|
%I #13 Apr 10 2019 12:24:03
%S 1,2,3,5,9,15,19,37,73,82,102,110,130,142,177,235,327,363,473,543,723,
%T 747,993,1023,1033,1291,2581,2889,3843,3903,5203,5973,6153,7029,7239,
%U 7365,8345,10013,10373,10593,12183,12313,13192,13240,13300,13480,13564,13677
%N a(1) = 1; a(n+1) is the smallest k > a(n) such that 2^k == 2^a(n) (mod a(n)).
%C The number of consecutive terms with the same parity: 1, 1, 7, 5, 28, 5, 16, 13, 47, 4, 70, 6, 56, 4, 32, 11, 17, 21, 22, 11, 2, 20, 67, 13, 22, 36, 8, 9,101, 47, 24, 4, 1, 7, 2, 79, 20, 71, 47, 92, 36, 57, 90, 38, 167, 215, 31, 17, 62, ... i.e.; 1 odd term, 1 even term, 7 odd terms, 5 even terms, 28 odd terms, 5 even terms, ...
%H Robert Israel, <a href="/A306829/b306829.txt">Table of n, a(n) for n = 1..3000</a>
%F a(n+1) = a(n) + ord_{A000265(a(n))}(2), where A000265(m) is the odd part of m.
%p A[1]:= 1:
%p for n from 2 to 100 do A[n]:= A[n-1] + numtheory:-order(2,A[n-1]/2^padic:-ordp(A[n-1],2)) od:
%p seq(A[n],n=1..100); # _Robert Israel_, Apr 10 2019
%o (PARI) odd(n) = n >> valuation(n, 2);
%o lista(nn) = {a = 1; print1(a, ", "); for (n=2, nn, a = a + znorder(Mod(2, odd(a))); print1(a, ", "););} \\ _Michel Marcus_, Mar 23 2019
%Y Cf. A000265, A002326.
%K nonn
%O 1,2
%A _Thomas Ordowski_, Mar 12 2019
%E More terms from _Amiram Eldar_, Mar 12 2019
|
