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%I #38 Jun 18 2024 15:28:42
%S 3,19,43,76,119,173,239,318,411,519,643,784,943,1121,1319,1538,1779,
%T 2043,2331,2644,2983,3349,3743,4166,4619,5103,5619,6168,6751,7369,
%U 8023,8714,9443,10211,11019,11868,12759,13693,14671,15694,16763
%N a(n) = (1/6)*(n^3 + 21*n^2 + 74*n + 18).
%C A floretion-generated sequence relating truncated triangle and pyramidal numbers. The following reasoning suggests that (a(n)) may not be the result of some "arbitrary" addition of these sequences--it may possess some geometric meaning of its own: The FAMP identity: "jesrightfor + jesleftfor = jesfor" holds and was used to find the relation a(n) = 2*A051936(n+4)_4 + A051937(n+4)_4 . In the above case, "jesfor" returns the truncated triangular numbers (times -1); "jesrightfor" returns the truncated pyramidal numbers; and (a(n)) is given by "jesleftfor" (times -1). All sequences result from a Force transform of the sequence c(n) = n + 5 (c was not chosen arbitrarily, for details see program code). Specifically, the sequence (a(n)) is the (ForType 1A) jesleftfor-transform of the sequence c(n) = n + 5 with respect to the floretion given in the program code.
%C Floretion Algebra Multiplication Program, FAMP Code: 1jesleftfor[A*B] with A = .25'i - .25i' - .25'ii' + .25'jj' + .25'kk' + .25'jk' + .25'kj' - .25e and B = + 'i + .5j' + .5k' + .5'ij' + .5'ik'; 1vesfor[A*B](n) = n + 5. ForType: 1A Alternative description: 1jesleftfor[A*B], ForType: 1A, LoopType: tes (first iteration after transforming the zero-sequence A000004).
%H Vincenzo Librandi, <a href="/A103145/b103145.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F a(n) = 2*A051936(n+4)_4 + A051937(n+4)_4 (for n = 0, 1, 2, 3) or a(m) = (1/6)*(m^3 + 9m^2 - 46m - 6) = 2*A051936(m) + A051937(m) (for m = 4, 5, 6).
%F G.f.: (3-2*x)*(1 + 3*x - 3*x^2)/(1-x)^4. - _Colin Barker_, Apr 30 2012
%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - _Vincenzo Librandi_, Jun 26 2012
%t CoefficientList[Series[(3-2*x)*(1+3*x-3*x^2)/(1-x)^4,{x,0,40}],x] (* or *) LinearRecurrence[ {4,-6,4,-1},{3,19,43,76},50] (* _Vincenzo Librandi_, Jun 26 2012 *)
%t Table[(n^3+21n^2+74n+18)/6,{n,0,50}] (* _Harvey P. Dale_, Jun 18 2024 *)
%o (Magma) I:=[3, 19, 43, 76]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..45]]; // _Vincenzo Librandi_, Jun 26 2012
%o (PARI) a(n) = (n^3+21*n^2+74*n+18)/6; \\ _Altug Alkan_, Sep 23 2018
%o (Python)
%o def A103145(n): return (n*(n*(n+21)+74)+18)//6 # _Chai Wah Wu_, Mar 07 2024
%Y Cf. A051936, A051937.
%K easy,nonn
%O 0,1
%A _Creighton Dement_, Mar 17 2005