A029747 -id:A029747 - 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

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

A364499 a(n) = A005940(n) - n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 2, 0, -2, 0, 4, 0, 12, 4, 12, 0, -6, -4, 2, 0, 14, 8, 22, 0, 24, 24, 48, 8, 96, 24, 50, 0, -20, -12, -2, -8, 18, 4, 24, 0, 36, 28, 62, 16, 130, 44, 88, 0, 72, 48, 96, 48, 192, 96, 170, 16, 286, 192, 316, 48, 564, 100, 180, 0, -48, -40, -28, -24, -4, -4, 28, -16, 18, 36, 90, 8, 198, 48, 110, 0, 62

Offset: 1

Antti Karttunen, Jul 28 2023

Compare to the scatter plot of A364563.
From Antti Karttunen, Aug 11 2023: (Start)
Can be computed as a certain kind of bitmask transformation of A364568 (analogous to the inverse Möbius transform that is appropriate for A156552-encoding of n).
See A364572, A364573 (and also A364576) for n (apart from those in A029747) where a(n) comes relatively close to the X-axis.
(End)

			A005940(528577) = 528581, therefore a(528577) = 528581 - 528577 = 4. (See A364576).
A005940(2109697) = 2109629, therefore a(2109697) = 2109629 - 2109697 = -68.

Antti Karttunen, Table of n, a(n) for n = 1..16384
Index entries for sequences related to binary expansion of n

Cf. A005940, A364500 [= gcd(n,a(n))], A364559, A364572, A364573, A364576.
Cf. A029747 (known positions of 0's), A364540 (positions of terms < 0), A364541 (of terms <= 0), A364542 (of terms >= 0), A364563 [= -a(A364543(n))].
Cf. also A364258, A364568.

nn = 81; Array[Set[a[#], #] &, 2]; Do[If[EvenQ[n], Set[a[n], 2 a[n/2]], Set[a[n], Times @@ Power @@@ Map[{Prime[PrimePi[#1] + 1], #2} & @@ # &, FactorInteger[a[(n + 1)/2]]]]], {n, 3, nn}]; Array[a[#] - # &, nn] (* Michael De Vlieger, Jul 28 2023 *)

A005940(n) = { my(p=2, t=1); n--; until(!n\=2, if((n%2), (t*=p), p=nextprime(p+1))); t };
A364499(n) = (A005940(n)-n);

A364499(n) = { my(m=1,p=2,x=0,z=1); n--; while(n, if(!(n%2), p=nextprime(1+p), x += m; z *= p); n>>=1; m <<=1); (z-x)-1; }; \\ Antti Karttunen, Aug 06 2023

from math import prod
from itertools import accumulate
from collections import Counter
from sympy import prime
def A364499(n): return prod(prime(len(a)+1)**b for a, b in Counter(accumulate(bin(n-1)[2:].split('1')[:0:-1])).items())-n # Chai Wah Wu, Aug 07 2023

a(n) = -A364559(A005940(n)).
For all n >= 1, a(2*n) = 2*a(n).
For all n >= 1, a(A029747(n)) = 0.

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

Original entry on oeis.org

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

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=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.

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).

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.

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

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

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

A246085 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, 11, 12, 13, 14, 16, 18, 21, 24, 27, 30, 33, 36, 40, 44, 48, 54, 63, 72, 81, 90, 99, 108, 120, 132, 144, 162, 189, 216, 243, 270, 297, 324, 360, 396, 432, 486, 567, 648, 729, 810, 891, 972, 1080, 1188, 1296, 1458, 1701, 1944, 2187, 2430, 2673, 2916, 3240, 3564, 3888, 4374, 5103, 5832, 6561, 7290, 8019, 8748, 9720, 10692

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

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

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 +8*x^10 +6*x^11 +4*x^12 +2*x^13 +x^14 +x^22 +2*x^23) / (1 -3*x^10) + O(x^100)) \\ Colin Barker, Nov 22 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-10) 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 +9*x^8 +10*x^9 +8*x^10 +6*x^11 +4*x^12 +2*x^13 +x^14 +x^22 +2*x^23) / (1 -3*x^10). - Colin Barker, Nov 22 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

A246093 Paradigm shift sequence for (3,3) 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, 21, 24, 27, 30, 33, 36, 40, 44, 48, 52, 56, 60, 65, 72, 81, 90, 99, 108, 120, 132, 144, 160, 176, 192, 208, 224, 243, 270, 297, 324, 360, 396, 432, 480, 528, 576, 640, 704, 768, 832, 896, 972, 1080, 1188, 1296, 1440, 1584, 1728, 1920, 2112, 2304, 2560, 2816, 3072, 3328, 3584, 3888, 4320, 4752, 5184

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

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

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) = 4*a(n-15) for all n >= 48.

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

PARI

Formula

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

PARI

Formula

A364499 a(n) = A005940(n) - n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

PARI

Python

Formula

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

PARI

Formula

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

PARI

Formula

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

PARI

Formula

A246085 Paradigm shift sequence for (1,3) production scheme with replacement.

Views

Author

Keywords