A333218 -id:A333218 - cp's OEIS frontend

A348615 Number of non-alternating permutations of {1...n}.

0, 0, 0, 2, 14, 88, 598, 4496, 37550, 347008, 3527758, 39209216, 473596070, 6182284288, 86779569238, 1303866853376, 20884006863710, 355267697410048, 6397563946377118, 121586922638606336, 2432161265800164950, 51081039175603191808, 1123862030028821404198

Offset: 0

Gus Wiseman, Nov 03 2021

nonn

A sequence is alternating if it is alternately strictly increasing and strictly decreasing, starting with either.

Also permutations of {1...n} matching the consecutive patterns (1,2,3) or (3,2,1). Matching only one of these gives A065429.

			The a(4) = 14 permutations:
  (1,2,3,4)  (3,1,2,4)
  (1,2,4,3)  (3,2,1,4)
  (1,3,4,2)  (3,4,2,1)
  (1,4,3,2)  (4,1,2,3)
  (2,1,3,4)  (4,2,1,3)
  (2,3,4,1)  (4,3,1,2)
  (2,4,3,1)  (4,3,2,1)

Wikipedia, Alternating permutation

The complement is counted by A001250, ranked by A333218.
The complementary version for compositions is A025047, ranked by A345167.
A directed version is A065429, complement A049774.
The version for compositions is A345192, ranked by A345168.
The version for ordered factorizations is A348613, complement A348610.
A345165 counts partitions w/o an alternating permutation, ranked by A345171.
A345170 counts partitions w/ an alternating permutation, ranked by A345172.
A348379 counts factorizations with an alternating permutation.
A348380 counts factorizations without an alternating permutation.
Cf. A056986, A102726, A325534, A325535, A344614, A344653, A344654, A347050, A347706, A348377, A348609.

b:= proc(u, o) option remember;
      `if`(u+o=0, 1, add(b(o-1+j, u-j), j=1..u))
    end:
a:= n-> n!-`if`(n<2, 1, 2)*b(n, 0):
seq(a(n), n=0..30);  # Alois P. Heinz, Nov 04 2021

wigQ[y_]:=Or[Length[y]==0,Length[Split[y]] ==Length[y]&&Length[Split[Sign[Differences[y]]]]==Length[y]-1];
Table[Length[Select[Permutations[Range[n]],!wigQ[#]&]],{n,0,6}]

from itertools import accumulate, count, islice
def A348615_gen(): # generator of terms
    yield from (0,0)
    blist, f = (0,2), 1
    for n in count(2):
        f *= n
        yield f - (blist := tuple(accumulate(reversed(blist),initial=0)))[-1]
A348615_list = list(islice(A348615_gen(),40)) # Chai Wah Wu, Jun 09-11 2022

a(n) = n! - A001250(n).

A333766 Maximum part of the n-th composition in standard order. a(0) = 0.

Original entry on oeis.org

0, 1, 2, 1, 3, 2, 2, 1, 4, 3, 2, 2, 3, 2, 2, 1, 5, 4, 3, 3, 3, 2, 2, 2, 4, 3, 2, 2, 3, 2, 2, 1, 6, 5, 4, 4, 3, 3, 3, 3, 4, 3, 2, 2, 3, 2, 2, 2, 5, 4, 3, 3, 3, 2, 2, 2, 4, 3, 2, 2, 3, 2, 2, 1, 7, 6, 5, 5, 4, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 3, 5, 4, 3, 3, 3, 2, 2

Offset: 0

Gus Wiseman, Apr 05 2020

nonn

One plus the longest run of 0's in the binary expansion of n.

A composition of n is a finite sequence of positive integers summing to n. The k-th composition in standard order (row k of A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

			The 100th composition in standard order is (1,3,3), so a(100) = 3.

Positions of ones are A000225.
Positions of terms <= 2 are A003754.
The version for prime indices is A061395.
Positions of terms > 1 are A062289.
Positions of first appearances are A131577.
The minimum part is given by A333768.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Compositions without 1's are A022340.
- Sum is A070939.
- Product is A124758.
- Runs are counted by A124767.
- Strict compositions are A233564.
- Constant compositions are A272919.
- Runs-resistance is A333628.
- Weakly decreasing compositions are A114994.
- Weakly increasing compositions are A225620.
- Strictly decreasing compositions are A333255.
- Strictly increasing compositions are A333256.
Cf. A029931, A048793, A087117, A228351, A328594, A333217, A333218, A333219, A333632, A333767.

stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Table[If[n==0,0,Max@@stc[n]],{n,0,100}]

For n > 0, a(n) = A087117(n) + 1.

A335458 Number of normal patterns contiguously matched by the n-th composition in standard order (A066099).

Original entry on oeis.org

1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 5, 3, 5, 5, 5, 2, 3, 3, 5, 3, 5, 5, 7, 3, 5, 5, 8, 5, 8, 7, 6, 2, 3, 3, 5, 3, 4, 5, 7, 3, 5, 4, 7, 5, 7, 8, 9, 3, 5, 5, 8, 4, 8, 7, 11, 5, 8, 7, 11, 7, 11, 9, 7, 2, 3, 3, 5, 3, 4, 5, 7, 3, 5, 5, 7, 5, 7, 8, 9, 3, 5, 5, 8, 5, 7

Offset: 0

Gus Wiseman, Jun 21 2020

nonn

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

We define a (normal) pattern to be a finite sequence covering an initial interval of positive integers. Patterns are counted by A000670 and ranked by A333217. A sequence S is said to match a pattern P if there is a not necessarily contiguous subsequence of S whose parts have the same relative order as P. For example, (3,1,1,3) matches (1,1,2), (2,1,1), and (2,1,2), but avoids (1,2,1), (1,2,2), and (2,2,1).

			The a(180) = 7 patterns are: (), (1), (1,2), (2,1), (1,2,3), (2,1,2), (2,1,2,3).

Wikipedia, Permutation pattern
Gus Wiseman, Sequences counting and ranking compositions by the patterns they match or avoid.
Gus Wiseman, Statistics, classes, and transformations of standard compositions

The non-contiguous version is A335454.
Summing over indices with binary length n gives A335457(n).
The nonempty version is A335474.
Patterns are counted by A000670 and ranked by A333217.
The n-th composition has A124771(n) distinct consecutive subsequences.
Knapsack compositions are counted by A325676 and ranked by A333223.
The n-th composition has A333257(n) distinct subsequence-sums.
The n-th composition has A334299(n) distinct subsequences.
Minimal avoided patterns are counted by A335465.
Cf. A108917, A124767, A124770, A181796, A269134, A333218, A333222, A333628, A334030, A335456.

stc[n_]:=Reverse[Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]];
mstype[q_]:=q/.Table[Union[q][[i]]->i,{i,Length[Union[q]]}];
Table[Length[Union[mstype/@ReplaceList[stc[n],{_,s___,_}:>{s}]]],{n,0,30}]

a(n) = A335474(n) + 1.

A124758 Product of the parts of the compositions in standard order.

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 2, 1, 4, 3, 4, 2, 3, 2, 2, 1, 5, 4, 6, 3, 6, 4, 4, 2, 4, 3, 4, 2, 3, 2, 2, 1, 6, 5, 8, 4, 9, 6, 6, 3, 8, 6, 8, 4, 6, 4, 4, 2, 5, 4, 6, 3, 6, 4, 4, 2, 4, 3, 4, 2, 3, 2, 2, 1, 7, 6, 10, 5, 12, 8, 8, 4, 12, 9, 12, 6, 9, 6, 6, 3, 10, 8, 12, 6, 12, 8, 8, 4, 8, 6, 8, 4, 6, 4, 4, 2, 6, 5, 8, 4, 9, 6

Offset: 0

Franklin T. Adams-Watters, Nov 06 2006

The standard order of compositions is given by A066099.

A composition of n is a finite sequence of positive integers summing to n. The k-th composition in standard order (row k of A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. - Gus Wiseman, Apr 03 2020

			Composition number 11 is 2,1,1; 2*1*1 = 2, so a(11) = 2.
The table starts:
  1
  1
  2 1
  3 2 2 1
  4 3 4 2 3 2 2 1
  5 4 6 3 6 4 4 2 4 3 4 2 3 2 2 1
The 146-th composition in standard order is (3,3,2), with product 18, so a(146) = 18. - _Gus Wiseman_, Apr 03 2020

Alois P. Heinz, Rows n = 0..14, flattened
George Beck and Karl Dilcher, A Matrix Related to Stern Polynomials and the Prouhet-Thue-Morse Sequence, arXiv:2106.10400 [math.CO], 2021.

Cf. A066099, A118851, A011782 (row lengths), A001906 (row sums).
The lengths of standard compositions are given by A000120.
The version for prime indices is A003963.
The version for binary indices is A096111.
Taking the sum instead of product gives A070939.
The sum of binary indices is A029931.
The sum of prime indices is A056239.
Taking GCD instead of product gives A326674.
Positions of first appearances are A331579.
Cf. A001519, A011782, A048793, A114994, A124767, A228351, A272919, A329369 (similar recurrence), A333218, A333219.

stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Table[Times@@stc[n],{n,0,100}] (* Gus Wiseman, Apr 03 2020 *)

For a composition b(1),...,b(k), a(n) = Product_{i=1}^k b(i).
a(A164894(n)) = a(A246534(n)) = n!. - Gus Wiseman, Apr 03 2020
a(A233249(n)) = a(A333220(n)) = A003963(n). - Gus Wiseman, Apr 03 2020
From Mikhail Kurkov, Jul 11 2021: (Start)
Some conjectures:
a(2n+1) = a(n) for n >= 0.
a(2n) = (1 + 1/A001511(n))*a(n) = 2*a(n) + a(n - 2^f(n)) - a(2n - 2^f(n)) for n > 0 with a(0)=1 where f(n) = A007814(n).
From the 1st formula for a(2n) we get a(4n+2) = 2*a(n), a(4n) = 2*a(2n) - a(n).
Sum_{k=0..2^n - 1} a(k) = A001519(n+1) for n >= 0.
a((4^n - 1)/3) = A011782(n) for n >= 0.
a(2^m*(2^n - 1)) = m + 1 for n > 0, m >= 0. (End)

A164894 Base-10 representation of the binary string formed by appending 10, 100, 1000, 10000, ..., etc., to 1.

Original entry on oeis.org

1, 6, 52, 840, 26896, 1721376, 220336192, 56406065280, 28879905423616, 29573023153783296, 60565551418948191232, 248076498612011791288320, 2032242676629600594233921536, 33296264013899376135928570454016, 1091051979207454757222107396637212672

Offset: 1

Gil Broussard, Aug 29 2009

These numbers are half the sum of powers of 2 indexed by differences of a triangular number and each smaller triangular number (e.g., 21 - 15 = 6, 21 - 10 = 11, ..., 21 - 0 = 21).
This suggests another way to think about these numbers: consider the number triangle formed by the characteristic function of the triangular numbers (A010054), join together the first n rows (the very first row is row 0) as a single binary string and that gives the (n + 1)th term of this sequence. - Alonso del Arte, Nov 15 2013
Numbers k such that the k-th composition in standard order (row k of A066099) is an initial interval. - Gus Wiseman, Apr 02 2020

			a(1) = 1, also 1 in binary.
a(2) = 6, or 110 in binary.
a(3) = 52, or 110100 in binary.
a(4) = 840, or 1101001000 in binary.

Harvey P. Dale, Table of n, a(n) for n = 1..81

The version for prime (rather than binary) indices is A002110.
The non-strict generalization is A225620.
The reversed version is A246534.
Standard composition numbers of permutations are A333218.
Standard composition numbers of strict increasing compositions are A333255.
Cf. A000120, A029931, A048793, A066099, A070939, A124768, A233564, A272919, A333217, A333220, A333379.

Table[Sum[2^((n^2 + n)/2 - (k^2 + k)/2 - 1), {k, 0, n - 1}], {n, 25}] (* Alonso del Arte, Nov 14 2013 *)
Module[{nn=15,t},t=Table[10^n,{n,0,nn}];Table[FromDigits[Flatten[IntegerDigits/@Take[t,k]],2],{k,nn}]] (* Harvey P. Dale, Jan 16 2024 *)

def a(n): return int("".join("1"+"0"*i for i in range(n)), 2)
print([a(n) for n in range(1, 16)]) # Michael S. Branicky, Jul 05 2021

def A164894(n): return sum(1<<(k*((n<<1)-k-1)>>1)+n-1 for k in range(n)) # Chai Wah Wu, Jul 11 2025

a(n) = Sum_{k=0..n-1} 2^((n^2 + n)/2 - (k^2 + k)/2 - 1). - Alonso del Arte, Nov 15 2013
Intersection of A333255 and A333217. - Gus Wiseman, Apr 02 2020
a(n) = Sum_{k=0..n-1} 2^(k*(2*n-k-1)/2+n-1). - Chai Wah Wu, Jul 11 2025

A246534 a(n) = Sum_{k=1..n} 2^(T(k)-1), where T(k)=k(k+1)/2 = A000217(k).

Original entry on oeis.org

0, 1, 5, 37, 549, 16933, 1065509, 135283237, 34495021605, 17626681066021, 18032025190548005, 36911520172609651237, 151152638972001256489509, 1238091191924352276155613733, 20283647694843594776223406899749, 664634281540152780046679753547072037

Offset: 0

M. F. Hasler, Aug 28 2014

nonn

Similar to A181388, this occurs as binary encoding of a straight n-omino lying on the y-axis, when the grid points of the first quadrant (N x N, N={0,1,2,...}) are given the weight 2^k, with k=0, 1,2, 3,4,5, ... filled in by antidiagonals.

Numbers k such that the k-th composition in standard order (row k of A066099) is a reversed initial interval. - Gus Wiseman, Apr 02 2020

			Label the cells of an infinite square matrix with 0,1,2,3,... along antidiagonals:
  0 1 3 6 10 ...
  2 4 7 ...
  5 8 ...
  9 ...
  ....
Now any subset of these cells can be represented by the sum of 2 raised to the power written in the given cells. In particular, the subset consisting of the first cell in the first 1, 2, 3, ... rows is represented by 2^0, 2^0+2^2, 2^0+2^2+2^5, ...

The version for prime (rather than binary) indices is A002110.
The non-strict generalization is A114994.
The non-reversed version is A164894.
Intersection of A333256 and A333217.
Partial sums of A036442.
Cf. A000120, A029931, A048793, A066099, A070939, A124766, A228351, A233564, A272919, A333218.

stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
normQ[m_]:=Or[m=={},Union[m]==Range[Max[m]]];
Select[Range[0,1000],normQ[stc[#]]&&Greater@@stc[#]&] (* Gus Wiseman, Apr 02 2020 *)

t=0;vector(20,n,t+=2^(n*(n+1)/2-1)) \\ yields the vector starting with a[1]=1

t=0;vector(20,n,if(n>1,t+=2^(n*(n-1)/2-1))) \\ yields the vector starting with 0

a = 0
for n in range(1,17): print(a, end =', '); a += 1<<(n-1)*(n+2)//2 # Ya-Ping Lu, Jan 23 2024

A333221 Irregular triangle read by rows where row n lists the set of STC-numbers of permutations of the prime indices of n.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 8, 7, 10, 9, 12, 16, 11, 13, 14, 32, 17, 24, 18, 20, 15, 64, 21, 22, 26, 128, 19, 25, 28, 34, 40, 33, 48, 256, 23, 27, 29, 30, 36, 65, 96, 42, 35, 49, 56, 512, 37, 38, 41, 44, 50, 52, 1024, 31, 66, 80, 129, 192, 68, 72, 43, 45, 46, 53, 54, 58

Offset: 1

Gus Wiseman, Mar 17 2020

This is a permutation of the nonnegative integers.
The k-th composition in standard order (row k of A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. We define the composition with STC-number k to be the k-th composition in standard order.
A prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798.

			Reading by columns gives:
  0  1  2  3  4  5  8  7  10  9   16  11  32  17  18  15  64  21  128  19
                 6            12      13      24  20          22       25
                                      14                      26       28
  34  33  256  23  36  65  42  35  512  37  1024  31  66  129  68  43
  40  48       27      96      49       38            80  192  72  45
               29              56       41                         46
               30                       44                         53
                                        50                         54
                                        52                         58
The sequence of terms together with the corresponding compositions begins:
     0: ()           24: (1,4)          27: (1,2,1,1)
     1: (1)          18: (3,2)          29: (1,1,2,1)
     2: (2)          20: (2,3)          30: (1,1,1,2)
     3: (1,1)        15: (1,1,1,1)      36: (3,3)
     4: (3)          64: (7)            65: (6,1)
     5: (2,1)        21: (2,2,1)        96: (1,6)
     6: (1,2)        22: (2,1,2)        42: (2,2,2)
     8: (4)          26: (1,2,2)        35: (4,1,1)
     7: (1,1,1)     128: (8)            49: (1,4,1)
    10: (2,2)        19: (3,1,1)        56: (1,1,4)
     9: (3,1)        25: (1,3,1)       512: (10)
    12: (1,3)        28: (1,1,3)        37: (3,2,1)
    16: (5)          34: (4,2)          38: (3,1,2)
    11: (2,1,1)      40: (2,4)          41: (2,3,1)
    13: (1,2,1)      33: (5,1)          44: (2,1,3)
    14: (1,1,2)      48: (1,5)          50: (1,3,2)
    32: (6)         256: (9)            52: (1,2,3)
    17: (4,1)        23: (2,1,1,1)    1024: (11)

Row lengths are A008480.
Column k = 1 is A233249.
Column k = -1 is A333220.
A related triangle for partitions is A215366.
Cf. A000120, A029931, A048793, A056239, A066099, A070939, A112798, A114994, A225620, A228351, A333218, A333219.

primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
fbi[q_]:=If[q=={},0,Total[2^q]/2];
Table[Sort[fbi/@Accumulate/@Permutations[primeMS[n]]],{n,30}]

A057335 a(0) = 1, and for n > 0, a(n) = A000040(A000120(n)) * a(floor(n/2)); essentially sequence A055932 generated using A000120, hence sorted by number of factors.

Original entry on oeis.org

1, 2, 4, 6, 8, 12, 18, 30, 16, 24, 36, 60, 54, 90, 150, 210, 32, 48, 72, 120, 108, 180, 300, 420, 162, 270, 450, 630, 750, 1050, 1470, 2310, 64, 96, 144, 240, 216, 360, 600, 840, 324, 540, 900, 1260, 1500, 2100, 2940, 4620, 486, 810, 1350, 1890, 2250, 3150, 4410

Offset: 0

Alford Arnold, Aug 27 2000

Note that for n>0 the prime divisors of a(n) are consecutive primes starting with 2. All of the least prime signatures (A025487) are included; with the other values forming A056808.
Using the formula, terms of b(n)= a(n)/A057334(n) are: 1, 1, 2, 2, 4, 4, 6, 6, 8, ..., indeed a(n) repeated. - Michel Marcus, Feb 09 2014
a(n) is the unique normal number whose unsorted prime signature is the k-th composition in standard order (graded reverse-lexicographic). This composition (row k of A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. A number is normal if its prime indices cover an initial interval of positive integers. Unsorted prime signature is the sequence of exponents in a number's prime factorization. - Gus Wiseman, Apr 19 2020

			From _Gus Wiseman_, Apr 19 2020: (Start)
The sequence of terms together with their prime indices begins:
      1: {}
      2: {1}
      4: {1,1}
      6: {1,2}
      8: {1,1,1}
     12: {1,1,2}
     18: {1,2,2}
     30: {1,2,3}
     16: {1,1,1,1}
     24: {1,1,1,2}
     36: {1,1,2,2}
     60: {1,1,2,3}
     54: {1,2,2,2}
     90: {1,2,2,3}
    150: {1,2,3,3}
    210: {1,2,3,4}
     32: {1,1,1,1,1}
     48: {1,1,1,1,2}
For example, the 27th composition in standard order is (1,2,1,1), and the normal number with prime signature (1,2,1,1) is 630 = 2*3*3*5*7, so a(27) = 630.
(End)

Michael De Vlieger, Table of n, a(n) for n = 0..10000

Cf. A000120, A057334, A055932 and A056808.
Cf. A324939.
Unsorted prime signature is A124010.
Numbers whose prime signature is aperiodic are A329139.
The reversed version is A334031.
A partial inverse is A334032.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Sum is A070939.
- Strict compositions are A233564.
- Constant compositions are A272919.
- Aperiodic compositions are A328594.
- Normal compositions are A333217.
- Permutations are A333218.
- Heinz number is A333219.
Cf. A005867, A029931, A048793, A052409, A056239, A066099, A112798, A124767, A228351, A233249, A333220, A345974.
Related to A019565 via A122111 and to A000079 via A336321.

Table[Times @@ Map[If[# == 0, 1, Prime@ #] &, Accumulate@ IntegerDigits[n, 2]], {n, 0, 54}] (* Michael De Vlieger, May 23 2017 *)

mg(n) = if (n==0, 1, prime(hammingweight(n))); \\ A057334
lista(nn) = {my(v = vector(nn)); v[1] = 1; for (i=2, nn, v[i] = mg(i-1)*v[(i+1)\2];); v;} \\ Michel Marcus, Feb 09 2014

A057335(n) = if(0==n,1,prime(hammingweight(n))*A057335(n\2)); \\ Antti Karttunen, Jul 20 2020

a(n) = A057334(n) * a (repeated).
A334032(a(n)) = n; a(A334032(n)) = A071364(n). - Gus Wiseman, Apr 19 2020
a(n) = A122111(A019565(n)); A019565(n) = A122111(a(n)). - Peter Munn, Jul 18 2020
a(n) = A336321(2^n). - Peter Munn, Mar 04 2022
Sum_{n>=0} 1/a(n) = Sum_{n>=0} 1/A005867(n) = 2.648101... (A345974). - Amiram Eldar, Jun 26 2025

More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Mar 29 2003

New primary name from Antti Karttunen, Jul 20 2020

A022340 Even Fibbinary numbers (A003714); also 2*Fibbinary(n).

Original entry on oeis.org

0, 2, 4, 8, 10, 16, 18, 20, 32, 34, 36, 40, 42, 64, 66, 68, 72, 74, 80, 82, 84, 128, 130, 132, 136, 138, 144, 146, 148, 160, 162, 164, 168, 170, 256, 258, 260, 264, 266, 272, 274, 276, 288, 290, 292, 296, 298, 320, 322, 324, 328, 330, 336, 338, 340, 512

Offset: 0

Marc LeBrun

nonn

Positions of ones in binomial(3k+2,k+1)/(3k+2) modulo 2 (A085405). - Paul D. Hanna, Jun 29 2003
Construction: start with strings S(0)={0}, S(1)={2}; for k>=2, concatenate all prior strings excluding S(k-1) and add 2^k to each element in the resulting string to obtain S(k); this sequence is the concatenation of all such generated strings: {S(0),S(1),S(2),...}. Example: for k=5, concatenate {S(0),S(1),S(2),S(3)} = {0, 2, 4, 8,10}; add 2^5 to each element to obtain S(5)={32,34,38,40,42}. - Paul D. Hanna, Jun 29 2003
From Gus Wiseman, Apr 08 2020: (Start)
The k-th composition in standard order (row k of A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions. This sequence lists all numbers k such that the k-th composition in standard order has no ones. For example, the sequence together with the corresponding compositions begins:
()          80: (2,5)        260: (6,3)
(2)         82: (2,3,2)      264: (5,4)
(3)         84: (2,2,3)      266: (5,2,2)
(4)        128: (8)          272: (4,5)
(2,2)      130: (6,2)        274: (4,3,2)
(5)        132: (5,3)        276: (4,2,3)
(3,2)      136: (4,4)        288: (3,6)
(2,3)      138: (4,2,2)      290: (3,4,2)
(6)        144: (3,5)        292: (3,3,3)
(4,2)      146: (3,3,2)      296: (3,2,4)
(3,3)      148: (3,2,3)      298: (3,2,2,2)
(2,4)      160: (2,6)        320: (2,7)
(2,2,2)    162: (2,4,2)      322: (2,5,2)
(7)        164: (2,3,3)      324: (2,4,3)
(5,2)      168: (2,2,4)      328: (2,3,4)
(4,3)      170: (2,2,2,2)    330: (2,3,2,2)
(3,4)      256: (9)          336: (2,2,5)
(3,2,2)    258: (7,2)        338: (2,2,3,2)
(End)

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Equals 2 * A003714.
Cf. A006013, A001045, A085405, A085407.
Compositions with no ones are counted by A212804.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Compositions without terms > 2 are A003754.
- Compositions without ones are A022340 (this sequence).
- Sum is A070939.
- Compositions with no twos are A175054.
- Strict compositions are A233564.
- Constant compositions are A272919.
- Normal compositions are A333217.
- Runs-resistance is A333628.
Cf. A066099, A124767, A228351, A318928, A333218.

a022340 = (* 2) . a003714 -- Reinhard Zumkeller, Feb 03 2015

f[n_Integer] := Block[{k = Ceiling[ Log[ GoldenRatio, n*Sqrt[5]]], t = n, fr = {}}, While[k > 1, If[t >= Fibonacci[k], AppendTo[fr, 1]; t = t - Fibonacci[k], AppendTo[fr, 0]]; k-- ]; FromDigits[fr, 2]]; Select[f /@ Range[0, 95], EvenQ[ # ] &] (* Robert G. Wilson v, Sep 18 2004 *)
Select[Range[2, 512, 2], BitAnd[#, 2#] == 0 &] (* Alonso del Arte, Jun 18 2012 *)

from itertools import count, islice
def A022340_gen(startvalue=0): # generator of terms >= startvalue
    return filter(lambda n:not n&(n>>1),count(max(0,startvalue+(startvalue&1)),2))
A022340_list = list(islice(A022340_gen(),30)) # Chai Wah Wu, Sep 07 2022

def A022340(n):
    tlist, s = [1,2], 0
    while tlist[-1]+tlist[-2] <= n: tlist.append(tlist[-1]+tlist[-2])
    for d in tlist[::-1]:
        if d <= n:
            s += 1
            n -= d
        s <<= 1
    return s # Chai Wah Wu, Apr 24 2025

For n>0, a(F(n))=2^n, a(F(n)-1)=A001045(n+2)-1, where F(n) is the n-th Fibonacci number with F(0)=F(1)=1.

a(n) + a(n)/2 = a(n) XOR a(n)/2, see A106409. - Reinhard Zumkeller, May 02 2005

Edited by Ralf Stephan, Sep 01 2004

A335515 Number of patterns of length n matching the pattern (1,2,3).

Original entry on oeis.org

0, 0, 0, 1, 19, 257, 3167, 38909, 498235, 6811453, 100623211, 1612937661, 28033056683, 526501880989, 10639153638795, 230269650097469, 5315570416909995, 130370239796988957, 3385531348514480651, 92801566389186549245, 2677687663571344712043, 81124824154544921317597

Offset: 0

Gus Wiseman, Jun 19 2020

nonn

We define a pattern to be a finite sequence covering an initial interval of positive integers. Patterns are counted by A000670 and ranked by A333217. A sequence S is said to match a pattern P if there is a not necessarily contiguous subsequence of S whose parts have the same relative order as P. For example, (3,1,1,3) matches (1,1,2), (2,1,1), and (2,1,2), but avoids (1,2,1), (1,2,2), and (2,2,1).

			The a(3) = 1 through a(4) = 19 patterns:
  (1,2,3)  (1,1,2,3)
           (1,2,1,3)
           (1,2,2,3)
           (1,2,3,1)
           (1,2,3,2)
           (1,2,3,3)
           (1,2,3,4)
           (1,2,4,3)
           (1,3,2,3)
           (1,3,2,4)
           (1,3,4,2)
           (1,4,2,3)
           (2,1,2,3)
           (2,1,3,4)
           (2,3,1,4)
           (2,3,4,1)
           (3,1,2,3)
           (3,1,2,4)
           (4,1,2,3)

Andrew Howroyd, Table of n, a(n) for n = 0..200
Wikipedia, Permutation pattern
Gus Wiseman, Sequences counting and ranking compositions by the patterns they match or avoid.

The complement A226316 is the avoiding version.
Compositions matching this pattern are counted by A335514 and ranked by A335479.
Permutations of prime indices matching this pattern are counted by A335520.
Patterns are counted by A000670 and ranked by A333217.
Patterns matching the pattern (1,1) are counted by A019472.
Permutations matching (1,2,3) are counted by A056986.
Combinatory separations are counted by A269134.
Patterns matched by standard compositions are counted by A335454.
Minimal patterns avoided by a standard composition are counted by A335465.
Cf. A034691, A158005, A158009, A238279, A333218, A333379, A333942.

allnorm[n_]:=If[n<=0,{{}},Function[s,Array[Count[s,y_/;y<=#]+1&,n]]/@Subsets[Range[n-1]+1]];
Table[Length[Select[Join@@Permutations/@allnorm[n],MatchQ[#,{_,x_,_,y_,_,z_,_}/;x

seq(n)=Vec( serlaplace(1/(2-exp(x + O(x*x^n)))) - 1/2 - 1/(1+sqrt(1-8*x+8*x^2 + O(x*x^n))), -(n+1)) \\ Andrew Howroyd, Jan 28 2024

a(n) = A000670(n) - A226316(n). - Andrew Howroyd, Jan 28 2024

a(9) onwards from Andrew Howroyd, Jan 28 2024

cp's OEIS Frontend

A348615 Number of non-alternating permutations of {1...n}.

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A333766 Maximum part of the n-th composition in standard order. a(0) = 0.

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A335458 Number of normal patterns contiguously matched by the n-th composition in standard order (A066099).

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A124758 Product of the parts of the compositions in standard order.

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A164894 Base-10 representation of the binary string formed by appending 10, 100, 1000, 10000, ..., etc., to 1.

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A246534 a(n) = Sum_{k=1..n} 2^(T(k)-1), where T(k)=k(k+1)/2 = A000217(k).

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PARI

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A333221 Irregular triangle read by rows where row n lists the set of STC-numbers of permutations of the prime indices of n.

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