OFFSET
1,2
COMMENTS
Properties of the sequence:
Of the first 1000 terms, 70.5% are powers of 2 (see the table below). We observe repeated terms a(n) = a(n+1) for n = 2, 4, 7, 10, 18, 19, 22, 26, 33, 34, 40, 44, 46, 49, 52, 55, ....
The following table lists statistics of a(n) for n=1..1000.
-------------------------------
| a(n) | frequency | % |
-------------------------------
| 1 | 1 | 0.1% |
| 2 | 61 | 6.1% |
| 3 | 9 | 0.9% |
| 4 | 235 | 23.5% |
| 5 | 2 | 0.2% |
| 6 | 72 | 7.2% |
| 7 | 1 | 0.1% |
| 8 | 266 | 26.6% |
| 9 | 12 | 1.2% |
| 10 | 6 | 0.6% |
| 12 | 116 | 11.6% |
| 14 | 1 | 0.1% |
| 16 | 130 | 13.0% |
| 18 | 10 | 1.0% |
| 20 | 11 | 1.1% |
| 24 | 45 | 4.5% |
| 27 | 1 | 0.1% |
| 32 | 12 | 1.2% |
| 36 | 5 | 0.5% |
| 40 | 1 | 0.1% |
| 48 | 2 | 0.2% |
| 54 | 1 | 0.1% |
-------------------------------
| TOTAL | 1000 | 100.0% |
-------------------------------
Based on the results in the table and the computing of Jon E. Schoenfield through n=3500, is it possible to determine an approximation of the probability p(a(n)= power of 2)?
Conjecture: the probability that a(n) is a power of 2 is such that 0.703 < p(a(n)=2^p) < 0.705.
The integers n such that a(n)=2 are 2, 3, 6, 9, 15, 21, 30, 36, 51, 54, 69, ... Is this A040040? - Michel Marcus, Jan 22 2015
LINKS
Michel Lagneau, Table of n, a(n) for n = 1..1000
EXAMPLE
a(7) = 4 because the 4 squares of the form x^2 + x + 7^2 are 49, 121, 289, 2401 for x = 0, 8, 15, 48, respectively.
a(8) = 4 because the 4 squares of the form x^2 + x + 8^2 are 64, 196, 484, 4096 for x = 0, 11, 20, 63, respectively.
MATHEMATICA
lst={}; Do[k=0; Do[If[IntegerQ[Sqrt[x^2+x+n^2]], k=k+1], {x, 0, n^2}]; AppendTo[lst, k], {n, 1, 100}]; lst
PROG
(PARI) a(n) = sum(x=0, n^2, issquare(x^2 + x + n^2)); \\ Michel Marcus, Jan 21 2015
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Jan 18 2015
STATUS
approved