A333381 -id:A333381 - cp's OEIS frontend

A064113 Indices k such that (1/3)*(prime(k)+prime(k+1)+prime(k+2)) is a prime.

2, 15, 36, 39, 46, 54, 55, 73, 102, 107, 110, 118, 129, 160, 164, 184, 187, 194, 199, 218, 239, 271, 272, 291, 339, 358, 387, 419, 426, 464, 465, 508, 520, 553, 599, 605, 621, 629, 633, 667, 682, 683, 702, 709, 710, 733, 761, 791, 813, 821, 822, 829, 830

Offset: 1

Jason Earls, Sep 08 2001

n such that d(n) = d(n+1), where d(n) = prime(n+1) - prime(n) = A001223(n).
Of interest because when I generalize it to d(n) = d(n+2), d(n) = d(n+3), etc. I am unable to find any positive number k such that d(n) = d(n+k) has no solution.
From Lei Zhou, Dec 06 2005: (Start)
When (1/3)*(prime(k) + prime(k+1) + prime(k+2)) is prime, then it is equal to prime(k+1).
Also, indices k such that (prime(k)+prime(k+2))/2 = prime(k+1).
The Mathematica program is based on the alternative definition. (End)
Inflection and undulation points of the primes, i.e., positions of zeros in A036263, the second differences of the primes. - Gus Wiseman, Mar 24 2020

			a(2) = 15 because (p(15)+p(16)+p(17)) = 1/3(47 + 53 + 59) = 53 (prime average of three successive primes).
Splitting the prime gaps into anti-runs gives: (1,2), (2,4,2,4,2,4,6,2,6,4,2,4,6), (6,2,6,4,2,6,4,6,8,4,2,4,2,4,14,4,6,2,10,2,6), (6,4,6), ... Then a(n) is the n-th partial sum of the lengths of these anti-runs. - _Gus Wiseman_, Mar 24 2020

Harry J. Smith, Table of n, a(n) for n=1..1000
Thomas Bloom, Let d_n = p_(n+1)-p_n. The set of n such that d_(n+1) >= d_n has density 1/2, and similarly for d_(n+1) <= d_n. Furthermore, there are infinitely many n such that d_(n+1) = d_n, Erdős Problems.
Terence Tao, Erdős problem database, see no. 218.
Wikipedia, Inflection point

Indices of zeros in A036263 (second differences of primes).
Indices (A000720 = primepi) of balanced primes A006562, minus 1.
Cf. A001223, A024675, A075540, A075541.
Cf. A262138.
Complement of A333214.
First differences are A333216.
The version for strict ascents is A258025.
The version for strict descents is A258026.
The version for weak ascents is A333230.
The version for weak descents is A333231.
A triangle for anti-runs of compositions is A106356.
Lengths of maximal runs of prime gaps are A333254.
Anti-runs of compositions in standard order are A333381.
Cf. A000040, A084758, A124762, A124767, A238279.

import Data.List (elemIndices)
a064113 n = a064113_list !! (n-1)
a064113_list = map (+ 1) $ elemIndices 0 a036263_list
-- Reinhard Zumkeller, Jan 20 2012

ct = 0; Do[If[(Prime[k] + Prime[k + 2] - 2*Prime[k + 1]) == 0, ct++; n[ct] = k], {k, 1, 2000}]; Table[n[k], {k, 1, ct}] (* Lei Zhou, Dec 06 2005 *)
Join@@Position[Differences[Array[Prime,100],2],0] (* Gus Wiseman, Mar 24 2020 *)

d(n) = prime(n+1)-prime(n); j=[]; for(n=1,1500, if(d(n)==d(n+1), j=concat(j,n))); j

{ n=0; for (m=1, 10^9, if (d(m)==d(m+1), write("b064113.txt", n++, " ", m); if (n==1000, break)) ) } \\ Using d(n) above. - Harry J. Smith, Sep 07 2009

[n | n<-[1..888], !A036263(n)] \\ M. F. Hasler, Oct 15 2024

\\ More efficient for larges range of n:
A064113_upto(N, n=1, L=List(), q=prime(n+1), d=q-prime(n))={forprime(p=1+q,, if(d==d=p-q, listput(L,n); #LM. F. Hasler, Oct 15 2024

from itertools import count, islice
from sympy import prime, nextprime
def A064113_gen(startvalue=1): # generator of terms >= startvalue
    c = max(startvalue,1)
    p = prime(c)
    q = nextprime(p)
    r = nextprime(q)
    for k in count(c):
        if p+r==(q<<1):
            yield k
        p, q, r = q, r, nextprime(r)
A064113_list = list(islice(A064113_gen(),20)) # Chai Wah Wu, Feb 27 2024

A036263(a(n)) = 0; A122535(n) = A000040(a(n)); A006562(n) = A000040(a(n) + 1); A181424(n) = A000040(a(n) + 2). - Reinhard Zumkeller, Jan 20 2012
A262138(2*a(n)) = 0. - Reinhard Zumkeller, Sep 12 2015
a(n) = A000720(A006562(n)) - 1, where A000720 = (prime)pi, A006562 = balanced primes. - M. F. Hasler, Oct 15 2024

A124762 Number of levels for compositions in standard order.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 1, 1, 0, 0, 1, 3, 0, 0, 0, 1, 0, 1, 0, 2, 0, 0, 1, 1, 1, 1, 2, 4, 0, 0, 0, 1, 1, 0, 0, 2, 0, 0, 2, 2, 0, 0, 1, 3, 0, 0, 0, 1, 0, 1, 0, 2, 1, 1, 2, 2, 2, 2, 3, 5, 0, 0, 0, 1, 0, 0, 0, 2, 0, 1, 1, 1, 0, 0, 1, 3, 0, 0, 0, 1, 1, 2, 1, 3, 0, 0, 1, 1, 1, 1, 2, 4, 0, 0, 0, 1, 1, 0, 0, 2, 0

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. This gives a bijective correspondence between nonnegative integers and integer compositions. This sequence gives the number of adjacent equal terms in the n-th composition in standard order. Alternatively, a(n) is one fewer than the number of maximal anti-runs in the same composition, where anti-runs are sequences without any adjacent equal terms. For example, the 1234567th composition in standard order is (3,2,1,2,2,1,2,5,1,1,1) with anti-runs ((3,2,1,2),(2,1,2,5,1),(1),(1)), so a(1234567) = 4 - 1 = 3. - Gus Wiseman, Apr 08 2020

			Composition number 11 is 2,1,1; 2>1=1, so a(11) = 1.
The table starts:
  0
  0
  0 1
  0 0 0 2
  0 0 1 1 0 0 1 3
  0 0 0 1 0 1 0 2 0 0 1 1 1 1 2 4
  0 0 0 1 1 0 0 2 0 0 2 2 0 0 1 3 0 0 0 1 0 1 0 2 1 1 2 2 2 2 3 5

Cf. A066099, A124760, A124761, A124763, A124764, A011782 (row lengths), A059570 (row sums).
Anti-runs summing to n are counted by A003242(n).
A triangle counting maximal anti-runs of compositions is A106356.
A triangle counting maximal runs of compositions is A238279.
Partitions whose first differences are an anti-run are A238424.
All of the following pertain to compositions in standard order (A066099):
- Weakly decreasing runs are counted by A124765.
- Weakly increasing runs are counted by A124766.
- Equal runs are counted by A124767.
- Strictly increasing runs are counted by A124768.
- Strictly decreasing runs are counted by A124769.
- Strict compositions are A233564.
- Constant compositions are A272919.
- Normal compositions are A333217.
- Adjacent unequal pairs are counted by A333382.
- Anti-runs are A333489.
Cf. A000120, A029931, A048793, A059893, A070939, A114994, A225620, A228351.

stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Table[Length[Select[Partition[stc[n],2,1],SameQ@@#&]],{n,0,100}] (* Gus Wiseman, Apr 08 2020 *)

For a composition b(1),...,b(k), a(n) = Sum_{1<=i=1

For n > 0, a(n) = A333381(n) - 1. - Gus Wiseman, Apr 08 2020

A345167 Numbers k such that the k-th composition in standard order is alternating.

Original entry on oeis.org

0, 1, 2, 4, 5, 6, 8, 9, 12, 13, 16, 17, 18, 20, 22, 24, 25, 32, 33, 34, 38, 40, 41, 44, 45, 48, 49, 50, 54, 64, 65, 66, 68, 70, 72, 76, 77, 80, 81, 82, 88, 89, 96, 97, 98, 102, 108, 109, 128, 129, 130, 132, 134, 140, 141, 144, 145, 148, 152, 153, 160, 161, 162

Offset: 1

Gus Wiseman, Jun 15 2021

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.

A sequence is alternating if it is alternately strictly increasing and strictly decreasing, starting with either. For example, the partition (3,2,2,2,1) has no alternating permutations, even though it does have the anti-run permutations (2,3,2,1,2) and (2,1,2,3,2).

			The terms together with their binary indices begin:
      1: (1)         25: (1,3,1)       66: (5,2)
      2: (2)         32: (6)           68: (4,3)
      4: (3)         33: (5,1)         70: (4,1,2)
      5: (2,1)       34: (4,2)         72: (3,4)
      6: (1,2)       38: (3,1,2)       76: (3,1,3)
      8: (4)         40: (2,4)         77: (3,1,2,1)
      9: (3,1)       41: (2,3,1)       80: (2,5)
     12: (1,3)       44: (2,1,3)       81: (2,4,1)
     13: (1,2,1)     45: (2,1,2,1)     82: (2,3,2)
     16: (5)         48: (1,5)         88: (2,1,4)
     17: (4,1)       49: (1,4,1)       89: (2,1,3,1)
     18: (3,2)       50: (1,3,2)       96: (1,6)
     20: (2,3)       54: (1,2,1,2)     97: (1,5,1)
     22: (2,1,2)     64: (7)           98: (1,4,2)
     24: (1,4)       65: (6,1)        102: (1,3,1,2)

Wikipedia, Alternating permutation

These compositions are counted by A025047, complement A345192.
The complement is A345168.
Partitions with a permutation of this type: A345170, complement A345165.
Factorizations with a permutation of this type: A348379.
A001250 counts alternating permutations, complement A348615.
A003242 counts anti-run compositions.
A345164 counts alternating permutations of prime indices.
A345194 counts alternating patterns, with twins A344605.
Statistics of standard compositions:
- Length is A000120.
- Constant runs are A124767.
- Heinz number is A333219.
- Number of maximal anti-runs is A333381.
- Runs-resistance is A333628.
- Number of distinct parts is A334028.
Classes of standard compositions:
- Weakly decreasing compositions (partitions) are A114994.
- Weakly increasing compositions (multisets) are A225620.
- Anti-runs are A333489.
- Non-alternating anti-runs are A345169.
Cf. A025048, A025049, A059893, A106356, A238279, A335448, A344604, A344615, A344653, A344742, A345163, A348377.

stc[n_]:=Differences[Prepend[Join@@Position[ Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
wigQ[y_]:=Or[Length[y]==0,Length[Split[y]] ==Length[y]&&Length[Split[Sign[Differences[y]]]]==Length[y]-1];
Select[Range[0,100],wigQ@*stc]

A374249 Numbers k such that the k-th composition in standard order has its equal parts contiguous.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 26, 28, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 47, 48, 50, 52, 56, 58, 60, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 78, 79, 80, 81, 83, 84, 85

Offset: 1

Gus Wiseman, Jul 13 2024

nonn

These are compositions avoiding the patterns (1,2,1) and (2,1,2).

			The terms together with their standard compositions begin:
   0: ()
   1: (1)
   2: (2)
   3: (1,1)
   4: (3)
   5: (2,1)
   6: (1,2)
   7: (1,1,1)
   8: (4)
   9: (3,1)
  10: (2,2)
  11: (2,1,1)
  12: (1,3)
  14: (1,1,2)
  15: (1,1,1,1)
  16: (5)
See A374253 for the complement: 13, 22, 25, 27, 29, ...

Gus Wiseman, Statistics, classes, and transformations of standard compositions

The strict (also anti-run) case is A233564, counted by A032020.
Compositions of this type are counted by A274174.
Permutations of prime indices of this type are counted by A333175.
The complement is A374253 (anti-run A374254), counted by A335548.
A003242 counts anti-run compositions, ranks A333489.
A011782 counts compositions.
A066099 lists compositions in standard order.
A124767 counts runs in standard compositions, anti-runs A333381.
A333755 counts compositions by number of runs.
A335454 counts patterns matched by standard compositions.
A335462 counts (1,2,1)- and (2,1,2)-matching permutations of prime indices.
- A335470 counts (1,2,1)-matching compositions, ranks A335466.
- A335471 counts (1,2,1)-avoiding compositions, ranks A335467.
- A335472 counts (2,1,2)-matching compositions, ranks A335468.
- A335473 counts (2,1,2)-avoiding compositions, ranks A335469.
Cf. A106356, A124762, A238130, A238279, A261982, A272919, A333382, A335450, A335460, A335524, A335525.

stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Select[Range[0,100],UnsameQ@@First/@Split[stc[#]]&]

Equals A335467 /\ A335469.

A333382 Number of adjacent unequal parts in the n-th composition in standard-order.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 2, 1, 0, 0, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 0, 0, 1, 1, 1, 0, 2, 2, 1, 1, 2, 0, 1, 2, 3, 2, 1, 1, 2, 2, 2, 2, 2, 3, 2, 1, 2, 1, 2, 1, 2, 1, 0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 3, 2, 1, 1, 2, 2, 2, 1, 1, 2

Offset: 0

Gus Wiseman, Mar 24 2020

nonn

For n > 0, a(n) is one fewer than the number of maximal runs of the n-th composition in standard-order.

			The 46th composition in standard order is (2,1,1,2), with maximal runs ((2),(1,1),(2)), so a(46) = 3 - 1 = 2.

Wikipedia, Longest increasing subsequence

Indices of first appearances (not counting 0) are A113835.
Partitions whose 0-appended first differences are a run are A007862.
Partitions whose first differences are a run are A049988.
A triangle counting maximal anti-runs of compositions is A106356.
A triangle counting maximal runs of compositions is A238279.
All of the following pertain to compositions in standard order (A066099):
- Adjacent equal pairs are counted by A124762.
- Weakly decreasing runs are counted by A124765.
- Weakly increasing runs are counted by A124766.
- Equal runs are counted by A124767.
- Strictly increasing runs are counted by A124768.
- Strictly decreasing runs are counted by A124769.
- Strict compositions are ranked by A233564.
- Constant compositions are ranked by A272919.
- Normal compositions are ranked by A333217.
- Anti-runs are ranked by A333489.
- Anti-runs are counted by A333381.
Cf. A000005, A000120, A003242, A029931, A048793, A059893, A070939, A114994, A225620, A228351, A238424.

stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Table[Length[Select[Partition[stc[n],2,1],UnsameQ@@#&]],{n,0,100}]

For n > 0, a(n) = A124767(n) - 1.

A333627 The a(n)-th composition in standard order is the sequence of run-lengths of the n-th composition in standard order.

Original entry on oeis.org

0, 1, 1, 2, 1, 3, 3, 4, 1, 3, 2, 6, 3, 7, 5, 8, 1, 3, 3, 6, 3, 5, 7, 12, 3, 7, 6, 14, 5, 11, 9, 16, 1, 3, 3, 6, 2, 7, 7, 12, 3, 7, 4, 10, 7, 15, 13, 24, 3, 7, 7, 14, 7, 13, 15, 28, 5, 11, 10, 22, 9, 19, 17, 32, 1, 3, 3, 6, 3, 7, 7, 12, 3, 5, 6, 14, 7, 15, 13

Offset: 0

Gus Wiseman, Mar 30 2020

nonn

			The standard compositions and their run-lengths:
       0 ~ () -> () ~ 0
      1 ~ (1) -> (1) ~ 1
      2 ~ (2) -> (1) ~ 1
     3 ~ (11) -> (2) ~ 2
      4 ~ (3) -> (1) ~ 1
     5 ~ (21) -> (11) ~ 3
     6 ~ (12) -> (11) ~ 3
    7 ~ (111) -> (3) ~ 4
      8 ~ (4) -> (1) ~ 1
     9 ~ (31) -> (11) ~ 3
    10 ~ (22) -> (2) ~ 2
   11 ~ (211) -> (12) ~ 6
    12 ~ (13) -> (11) ~ 3
   13 ~ (121) -> (111) ~ 7
   14 ~ (112) -> (21) ~ 5
  15 ~ (1111) -> (4) ~ 8
     16 ~ (5) -> (1) ~ 1
    17 ~ (41) -> (11) ~ 3
    18 ~ (32) -> (11) ~ 3
   19 ~ (311) -> (12) ~ 6

Claude Lenormand, Deux transformations sur les mots, Preprint, 5 pages, Nov 17 2003.

Positions of first appearances are A333630.
All of the following pertain to compositions in standard order (A066099):
- The length is A000120.
- The partial sums from the right are A048793.
- The sum is A070939.
- Adjacent equal pairs are counted by A124762.
- Equal runs are counted by A124767.
- Strict compositions are ranked by A233564.
- The partial sums from the left are A272020.
- Constant compositions are ranked by A272919.
- Normal compositions are ranked by A333217.
- Heinz number is A333219.
- Anti-runs are counted by A333381.
- Adjacent unequal pairs are counted by A333382.
- Runs-resistance is A333628.
- First appearances of run-resistances are A333629.
Cf. A029931, A098504, A114994, A225620, A228351, A238279, A242882, A318928, A329744, A329747, A333489.

stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Table[Total[2^(Accumulate[Reverse[Length/@Split[stc[n]]]])]/2,{n,0,30}]

A000120(n) = A070939(a(n)).

A000120(a(n)) = A124767(n).

A124765 Number of monotonically decreasing runs for compositions in standard order.

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 2, 1, 1, 2, 2, 2, 1, 2, 2, 2, 2, 3, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 1, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 3, 2, 3

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. a(n) is the number of maximal weakly decreasing runs in this composition. Alternatively, a(n) is one plus the number of strict ascents in the same composition. For example, the weakly decreasing runs of the 1234567th composition are ((3,2,1),(2,2,1),(2),(5,1,1,1)), so a(1234567) = 4. The 3 strict ascents together with the weak descents are: 3 >= 2 >= 1 < 2 >= 2 >= 1 < 2 < 5 >= 1 >= 1 >= 1. - Gus Wiseman, Apr 08 2020

			Composition number 11 is 2,1,1; the decreasing runs are 2,1,1; so a(11) = 1.
The table starts:
  0
  1
  1 1
  1 1 2 1
  1 1 1 1 2 2 2 1
  1 1 1 1 2 1 2 1 2 2 2 2 2 2 2 1
  1 1 1 1 1 1 2 1 2 2 1 1 2 2 2 1 2 2 2 2 3 2 3 2 2 2 2 2 2 2 2 1

Antti Karttunen, Table of n, a(n) for n = 0..16383

Cf. A066099, A124760, A011782 (row lengths).
Compositions of n with k strict ascents are A238343.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Sum is A070939.
- Weakly decreasing compositions are A114994.
- Weakly decreasing runs are counted by A124765.
- Weakly increasing runs are counted by A124766.
- Equal runs are counted by A124767.
- Strictly increasing runs are counted by A124768.
- Strictly decreasing runs are counted by A124769.
- Weakly increasing compositions are A225620.
- Reverse is A228351 (triangle).
- Strict compositions are A233564.
- Constant compositions are A272919.
- Normal compositions are A333217.
- Strictly decreasing compositions are A333255.
- Strictly increasing compositions are A333256.
- Anti-runs are counted by A333381.
- Adjacent unequal pairs are counted by A333382.
- Anti-runs are A333489.
- Runs-resistance is A333628.
Cf. A124761, A124763, A124764, A233249, A333213, A333380.

stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Table[Length[Split[stc[n],GreaterEqual]],{n,0,100}] (* Gus Wiseman, Apr 08 2020 *)

a(0) = 0, a(n) = A124760(n) + 1 for n > 0.

A353847 Composition run-sum transformation in terms of standard composition numbers. The a(k)-th composition in standard order is the sequence of run-sums of the k-th composition in standard order. Takes each index of a row of A066099 to the index of the row consisting of its run-sums.

Original entry on oeis.org

0, 1, 2, 2, 4, 5, 6, 4, 8, 9, 8, 10, 12, 13, 10, 8, 16, 17, 18, 18, 20, 17, 22, 20, 24, 25, 24, 26, 20, 21, 18, 16, 32, 33, 34, 34, 32, 37, 38, 36, 40, 41, 32, 34, 44, 45, 42, 40, 48, 49, 50, 50, 52, 49, 54, 52, 40, 41, 40, 42, 36, 37, 34, 32, 64, 65, 66, 66

Offset: 0

Gus Wiseman, May 30 2022

nonn

Every sequence can be uniquely split into a sequence of non-overlapping runs. For example, the runs of (2,2,1,1,1,3,2,2) are ((2,2),(1,1,1),(3),(2,2)), with sums (4,3,3,4).

			As a triangle:
   0
   1
   2  2
   4  5  6  4
   8  9  8 10 12 13 10  8
  16 17 18 18 20 17 22 20 24 25 24 26 20 21 18 16
These are the standard composition numbers of the following compositions (transposed):
  ()  (1)  (2)  (3)    (4)      (5)
           (2)  (2,1)  (3,1)    (4,1)
                (1,2)  (4)      (3,2)
                (3)    (2,2)    (3,2)
                       (1,3)    (2,3)
                       (1,2,1)  (4,1)
                       (2,2)    (2,1,2)
                       (4)      (2,3)
                                (1,4)
                                (1,3,1)
                                (1,4)
                                (1,2,2)
                                (2,3)
                                (2,2,1)
                                (3,2)
                                (5)

Mathematics Stack Exchange, What is a sequence run? (answered 2011-12-01)

Standard compositions are listed by A066099.
The version for partitions is A353832.
The run-sums themselves are listed by A353932, with A353849 distinct terms.
A005811 counts runs in binary expansion.
A300273 ranks collapsible partitions, counted by A275870.
A353838 ranks partitions with all distinct run-sums, counted by A353837.
A353851 counts compositions with all equal run-sums, ranked by A353848.
A353840-A353846 pertain to partition run-sum trajectory.
A353852 ranks compositions with all distinct run-sums, counted by A353850.
A353853-A353859 pertain to composition run-sum trajectory.
A353860 counts collapsible compositions.
A353863 counts run-sum-complete partitions.
Cf. A003242. A175413, A181819, A238279, A274174, A333381, A333489, A333755, A353835, A353839, A353864.

stc[n_]:=Differences[Prepend[Join@@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
stcinv[q_]:=Total[2^(Accumulate[Reverse[q]])]/2;
Table[stcinv[Total/@Split[stc[n]]],{n,0,100}]

A373948 Run-compression encoded as a transformation of compositions in standard order.

Original entry on oeis.org

0, 1, 2, 1, 4, 5, 6, 1, 8, 9, 2, 5, 12, 13, 6, 1, 16, 17, 18, 9, 20, 5, 22, 5, 24, 25, 6, 13, 12, 13, 6, 1, 32, 33, 34, 17, 4, 37, 38, 9, 40, 41, 2, 5, 44, 45, 22, 5, 48, 49, 50, 25, 52, 13, 54, 13, 24, 25, 6, 13, 12, 13, 6, 1, 64, 65, 66, 33, 68, 69, 70, 17, 72

Offset: 0

Gus Wiseman, Jun 24 2024

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 the (run-) compression of a sequence to be the anti-run obtained by reducing each run of repeated parts to a single part. Alternatively, compression removes all parts equal to the part immediately to their left. For example, (1,1,2,2,1) has compression (1,2,1).
For the present sequence, the a(n)-th composition in standard order is obtained by compressing the n-th composition in standard order.

			The standard compositions and their compressions begin:
   0: ()        -->  0: ()
   1: (1)       -->  1: (1)
   2: (2)       -->  2: (2)
   3: (1,1)     -->  1: (1)
   4: (3)       -->  4: (3)
   5: (2,1)     -->  5: (2,1)
   6: (1,2)     -->  6: (1,2)
   7: (1,1,1)   -->  1: (1)
   8: (4)       -->  8: (4)
   9: (3,1)     -->  9: (3,1)
  10: (2,2)     -->  2: (2)
  11: (2,1,1)   -->  5: (2,1)
  12: (1,3)     --> 12: (1,3)
  13: (1,2,1)   --> 13: (1,2,1)
  14: (1,1,2)   -->  6: (1,2)
  15: (1,1,1,1) -->  1: (1)

Gus Wiseman, Statistics, classes, and transformations of standard compositions

Positions of 1's are A000225.
The image is A333489, counted by A003242.
Sum of standard composition for a(n) is given by A373953, length A124767.
A037201 gives compression of first differences of primes, halved A373947.
A066099 lists the parts of all compositions in standard order.
A114901 counts compositions with no isolated parts.
A116861 counts partitions by compressed sum, by length A116608.
A240085 counts compositions with no unique parts.
A333755 counts compositions by compressed length.
A373949 counts compositions by compressed sum, opposite A373951.
Cf. A106356, A124762, A238130, A238279, A272919, A333381, A333382, A333627, A373952, A373954.

stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
stcinv[q_]:=Total[2^(Accumulate[Reverse[q]])]/2;
Table[stcinv[First/@Split[stc[n]]],{n,0,30}]

A029837(a(n)) = A373953(n).

A000120(a(n)) = A124767(n).

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A064113 Indices k such that (1/3)*(prime(k)+prime(k+1)+prime(k+2)) is a prime.

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A373949 Triangle read by rows where T(n,k) is the number of integer compositions of n such that replacing each run of repeated parts with a single part (run-compression) yields a composition of k.

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A124762 Number of levels for compositions in standard order.

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A345167 Numbers k such that the k-th composition in standard order is alternating.

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A374249 Numbers k such that the k-th composition in standard order has its equal parts contiguous.

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A333382 Number of adjacent unequal parts in the n-th composition in standard-order.

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A333627 The a(n)-th composition in standard order is the sequence of run-lengths of the n-th composition in standard order.

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