This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A106272 #33 Sep 02 2024 08:39:50
%S A106272 1,-1,-1,-6,-15,-48,-147,-477,-1577,-5339,-18373,-64125,-226385,
%T A106272 -807025,-2900825,-10501870,-38258495,-140146660,-515897195,
%U A106272 -1907409850,-7080017615,-26373676870,-98562581255,-369433290520,-1388466728579,-5231379691972
%N A106272 Antidiagonal sums of number triangle A106270.
%C A106272 To prove _R. J. Mathar_'s conjecture, let A(x) be the g.f. of the current sequence. We note first that
%C A106272 Sum_{n >= 3} (n+1)*a(n)*x^n = (x*A(x))' + (-1 + 2*x + 3*x^2),
%C A106272 Sum_{n >= 3} 2*(1-2*n)*a(n-1)*x^n = 2*x*A(x) - 4*x*(x*A(x))' + (2*x - 6*x^2),
%C A106272 Sum_{n >= 3} -(n+1)*a(n-2)*x^n = -(x^3*A(x))' + 3*x^2, and
%C A106272 Sum_{n >= 3} 2*(2*n-1)*a(n-3)*x^n = 4*x*(x^3*A(x))' - 2*x^3*A(x).
%C A106272 Adding these equations (side by side), we get
%C A106272 Sum_{n >= 3} ((n+1)*a(n) + 2*(1-2*n)*a(n-1) - (n+1)*a(n-2) + 2*(2*n-1)*a(n-3))*x^n = 0,
%C A106272 which proves the conjecture. - _Petros Hadjicostas_, Jul 15 2019
%F A106272 G.f.: c(x)*sqrt(1 - 4*x)/(1 - x^2), where c(x) is the g.f. of A000108.
%F A106272 a(n) = Sum_{k = 0..floor(n/2)} 2*0^(n-2k) - C(n-2k).
%F A106272 Conjecture: (n+1)*a(n) + 2*(1-2*n)*a(n-1) - (n+1)*a(n-2) + 2*(2*n-1)*a(n-3) = 0. - _R. J. Mathar_, Nov 09 2012
%o A106272 (PARI) c(x) = (1-sqrt(1-4*x))/(2*x);
%o A106272 my(x='x+O('x^35)); Vec(c(x)*sqrt(1 - 4*x)/(1 - x^2)) \\ _Michel Marcus_, Jul 16 2019
%Y A106272 Cf. A000108, A106270.
%K A106272 easy,sign
%O A106272 0,4
%A A106272 _Paul Barry_, Apr 28 2005