A175631 a(n) = (n-th pentagonal number) modulo (n-th triangular number).
0, 2, 0, 2, 5, 9, 14, 20, 27, 35, 44, 54, 65, 77, 90, 104, 119, 135, 152, 170, 189, 209, 230, 252, 275, 299, 324, 350, 377, 405, 434, 464, 495, 527, 560, 594, 629, 665, 702, 740, 779, 819, 860, 902, 945, 989, 1034, 1080, 1127, 1175, 1224, 1274, 1325, 1377, 1430
Offset: 1
Keywords
Examples
a(1)=0 because (1(3-1)/2) mod (1(1+1)/2) = 1 mod 1 = 0, a(2)=2 because (2(6-1)/2) mod (2(2+1)/2) = 5 mod 3 = 2.
Links
- G. C. Greubel, Table of n, a(n) for n = 1..1000
- Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
Crossrefs
Programs
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Magma
[n lt 4 select 1+(-1)^n else n*(n-3)/2: n in [1..60]]; // G. C. Greubel, Jan 30 2022
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Mathematica
Table[Mod[n(3n-1)/2, n(n+1)/2],{n,100}] Module[{nn=60},Mod[#[[1]],#[[2]]]&/@Thread[{PolygonalNumber[ 5,Range[ nn]],Accumulate[ Range[nn]]}]] (* Harvey P. Dale, Nov 19 2022 *) -
Sage
def A175631(n): return 1+(-1)^n if (n<4) else 9*binomial(n/3, 2) [A175631(n) for n in (1..60)] # G. C. Greubel, Jan 30 2022
Formula
For n>=3, a(n) = A000096(n-2).
From Chai Wah Wu, Oct 12 2018: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 5.
G.f.: x^2*(2 - 6*x + 8*x^2 - 3*x^3)/(1 - x)^3. (End)
E.g.f.: (x/2)*(2 + 3*x - (2 - x)*exp(x)). - G. C. Greubel, Jan 30 2022