cp's OEIS Frontend

A135392 A triangular sequence from a general proportionality to modular function polynomial triangular function.

%I A135392 #7 Jan 24 2013 14:52:54
%S A135392 -1,-1,1,1,-1,2,3,1,-1,3,6,4,1,-1,4,10,10,5,1,-1,5,15,20,15,6,1,-1,6,
%T A135392 21,35,35,21,7,1,-1,7,28,56,70,56,28,8,1,-1,8,36,84,126,126,84,36,9,1,
%U A135392 -1,9,45,120,210,252,210,120,45,10,1,-1,10,55,165,330,462,462,330,165,55,11,1
%N A135392 A triangular sequence from a general proportionality to modular function polynomial triangular function.
%C A135392 Proportionality function: 1) f(q*x)=f(x)/q
%C A135392 Modified Modular function to fit this form: 2) f((a*x+b)/(c*x+d))=f(x)/(c*x+d)^(2*n)
%C A135392 such that: q=(c*x+d)^(2*n)=(a*x+b)/(c*x+d)
%C A135392 gives the general polynomial: p(x,n)=(c*x + d)^(2*n + 1) - (a*x + b)=0
%C A135392 The integer form when: a = 1; c = 1; b = Integer; d = Integer; has a very general Pascal triangle like form.
%C A135392 The only problem is the polynomial here skips the linear level: a0*x+b0->{a0,b0}
%C A135392 But more generally stars at that level.
%C A135392 Two quantities are said to be directly proportional if, by whatever positive real number p one of them is multiplied, the other changes by the same factor p. But is it necessary to go through all p's? What more is required than insisting, say, that if one of the quantities doubles, then the second doubles, too? Given a positive real number p different from 1 and a real-valued function f on positive real numbers satisfying f(px)=pf(x) for all x, we study conditions under which f(x)=f(1)x for all x. Similarly, we consider what we need to assume on a function f satisfying f(px)=f(x)/p for all x to ensure that f(x)=f(1)/x for all x.
%D A135392 Markku Halmetoja, et al.,"On Direct and Inverse Proportionality", American Math Monthly, volume 114, number 10, Dec 2007, pages 871-881.
%D A135392 McKean and Moll, Elliptic Curves, Function Theory,Geometry, Arithmetic, Cambridge University Press, New York, 199, page 172
%F A135392 p(x,n)=(c*x + d)^(2*n + 1) - (a*x + b); a = 1; c = 1; b = 2; d = 1; t(n,m)=CoefficientList[p(x,n),x];
%e A135392 {-1},
%e A135392 {-1, 1, 1},
%e A135392 {-1, 2, 3, 1},
%e A135392 {-1, 3, 6, 4, 1},
%e A135392 {-1, 4, 10, 10,5, 1},
%e A135392 {-1, 5, 15, 20, 15, 6, 1},
%e A135392 {-1, 6, 21, 35, 35, 21, 7, 1},
%e A135392 {-1, 7, 28, 56, 70, 56, 28, 8, 1},
%e A135392 {-1, 8, 36, 84, 126, 126, 84, 36, 9, 1},
%e A135392 {-1, 9, 45, 120, 210, 252, 210, 120, 45, 10, 1},
%e A135392 {-1, 10, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1}
%t A135392 f[x_, n_] = (c*x + d)^(2*n + 1) - (a*x + b) a = 1; c = 1; b = 2; d = 1; Table[CoefficientList[f[x, n], x], {n, 0, 5, 1/2}] Flatten[%]
%Y A135392 Cf. A115346, A115514.
%K A135392 tabf,uned,sign
%O A135392 1,6
%A A135392 _Roger L. Bagula_, Dec 10 2007