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A103145
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a(n) = (1/6)*(n^3 + 21*n^2 + 74*n + 18).
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1
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3, 19, 43, 76, 119, 173, 239, 318, 411, 519, 643, 784, 943, 1121, 1319, 1538, 1779, 2043, 2331, 2644, 2983, 3349, 3743, 4166, 4619, 5103, 5619, 6168, 6751, 7369, 8023, 8714, 9443, 10211, 11019, 11868, 12759, 13693, 14671, 15694, 16763
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OFFSET
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0,1
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COMMENTS
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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.
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).
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LINKS
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FORMULA
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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).
G.f.: (3-2*x)*(1 + 3*x - 3*x^2)/(1-x)^4. - Colin Barker, Apr 30 2012
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MATHEMATICA
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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 *)
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PROG
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(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
(PARI) a(n) = (n^3+21*n^2+74*n+18)/6; \\ Altug Alkan, Sep 23 2018
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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