A193286 -id:A193286 - cp's OEIS frontend

A246076 Paradigm shift sequence for the (-2,5) production scheme with replacement.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 18, 20, 22, 24, 26, 28, 30, 33, 36, 40, 44, 48, 52, 56, 60, 66, 72, 80, 88, 96, 104, 112, 120, 132, 144, 160, 176, 192, 208, 224, 240, 264, 288, 320, 352, 384, 416, 448, 480, 528, 576, 640, 704, 768, 832, 896, 960, 1056, 1152, 1280, 1408, 1536, 1664, 1792

Offset: 1

Jonathan T. Rowell, Aug 13 2014

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=-2 steps), or implement the current bundled action (which requires q=5 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?"
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.
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.
The ideal number of implementations per bundle, as measured by the geometric growth rate (p+zq root of z), is z = 2.
All solutions will be of the form a(n) = (qm+r) * m^b * (m+1)^d.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,2).

Paradigm shift sequences with q=5: A103969, A246074, A246075, A246076, A246079, A246083, A246087, A246091, A246095, A246099, A246103.

Paradigm shift sequences with p<0: A103969, A246074, A246075, A246076, A246079, A029750, A246078, A029747, A246077, A029744, A029747, A131577.

Vec(x*(1 +2*x +3*x^2 +4*x^3 +5*x^4 +6*x^5 +7*x^6 +8*x^7 +7*x^8 +6*x^9 +5*x^10 +4*x^11 +3*x^12 +2*x^13 +x^14 +x^23) / (1 -2*x^8) + O(x^100)) \\ Colin Barker, Nov 18 2016

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) = 2*a(n-8) for all n >= 25.
G.f.: x*(1 +2*x +3*x^2 +4*x^3 +5*x^4 +6*x^5 +7*x^6 +8*x^7 +7*x^8 +6*x^9 +5*x^10 +4*x^11 +3*x^12 +2*x^13 +x^14 +x^23) / (1 -2*x^8). - Colin Barker, Nov 18 2016

A246078 Paradigm shift sequence for (-1,4) production scheme with replacement.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 16, 18, 20, 22, 24, 27, 30, 33, 36, 40, 44, 48, 54, 60, 66, 72, 81, 90, 99, 108, 120, 132, 144, 162, 180, 198, 216, 243, 270, 297, 324, 360, 396, 432, 486, 540, 594, 648, 729, 810, 891, 972, 1080, 1188, 1296, 1458, 1620, 1782, 1944, 2187, 2430, 2673, 2916, 3240, 3564, 3888, 4374

Offset: 1

Jonathan T. Rowell, Aug 13 2014

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=-1 steps), or implement the current bundled action (which requires q=4 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?"
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.
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.
The ideal number of implementations per bundle, as measured by the geometric growth rate (p+zq root of z), is z = 3.
All solutions will be of the form a(n) = (qm+r) * m^b * (m+1)^d.
For large n, the sequence is recursively defined.

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

Paradigm shift sequences with q=4: A029750, A103969, A246074, A246078, A246082, A246086, A246090, A246094, A246098, A246102.

Paradigm shift sequences with p<0: A103969, A246074, A246075, A246076, A246079, A029750, A246078, A029747, A246077, A029744, A029747, A131577.

CoefficientList[Series[x (1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4 + 6 x^5 + 7 x^6 + 8 x^7 + 9 x^8 + 10 x^9 + 11 x^10 + 9 x^11 + 7 x^12 + 5 x^13 + 4 x^14 + 3 x^15 + 2 x^16 + x^17 + x^23 + 2 x^24)/(1 - 3 x^11), {x, 0, 71}], x] (* Michael De Vlieger, Nov 18 2016 *)

Vec(x*(1 +2*x +3*x^2 +4*x^3 +5*x^4 +6*x^5 +7*x^6 +8*x^7 +9*x^8 +10*x^9 +11*x^10 +9*x^11 +7*x^12 +5*x^13 +4*x^14 +3*x^15 +2*x^16 +x^17 +x^23 +2*x^24) / (1 -3*x^11) + O(x^100)) \\ Colin Barker, Nov 18 2016

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-11) for all n >= 26.
G.f.: x*(1 +2*x +3*x^2 +4*x^3 +5*x^4 +6*x^5 +7*x^6 +8*x^7 +9*x^8 +10*x^9 +11*x^10 +9*x^11 +7*x^12 +5*x^13 +4*x^14 +3*x^15 +2*x^16 +x^17 +x^23 +2*x^24) / (1 -3*x^11). - Colin Barker, Nov 18 2016

A246091 Paradigm shift sequence for (2,5) production scheme with replacement.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 26, 28, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 64, 68, 72, 76, 80, 84, 90, 99, 108, 117, 126, 135, 144, 153, 162, 171, 180, 192, 204, 216, 228, 240, 256, 272, 297, 324, 351, 378, 405, 432, 459, 486, 513, 540, 576, 612, 648, 684, 720, 768, 816, 891, 972, 1053, 1134, 1215, 1296, 1377, 1458, 1539, 1620

Offset: 1

Jonathan T. Rowell, Aug 13 2014

nonn

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=2 steps), or implement the current bundled action (which requires q=5 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?"
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.
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.
The ideal number of implementations per bundle, as measured by the geometric growth rate (p+zq root of z), is z = 3.
All solutions will be of the form a(n) = (qm+r) * m^b * (m+1)^d.

Paradigm shift sequences with q=5: A103969, A246074, A246075, A246076, A246079, A246083, A246087, A246091, A246095, A246099, A246103.

Paradigm shift sequences with p=2: A193286, A246088, A246089, A246090, A246091.

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) ).

Recursive: a(n) = 3*a(n-17) for all n >= 62.

A246077 Paradigm shift sequence for (-1,-3) production scheme with replacement.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 16, 18, 21, 24, 28, 32, 36, 42, 48, 56, 64, 72, 84, 96, 112, 128, 144, 168, 192, 224, 256, 288, 336, 384, 448, 512, 576, 672, 768, 896, 1024, 1152, 1344, 1536, 1792, 2048, 2304, 2688, 3072, 3584, 4096, 4608, 5376, 6144, 7168, 8192, 9216

Offset: 1

Jonathan T. Rowell, Aug 13 2014

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=-1 steps), or implement the current bundled action (which requires q=3 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?"
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.
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.
The ideal number of implementations per bundle, as measured by the geometric growth rate (p+zq root of z), is z = 2.
All solutions will be of the form a(n) = (qm+r) * m^b * (m+1)^d.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,2).

Paradigm shifts with q=3: A029747, A029750, A246077, A246081, A246085, A246089, A246093, A246097, A246101.

Paradigm shifts with q<0: A103969, A246074, A246075, A246076, A246079, A029750, A246078, A029747, A246077, A029744, A029747, A131577.

Vec(x*(1 +x^2) * (1 +2*x +2*x^2 +2*x^3 +3*x^4 +2*x^5 +x^8-x^10 +x^12) / (1 -2*x^5) + O(x^100)) \\ Colin Barker, Nov 19 2016

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) = 2*a(n-5) for all n >= 16.
G.f.: x*(1 +x^2) * (1 +2*x +2*x^2 +2*x^3 +3*x^4 +2*x^5 +x^8-x^10 +x^12) / (1 -2*x^5). - Colin Barker, Nov 19 2016

A193455 Paradigm shift sequence with procedure length p=3.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 16, 20, 25, 30, 36, 42, 49, 64, 80, 100, 125, 150, 180, 216, 256, 320, 400, 500, 625, 750, 900, 1080, 1296, 1600, 2000, 2500, 3125, 3750, 4500, 5400, 6480, 8000, 10000, 12500, 15625, 18750, 22500, 27000, 32400, 40000, 50000

Offset: 1

Jonathan T. Rowell, Jul 26 2011

This sequence is the solution to the following problem: "Suppose you have the choice of using one of three production options: apply a simple action, bundle all existing actions (which requires p=3 steps), or apply the current bundled action. The first use of a novel bundle erases (or makes obsolete) all prior actions.  How many total actions (simple) can be applied in n time steps?"
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 Procedural Lengths p=1 and 2, respectively.
The optimal number of pastes per copy, as measured by the geometric growth rate (p+z root of z), is z = 5. [Non-integer maximum between 4 and 5.]
The function a(n) = maximum value of the product of the terms k_i, where Sum (k_i) = n+ 3 - 3*i_max
All solutions will be of the form a(n) = m^b * (m+1)^d

			For n=20, C = floor(26/8) = 3, R = (23 mod 3) = 2, m = floor (23-9/3) = floor(14/3)=4; therefore a(20) = 4^(3-2)*5^(2) = 4*5^2 = 100.
For n=25, the same general formula is used, but C=4 (instead of 3). R=28 mod 4 =0, m = floor(28-12/4)=4; therefore a(25) = 4^4 = 256.
For n=35, C = floor(41/8)=5, R = 1, b = max(0,2)=2, d=max(0,-2)=0; therefore a(35) = 4^2*5^(5-2)*6^0 = 2000.

Jonathan T. Rowell, Solution Sequences for the Keyboard Problem and its Generalizations, Journal of Integer Sequences, Vol. 18 (2015), Article 15.10.7.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,5).

Paradigm shift sequences: A000792 (p=0), A178715 (p=1), A193286 (p=2), A193455 (p=3), A193456 (p=4), A193457 (p=5).

def a(n):
    c=(n + 6)//8
    if n<25:
        if n<10: return n
        r=(n + 3)%c
        m=(n + 3 - 3*c)//c
        return m**(c - r)*(m + 1)**r
    elif n==25: return 256
    else:
        r=(n + 6)%8
        b=max(0, 3 - r)
        d=max(0, r - 3)
        return 4**b*5**(c - (b + d))*6**d
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jun 27 2017

a(n) =
a(25)   = 256  [C = 4 below]
a(1:24) = m^(C-R) * (m+1)^R
                where C = floor((n+6)/8) [min C=1],
                  R = n+3 mod C, m = floor((n+3-3*C)/C)
  a(n>=26) = 4^b * 5^(C-(b+d)) * 6^d
                  where C = floor((n+6)/8), R = n+6 mod 8,
                  b = max(0,3-R), and d = max(0, R-3)
Recursive:  a(n) = 5*a(n-8) for all n >= 34

A193456 Paradigm shift sequence with procedure length p=4.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 16, 20, 25, 30, 36, 42, 49, 56, 64, 80, 100, 125, 150, 180, 216, 252, 294, 343, 400, 500, 625, 750, 900, 1080, 1296, 1512, 1764, 2058, 2500, 3125, 3750, 4500, 5400, 6480, 7776, 9072, 10584, 12500, 15625, 18750, 22500, 27000

Offset: 1

Jonathan T. Rowell, Jul 26 2011

nonn

This sequence is the solution to the following problem: "Suppose you have the choice of using one of three production options: apply a simple action, bundle all existing actions (which requires p =4 steps), or apply the current bundled action. The first use of a novel bundle erases (or makes obsolete) all prior actions.  How many total actions (simple) can be applied in n time steps?"
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 Procedural Lengths p=1 and 2, respectively.
The optimal number of pastes per copy, as measured by the geometric growth rate (p+z root of z), is z = 6. [Non-integer maximum between 5 and 6.]
The function a(n) = maximum value of the product of the terms k_i, where Sum (k_i) = n+ 4 - 4*i_max
All solutions will be of the form a(n) = m^b * (m+1)^d

			For n = 18, a(18) uses the general formula given for n in [1:67], but uses C=2 (rather than C=3).  m = floor(22/2)-4 = 7; R = 22 mod 2 = 0; therefore a(18) = 7^(2-0)*8^0 = 49
For n=37, a(37) has: C = floor(39/10) +1 = 3+1=4.  m = floor(41/4)-4 = 10-4=6, R = 41 mod 4 = 1; therefore, a(37) = 6^(4-1)*7^(1) = 6^3 *7 = 1512.

Jonathan T. Rowell, Solution Sequences for the Keyboard Problem and its Generalizations, Journal of Integer Sequences, Vol. 18 (2015), Article 15.10.7.

Paradigm shift sequences: A000792, A178715, A193286, A193455, A193456, and A193457 for p=0,1,...,5.

a(n) =
      a(8:10) = 8; 9; 10     [C=1 below]
      a(18:20) = 49; 56; 88  [C=2 below]
      a(28:29) = 294; 343    [C=3 below]
      a(38:39) = 1764; 2058  [C=4 below]
      a(48) = 10584          [C=5 below]
      a(58) = 63504          [C=6 below]
       a(1:67) = m^(C-R) * (m+1)^R
                where C = floor((n+2)/10) +1 [min C=1]
                m = floor ((n+4)/C)-4, and R = n+4 mod C
      a(n>=68) = 5^b * 6^(C-b-d) * 7^d
                where C = floor((n+2)/10) +1
                R = n+2 mod 10
                b = max(0, 8-R); d = max(0, R-8)
Recursive: for n>=69, a(n)=6*a(n-10)

A193457 Paradigm shift sequence with procedure length p=5.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 16, 20, 25, 30, 36, 42, 49, 56, 64, 72, 81, 100, 125, 150, 180, 216, 252, 294, 343, 392, 448, 512, 625, 750, 900, 1080, 1296, 1512, 1764, 2058, 2401, 2744, 3136, 3750, 4500, 5400, 6480, 7776, 9072, 10584, 12348, 14406

Offset: 1

Jonathan T. Rowell, Jul 27 2011

nonn

This sequence is the solution to the following problem: "Suppose you have the choice of using one of three production options: apply a simple action, bundle all existing actions (which requires p=5 steps), or apply the current bundled action. The first use of a novel bundle erases (or makes obsolete) all prior actions.  How many total actions (simple) can be applied in n time steps?"
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 Procedural Lengths p=1 and 2, respectively.
The optimal number of pastes per copy, as measured by the geometric growth rate (p+z root of z), is z = 6. [Non-integer maximum between 6 and 7.]
The function a(n) = maximum value of the product of the terms k_i, where Sum (k_i) = n+ 5 - 5*i_max
All solutions will be of the form a(n) = m^b * (m+1)^d

			For n=30, C=floor(38/11)=3, m=floor(35/3)-5 = 11-5 = 6, and R= (35 mod 3) = 2; therefore a(30) = 6^(3-2)*7^2 = 6*7^2 =294.
For n=13, use the general formula with C=2 (rather than C=1), with R = (18 mod 2) = 0, m=floor(18/2)-5=9-5=4; therefore a(13)=4^2*5^0=16.
For n=80, C = floor(88/11)=8, R=(88 mod 11) = 0, b = max(0,3)=3, and d=max(0,-3)=0; therefore a(80) = 5^3*6^(8-3)*7^0 = 5^3*6^5 = 972000

Jonathan T. Rowell, Solution Sequences for the Keyboard Problem and its Generalizations, Journal of Integer Sequences, Vol. 18 (2015), Article 15.10.7.

A000792 (n>=1), A178715, A193286, A193455, A193456, and A193457 are paradigm shift sequences for p=0,1,...5 respectively.

a(n) =
      a(13) = 16              [C=2 below]
      a(24) = 100             [C=3 below]
      a(46) = 3750            [C=5 below]
      a(57) = 22500           [C=6 below]
      a(68) = 135000          [C=7 below]
      a(1:68) = m^(C-R) * (m+1)^R
                where C = floor((n+8)/11) [min C=1]
                m = floor ((n+5)/C)-5, and R = n+5 mod C
      a(n>=69) = 5^b * 6^(C-b-d) * 7^d
                where C = floor((n+8)/11)
                R = n+8 mod 11
                b = max(0, 3-R); d = max(0, R-3)
Recursive: for n>=80, a(n)=6*a(n-11)

A246082 Paradigm shift sequence for (0,4) production scheme with replacement.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 18, 20, 22, 24, 27, 30, 33, 36, 39, 42, 45, 48, 54, 60, 66, 72, 81, 90, 99, 108, 117, 126, 135, 144, 162, 180, 198, 216, 243, 270, 297, 324, 351, 378, 405, 432, 486, 540, 594, 648, 729, 810, 891, 972, 1053, 1134, 1215, 1296, 1458, 1620, 1782

Offset: 1

Jonathan T. Rowell, Aug 13 2014

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=4 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?"
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.
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.
The ideal number of implementations per bundle, as measured by the geometric growth rate (p+zq root of z), is z = 3.
All solutions will be of the form a(n) = (qm+r) * m^b * (m+1)^d.
For large n, the sequence is recursively defined.

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

Paradigm shift sequences with q=4: A029750, A103969, A246074, A246078, A246082, A246086, A246090, A246094, A246098, A246102.

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

Vec(x*(1+x)^2 * (1+x^2)^2 * (1+2*x^4+3*x^8+x^12) / (1-3*x^12) + O(x^100)) \\ Colin Barker, Nov 19 2016

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-12) for all n >= 20.
G.f.: x*(1+x)^2 * (1+x^2)^2 * (1+2*x^4+3*x^8+x^12) / (1-3*x^12). - Colin Barker, Nov 19 2016

A246086 Paradigm shift sequence for (1,4) production scheme with replacement.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 20, 22, 24, 27, 30, 33, 36, 39, 42, 45, 48, 52, 56, 60, 66, 72, 81, 90, 99, 108, 117, 126, 135, 144, 156, 168, 180, 198, 216, 243, 270, 297, 324, 351, 378, 405, 432, 468, 504, 540, 594, 648, 729, 810, 891, 972, 1053, 1134, 1215, 1296, 1404, 1512, 1620, 1782, 1944, 2187, 2430, 2673, 2916, 3159, 3402, 3645, 3888, 4212

Offset: 1

Jonathan T. Rowell, Aug 13 2014

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=1 steps), or implement the current bundled action (which requires q=4 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?"
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.
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.
The ideal number of implementations per bundle, as measured by the geometric growth rate (p+zq root of z), is z = 3.
All solutions will be of the form a(n) = (qm+r) * m^b * (m+1)^d.

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

Paradigm shift sequences with q=4: A029750, A103969, A246074, A246078, A246082, A246086, A246090, A246094, A246098, A246102.

Paradigm shift sequences with p=1: A178715, A246084, A246085, A246086, A246087.

Vec(x*(1 +2*x +3*x^2 +4*x^3 +5*x^4 +6*x^5 +7*x^6 +8*x^7 +9*x^8 +10*x^9 +11*x^10 +12*x^11 +13*x^12 +11*x^13 +9*x^14 +7*x^15 +5*x^16 +3*x^17 +2*x^18 +x^19 +x^29 +2*x^30) / (1-3*x^13) + O(x^100)) \\ Colin Barker, Nov 19 2016

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-13) for all n >= 32.
G.f.: x*(1 +2*x +3*x^2 +4*x^3 +5*x^4 +6*x^5 +7*x^6 +8*x^7 +9*x^8 +10*x^9 +11*x^10 +12*x^11 +13*x^12 +11*x^13 +9*x^14 +7*x^15 +5*x^16 +3*x^17 +2*x^18 +x^19 +x^29 +2*x^30) / (1-3*x^13). - Colin Barker, Nov 19 2016

A246094 Paradigm shift sequence for (3,4) production scheme with replacement.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 24, 27, 30, 33, 36, 39, 42, 45, 48, 52, 56, 60, 64, 68, 72, 76, 81, 90, 99, 108, 117, 126, 135, 144, 156, 168, 180, 192, 208, 224, 240, 256, 272, 297, 324, 351, 378, 405, 432, 468, 504, 540, 576, 624, 672, 720, 768, 832, 896, 972, 1053, 1134, 1215, 1296, 1404, 1512, 1620, 1728, 1872, 2016, 2160, 2304, 2496, 2688, 2916, 3159, 3402, 3645, 3888, 4212, 4536, 4860, 5184, 5616, 6048, 6480, 6912, 7488

Offset: 1

Jonathan T. Rowell, Aug 13 2014

nonn

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=3 steps), or implement the current bundled action (which requires q=4 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?"
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.
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.
The ideal number of implementations per bundle, as measured by the geometric growth rate (p+zq root of z), is z = 3.
All solutions will be of the form a(n) = (qm+r) * m^b * (m+1)^d.

Paradigm shift sequences with q=4: A029750, A103969, A246074, A246078, A246082, A246086, A246090, A246094, A246098, A246102.

Paradigm shift sequences with p=3: A193455, A246092, A246093, A246094, A246095.

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) ).

Recursive: a(n) = 3*a(n-15) for all n >= 72.

cp's OEIS Frontend

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A246091 Paradigm shift sequence for (2,5) production scheme with replacement.

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A193455 Paradigm shift sequence with procedure length p=3.

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A193457 Paradigm shift sequence with procedure length p=5.

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A246082 Paradigm shift sequence for (0,4) production scheme with replacement.

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A246086 Paradigm shift sequence for (1,4) production scheme with replacement.

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