cp's OEIS Frontend

A135503 a(n) = n*(n^2 - 1)/2.

%I A135503 #71 Mar 19 2025 06:02:50
%S A135503 0,0,3,12,30,60,105,168,252,360,495,660,858,1092,1365,1680,2040,2448,
%T A135503 2907,3420,3990,4620,5313,6072,6900,7800,8775,9828,10962,12180,13485,
%U A135503 14880,16368,17952,19635,21420,23310,25308,27417,29640,31980,34440
%N A135503 a(n) = n*(n^2 - 1)/2.
%C A135503 Previous name was: Integer values of sqrt(b) solving sqrt(d) + sqrt(b) = sqrt(c) with d^2 + b = c.
%C A135503 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)
%C A135503 -> 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).
%C A135503 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).
%C A135503 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.
%C A135503 A shifted version of A027480. - _R. J. Mathar_, Apr 07 2009
%C A135503 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
%C A135503 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
%H A135503 G. C. Greubel, <a href="/A135503/b135503.txt">Table of n, a(n) for n = 0..1000</a>
%H A135503 C. D. Bennet, A. M. W. Glass and G. J. Székely, <a href="http://www.jstor.org/stable/4145241">Fermat's Last Theorem for Rational Exponents</a>, Am. Math. Monthly 111 (2004), 322-329. - _R. J. Mathar_, Apr 21 2009
%H A135503 Terrel Trotter, <a href="https://web.archive.org/web/20070106085340/http://www.trottermath.net/simpleops/pmp.html">Perimeter-Magic Polygons</a>, Journal of Recreational Mathematics Vol. 7, No. 1, 1974, pp. 14-20.
%F A135503 a(n) = 3*A000292(n-1).
%F A135503 From _R. J. Mathar_ Feb 20 2008: (Start)
%F A135503 O.g.f.: 3*x^2/(-1+x)^4.
%F A135503 a(n) = n*(n^2 - 1)/2 = A007531(n+1)/2. (End)
%F A135503 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
%F A135503 a(n) = A006003(n+1) - A000326(n+1). - _J. M. Bergot_, Dec 04 2014
%F A135503 E.g.f.: (1/2)* x^2 *(3 + x)*exp(x). - _G. C. Greubel_, Oct 15 2016
%F A135503 From _Miquel Cerda_, Dec 25 2016: (Start)
%F A135503 a(n) = A000578(n) - A006003(n).
%F A135503 a(n) = A004188(n) - A000578(n).
%F A135503 a(n) = A007588(n) - A004188(n). (End)
%F A135503 a(n) = A002411(n) - A000217(n). - _Justin Gaetano_, Feb 20 2018
%F A135503 From _Amiram Eldar_, Jan 09 2021: (Start)
%F A135503 Sum_{n>=2} 1/a(n) = 1/2.
%F A135503 Sum_{n>=2} (-1)^n/a(n) = 4*log(2) - 5/2. (End)
%e A135503 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.
%t A135503 Array[# (#^2 - 1)/2 &, 42, 0] (* _Michael De Vlieger_, Feb 20 2018 *)
%o A135503 (PARI) flt2(n,p) = { local(a,b); for(a=0,n, b = (a^3-a)/2; print1(b", ") ) }
%Y A135503 Cf. A000217, A000578, A002411, A004188, A006003, A007531, A007588, A027480, A342758.
%K A135503 nonn,easy
%O A135503 0,3
%A A135503 _Cino Hilliard_, Feb 09 2008
%E A135503 Edited by _R. J. Mathar_, Apr 21 2009
%E A135503 New name using _R. J. Mathar_'s formula, _Joerg Arndt_, Dec 05 2014