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

1,6

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.

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

Links

Table of n, a(n) for n=1..76.

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];

Example

{-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}

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[%]

Crossrefs

Cf. A115346, A115514.

Sequence in context: A265848 A139438 A377842 * A071947 A139343 A059247

Adjacent sequences: A135389 A135390 A135391 * A135393 A135394 A135395

Keyword

tabf,uned,sign

Author

Roger L. Bagula, Dec 10 2007

Status

approved