A365938 - OEIS

%I #17 Sep 25 2023 23:54:19

%S 1,0,1,1,0,1,1,1,0,1,0,1,1,0,0,0,0,1,0,1,0,1,1,0,0,1,0,0,1,0,0,1,1,0,

%T 1,0,0,1,1,0,1,0,1,0,1,0,1,1,0,1,0,1,0,1,1,0,0,1,0,0,0,1,0,1,0,1,0,0,

%U 0,1,1,1,1,0,1,0,0,1,0,0,0,1,0,0,1,0,0,0,0,1,0,0,1,0,1,1,0,1,0,1,1

%N Form the "Gilbreath Transform" array for the Euler phi function (A000010); sequence gives the third column, divided by 2.

%H Paolo Xausa, <a href="/A365938/b365938.txt">Table of n, a(n) for n = 1..10000</a>

%e The first 4 rows of the Gilbreath Transform array for phi(n) are:

%e 1, 1, 2, 2, 4, 2, 6, 4, 6, 4,10, 4,12, 6, 8, 8,16, 6,18, 8,12,10,22, 8,20

%e 0, 1, 0, 2, 2, 4, 2, 2, 2, 6, 6, 8, 6, 2, 0, 8,10,12,10, 4, 2,12,14,12, 8

%e 1, 1, 2, 0, 2, 2, 0, 0, 4, 0, 2, 2, 4, 2, 8, 2, 2, 2, 6, 2,10, 2, 2, 4, 2

%e 0, 1, 2, 2, 0, 2, 0, 4, 4, 2, 0, 2, 2, 6, 6, 0, 0, 4, 4, 8, 8, 0, 2, 2, 2

%e The third column is 2, 0, 2, 2, ..., which when halved gives the current sequence 1, 0, 1, 1, ...

%e I conjecture that the first column is (1,0) repeated, and the second column is all 1's.

%t A365938list[nmax_]:=Module[{d=EulerPhi[Range[3,nmax+2]]},Join[{1},Table[First[d=Abs[Differences[d]]],nmax-1]/2]];A365938list[200] (*  _Paolo Xausa_, Sep 25 2023 *)

%Y Cf. A000010, A036262, A361897.

%K nonn

%O 1

%A _N. J. A. Sloane_, Sep 23 2023, following a suggestion from _Sean A. Irvine_