

A309274


Ackermann Coding (BIT predicate) of transitive hereditarily finite sets.


2



0, 1, 3, 7, 11, 15, 23, 31, 39, 47, 55, 63, 71, 79, 87, 95, 103, 111, 119, 127, 135, 143, 151, 159, 167, 175, 183, 191, 199, 207, 215, 223, 231, 239, 247, 255, 267, 271, 287, 303, 319, 335, 351, 367, 383, 399, 415, 431, 447, 463, 479, 495, 511, 523, 527, 543
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OFFSET

1,3


COMMENTS

If the representation of a(n) in base 2 contains the kth bit (2^k), then it must contain the bits of k.
A034797 is a subsequence, and can be seen as a recursive variant of this sequence.  Rémy Sigrist, Jul 25 2019


LINKS



EXAMPLE

23 is in the sequence because 23 = 2^4 + 2^2 + 2^1 + 2^0 encodes the transitive set {0,1,{1},{{1}}} (remember that 0 is the empty set and 1 is {0}).


MATHEMATICA

b[n_] := (Flatten @ Position[Reverse[IntegerDigits[n, 2]], 1]  1);
okQ[n_] := With[{bb = b[n]}, AllTrue[b /@ bb, Intersection[bb, #] == #&]];


PROG

(PARI) is(n) = { for (b=0, #binary(n), if (bittest(n, b), if (bitand(n, b)!=b, return (0)))); return (1) } \\ Rémy Sigrist, Jul 25 2019


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



