A174835 -id:A174835 - cp's OEIS frontend

A174837 Sequence built as it follows in comments.

9, 2, 45, 2, 11, 2, 225, 2, 11, 2, 56, 2, 11, 2, 1125, 2, 11, 2, 56, 2, 11, 281, 2, 11, 2, 56, 2, 11, 2, 5625

Offset: 0

Richard Choulet, Mar 30 2010

If we denote "A" the finite sequence between a(2^(n-1)-2) and a(2^n-2), the subsequence of a between

a(2^(n)-2) and a(2^(n+1)-2) is given by: " A - a(3*2^(n-1)-2) - A" for every n>=2.

			a(4)=a(2*3-2)=(9*4-1)/4=11. a(14)=a(2^4-2)==9*5^3=125*9=1125.
Between a(2) and a(6) the subsequence is "2, 11, 2"; then between a(6) and a(14) the subsequence of a is:
"2, 11, 2, a(10)=56, 2, 11, 2".
It seems that this new sequence gives the number of 2 in the sets of 2 of the sequence A174835.

Cf. A174835.

a(2n+1)=2. a(2^(n+1)-2)=9*5^n. a(3*2^n-2)=(9*5^n-1)/4.

A174978 For definition see comments lines.

Original entry on oeis.org

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, 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, 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

Offset: 0

Richard Choulet, Apr 03 2010

It is easier to explain the rule of recurrence when the numbers are written as follows:
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,
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,
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,
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.
At first a(2^(n+1)-1) = (3*5^n+5)/4 (n>=0). Let A be the sequence defined as follows:
A(0)=2; W(A(0))=5; A(1)=A(0),W(A(0))=2, 5; W(A(1))=2, 20.
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)).
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;
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.
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.
If we denote by <> 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 <>.
It seems that this sequence gives the numbers of 1 in the successive sets of 1 in the sequence A174835.

			a(1)=a(2^1-1)=(3*5^0+5)/4=2. a(3)=a(2^2-1)=(3*5+5)/4=5.
a(7)=a(2^3-1)=(75+5)/4=20. a(15)=a(2^4-1)=(3*125+5)/4=380/4=95.
Between 20 and 95 the subsequence of a is: 2, 5, 2, 5, 2, 2, 1.
Then with the definition, the subsequence of a, between 95 and 470 is:
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.

Cf. A174835, A174837.

cp's OEIS Frontend

A174837 Sequence built as it follows in comments.

Views

Author

Keywords

Comments

Examples

Crossrefs

Formula