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A175631 a(n) = (n-th pentagonal number) modulo (n-th triangular number).

%I A175631 #13 Nov 19 2022 12:35:38
%S A175631 0,2,0,2,5,9,14,20,27,35,44,54,65,77,90,104,119,135,152,170,189,209,
%T A175631 230,252,275,299,324,350,377,405,434,464,495,527,560,594,629,665,702,
%U A175631 740,779,819,860,902,945,989,1034,1080,1127,1175,1224,1274,1325,1377,1430
%N A175631 a(n) = (n-th pentagonal number) modulo (n-th triangular number).
%H A175631 G. C. Greubel, <a href="/A175631/b175631.txt">Table of n, a(n) for n = 1..1000</a>
%H A175631 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A175631 For n>=3, a(n) = A000096(n-2).
%F A175631 From _Chai Wah Wu_, Oct 12 2018: (Start)
%F A175631 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 5.
%F A175631 G.f.: x^2*(2 - 6*x + 8*x^2 - 3*x^3)/(1 - x)^3. (End)
%F A175631 E.g.f.: (x/2)*(2 + 3*x - (2 - x)*exp(x)). - _G. C. Greubel_, Jan 30 2022
%e A175631 a(1)=0 because (1(3-1)/2) mod (1(1+1)/2) = 1 mod 1 = 0,
%e A175631 a(2)=2 because (2(6-1)/2) mod (2(2+1)/2) = 5 mod 3 = 2.
%t A175631 Table[Mod[n(3n-1)/2, n(n+1)/2],{n,100}]
%t A175631 Module[{nn=60},Mod[#[[1]],#[[2]]]&/@Thread[{PolygonalNumber[ 5,Range[ nn]],Accumulate[ Range[nn]]}]] (* _Harvey P. Dale_, Nov 19 2022 *)
%o A175631 (Magma) [n lt 4 select 1+(-1)^n else n*(n-3)/2: n in [1..60]]; // _G. C. Greubel_, Jan 30 2022
%o A175631 (Sage)
%o A175631 def A175631(n): return 1+(-1)^n if (n<4) else 9*binomial(n/3, 2)
%o A175631 [A175631(n) for n in (1..60)] # _G. C. Greubel_, Jan 30 2022
%Y A175631 Cf. A000096 (n(n+3)/2), A000217 (triangular numbers), A000326 (pentagonal numbers), A175630 (n-th pentagonal number mod (n+2)).
%Y A175631 Cf. A080956, A132315, A132316, A132317.
%K A175631 nonn
%O A175631 1,2
%A A175631 _Zak Seidov_, Jul 29 2010