A246080 -id:A246080 - cp's OEIS frontend

A246083 Paradigm shift sequence for (0,5) production scheme with replacement.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 22, 24, 26, 28, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 66, 72, 78, 84, 90, 99, 108, 117, 126, 135, 144, 153, 162, 171, 180, 198, 216, 234, 252, 270, 297, 324, 351, 378, 405, 432, 459, 486, 513, 540, 594, 648, 702, 756, 810, 891, 972, 1053, 1134, 1215, 1296, 1377, 1458

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=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.
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 (0,0,0,0,0,0,0,0,0,0,0,0,0,0,3).

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

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

Vec(x*(1 +x +x^2 +x^3 +x^4)^2 * (1 +2*x^5 +3*x^10 +x^15) / (1 -3*x^15) + 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-15) for all n >= 25.
G.f.: x*(1 +x +x^2 +x^3 +x^4)^2 * (1 +2*x^5 +3*x^10 +x^15) / (1 -3*x^15). - Colin Barker, Nov 18 2016

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

A246081 Paradigm shift sequence for (0,3) production scheme with replacement.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 16, 18, 21, 24, 27, 30, 33, 36, 42, 48, 54, 63, 72, 81, 90, 99, 108, 126, 144, 162, 189, 216, 243, 270, 297, 324, 378, 432, 486, 567, 648, 729, 810, 891, 972, 1134, 1296, 1458, 1701, 1944, 2187, 2430, 2673, 2916, 3402, 3888, 4374, 5103, 5832, 6561, 7290, 8019, 8748

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=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 = 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 (0,0,0,0,0,0,0,0,3).

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

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

Vec(x*(1+x+x^2)^2 * (1-x+x^3) * (1+x+x^2+2*x^3+x^4+x^6) / (1-3*x^9) + 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-9) for all n >= 15.
G.f.: x*(1+x+x^2)^2 * (1-x+x^3) * (1+x+x^2+2*x^3+x^4+x^6) / (1-3*x^9). - Colin Barker, Nov 19 2016

A246084 Paradigm shift sequence for (1,2) production scheme with replacement.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 15, 18, 21, 24, 28, 32, 36, 45, 54, 63, 72, 84, 96, 112, 135, 162, 189, 216, 252, 288, 336, 405, 486, 567, 648, 756, 864, 1008, 1215, 1458, 1701, 1944, 2268, 2592, 3024, 3645, 4374, 5103, 5832, 6804, 7776, 9072, 10935, 13122, 15309, 17496, 20412, 23328, 27216, 32805, 39366, 45927

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=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?"
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 (0,0,0,0,0,0,3).

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

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 +5*x^7 +3*x^8 +x^9 +x^15 +2*x^16 +4*x^24) / (1 -3*x^7) + 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-7) 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 +5*x^7 +3*x^8 +x^9 +x^15 +2*x^16 +4*x^24) / (1 -3*x^7). - Colin Barker, Nov 19 2016

A246092 Paradigm shift sequence for (3,2) production scheme with replacement.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 18, 21, 24, 28, 32, 36, 40, 45, 50, 55, 63, 72, 84, 96, 112, 128, 144, 160, 180, 200, 225, 252, 288, 336, 384, 448, 512, 576, 640, 720, 800, 900, 1008, 1152, 1344, 1536, 1792, 2048, 2304, 2560, 2880, 3200, 3600, 4032, 4608, 5376, 6144, 7168, 8192, 9216, 10240, 11520, 12800, 14400, 16128, 18432, 21504, 24576, 28672

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=3 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?"
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 = 4.
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 (0,0,0,0,0,0,0,0,0,0,4).

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

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

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 +8*x^11 +5*x^12 +3*x^13 +2*x^14 +x^15 +x^21 +2*x^22 +3*x^23 +3*x^24 +5*x^34) / (1 -4*x^11) + 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) = 4*a(n-11) for all n >= 36.
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 +8*x^11 +5*x^12 +3*x^13 +2*x^14 +x^15 +x^21 +2*x^22 +3*x^23 +3*x^24 +5*x^34) / (1 -4*x^11). - Colin Barker, Nov 19 2016

A246096 Paradigm shift sequence for (4,2) production scheme with replacement.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 18, 21, 24, 28, 32, 36, 40, 45, 50, 55, 60, 66, 72, 84, 96, 112, 128, 144, 160, 180, 200, 225, 250, 275, 300, 336, 384, 448, 512, 576, 640, 720, 800, 900, 1000, 1125, 1250, 1375, 1536, 1792, 2048, 2304, 2560, 2880, 3200, 3600, 4000, 4500, 5000, 5625, 6250, 7168, 8192, 9216, 10240, 11520, 12800, 14400, 16000, 18000, 20000, 22500, 25000

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

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

Paradigm shift sequences with p=4: A193456, A246096, A246097, A246098, A246099.

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) = 4*a(n-12) for all n >= 67.

A246100 Paradigm shift sequence for (5,2) 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, 21, 24, 28, 32, 36, 40, 45, 50, 55, 60, 66, 72, 78, 84, 96, 112, 128, 144, 160, 180, 200, 225, 250, 275, 300, 330, 360, 396, 448, 512, 576, 640, 720, 800, 900, 1000, 1125, 1250, 1375, 1500, 1650, 1800, 2048, 2304, 2560, 2880, 3200, 3600, 4000, 4500, 5000, 5625, 6250, 6875, 7500, 8250, 9216, 10240, 11520, 12800, 14400, 16000, 18000, 20000, 22500, 25000, 28125, 31250, 34375, 37500, 41250, 46080, 51200, 57600, 64000, 72000, 80000, 90000, 100000, 112500, 125000, 140625, 156250

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=5 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?"
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 = 5.
All solutions will be of the form a(n) = (qm+r) * m^b * (m+1)^d.

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

Paradigm shift sequences with p=5: A193457, A246100, A246101, A246102, A246103.

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) = 5*a(n-15) for all n >= 75.

A246088 Paradigm shift sequence for (2,2) production scheme with replacement.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 15, 18, 21, 24, 28, 32, 36, 40, 45, 54, 63, 72, 84, 96, 112, 128, 144, 162, 189, 216, 252, 288, 336, 384, 448, 512, 576, 648, 756, 864, 1008, 1152, 1344, 1536, 1792, 2048, 2304, 2592, 3024, 3456, 4032, 4608, 5376, 6144, 7168, 8192, 9216, 10368, 12096, 13824, 16128, 18432, 21504, 24576, 28672, 32768, 36864, 41472

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=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?"
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 = 4.
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 (0,0,0,0,0,0,0,0,0,4).

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

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

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 +7*x^10 +4*x^11 +3*x^12 +2*x^13 +x^14 +x^20 +6*x^21 +3*x^22 +2*x^29 +9*x^30) / ((1 -2*x^5) * (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) = 4*a(n-10) 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 +7*x^10 +4*x^11 +3*x^12 +2*x^13 +x^14 +x^20 +6*x^21 +3*x^22 +2*x^29 +9*x^30) / ((1 -2*x^5) * (1 +2*x^5)). - Colin Barker, Nov 19 2016

cp's OEIS Frontend

A246083 Paradigm shift sequence for (0,5) production scheme with replacement.

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

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A246081 Paradigm shift sequence for (0,3) production scheme with replacement.

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A246084 Paradigm shift sequence for (1,2) production scheme with replacement.

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A246092 Paradigm shift sequence for (3,2) production scheme with replacement.

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A246096 Paradigm shift sequence for (4,2) production scheme with replacement.

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A246100 Paradigm shift sequence for (5,2) production scheme with replacement.

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A246088 Paradigm shift sequence for (2,2) production scheme with replacement.

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