OFFSET
1,1
COMMENTS
Bisection of A008438.
a(n) is also the total number of cells in the n-th branch of the third quadrant of the spiral formed by the parts of the symmetric representation of sigma(4n-1), see example. For the quadrants 1, 2, 4 see A112610, A239052, A193553. The spiral has been obtained according to the following way: A196020 --> A236104 --> A235791 --> A237591 --> A237593 --> A237270.
We can find the spiral (mentioned above) on the terraces of the pyramid described in A244050. - Omar E. Pol, Dec 06 2016
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000
FORMULA
a(n) = 4*A097723(n-1). - Joerg Arndt, Mar 09 2014
Sum_{k=1..n} a(k) = (Pi^2/4) * n^2 + O(n*log(n)). - Amiram Eldar, Dec 17 2022
EXAMPLE
Illustration of initial terms:
-----------------------------------------------------
. Branches of the spiral
. in the third quadrant n a(n)
-----------------------------------------------------
. _ _ _ _
. | | | | | | | |
. | | | | | | |_|_ _
. | | | | | | 2 |_ _| 1 4
. | | | | |_|_ 2
. | | | | 4 |_
. | | |_|_ _ |_ _ _ _
. | | 6 |_ |_ _ _ _| 2 8
. |_|_ _ _ |_ 4
. 8 | |_ _ |
. |_ | |_ _ _ _ _ _
. |_ |_ |_ _ _ _ _ _| 3 12
. 8 |_ _| 6
. |
. |_ _ _ _ _ _ _ _
. |_ _ _ _ _ _ _ _| 4 24
. 8
.
For n = 4 the sum of divisors of 4*n-1 is 1 + 3 + 5 + 15 = A000203(15) = 24. On the other hand the parts of the symmetric representation of sigma(15) are [8, 8, 8] and the sum of them is 8 + 8 + 8 = 24, equaling the sum of divisors of 15, so a(4) = 24.
MAPLE
MATHEMATICA
DivisorSigma[1, 4*Range[60]-1] (* Harvey P. Dale, Dec 06 2016 *)
Table[DivisorSigma[1, 4 n - 1], {n, 100}] (* Vincenzo Librandi, Dec 07 2016 *)
PROG
(Magma) [SumOfDivisors(4*n-1): n in [1..60]]; // Vincenzo Librandi, Dec 07 2016
(PARI) a(n) = sigma(4*n-1); \\ Michel Marcus, Dec 07 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Omar E. Pol, Mar 09 2014
STATUS
approved