cp's OEIS Frontend

A174978 For definition see comments lines.

%I A174978 #10 Dec 10 2016 03:12:58
%S A174978 1,2,1,5,2,2,1,20,2,5,2,5,2,2,1,95,2,5,2,20,2,5,2,20,2,5,2,5,2,2,1,
%T A174978 470,2,5,2,20,2,5,2,95,2,5,2,20,2,5,2,95,2,5,2,20,2,5,2,20,2,5,2,5,2,
%U A174978 2,1,2345,2,5,2,20,2,5,2,95,2,5,2,20,2,5,2,470,2,5,2,20,2,5,2,95,2,5,2,20,2
%N A174978 For definition see comments lines.
%C A174978 It is easier to explain the rule of recurrence when the numbers are written as follows:
%C A174978 1,
%C A174978 2, 1,
%C A174978 5, 2, 2, 1,
%C A174978 20, 2, 5, 2, 5, 2, 2, 1,
%C A174978 95, 2, 5, 2, 20, 2, 5, 2, 20, 2, 5, 2, 5, 2, 2, 1,
%C A174978 470, 2, 5, 2, 20, 2, 5, 2, 95, 2, 5, 2, 20, 2, 5, 2, 95, 2, 5, 2, 20, 2, 5, 2, 20, 2, 5, 2, 5, 2, 2, 1,
%C A174978 2345, 2, 5, 2, 20, 2, 5, 2, 95, 2, 5, 2, 20, 2, 5, 2, 470, 2, 5, 2, 20, 2, 5, 2, 95, 2, 5, 2, 20, 2, 5, 2, 470,
%C A174978 2, 5, 2, 20, 2, 5, 2, 95, 2, 5, 2, 20, 2, 5, 2, 95, 2, 5, 2, 20, 2, 5, 2, 20, 2, 5, 2, 5, 2, 2, 1.
%C A174978 At first a(2^(n+1)-1) = (3*5^n+5)/4 (n>=0). Let A be the sequence defined as follows:
%C A174978 A(0)=2; W(A(0))=5; A(1)=A(0),W(A(0))=2, 5; W(A(1))=2, 20.
%C A174978 More generally with A(n)=B(n), {3*5^n+5)/4; we define W(A(n))=B(n), (3*5^(n+1)+5)/4 and A(n+1)=A(n), W(A(n)).
%C A174978 Here we obtain A(1)=2, 5; W(A(1))=2, 20; A(2)=2, 5, 2, 20; W(A(2))=2, 5, 2, 95; A(3)=2, 5, 2, 20, 2, 5, 2, 95;
%C A174978 W(A(3))=2, 5, 2, 20, 2, 5, 2, 470; A(4)=2, 5, 2, 20, 2, 5, 2, 95, 2, 5, 2, 20, 2, 5, 2, 470, etc.
%C A174978 In fact: B(1)=2; B(2)=2, 5, 2; B(3)=2, 5, 2, 20, 2, 5, 2; B(4)=2, 5, 2, 20, 2, 5, 2, 95, 2, 5, 2, 20, 2, 5, 2, etc.
%C A174978 If we denote by <<A|UA|>> the subsequence of a between a(2^(n+1)-1) and a(2^(n+2)-1), the subsequence of a between a(2^(n+2)-1) and a(2^(n+3)-1) is given by <<A|A(n+1), A(n+1), UA|>>.
%C A174978 It seems that this sequence gives the numbers of 1 in the successive sets of 1 in the sequence A174835.
%e A174978 a(1)=a(2^1-1)=(3*5^0+5)/4=2. a(3)=a(2^2-1)=(3*5+5)/4=5.
%e A174978 a(7)=a(2^3-1)=(75+5)/4=20. a(15)=a(2^4-1)=(3*125+5)/4=380/4=95.
%e A174978 Between 20 and 95 the subsequence of a is: 2, 5, 2, 5, 2, 2, 1.
%e A174978 Then with the definition, the subsequence of a, between 95 and 470 is:
%e A174978 A(2), A(2), 2, 5, 2, 5, 2, 2, 1, i.e., 2, 5, 2, 20, 2, 5, 2, 20, 2, 5, 2, 5, 2, 2, 1.
%Y A174978 Cf. A174835, A174837.
%K A174978 easy,nonn,uned
%O A174978 0,2
%A A174978 _Richard Choulet_, Apr 03 2010