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A322141
a(n) is also the sum of the even-indexed terms of the n-th row of the triangle A237591.
3
0, 0, 1, 1, 2, 1, 2, 2, 2, 3, 4, 3, 4, 5, 4, 4, 5, 5, 6, 5, 6, 7, 8, 7, 7, 8, 9, 8, 9, 8, 9, 9, 10, 11, 10, 10, 11, 12, 13, 12, 13, 12, 13, 14, 13, 14, 15, 14, 14, 14, 15, 16, 17, 16, 17, 16, 17, 18, 19, 18, 19, 20, 19, 19, 20, 19, 20, 21, 22, 21, 22, 20, 21
OFFSET
1,5
FORMULA
a(n) = n - A240542(n).
EXAMPLE
Illustration of initial terms in two ways:
.
n a(n)
1 0
2 0 _ _
3 1 |_| _|_|
4 1 _|_| _|_|
5 2 |_ _| _|_ _|
6 1 _|_| _|_|
7 2 |_ _| _|_ _|
8 2 _|_ _| _|_ _|
9 2 |_ _| _ _|_ _|
10 3 _|_ _| |_| _|_ _|_|
11 4 |_ _ _| |_| _|_ _ _|_|
12 3 _|_ _| |_| _|_ _|_|
13 4 |_ _ _| _|_| _|_ _ _|_|
14 5 _|_ _ _| |_ _| _|_ _ _|_ _|
15 4 |_ _ _| |_| _|_ _ _|_|
16 4 |_ _ _| |_| |_ _ _|_|
...
Figure 1. Figure 2.
.
Figure 1 shows the illustration of initial terms taken from the isosceles triangle of A237593 (see link). For n = 16 there are (3 + 1) = 4 cells in the 16th row of the diagram, so a(16) = 4.
Figure 2 shows the illustration of initial terms taken from an octant of the pyramid described in A244050 and A245092 (see link). For n = 16 there are (3 + 1) = 4 cells in the 16th row of the diagram, so a(16) = 4.
Note that if we fold each level (or row) of that isosceles triangle of A237593 we essentially obtain the structure of the pyramid described in A245092 whose terraces at the n-th level have a total area equal to sigma(n) = A000203(n).
PROG
(PARI) row235791(n) = vector((sqrtint(8*n+1)-1)\2, i, 1+(n-(i*(i+1)/2))\i);
row237591(n) = {my(orow = concat(row235791(n), 0)); vector(#orow -1, i, orow[i] - orow[i+1]); }
a003056(n) = floor((sqrt(1+8*n)-1)/2);
a(n) = my(row=row237591(n)); sum(k=1, a003056(n), if (!(k%2), row[k])); \\ Michel Marcus, Dec 22 2020
KEYWORD
nonn
AUTHOR
Omar E. Pol, Dec 21 2020
STATUS
approved