A103969 -id:A103969 - cp's OEIS frontend

A005940 The Doudna sequence: write n-1 in binary; power of prime(k) in a(n) is # of 1's that are followed by k-1 0's.

1, 2, 3, 4, 5, 6, 9, 8, 7, 10, 15, 12, 25, 18, 27, 16, 11, 14, 21, 20, 35, 30, 45, 24, 49, 50, 75, 36, 125, 54, 81, 32, 13, 22, 33, 28, 55, 42, 63, 40, 77, 70, 105, 60, 175, 90, 135, 48, 121, 98, 147, 100, 245, 150, 225, 72, 343, 250, 375, 108, 625, 162, 243, 64, 17, 26, 39

Offset: 1

N. J. A. Sloane and J. H. Conway

A permutation of the natural numbers. - Robert G. Wilson v, Feb 22 2005
Fixed points: A029747. - Reinhard Zumkeller, Aug 23 2006
The even bisection, when halved, gives the sequence back. - Antti Karttunen, Jun 28 2014
From Antti Karttunen, Dec 21 2014: (Start)
This irregular table can be represented as a binary tree. Each child to the left is obtained by applying A003961 to the parent, and each child to the right is obtained by doubling the parent:
                                      1
                                      |
                   ...................2...................
                  3                                       4
        5......../ \........6                   9......../ \........8
       / \                 / \                 / \                 / \
      /   \               /   \               /   \               /   \
     /     \             /     \             /     \             /     \
    7       10         15       12         25       18         27       16
  11 14   21  20     35  30   45  24     49  50   75  36    125  54   81  32
etc.
Sequence A163511 is obtained by scanning the same tree level by level, from right to left. Also in binary trees A253563 and A253565 the terms on level of the tree are some permutation of the terms present on the level n of this tree. A252464(n) gives the distance of n from 1 in all these trees.
A252737(n) gives the sum and A252738(n) the product of terms on row n (where 1 is on row 0, 2 on row 1, 3 and 4 on row 2, etc.). A252745(n) gives the number of nodes on level n whose left child is larger than the right child, A252750 the difference between left and right child for each node from node 2 onward.
(End)
-A008836(a(1+n)) gives the corresponding numerator for A323505(n). - Antti Karttunen, Jan 19 2019
(a(2n+1)-1)/2 [= A244154(n)-1, for n >= 0] is a permutation of the natural numbers. - George Beck and Antti Karttunen, Dec 08 2019
From Peter Munn, Oct 04 2020: (Start)
Each term has the same even part (equivalently, the same 2-adic valuation) as its index.
Using the tree depicted in Antti Karttunen's 2014 comment:
Numbers are on the right branch (4 and descendants) if and only if divisible by the square of their largest prime factor (cf. A070003).
Numbers on the left branch, together with 2, are listed in A102750.
(End)
According to Kutz (1981), he learned of this sequence from American mathematician Byron Leon McAllister (1929-2017) who attributed the invention of the sequence to a graduate student by the name of Doudna (first name Paul?) in the mid-1950's at the University of Wisconsin. - Amiram Eldar, Jun 17 2021
From David James Sycamore, Sep 23 2022: (Start)
Alternative (recursive) definition: If n is a power of 2 then a(n)=n. Otherwise, if 2^j is the greatest power of 2 not exceeding n, and if k = n - 2^j, then a(n) is the least m*a(k) that has not occurred previously, where m is an odd prime.
Example: Use recursion with n = 77 = 2^6 + 13. a(13) = 25 and since 11 is the smallest odd prime m such that m*a(13) has not already occurred (see a(27), a(29),a(45)), then a(77) = 11*25 = 275. (End)
The odd bisection, when transformed by replacing all prime(k)^e in a(2*n - 1) with prime(k-1)^e, returns a(n), and thus gives the sequence back. - David James Sycamore, Sep 28 2022

			From _N. J. A. Sloane_, Aug 22 2022: (Start)
Let c_i = number of 1's in binary expansion of n-1 that have i 0's to their right, and let p(j) = j-th prime.  Then a(n) = Product_i p(i+1)^c_i.
If n=9, n-1 is 1000, c_3 = 1, a(9) = p(4)^1 = 7.
If n=10, n-1 = 1001, c_0 = 1, c_2 = 1, a(10) = p(1)*p(3) = 2*5 = 10.
If n=11, n-1 = 1010, c_1 = 1, c_2 = 1, a(11) = p(2)*p(3) = 15. (End)

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Antti Karttunen, Table of n, a(n) for n = 1..8192 (terms 1..1024 from Reinhard Zumkeller)
Michael De Vlieger, 6 row Doudna Tree diagram as mentioned in Comments.
Michael De Vlieger, Annotated fan-style binary tree showing 10 levels, with a color function where 2^m is shown in medium blue in row m, k < 2^m is darker blue, and k > 2^m is brighter green, with records in each row shown in red.
Ronald E. Kutz, Two unusual sequences, Two-Year College Mathematics Journal, Vol. 12, No. 5 (1981), pp. 316-319.
Index entries for sequences that are permutations of the natural numbers

Cf. A103969. Inverse is A005941 (A156552).
Cf. A125106. [From Franklin T. Adams-Watters, Mar 06 2010]
Cf. A252737 (gives row sums), A252738 (row products), A332979 (largest on row).
Related permutations of positive integers: A163511 (via A054429), A243353 (via A006068), A244154, A253563 (via A122111), A253565, A332977, A334866 (via A225546).
A000120, A003602, A003961, A006519, A053645, A070939, A246278, A250246, A252753, A253552 are used in a formula defining this sequence.
Formulas for f(a(n)) are given for f = A000265, A003963, A007949, A055396, A056239.
Numbers that occur at notable sets of positions in the binary tree representation of the sequence: A000040, A000079, A002110, A070003, A070826, A102750.
Cf. also A000142, A001511, A002450, A112798, A252463, A252464, A252745, A252750, A324054, A324106, A323505, A323508.
Cf. A106737, A290077, A323915, A324052, A324054, A324055, A324056, A324057, A324058, A324114, A324335, A324340, A324348, A324349 for various number-theoretical sequences applied to (i.e., permuted by) this sequence.
k-adic valuation: A007814 (k=2), A337821 (k=3).
Positions of multiples of 3: A091067.
Primorial deflation: A337376 / A337377.
Sum of prime indices of a(n) is A161511, reverse version A359043.
A048793 lists binary indices, ranked by A019565.
A066099 lists standard comps, partial sums A358134 (ranked by A358170).
Cf. A001222, A029837, A059893, A241916, A242628, A253566, A359042.

a005940 n = f (n - 1) 1 1 where
   f 0 y _          = y
   f x y i | m == 0 = f x' y (i + 1)
           | m == 1 = f x' (y * a000040 i) i
           where (x',m) = divMod x 2
-- Reinhard Zumkeller, Oct 03 2012
(Scheme, with memoization-macro definec from Antti Karttunen's IntSeq-library)
(define (A005940 n) (A005940off0 (- n 1))) ;; The off=1 version, utilizing any one of three different offset-0 implementations:
(definec (A005940off0 n) (cond ((< n 2) (+ 1 n)) (else (* (A000040 (- (A070939 n) (- (A000120 n) 1))) (A005940off0 (A053645 n))))))
(definec (A005940off0 n) (cond ((<= n 2) (+ 1 n)) ((even? n) (A003961 (A005940off0 (/ n 2)))) (else (* 2 (A005940off0 (/ (- n 1) 2))))))
(define (A005940off0 n) (let loop ((n n) (i 1) (x 1)) (cond ((zero? n) x) ((even? n) (loop (/ n 2) (+ i 1) x)) (else (loop (/ (- n 1) 2) i (* x (A000040 i)))))))
;; Antti Karttunen, Jun 26 2014

f := proc(n,i,x) option remember ; if n = 0 then x; elif type(n,'even') then procname(n/2,i+1,x) ; else procname((n-1)/2,i,x*ithprime(i)) ; end if; end proc:
A005940 := proc(n) f(n-1,1,1) ; end proc: # R. J. Mathar, Mar 06 2010

f[n_] := Block[{p = Partition[ Split[ Join[ IntegerDigits[n - 1, 2], {2}]], 2]}, Times @@ Flatten[ Table[q = Take[p, -i]; Prime[ Count[ Flatten[q], 0] + 1]^q[[1, 1]], {i, Length[p]}] ]]; Table[ f[n], {n, 67}] (* Robert G. Wilson v, Feb 22 2005 *)
Table[Times@@Prime/@(Join@@Position[Reverse[IntegerDigits[n,2]],1]-Range[DigitCount[n,2,1]]+1),{n,0,100}] (* Gus Wiseman, Dec 28 2022 *)

A005940(n) = { my(p=2, t=1); n--; until(!n\=2, n%2 && (t*=p) || p=nextprime(p+1)); t } \\ M. F. Hasler, Mar 07 2010; update Aug 29 2014

a(n)=my(p=2, t=1); for(i=0,exponent(n), if(bittest(n,i), t*=p, p=nextprime(p+1))); t \\ Charles R Greathouse IV, Nov 11 2021

from sympy import prime
import math
def A(n): return n - 2**int(math.floor(math.log(n, 2)))
def b(n): return n + 1 if n<2 else prime(1 + (len(bin(n)[2:]) - bin(n)[2:].count("1"))) * b(A(n))
print([b(n - 1) for n in range(1, 101)]) # Indranil Ghosh, Apr 10 2017

from math import prod
from itertools import accumulate
from collections import Counter
from sympy import prime
def A005940(n): return prod(prime(len(a)+1)**b for a, b in Counter(accumulate(bin(n-1)[2:].split('1')[:0:-1])).items()) # Chai Wah Wu, Mar 10 2023

From Reinhard Zumkeller, Aug 23 2006, R. J. Mathar, Mar 06 2010: (Start)
a(n) = f(n-1, 1, 1)
  where f(n, i, x) = x if n = 0,
                   = f(n/2, i+1, x) if n > 0 is even
                   = f((n-1)/2, i, x*prime(i)) otherwise. (End)
From Antti Karttunen, Jun 26 2014: (Start)
Define a starting-offset 0 version of this sequence as:
  b(0)=1, b(1)=2, [base cases]
and then compute the rest either with recurrence:
  b(n) = A000040(1+(A070939(n)-A000120(n))) * b(A053645(n)).
or
  b(2n) = A003961(b(n)), b(2n+1) = 2 * b(n). [Compare this to the similar recurrence given for A163511.]
Then define a(n) = b(n-1), where a(n) gives this sequence A005940 with the starting offset 1.
Can be also defined as a composition of related permutations:
a(n+1) = A243353(A006068(n)).
a(n+1) = A163511(A054429(n)). [Compare the scatter plots of this sequence and A163511 to each other.]
This permutation also maps between the partitions as enumerated in the lists A125106 and A112798, providing identities between:
A161511(n) = A056239(a(n+1)). [The corresponding sums ...]
A243499(n) = A003963(a(n+1)). [... and the products of parts of those partitions.]
(End)
From Antti Karttunen, Dec 21 2014  - Jan 04 2015: (Start)
A002110(n) =  a(1+A002450(n)). [Primorials occur at (4^n - 1)/3 in the offset-0 version of the sequence.]
a(n) = A250246(A252753(n-1)).
a(n) = A122111(A253563(n-1)).
For n >= 1, A055396(a(n+1)) = A001511(n).
For n >= 2, a(n) = A246278(1+A253552(n)).
(End)
From Peter Munn, Oct 04 2020: (Start)
A000265(a(n)) = a(A000265(n)) = A003961(a(A003602(n))).
A006519(a(n)) = a(A006519(n)) = A006519(n).
a(n) = A003961(a(A003602(n))) * A006519(n).
A007814(a(n)) = A007814(n).
A007949(a(n)) = A337821(n) = A007814(A003602(n)).
a(n) = A225546(A334866(n-1)).
(End)
a(2n) = 2*a(n), or generally a(2^k*n) = 2^k*a(n). - Amiram Eldar, Oct 03 2022
If n-1 = Sum_{i} 2^(q_i-1), then a(n) = Product_{i} prime(q_i-i+1). These are the Heinz numbers of the rows of A125106. If the offset is changed to 0, the inverse is A156552. - Gus Wiseman, Dec 28 2022

More terms from Robert G. Wilson v, Feb 22 2005
Sign in a formula switched and Maple program added by R. J. Mathar, Mar 06 2010
Binary tree illustration and keyword tabf added by Antti Karttunen, Dec 21 2014

A029747 Numbers of the form 2^k times 1, 3 or 5.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 10, 12, 16, 20, 24, 32, 40, 48, 64, 80, 96, 128, 160, 192, 256, 320, 384, 512, 640, 768, 1024, 1280, 1536, 2048, 2560, 3072, 4096, 5120, 6144, 8192, 10240, 12288, 16384, 20480, 24576, 32768, 40960, 49152, 65536, 81920, 98304, 131072, 163840, 196608

Offset: 1

N. J. A. Sloane

Fixed points of the Doudna sequence: A005940(a(n)) = A005941(a(n)) = a(n). - Reinhard Zumkeller, Aug 23 2006
Subsequence of A103969. - R. J. Mathar, Mar 06 2010
Question: Is there a simple proof that A005940(c) = c would never allow an odd composite c as a solution? See also my comments in A163511 and in A335431 concerning similar problems, also A364551 and A364576. - Antti Karttunen, Jul 28 & Aug 11 2023

			128 = 2^7 * 1 is in the sequence as well as 160 = 2^5 * 5. - _David A. Corneth_, Sep 18 2020

David A. Corneth, Table of n, a(n) for n = 1..9963 (terms <= 10^1000)
Index entries for linear recurrences with constant coefficients, signature (0,0,2).

Cf. A122132, A000079, A007283, A020714, A130793.
Cf. A005940, A005941, A163511, A335431, A364499, A364551, A364576.
Subsequence of the following sequences: A103969, A253789, A364541, A364542, A364544, A364546, A364548, A364550, A364560, A364565.
Even terms form a subsequence of A320674.

m = 200000; Select[Union @ Flatten @ Outer[Times, {1, 3, 5}, 2^Range[0, Floor[Log2[m]]]], # < m &] (* Amiram Eldar, Oct 15 2020 *)

is(n) = n>>valuation(n, 2) <= 5 \\ David A. Corneth, Sep 18 2020

def A029747(n):
    if n<3: return n
    a, b = divmod(n,3)
    return 1<Chai Wah Wu, Apr 02 2025

a(n) = if n < 6 then n else 2*a(n-3). - Reinhard Zumkeller, Aug 23 2006
G.f.: (1+x+x^2)^2/(1-2*x^3). - R. J. Mathar, Mar 06 2010
Sum_{n>=1} 1/a(n) = 46/15. - Amiram Eldar, Oct 15 2020

Edited by David A. Corneth and Peter Munn, Sep 18 2020

A005941 Inverse of the Doudna sequence A005940.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 9, 8, 7, 10, 17, 12, 33, 18, 11, 16, 65, 14, 129, 20, 19, 34, 257, 24, 13, 66, 15, 36, 513, 22, 1025, 32, 35, 130, 21, 28, 2049, 258, 67, 40, 4097, 38, 8193, 68, 23, 514, 16385, 48, 25, 26, 131, 132, 32769, 30, 37, 72, 259, 1026, 65537, 44, 131073, 2050, 39, 64

Offset: 1

N. J. A. Sloane

nonn

a(2^k) = 2^k. - Robert G. Wilson v, Feb 22 2005
Fixed points: A029747. - Reinhard Zumkeller, Aug 23 2006
Question: Is there a simple proof that a(c) = c would never allow an odd composite c as a solution? See also A364551. - Antti Karttunen, Jul 30 2023

J. H. Conway, personal communication.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Cf. A103969. Inverse of A005940. One more than A156552.
Cf. A364559 [= a(n)-n], A364557 (Möbius transform), A364558.
Cf. A029747 [known positions where a(n) = n], A364560 [where a(n) <= n], A364561 [where a(n) <= n and n is odd], A364562 [where a(n) > n], A364548 [where n divides a(n)], A364549 [where odd n divides a(n)], A364550 [where a(n) divides n], A364551 [where a(n) divides n and n is odd].

A005941 := proc(n)
    local k ;
    for k from 1 do
    if A005940(k) = n then # code reuse
        return k;
    end if;
    end do ;
end proc: # R. J. Mathar, Mar 06 2010

f[n_] := Block[{p = Partition[ Split[ Join[ IntegerDigits[n - 1, 2], {2}]], 2]}, Times @@ Flatten[ Table[q = Take[p, -i]; Prime[ Count[ Flatten[q], 0] + 1]^q[[1, 1]], {i, Length[p]}] ]]; t = Table[ f[n], {n, 10^5}]; Flatten[ Table[ Position[t, n, 1, 1], {n, 64}]] (* Robert G. Wilson v, Feb 22 2005 *)

A005941(n) = { my(f=factor(n), p, p2=1, res=0); for(i=1, #f~, p = 1 << (primepi(f[i, 1])-1); res += (p * p2 * (2^(f[i, 2])-1)); p2 <<= f[i, 2]); (1+res) }; \\ (After David A. Corneth's program for A156552) - Antti Karttunen, Jul 30 2023

from sympy import primepi, factorint
def A005941(n): return sum((1<Chai Wah Wu, Mar 11 2023

(define (A005941 n) (+ 1 (A156552 n))) ;; Antti Karttunen, Jun 26 2014

a(n) = h(g(n,1,1), 0) / 2 + 1 with h(n, m) = if n=0 then m else h(floor(n/2), 2*m + n mod 2) and g(n, i, x) = if n=1 then x else (if n mod prime(i) = 0 then g(n/prime(i), i, 2*x+1) else g(n, i+1, 2*x)). - Reinhard Zumkeller, Aug 23 2006

a(n) = 1 + A156552(n). - Antti Karttunen, Jun 26 2014

More terms from Robert G. Wilson v, Feb 22 2005

a(61) inserted by R. J. Mathar, Mar 06 2010

A246074 Paradigm Shift Sequence for a (-4,5) production scheme with replacement.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 16, 18, 20, 22, 24, 28, 32, 36, 40, 44, 48, 56, 64, 72, 80, 88, 96, 112, 128, 144, 160, 176, 192, 224, 256, 288, 320, 352, 384, 448, 512, 576, 640, 704, 768, 896, 1024, 1152, 1280, 1408, 1536, 1792, 2048, 2304, 2560, 2816, 3072, 3584, 4096, 4608, 5120, 5632, 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=-4 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 followinging 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.
The paradigm shift sequence for the R(-4,5) scheme is also the solution to the R(-2,4) scheme.

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

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

Paradigm shift sequences with negative bundling steps: A103969, A246074, A246075, A246076, A246079, A029750, A246078, A029747, A246077, A029744, A029747, A131577.

Join[{1, 2, 3, 4, 5}, LinearRecurrence[{0, 0, 0, 0, 0, 2}, {6, 7, 8, 9, 10, 11}, 64]] (* Jean-François Alcover, Sep 25 2017 *)

Vec(x*(1+x)^2 * (1-x+x^2)^2 * (1+x+x^2)^2 / (1-2*x^6) + 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-6) for all n >= 12.
G.f.: x*(1+x)^2 * (1-x+x^2)^2 * (1+x+x^2)^2 / (1-2*x^6). - Colin Barker, Nov 18 2016

A246079 Paradigm shift sequence for (-1,5) 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, 20, 22, 24, 26, 28, 30, 33, 36, 39, 42, 45, 48, 52, 56, 60, 66, 72, 78, 84, 90, 99, 108, 117, 126, 135, 144, 156, 168, 180, 198, 216, 234, 252, 270, 297, 324, 351, 378, 405, 432, 468, 504, 540, 594, 648, 702, 756, 810, 891, 972, 1053, 1134, 1215, 1296, 1404

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=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.

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

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.

Join[Range[18], LinearRecurrence[PadLeft[{3}, 14], {20, 22, 24, 26, 28, 30, 33, 36, 39, 42, 45, 48, 52, 56}, 55]] (* Jean-François Alcover, Sep 25 2017 *)

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 +14*x^13 +12*x^14 +10*x^15 +8*x^16 +6*x^17 +5*x^18 +4*x^19 +3*x^20 +2*x^21 +x^22 +x^30 +2*x^31) / (1 -3*x^14) + 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-14) for all n >= 33.
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 +14*x^13 +12*x^14 +10*x^15 +8*x^16 +6*x^17 +5*x^18 +4*x^19 +3*x^20 +2*x^21 +x^22 +x^30 +2*x^31) / (1 -3*x^14). - Colin Barker, Nov 18 2016

A246083 Paradigm shift sequence for (0,5) 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, 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

A246087 Paradigm shift sequence for (1,5) 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, 24, 26, 28, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 64, 68, 72, 78, 84, 90, 99, 108, 117, 126, 135, 144, 153, 162, 171, 180, 192, 204, 216, 234, 252, 270, 297, 324, 351, 378, 405, 432, 459, 486, 513, 540, 576, 612, 648, 702, 756, 810, 891, 972, 1053, 1134, 1215, 1296

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=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.

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

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

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

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 + 12 x^11 + 13 x^12 + 14 x^13 + 15 x^14 + 16 x^15 + 14 x^16 + 12 x^17 + 10 x^18 + 8 x^19 + 6 x^20 + 4 x^21 + 3 x^22 + 2 x^23 x^24 + x^36 + 2 x^37)/(1 - 3 x^16), {x, 0, 80}], 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 +12*x^11 +13*x^12 +14*x^13 +15*x^14 +16*x^15 +14*x^16 +12*x^17 +10*x^18 +8*x^19 +6*x^20 +4*x^21 +3*x^22 +2*x^23 +x^24 +x^36 +2*x^37) / (1 -3*x^16) + 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-16) for all n >= 39.
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 +14*x^13 +15*x^14 +16*x^15 +14*x^16 +12*x^17 +10*x^18 +8*x^19 +6*x^20 +4*x^21 +3*x^22 +2*x^23 +x^24 +x^36 +2*x^37) / (1 -3*x^16). - Colin Barker, Nov 18 2016

A246095 Paradigm shift sequence for (3,5) 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, 25, 26, 28, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 64, 68, 72, 76, 80, 84, 88, 92, 99, 108, 117, 126, 135, 144, 153, 162, 171, 180, 192, 204, 216, 228, 240, 256, 272, 288, 304, 324, 351, 378, 405, 432, 459, 486, 513, 540, 576, 612, 648, 684, 720, 768, 816, 864, 912, 972

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=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=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-18) for all n >= 66.

A246099 Paradigm shift sequence for (4,5) 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, 25, 26, 27, 28, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100, 108, 117, 126, 135, 144, 153, 162, 171, 180, 192, 204, 216, 228, 240, 256, 272, 288, 304, 320, 336, 352, 378, 405, 432, 459, 486, 513, 540, 576, 612, 648, 684, 720, 768, 816, 864, 912, 960, 1024, 1088, 1152, 1216, 1296, 1377, 1458, 1539, 1620, 1728, 1836, 1944, 2052, 2160, 2304, 2448, 2592, 2736, 2880, 3072, 3264, 3456, 3648, 3888, 4131, 4374, 4617, 4864, 5184, 5508, 5832, 6156, 6480, 6912, 7344, 7776, 8208, 8640, 9216, 9792, 10368, 10944, 11664, 12393, 13122, 13851, 14592, 15552, 16524, 17496, 18468, 19456, 20736, 22032, 23328

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=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.
The optimal action string that generates this integer sequence (x=a^k PV^m ...) has inversions in the minimum number of bundling procedures P. These inversions are persistent as they are present when the recursive formula for the output first holds.

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

Paradigm shift sequences for q=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) = 3*a(n-19) for all n >= 164.

A246103 Paradigm shift sequence for (5,5) 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, 25, 26, 27, 28, 29, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100, 105, 110, 117, 126, 135, 144, 153, 162, 171, 180, 192, 204, 216, 228, 240, 256, 272, 288, 304, 320, 336, 352, 368, 384, 405, 432, 459, 486, 513, 540, 576, 612, 648, 684, 720, 768, 816, 864, 912, 960, 1024, 1088, 1152, 1216, 1280, 1344, 1408, 1472, 1539, 1620, 1728, 1836, 1944, 2052, 2160, 2304, 2448, 2592, 2736, 2880, 3072, 3264, 3456, 3648, 3840, 4096, 4352, 4608, 4864, 5120, 5376, 5632, 5888, 6156

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=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 = 4.
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=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) = 4*a(n-25) for all n >= 100.

cp's OEIS Frontend

A005940 The Doudna sequence: write n-1 in binary; power of prime(k) in a(n) is # of 1's that are followed by k-1 0's.

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A029747 Numbers of the form 2^k times 1, 3 or 5.

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A005941 Inverse of the Doudna sequence A005940.

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A246074 Paradigm Shift Sequence for a (-4,5) production scheme with replacement.

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A246079 Paradigm shift sequence for (-1,5) production scheme with replacement.

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

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