++, r = 2*r + 1 - Mod[m, 2]; m = Quotient[m, 2]]; r + 2^i];
SC = {0};
For[n = 1, n <= nmax, n++, If[a[n] == n, SC = Union[SC, {n}]]];
SC (* Jean-François Alcover, Dec 16 2020, after Maple *)

PARI

a(n) = my(v=binary(max(1,n-1))[^1]); n<<#v + bitneg(fromdigits(Vecrev(v),2)); \\ Kevin Ryde, Nov 30 2021

Python

a = lambda n: int(bin(n-1)[2:] + ''.join(str(1 ^ int(ch)) for ch in bin(n-1)[-1:2:-1]), 2) # Peter J. Taylor, Sep 24 2021

Formula

{ 0 } union fixed points of A059894. - Alois P. Heinz, Aug 24 2017

a(n) = a(n-1) + 2*4^(f(n-1) - 1) + 3*2^(f(n-1) - 1) - 1 if n = 2^k + 1, k > 0, otherwise a(n-1) + (2^(A007814(n-1) + 2) - 3)*2^f(A025480(n-2)) with a(1) = 0, a(2) = 1 where f(n) = A000523(n) for n > 0 with f(0) = 0. - Mikhail Kurkov, Sep 24 2021 [verification needed]