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.

A124437 Experience Points thresholds for levels in the pen and paper role-playing game "Das Schwarze Auge" (DSA, a.k.a. "The Dark Eye").

Original entry on oeis.org

0, 100, 300, 600, 1000, 1500, 2100, 2800, 3600, 4500, 5500, 6600, 7800, 9100, 10500, 12000, 13600, 15300, 17100, 19000, 21000, 23100, 25300, 27600, 30000, 32500, 35100, 37800, 40600, 43500, 46500, 49600, 52800, 56100, 59500, 63000, 66600
Offset: 1

Views

Author

Christoph D. Schmidt (snu(AT)power.ms), Dec 16 2006, Dec 21 2006

Keywords

Examples

			a(7)=2100 because a(6) + (7-1)*100 = (a(5) + 500) + 600 =(((((0+100) + 200) + 300) + 400) + 500) + 600.
		

References

  • "Das Schwarze Auge - Basisregelwerk" (basic rule book), Fantasy Productions Verlags- und Medienvertriebsgesellschaft mbH, Erkrath, Germany, 2005, ISBN 3890644406

Programs

  • C
    int folge(n){ if (n==1) return 0; return (folge(n-1)+(n-1)*100); }
    
  • Mathematica
    Table[50 n (n - 1), {n, 37}] (* or *)
    CoefficientList[Series[100 x^2/(1 - x)^3, {x, 0, 37}], x] (* Michael De Vlieger, Jul 11 2016 *)
    LinearRecurrence[{3,-3,1},{0,100,300},40] (* Harvey P. Dale, Jun 26 2017 *)
  • PARI
    a(n)=50*n*(n-1) \\ Charles R Greathouse IV, Jun 17 2017

Formula

a(n) = a(n-1) + (n-1)*100; a(1) = 0.
From Chai Wah Wu, Jul 11 2016: (Start)
a(n) = 50*n*(n-1).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3.
G.f.: 100*x^2/(1 - x)^3. (End)