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%I #26 Feb 27 2015 23:29:52
%S 1,2,2,3,3,2,4,4,2,4,4,3,6,4,2,4,8,4,4,4,2,6,6,3,6,4,4,8,4,2,6,12,4,4,
%T 4,2,6,12,4,5,5,4,8,4,4,8,8,4,6,6,2,8,8,2,4,4,4,12,12,6,6,8,4,4,4,4,
%U 16,8,2,4,8,8,12,6,2,6,12,4,4,8,4,8,8,3,9
%N a(n) is the number of squares of the form x^2 + x + n^2 for 0 <= x <= n^2.
%C Properties of the sequence:
%C 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, ....
%C The following table lists statistics of a(n) for n=1..1000.
%C -------------------------------
%C | a(n) | frequency | % |
%C -------------------------------
%C | 1 | 1 | 0.1% |
%C | 2 | 61 | 6.1% |
%C | 3 | 9 | 0.9% |
%C | 4 | 235 | 23.5% |
%C | 5 | 2 | 0.2% |
%C | 6 | 72 | 7.2% |
%C | 7 | 1 | 0.1% |
%C | 8 | 266 | 26.6% |
%C | 9 | 12 | 1.2% |
%C | 10 | 6 | 0.6% |
%C | 12 | 116 | 11.6% |
%C | 14 | 1 | 0.1% |
%C | 16 | 130 | 13.0% |
%C | 18 | 10 | 1.0% |
%C | 20 | 11 | 1.1% |
%C | 24 | 45 | 4.5% |
%C | 27 | 1 | 0.1% |
%C | 32 | 12 | 1.2% |
%C | 36 | 5 | 0.5% |
%C | 40 | 1 | 0.1% |
%C | 48 | 2 | 0.2% |
%C | 54 | 1 | 0.1% |
%C -------------------------------
%C | TOTAL | 1000 | 100.0% |
%C -------------------------------
%C 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)?
%C Conjecture: the probability that a(n) is a power of 2 is such that 0.703 < p(a(n)=2^p) < 0.705.
%C 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
%H Michel Lagneau, <a href="/A253900/b253900.txt">Table of n, a(n) for n = 1..1000</a>
%e 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.
%e 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.
%t 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
%o (PARI) a(n) = sum(x=0, n^2, issquare(x^2 + x + n^2)); \\ _Michel Marcus_, Jan 21 2015
%K nonn
%O 1,2
%A _Michel Lagneau_, Jan 18 2015
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