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%I #31 Sep 08 2022 08:45:06
%S 3,4,7,14,25,46,85,156,287,528,971,1786,3285,6042,11113,20440,37595,
%T 69148,127183,233926,430257,791366,1455549,2677172,4924087,9056808,
%U 16658067,30638962,56353837,103650866,190643665,350648368,644942899
%N a(n) = Sum_{k=0..n} S(k), where S(n) are the tribonacci generalized numbers A001644.
%H Robert Israel, <a href="/A073728/b073728.txt">Table of n, a(n) for n = 0..3399</a>
%H Daniel Birmajer, Juan B. Gil, Michael D. Weiner, <a href="http://arxiv.org/abs/1505.06339">Linear recurrence sequences with indices in arithmetic progression and their sums</a>, arXiv:1505.06339 [math.NT], 2015.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,1).
%F a(n) = a(n-1) + a(n-2) + a(n-3), a(0)=3, a(1)=4, a(2)=7.
%F G.f.: (3+x)/(1-x-x^2-x^3).
%F a(n) = 3*T(n+1) + T(n), where T(n) are the tribonacci numbers A000073.
%F a(n) = (S(n+3) - S(n+1))/2, where S(n) = A001644(n). - _Michael D. Weiner_, Mar 27 2015
%p A:= gfun[rectoproc]({a(n)=a(n-1)+a(n-2)+a(n-3), a(0)=3, a(1)=4, a(2)=7},a(n),remember):
%p seq(A(n),n=0..100); # _Robert Israel_, Mar 26 2015
%t CoefficientList[Series[(3+x)/(1-x-x^2-x^3), {x, 0, 40}], x]
%o (Magma) I:=[3,4,7]; [n le 3 select I[n] else Self(n-1)+Self(n-2) +Self(n-3): n in [1..40]]; // _Vincenzo Librandi_, Mar 27 2015
%o (PARI) my(x='x+O('x^40)); Vec((3+x)/(1-x-x^2-x^3)) \\ _G. C. Greubel_, Apr 09 2019
%o (Sage) ((3+x)/(1-x-x^2-x^3)).series(x, 40).coefficients(x, sparse=False) # _G. C. Greubel_, Apr 09 2019
%Y Cf. A001644, A000073.
%K easy,nonn
%O 0,1
%A Mario Catalani (mario.catalani(AT)unito.it), Aug 06 2002
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