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

-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, 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, -1, 9, 45, 120, 210, 252, 210, 120, 45, 10, 1, -1, 10, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1

Offset: 1

Author

Roger L. Bagula, Dec 10 2007

Keywords

• tabf
• uned
• sign

Comments

Proportionality function: 1) f(q*x)=f(x)/q

Modified Modular function to fit this form: 2) f((a*x+b)/(c*x+d))=f(x)/(c*x+d)^(2*n)

such that: q=(c*x+d)^(2*n)=(a*x+b)/(c*x+d)

gives the general polynomial: p(x,n)=(c*x + d)^(2*n + 1) - (a*x + b)=0

The integer form when: a = 1; c = 1; b = Integer; d = Integer; has a very general Pascal triangle like form.

The only problem is the polynomial here skips the linear level: a0*x+b0->{a0,b0}

But more generally stars at that level.

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.

Examples

			{-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, 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},
{-1, 9, 45, 120, 210, 252, 210, 120, 45, 10, 1},
{-1, 10, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1}
		

References

• Markku Halmetoja, et al.,"On Direct and Inverse Proportionality", American Math Monthly, volume 114, number 10, Dec 2007, pages 871-881.
• McKean and Moll, Elliptic Curves, Function Theory,Geometry, Arithmetic, Cambridge University Press, New York, 199, page 172

Crossrefs

Cf. A115346, A115514.

Programs

Mathematica

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[%]

Formula

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];