A029744 -id:A029744 - 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

A246352 Numbers n such that A048673(n) >= n.

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 25, 26, 27, 28, 30, 32, 33, 35, 36, 39, 40, 42, 44, 45, 48, 49, 50, 52, 54, 56, 57, 60, 63, 64, 66, 68, 69, 70, 72, 75, 76, 78, 80, 81, 84, 88, 90, 91, 92, 93, 96, 98, 99, 100, 102, 104, 105, 108, 110, 112, 114, 116, 117, 120, 124, 125, 126, 128, 130, 132

Offset: 1

Antti Karttunen, Aug 24 2014

nonn

Antti Karttunen, Table of n, a(n) for n = 1..10000

Complement: A246351
Union of A246282 and A048674.
Subsequence: A029744 (gives the positions of records in A048673).
Cf. A246372.

default(primelimit, 2^22);
A003961(n) = my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); \\ From Michel Marcus
A048673(n) = (A003961(n)+1)/2;
isA246352(n) = (A048673(n) >= n);
n = 0; i = 0; while(i < 10000, n++; if(isA246352(n), i++; write("b246352.txt", i, " ", n)));
(Scheme, with Antti Karttunen's IntSeq-library)
(define A246352 (MATCHING-POS 1 1 (lambda (n) (>= (A048673 n) n))))

A048985 Working in base 2, replace n with the concatenation of its prime divisors in increasing order (write answer in base 10).

Original entry on oeis.org

1, 2, 3, 10, 5, 11, 7, 42, 15, 21, 11, 43, 13, 23, 29, 170, 17, 47, 19, 85, 31, 43, 23, 171, 45, 45, 63, 87, 29, 93, 31, 682, 59, 81, 47, 175, 37, 83, 61, 341, 41, 95, 43, 171, 125, 87, 47, 683, 63, 173, 113, 173, 53, 191, 91, 343, 115, 93, 59, 349, 61, 95, 127, 2730

Offset: 1

N. J. A. Sloane

			15 = 3*5 -> 11.101 -> 11101 = 29, so a(15) = 29.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Patrick De Geest, Home Primes

Cf. A037276, A048986, A064841.
Cf. A193652, A029744 (record values and where they occur).
Cf. A027746.

-- import Data.List (unfoldr)
a048985 = foldr (\d v -> 2 * v + d) 0 . concatMap
   (unfoldr (\x -> if x == 0 then Nothing else Just $ swap $ divMod x 2))
   . reverse . a027746_row
-- Reinhard Zumkeller, Jul 16 2012

f[n_] := FromDigits[ Flatten[ IntegerDigits[ Flatten[ Table[ #1, {#2}] & @@@ FactorInteger@n], 2]], 2]; Array[f, 64] (* Robert G. Wilson v, Jun 02 2010 *)

from sympy import factorint
def a(n):
    if n == 1: return 1
    return int("".join(bin(p)[2:]*e for p, e in factorint(n).items()), 2)
print([a(n) for n in range(1, 65)]) # Michael S. Branicky, Oct 07 2022

More terms from Sam Alexander (pink2001x(AT)hotmail.com) and Michel ten Voorde

A246360 a(1) = 1, then A007051 ((3^n)+1)/2 interleaved with A057198 (5*3^(n-1)+1)/2.

Original entry on oeis.org

1, 2, 3, 5, 8, 14, 23, 41, 68, 122, 203, 365, 608, 1094, 1823, 3281, 5468, 9842, 16403, 29525, 49208, 88574, 147623, 265721, 442868, 797162, 1328603, 2391485, 3985808, 7174454, 11957423, 21523361, 35872268, 64570082, 107616803, 193710245, 322850408, 581130734

Offset: 1

Antti Karttunen, Aug 24 2014

Also record values in A048673.

Antti Karttunen, Table of n, a(n) for n = 1..64
Index entries for linear recurrences with constant coefficients, signature (1,3,-3).

Even bisection: A007051 from A007051(1) onward: [2, 5, 14, 41, ...]
Odd bisection: 1 followed by A057198.
A029744 gives the corresponding record positions in A048673.
A247284 gives the maximum values of A048673 between these records and A247283 gives the positions where they occur.
Subsequence of A246361.
Cf. A000244, A193652, A246347.

LinearRecurrence[{1, 3, -3}, {1, 2, 3, 5}, 40] (* Hugo Pfoertner, Sep 27 2022 *)

def A246360(n): return 1 if n==1 else (3+((n&1)<<1))*3**((n>>1)-1)+1>>1 # Chai Wah Wu, Sep 02 2025

(define (A246360 n) (cond ((<= n 1) n) ((even? n) (/ (+ 1 (A000244 (/ n 2))) 2)) (else (/ (+ 1 (* 5 (A000244 (/ (- n 3) 2)))) 2))))

a(1) = 1, a(2n) = (3^n+1)/2, a(2n+1) = (5 * 3^(n-1)+1)/2.
a(n) = A048673(A029744(n)).
a(n) = A087503(n-3) + 2 for n >= 3. - Peter Kagey, Nov 30 2019
G.f.: x -x^2*(-2-x+4*x^2) / ( (x-1)*(3*x^2-1) ). - R. J. Mathar, Sep 23 2014

A063759 Spherical growth series for modular group.

Original entry on oeis.org

1, 3, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96, 128, 192, 256, 384, 512, 768, 1024, 1536, 2048, 3072, 4096, 6144, 8192, 12288, 16384, 24576, 32768, 49152, 65536, 98304, 131072, 196608, 262144, 393216, 524288, 786432, 1048576, 1572864, 2097152

Offset: 0

N. J. A. Sloane, Aug 14 2001

Also number of sequences S of length n with entries in {1,..,q} where q = 3, satisfying the condition that adjacent terms differ in absolute value by exactly 1, see examples. - W. Edwin Clark, Oct 17 2008

			For n = 2 the a(2) = 4 sequences are (1,2),(2,1),(2,3),(3,2). - _W. Edwin Clark_, Oct 17 2008
From _Joerg Arndt_, Nov 23 2012: (Start)
There are a(6) = 16 such words of length 6:
[ 1]   [ 1 2 1 2 1 2 ]
[ 2]   [ 1 2 1 2 3 2 ]
[ 3]   [ 1 2 3 2 1 2 ]
[ 4]   [ 1 2 3 2 3 2 ]
[ 5]   [ 2 1 2 1 2 1 ]
[ 6]   [ 2 1 2 1 2 3 ]
[ 7]   [ 2 1 2 3 2 1 ]
[ 8]   [ 2 1 2 3 2 3 ]
[ 9]   [ 2 3 2 1 2 1 ]
[10]   [ 2 3 2 1 2 3 ]
[11]   [ 2 3 2 3 2 1 ]
[12]   [ 2 3 2 3 2 3 ]
[13]   [ 3 2 1 2 1 2 ]
[14]   [ 3 2 1 2 3 2 ]
[15]   [ 3 2 3 2 1 2 ]
[16]   [ 3 2 3 2 3 2 ]
(End)

P. de la Harpe, Topics in Geometric Group Theory, Univ. Chicago Press, 2000, p. 156.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Index entries for sequences related to modular groups
Index entries for linear recurrences with constant coefficients, signature (0,2)

Cf. A054886, A029744.

The sequence (ternary strings) seems to be related to A029744 and A090989.

import Data.List (transpose)
a063759 n = a063759_list !! n
a063759_list = concat $ transpose [a151821_list, a007283_list]
-- Reinhard Zumkeller, Dec 16 2013

CoefficientList[Series[(1+3*x+2*x^2)/(1-2*x^2),{x,0,40}],x](* Jean-François Alcover, Mar 21 2011 *)
Join[{1},Transpose[NestList[{Last[#],2First[#]}&,{3,4},40]][[1]]] (* Harvey P. Dale, Oct 22 2011 *)

a(n)=([0,1; 2,0]^n*[1;3])[1,1] \\ Charles R Greathouse IV, Feb 09 2017

G.f.: (1+3*x+2*x^2)/(1-2*x^2).
a(n) = 2*a(n-2), n>2. - Harvey P. Dale, Oct 22 2011
a(2*n) = A151821(n+1); a(2*n+1) = A007283(n). - Reinhard Zumkeller, Dec 16 2013

Information from A145751 included by Joerg Arndt, Dec 03 2012

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

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

cp's OEIS Frontend

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

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

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

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A246352 Numbers n such that A048673(n) >= n.

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A048985 Working in base 2, replace n with the concatenation of its prime divisors in increasing order (write answer in base 10).

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A246360 a(1) = 1, then A007051 ((3^n)+1)/2 interleaved with A057198 (5*3^(n-1)+1)/2.

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A063759 Spherical growth series for modular group.

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