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a(n) = Sum_{1 <= x_1, x_2, x_3 <= n} ( gcd(x_1, n)/gcd(x_1, x_2, x_3, n) )^2.
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1, 17, 91, 289, 701, 1547, 2647, 4769, 7705, 11917, 15731, 26299, 30421, 44999, 63791, 77473, 87857, 130985, 136459, 202589, 240877, 267427, 290951, 433979, 448201, 517157, 633187, 764983, 729989, 1084447, 951391, 1248929, 1431521, 1493569, 1855547, 2226745
FORMULA
a(n) = Sum_{d|n} phi(n/d) * (n/d)^2 * sigma_4(d^2)/sigma_2(d^2).
Multiplicative with a(p^e) = (p^(4*e+1)*(p+1)*(p^2+p+1) - p^(3*e+1)*(p^2+1) + p + 1)/((p^2+1)*(p^2+p+1)).
Dirichlet g.f.: zeta(s)*zeta(s-3)*zeta(s-4)/zeta(s-2)^2.
Sum_{k=1..n} a(k) ~ c * n^5 / 5, where c = zeta(2)*zeta(5)/zeta(3)^2 = 1.180448217... . (End)
MATHEMATICA
f[p_, e_] := (p^(4*e+1)*(p+1)*(p^2+p+1) - p^(3*e+1)*(p^2+1) + p + 1)/((p^2+1)*(p^2+p+1)); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 36] (* Amiram Eldar, May 24 2024 *)
PROG
(PARI) a(n) = sumdiv(n, d, eulerphi(n/d)*(n/d)^2*sigma(d^2, 4)/sigma(d^2, 2));
The y-coordinate of the point where x + y = n, x is an integer and x/y is as close as possible to phi (by absolute difference).
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2
1, 1, 1, 2, 2, 2, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 7, 7, 7, 8, 8, 8, 9, 9, 10, 10, 10, 11, 11, 11, 12, 12, 13, 13, 13, 14, 14, 15, 15, 15, 16, 16, 16, 17, 17, 18, 18, 18, 19, 19, 19, 20, 20, 21, 21, 21, 22, 22, 23, 23, 23, 24, 24, 24, 25, 25, 26, 26, 26, 27, 27, 28
COMMENTS
a(n) = k = n - either ceiling or floor of n/phi, according to which minimizes abs((n-k)/k - phi).
Each term is equal to or one greater than the previous term.
The average run length approaches 1+phi.
The 4 following statements are equivalent for any positive integer n and any function f(x) such that for any real x, f(x) equals a integer within the range (x-1,x+1):
a(n) != n-f(n/phi) xor A371627(n) != n-f(n/phi); Let s(n) = (phi*n - 1 - sqrt(1+(n^2)*(phi^-4)))/2.
Floor(s(n)) equals the number of times that a(n) swapped from being equal to n-floor(n/phi) to being equal to n-ceiling(n/phi) when n is extended to the reals.
This is true because s(n) is the solution to the equation n = (phi/4) * (phi(2r+1) + sqrt((2r+1)^2 / phi^4 + 4/phi)) solved for w. The equation gives the n-value of w-th swap from a(n) = n-floor(n/phi) to a(n) = n-ceiling(n/phi).
s(n) is asymptotic to n/phi - 1/2.
Floor(s(n)) != floor(n/phi - 1/2) <-> a(n) != round(n).
Floor(n/phi) equals the number of that a(n) swapped from being equal to n-ceiling(n/phi) to being equal to n-floor(n/phi) when n is extended to the reals.
FORMULA
Let s(n) = (phi*n - 1 - sqrt(1+(n^2) / phi^4))/2.
Floor(s(n))+floor(n/phi) is even -> a(n) = n-ceiling(n/phi) = (n mod 1) + floor(n/phi^2).
Floor(s(n))+floor(n/phi) is odd -> a(n) = n-floor(n/phi) = (n mod 1) + ceiling(n/phi^2).
a(n) = -a(-n).
EXAMPLE
For n=5, the possibilities are (0,5), (1,4), (2,3), (3,2), & (4,1). Of those, 3/2 is the closest to phi, so a(5)=3.
The x-coordinate of the point where x + y = n, x is an integer and x/y is as close as possible to 1/phi.
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2
0, 1, 1, 1, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 8, 8, 8, 9, 9, 10, 10, 10, 11, 11, 11, 12, 12, 13, 13, 13, 14, 14, 15, 15, 15, 16, 16, 16, 17, 17, 18, 18, 18, 19, 19, 19, 20, 20, 21, 21, 21, 22, 22, 23, 23, 23, 24, 24, 24, 25, 25, 26, 26, 26, 27, 27, 27
COMMENTS
Each term is equal to or one greater than the previous term.
The average run length approaches 1+phi.
a(n) = x = either ceiling or floor of n/phi^2, according as which minimizes abs(x/(n-x) - phi).
The 4 following statements are equivalent for any positive integer n and any function f(x) such that for all real x, x-1<f(x)<x+1 and f(x) is integral:
a(n) != n-f(n/phi) xor A371626(n) != n-f(n/phi);
EXAMPLE
For n=4, the possibilities are (0,4), (1,3), (2,2), and (3,1). 1/3 is the closest to 1/phi out of them so a(4)=1
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