cp's OEIS Frontend

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.

Showing 1-2 of 2 results.

A193588 A Fibonacci triangle: T(n,k) = Fib(k+2) for 0 <= k <= n.

Original entry on oeis.org

1, 1, 2, 1, 2, 3, 1, 2, 3, 5, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 13, 1, 2, 3, 5, 8, 13, 21, 1, 2, 3, 5, 8, 13, 21, 34, 1, 2, 3, 5, 8, 13, 21, 34, 55, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233
Offset: 0

Views

Author

Clark Kimberling, Jul 31 2011

Keywords

Comments

n-th row sum: A001911, Fib(n+3)-2;
n-th alternating row sum: A000045, F(n).
The augmentation (as defined at A193091) of A193588 is A193589.

Examples

			First 5 rows of A193588:
  1;
  1, 2;
  1, 2, 3;
  1, 2, 3, 5;
  1, 2, 3, 5, 8;
		

Crossrefs

Cf. A193588.

Programs

  • Mathematica
    (See A193589, the augmentation of A193588.)
    Table[Fibonacci[k+2],{n,0,20},{k,0,n}]//Flatten (* Harvey P. Dale, Nov 29 2017 *)
    Module[{nn=15,fibs},fibs=Fibonacci[Range[2,nn]];Table[Take[fibs,n],{n,nn-1}]]// Flatten (* Harvey P. Dale, Mar 02 2023 *)

Formula

a(n) = A115346(n) + 1. - Filip Zaludek, Nov 19 2016

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

Original entry on oeis.org

-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

Views

Author

Roger L. Bagula, Dec 10 2007

Keywords

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

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];
Showing 1-2 of 2 results.