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.

A110351 Least sum (n+1) + (n+2) + ... + (n+k) that is a multiple of the n-th triangular number, n(n+1)/2.

Original entry on oeis.org

2, 3, 30, 110, 30, 84, 1148, 1980, 585, 935, 462, 702, 2912, 105, 3120, 32504, 9027, 11457, 2660, 420, 2772, 26312, 16560, 19800, 44525, 52299, 17388, 20300, 10005, 11625, 460784, 61248, 11220, 2975, 24570, 43290, 220742, 29640, 4680, 188600
Offset: 1

Views

Author

Amarnath Murthy, Jul 21 2005

Keywords

Examples

			a(3) = 4 + 5 + 6 + 7 + 8 = 30 = 5*6, T(3) = 3*4/2 = 6.
		

Crossrefs

Programs

  • Maple
    a:=proc(n) local p,m: p:=proc(k) if type(sum(n+j,j=1..k)/(n*(n+1)/2),integer)=true then k else fi end: m:=[seq(p(k),k=1..1200)][1]: sum(n+j,j=1..m) end: seq(a(n),n=1..45); # Emeric Deutsch, Aug 08 2005
  • Mathematica
    Do[t = n*(n+1)/2; k = 1; While[Mod[k*n + k*(k+1)/2, t] > 0, k++ ]; Print[k*n + k*(k+1)/2], {n, 1, 50}] (* Ryan Propper, Aug 08 2005  *)

Extensions

More terms from Emeric Deutsch and Ryan Propper, Aug 08 2005

A110352 A110351(n)/T(n) where T(n) is the n-th triangular number.

Original entry on oeis.org

2, 1, 5, 11, 2, 4, 41, 55, 13, 17, 7, 9, 32, 1, 26, 239, 59, 67, 14, 2, 12, 104, 60, 66, 137, 149, 46, 50, 23, 25, 929, 116, 20, 5, 39, 65, 314, 40, 6, 230, 34, 36, 117, 4, 82, 494, 219, 229, 563, 54, 27, 146, 662, 97, 9, 49, 91, 797, 220, 228, 884, 11, 740, 614, 18, 122, 251
Offset: 1

Views

Author

Amarnath Murthy, Jul 21 2005

Keywords

Crossrefs

Programs

  • Maple
    A110351 := proc(n) local k ; for k from 1 to n*(n+1) do if (k*(2*n+k+1)) mod (n*(n+1)) =0 then RETURN(k*(n+(k+1)/2)) ; fi ; od ; RETURN(0) ; end: A110352 := proc(n) 2*A110351(n)/(n*(n+1)) ; end: for n from 1 to 100 do printf("%d, ",A110352(n)) ; od ; # R. J. Mathar, May 02 2007
  • Mathematica
    Do[t = n*(n + 1)/2; k = 1; While[Mod[k*n + k*(k + 1)/2, t] > 0, k++ ]; Print[(k*n + k*(k + 1)/2)/((n*(n + 1))/2)], {n, 1, 60}] (* Stefan Steinerberger, Feb 17 2006 *)

Extensions

More terms from Stefan Steinerberger, Feb 17 2006
More terms from R. J. Mathar, May 02 2007
Showing 1-2 of 2 results.