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A135503
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a(n) = n*(n^2 - 1)/2.
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9
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0, 0, 3, 12, 30, 60, 105, 168, 252, 360, 495, 660, 858, 1092, 1365, 1680, 2040, 2448, 2907, 3420, 3990, 4620, 5313, 6072, 6900, 7800, 8775, 9828, 10962, 12180, 13485, 14880, 16368, 17952, 19635, 21420, 23310, 25308, 27417, 29640, 31980, 34440
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OFFSET
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0,3
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COMMENTS
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Previous name was: Integer values of sqrt(b) solving sqrt(d) + sqrt(b) = sqrt(c) with d^2 + b = c.
Squaring the first equation and setting the result equal to the second, we need d + b + 2*sqrt(d*b) = d^2+b -> d + 2*sqrt(d*b) = d^2 -> d^2 - d = 2*sqrt(d*b)
-> d^2*(d-1)^2 = 4*d*b -> b = d*(d-1)^2/4 -> sqrt(b) = (d-1)*sqrt(d)/2. Setting d = (n+1)^2 yields sqrt(b) = A027480(n).
This is the case k = 2 for FLTR, Fermat's Last Theorem with rational exponents 1/k: Consider x + y = x + y. Then (x^k)^(1/k) + (y^k)^(1/k) = ((x+y)^k)^(1/k).
For k > 2, there are infinitely many solutions to d^(1/k) + b^(1/k) = c^(1/k). E.g., 8^(1/3) + 27^(1/3) = 125^(1/3) at k = 3. However, in conjunction with d^2 + b = c, I could not find any nontrivial solutions.
For n > 2, a(n) is the maximum value of the magic constant in a perimeter-magic n-gon of order n (see A342758). - Stefano Spezia, Mar 21 2021
a(n) is equal to the total number of P_3 edge-disjoint subgraphs of the complete graph on n vertices. - Samuel J. Bevins, May 09 2023
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LINKS
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FORMULA
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O.g.f.: 3*x^2/(-1+x)^4.
a(n) = n*(n^2 - 1)/2 = A007531(n+1)/2. (End)
G.f.: 3*x^2*G(0)/2, where G(k) = 1 + 1/(1 - x/(x + (k+1)/(k+4)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013
Sum_{n>=2} 1/a(n) = 1/2.
Sum_{n>=2} (-1)^n/a(n) = 4*log(2) - 5/2. (End)
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EXAMPLE
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For d = 9, b = 144, c = 225, 9^(1/2) + 144^(1/2) = 225^(1/2) and 9^2 + 144 = 225. So b^(1/2) = 12 is the 4th entry in the sequence.
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MATHEMATICA
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PROG
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(PARI) flt2(n, p) = { local(a, b); for(a=0, n, b = (a^3-a)/2; print1(b", ") ) }
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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