A103145 a(n) = (1/6)*(n^3 + 21*n^2 + 74*n + 18).

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

Offset: 0

Author

Creighton Dement, Mar 17 2005

Keywords

• easy
• nonn

Comments

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).

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Cf. A051936, A051937.

Programs

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

Mathematica

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 *)
Table[(n^3+21n^2+74n+18)/6,{n,0,50}] (* Harvey P. Dale, Jun 18 2024 *)

PARI

a(n) = (n^3+21*n^2+74*n+18)/6; \\ Altug Alkan, Sep 23 2018

Python

def A103145(n): return (n*(n*(n+21)+74)+18)//6 # Chai Wah Wu, Mar 07 2024

Formula

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

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Vincenzo Librandi, Jun 26 2012