A179820 a(n) = n-th triangular number mod (n+2).
0, 1, 3, 1, 4, 1, 5, 1, 6, 1, 7, 1, 8, 1, 9, 1, 10, 1, 11, 1, 12, 1, 13, 1, 14, 1, 15, 1, 16, 1, 17, 1, 18, 1, 19, 1, 20, 1, 21, 1, 22, 1, 23, 1, 24, 1, 25, 1, 26, 1, 27, 1, 28, 1, 29, 1, 30, 1, 31, 1, 32, 1, 33, 1, 34, 1, 35, 1, 36, 1, 37, 1, 38, 1, 39, 1, 40, 1, 41, 1, 42, 1, 43, 1, 44, 1, 45
Offset: 0
Links
- Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).
Programs
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Mathematica
Table[Mod[n(n+1)/2,n+2],{n,0,200}] LinearRecurrence[{0,2,0,-1},{0,1,3,1,4},110] (* or *) Join[{0,1},Riffle[Range[3,50],1]] (* Harvey P. Dale, Apr 02 2024 *)
Formula
a(0)=0, afterwards if n is odd then a(n)=1 else a(n)=(n+4)/2
a(0)=0, afterwards a(n)=1 for odd n and n/2+2 for even n.
a(n)= +2*a(n-2) -a(n-4), n>4. a(n) = (6+n*((-1)^n+1)+2*(-1)^n)/4, n>0. G.f.: -x*(-1-3*x+x^2+2*x^3) / ( (x-1)^2*(1+x)^2 ). [From R. J. Mathar, Aug 03 2010]