A165783 - OEIS

%I #10 May 14 2015 21:28:49

%S 2,3,6,4,9,15,18,5,12,27,8,15,30,27,42,6,15,17,54,16,30,21,17,32,31,

%T 10,78,28,27,87,90,7,18,99,33,49,12,29,45,56,81,123,10,39,15,16,13,50,

%U 72,45,150,74,16,159,54,50,42,63,15,33,165,26,150,8,21,195,26,53,102,207

%N a(n) = A002326(n-1) + A000120(A165781(n-1)).

%C Given a shift register : r(k)=r(k-1)+ X if r(k-1) is not divisible Y, else r(k)=r(k-1)/Y.

%C Gcd(r(0), X))=1, Gcd(X, Y)=1.

%C Then the length of the period orbit of such a register is L + digitsum (r(L)*(Y^L-1)/ X). Digitsum(z)in base X.

%C r(L) a point from period orbit, L minimal possible exponent such that (Y^L-1)/X)is a positive integer.

%C Number of period orbits is the order of the cyclic group connected to the register.

%C a(n) is the period length for Y=2, X=2*n-1, r(L)=1. [_Ctibor O. Zizka_, Nov 24 2009]

%H Ivan Neretin, <a href="/A165783/b165783.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = L + digitsum((2^L -1)/(2*n-1)). Digitsum(z)in base 2. [_Ctibor O. Zizka_, Nov 24 2009]

%e n=1, a(1)=1 + digitsum(1)= 2.

%e n=2, a(2)=2 + digitsum(1)=3.

%e n=3, a(3)= 4 + digitsum(3) = 6.

%e n=4, a(4)= 3 + digitsum(1)=4.

%e n=5, a(5)= 6 + digitsum(7)=9. [_Ctibor O. Zizka_, Nov 24 2009]

%p A002326 := proc(n) if n = 0 then 1; else numtheory[order](2,2*n+1) ; end if ; end proc:

%p A165781 := proc(n) (2^A002326(n)-1)/(2*n+1) ; end proc:

%p read("transforms") ; A165783 := proc(n) A002326(n-1)+wt(A165781(n-1) ) ; end proc:

%p seq(A165783(n),n=1..80) ; # _R. J. Mathar_, Nov 26 2009

%t Table[(b = MultiplicativeOrder[2, 2 n - 1]) + Plus @@ IntegerDigits[(2^b - 1)/(2 n - 1), 2], {n, 1, 70}] (* _Ivan Neretin_, May 09 2015 *)

%o (PARI) hamming(n)=my(v=binary(n));sum(i=1,#v,v[i])

%o a(n)=my(x=2*n+1,m=znorder(Mod(2,x)));m+hamming((1<<m)\x)

%Y Cf. A002326, A053446.

%K easy,nonn

%O 1,1

%A _Ctibor O. Zizka_, Sep 26 2009

%E Program and extension by _Charles R Greathouse IV_, Nov 24 2009

%E Definition corrected and comments merged by _R. J. Mathar_, Nov 26 2009