This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
For n=6: 6 has two ways -- (d=3; 3|6), 1+2+3=6; and (d=1; 1|6), 6=6 -- so a(6)=2.
N:= 1000: # to get a(1) to a(N) g:= add(x^(k*(2*k-1))/(1-x^(2*k-1)), k=1..floor(sqrt(N/2))): S:= series(g,x,N+1): seq(coeff(S,x,n),n=1..N); # Robert Israel, Dec 08 2015
a(n) = my(q = sqrt(2*n)); sumdiv(n, d, (d%2) && (d < q)); \\ Michel Marcus, Jul 04 2014
Illustration of initial terms:
n a(n) Diagram
_
1 1 _|1|_
2 2 _|1 _ 1|_
3 3 _|1 |1| 1|_
4 2 _|1 _| |_ 1|_
5 4 _|1 |1 _ 1| 1|_
6 3 _|1 _| |1| |_ 1|_
7 4 _|1 |1 | | 1| 1|_
8 2 _|1 _| _| |_ |_ 1|_
9 6 _|1 |1 |1 _ 1| 1| 1|_
10 3 _|1 _| | |1| | |_ 1|_
11 4 _|1 |1 _| | | |_ 1| 1|_
12 4 _|1 _| |1 | | 1| |_ 1|_
13 4 _|1 |1 | _| |_ | 1| 1|_
14 4 _|1 _| _| |1 _ 1| |_ |_ 1|_
15 7 _|1 |1 |1 | |1| | 1| 1| 1|_
16 2 |1 | | | | | | | | 1|
...
For n = 6 (above), the total number of steps in all double staircases that have at least one step in the 6th level of the structure is equal to 3, since there are two steps in the first double staircase, there are no steps in the second double staircase, and there is only one step in the third double staircase, so a(3) = 2 + 0 + 1 = 3.
From the theorem (see comments) for n = 6, let s(k) = A196020(6,k) be the total number of steps from level n to the top, in the k-th double staircase that has at least a step in the 6th level of the structure, otherwise s(k) = 0. We have that s(1) = 11, s(2) = 0 and s(3) = 1. So the alternating sum is 11 - 0 + 1 = 12, which equals sigma(6) = 1 + 2 + 3 + 6 = 12.
Note that to evaluate sigma(n), it is sufficient to have only the n-th level of the diagram, since the width of the base level of a double staircase equals the number of its steps. See below:
For n = 6 the 6th level of the above diagram looks like this:
_ _ _
|1 | |1| | 1|
.
Width of the 1st staircase: |<-------- 11 ------->|
.
Width of the 3rd staircase: --->|1|<---
.
The width of the first double staircase is 11, the width of the second double staircase does not count, and the width of the third double staircase is 1, so the alternating sum is 11 - 0 + 1 = 12 = sigma(6).
For n = 15 the alternating sum is 29 - 13 + 7 - 0 + 1 = 24 = sigma(15).
For n = 16 the alternating sum is 31 - 0 + 0 - 0 + 0 = 31 = sigma(16).
For more information about these alternating sums see A196020.
N:= 100: S := convert(series( add( x^(n*(n+1)/2)*(1 + x^n)/(1 - x^n), n = 1..floor(sqrt(2*N)) ), x, N+1 ), polynom): seq(coeff(S, x, n), n = 1..N); # Peter Bala, Jan 20 2021
A335616[n_]:=2DivisorSigma[0,n/2^IntegerExponent[n,2]]-Boole[IntegerQ[(Sqrt[8n+1]-1)/2]];Array[A335616,100] (* Paolo Xausa, Sep 03 2023 *)
a(3) = 4 because 3 = (-2)+(-1)+0+1+2+3 or 0+1+2 or 1+2 or 3; a(13) = 4 because 13 = (-12)+...+13 or (-5)+...+7 or 6+7 or 13.
From _Omar E. Pol_, Nov 28 2020: (Start)
Illustration of initial terms:
Diagram
n a(n) _ _
1 2 _|1 1|_
2 2 _|1 _ _ 1|_
3 4 _|1 |1 1| 1|_
4 2 _|1 _| |_ 1|_
5 4 _|1 |1 _ _ 1| 1|_
6 4 _|1 _| |1 1| |_ 1|_
7 4 _|1 |1 | | 1| 1|_
8 2 _|1 _| _| |_ |_ 1|_
9 6 _|1 |1 |1 _ _ 1| 1| 1|_
10 4 _|1 _| | |1 1| | |_ 1|_
11 4 _|1 |1 _| | | |_ 1| 1|_
12 4 _|1 _| |1 | | 1| |_ 1|_
13 4 _|1 |1 | _| |_ | 1| 1|_
14 4 _|1 _| _| |1 _ _ 1| |_ |_ 1|_
15 8 _|1 |1 |1 | |1 1| | 1| 1| 1|_
16 2 |1 | | | | | | | | 1|
...
a(n) is the number of horizontal toothpicks in the n-th level of the diagram. (End)
a(n)=2*sumdiv(n,d,d%2)
A054844(n) = (2*numdiv(n>>valuation(n, 2))); \\ Antti Karttunen, Sep 27 2018
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