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A320773
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Numbers (excluding squares) whose square root has a continued fraction with a period < 3.
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2
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2, 3, 5, 6, 8, 10, 11, 12, 15, 17, 18, 20, 24, 26, 27, 30, 35, 37, 38, 39, 40, 42, 48, 50, 51, 56, 63, 65, 66, 68, 72, 80, 82, 83, 84, 87, 90, 99, 101, 102, 104, 105, 110, 120, 122, 123, 132, 143, 145, 146, 147, 148, 150, 152, 156, 168, 170, 171, 182, 195, 197, 198, 200
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OFFSET
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1,1
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COMMENTS
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The Heron sequence of every number a(n) has the following relationship: numerator(h(k))^2 - a(n)*denominator(h(k))^2 = 1 for k > 1.
The Heron sequence of every number a(n) has the following relationship with the continued fraction f(s) convergent to sqrt(a(n)): h(k) = f(2^k-1).
Numbers k = m^2 + r with m > 0 and 0 < r <= 2m such that r is a divisor of 2m.
Continued fraction: k = [m; 2m/r, 2m, 2m/r, 2m, ...].
The number of terms that are between m^2 and (m+1)^2 is equal to the number of divisors of 2m, which is A099777(m).
Proof see link. The Maxima code below demonstrates the divisor property. Note that there is no divisor of 2m between m and 2m.
(End)
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LINKS
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EXAMPLE
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The continued fraction of sqrt(6) = 2 + 1/(2 + 1/(4 + 1/(2 + 1/(4 + 1/(2 + 1/(4 + ...)))))) = [2; 2, 4, 2, 4, 2, 4, ...] has repeating portion (2, 4) with period 2, so 6 is a term.
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MAPLE
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Digits:=40: nr:=0:
for z from 2 to 200 do
test:=true: c:=sqrt(z):
if (c=floor(c)) the test:=false: end if:
while (test=true) do
b[0]:=floor(c):
r[0]:=c:
for k from 1 to 2 do
r[k]:=evalf(1/(r[k-1]-b[k-1])):
b[k]:=floor(r[k]):
end do:
if (b[1]=2*b[0]) or (b[2]=2*b[0]) then nr:=nr+1: a[nr]:=z: printf("%4d", z): end if:
test:=false:
end do:
end do:
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MATHEMATICA
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Select[Range[200], !IntegerQ[Sqrt[#]] && Length@ContinuedFraction[Sqrt[#]][[-1]]<3 &] (* Amiram Eldar, Nov 01 2018 *)
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PROG
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(Maxima)
block([n: 2, m: 0, r: 0, k: 0, kmax: 10, v: ""],
while k<kmax do
(m: floor(sqrt(n)), r: n-m^2,
if mod(2*m, r)=0 then (k: k+1, print(n)),
if r= m then n: n+m else (if r= 2*m then n: n+2 else n: n+1)));
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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