A246080 Paradigm shift sequence for (0,2) production scheme with replacement.

1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 15, 18, 21, 24, 30, 36, 45, 54, 63, 72, 90, 108, 135, 162, 189, 216, 270, 324, 405, 486, 567, 648, 810, 972, 1215, 1458, 1701, 1944, 2430, 2916, 3645, 4374, 5103, 5832, 7290, 8748, 10935, 13122, 15309, 17496, 21870, 26244

Offset: 1

Author

Jonathan T. Rowell, Aug 13 2014

Keywords

• nonn
• easy

Comments

This sequence is the solution to the following problem: "Suppose you have the choice of using one of three production options: apply a simple incremental action, bundle existing output as an integrated product (which requires p=0 steps), or implement the current bundled action (which requires q=2 steps). The first use of a novel bundle erases (or makes obsolete) all prior actions. How large an output can be generated in n time steps?"

1. A production scheme with replacement R(p,q) eliminates existing output following a bundling action, while an additive scheme A(p,q) retains the output. The schemes correspond according to A(p,q)=R(p-q,q), with the replacement scheme serving as the default presentation.

2. This problem is structurally similar to the Copy and Paste Keyboard problem: Existing sequences (A178715 and A193286) should be regarded as Paradigm-Shift Sequences with production schemes R(1,1) and R(2,1) with replacement, respectively.

3. The ideal number of implementations per bundle, as measured by the geometric growth rate (p+zq root of z), is z = 3.

4. All solutions will be of the form a(n) = (qm+r) * m^b * (m+1)^d.

Links

• Colin Barker, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,3).

Crossrefs

Paradigm shift sequences with q=2: A029744, A029747, A246080, A246084, A246088, A246092, A246096, A246100.

Paradigm shift sequences with p=0: A000792, A246080, A246081, A246082, A246083.

Programs

PARI

Vec(x*(1+x)^2 * (1+2*x^2+3*x^4+x^6) / (1-3*x^6) + O(x^100)) \\ Colin Barker, Nov 19 2016

Formula

a(n) = (qd+r) * d^(C-R) * (d+1)^R, where r = (n-Cp) mod q, Q = floor( (R-Cp)/q ), R = Q mod (C+1), and d = floor (Q/(C+1) ).

a(n) = 3*a(n-6) for all n >= 10.

G.f.: x*(1+x)^2 * (1+2*x^2+3*x^4+x^6) / (1-3*x^6). - Colin Barker, Nov 19 2016