cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Previous Showing 71-80 of 314 results. Next

A374638 Numbers k such that the leaders of anti-runs in the k-th composition in standard order (A066099) are distinct.

Original entry on oeis.org

0, 1, 2, 4, 5, 6, 8, 9, 11, 12, 13, 16, 17, 18, 19, 20, 22, 24, 25, 26, 32, 33, 34, 35, 37, 38, 40, 41, 44, 45, 46, 48, 49, 50, 52, 53, 54, 64, 65, 66, 67, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82, 83, 88, 89, 91, 92, 93, 96, 97, 98, 100, 101, 102, 104
Offset: 1

Views

Author

Gus Wiseman, Aug 01 2024

Keywords

Comments

The leaders of anti-runs in a sequence are obtained by splitting it into maximal consecutive anti-runs (sequences with no adjacent equal terms) and taking the first term of each.
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.

Examples

			The terms together with corresponding compositions begin:
   0: ()
   1: (1)
   2: (2)
   4: (3)
   5: (2,1)
   6: (1,2)
   8: (4)
   9: (3,1)
  11: (2,1,1)
  12: (1,3)
  13: (1,2,1)
  16: (5)
  17: (4,1)
  18: (3,2)
  19: (3,1,1)
  20: (2,3)
  22: (2,1,2)
  24: (1,4)
  25: (1,3,1)
  26: (1,2,2)
		

Crossrefs

Positions of distinct (strict) rows in A374515.
Compositions of this type are counted by A374518.
For identical instead of distinct we have A374519, counted by A374517.
The complement is A374639.
Other types of runs (instead of anti-):
- For identical runs we have A374249, counted by A274174.
- For weakly increasing runs we have A374768, counted by A374632.
- For strictly increasing runs we have A374698, counted by A374687.
- For weakly decreasing runs we have A374701, counted by A374743.
- For strictly decreasing runs we have A374767, counted by A374761.
A065120 gives leaders of standard compositions.
A106356 counts compositions by number of maximal anti-runs.
A238279 counts compositions by number of maximal runs
A238424 counts partitions whose first differences are an anti-run.
All of the following pertain to compositions in standard order:
- Length is A000120.
- Sum is A029837(n+1).
- Parts are listed by A066099.
- Number of adjacent equal pairs is A124762, unequal A333382.
- Anti-runs are ranked by A333489, counted by A003242.
- Run-length transform is A333627, sum A070939.
- Run-compression transform is A373948, sum A373953, excess A373954.
Six types of maximal runs:

Programs

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

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

Views

Author

Alford Arnold, Aug 27 2000

Keywords

Comments

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

Examples

			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)
		

Crossrefs

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.
Related to A019565 via A122111 and to A000079 via A336321.

Programs

  • Mathematica
    Table[Times @@ Map[If[# == 0, 1, Prime@ #] &, Accumulate@ IntegerDigits[n, 2]], {n, 0, 54}] (* Michael De Vlieger, May 23 2017 *)
  • PARI
    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
    
  • PARI
    A057335(n) = if(0==n,1,prime(hammingweight(n))*A057335(n\2)); \\ Antti Karttunen, Jul 20 2020

Formula

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

Extensions

More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Mar 29 2003
New primary name from Antti Karttunen, Jul 20 2020

A073642 Replace 2^k in the binary representation of n with k (i.e., if n = 2^b + 2^c + 2^d + ... then a(n) = b + c + d + ...).

Original entry on oeis.org

0, 0, 1, 1, 2, 2, 3, 3, 3, 3, 4, 4, 5, 5, 6, 6, 4, 4, 5, 5, 6, 6, 7, 7, 7, 7, 8, 8, 9, 9, 10, 10, 5, 5, 6, 6, 7, 7, 8, 8, 8, 8, 9, 9, 10, 10, 11, 11, 9, 9, 10, 10, 11, 11, 12, 12, 12, 12, 13, 13, 14, 14, 15, 15, 6, 6, 7, 7, 8, 8, 9, 9, 9, 9, 10, 10, 11, 11, 12, 12, 10, 10, 11, 11, 12, 12, 13, 13
Offset: 0

Views

Author

Benoit Cloitre, Aug 29 2002

Keywords

Comments

For n >= 1, a(n) is the n-th row sum of the irregular triangle A133457. - Vladimir Shevelev, Dec 14 2015
For n >= 0, 2^a(n) is the number of partitions of n whose dimension (given by the hook-length formula) is an odd integer. See the MacDonald reference. - Arvind Ayyer, May 12 2016

Examples

			9 = 2^3 + 2^0, hence a(9) = 3 + 0 = 3;
25 = 2^4 + 2^3 + 2^0, hence a(25) = 4 + 3 + 0 = 7.
		

Crossrefs

Programs

  • Haskell
    a073642 = sum . zipWith (*) [0..] . a030308_row
    -- Reinhard Zumkeller, Jun 01 2013
    
  • Maple
    A073642 := proc(n)
            local bdgs ;
            bdgs := convert(n,base,2) ;
            add( op(i,bdgs)*(i-1), i=1..nops(bdgs)) ;
    end proc: # R. J. Mathar, Nov 17 2011
  • Mathematica
    Total[Flatten[Position[Rest[Reverse[IntegerDigits[#, 2]]], 1]]] & /@ Range[0, 87] (* Jayanta Basu, Jul 03 2013 *)
  • PARI
    a(n)=sum(i=1,length(binary(n)), component(binary(n),i)*(length(binary(n))-i))
    
  • PARI
    a(n) = my(b=binary(n)); b*-[-#b+1..0]~; \\ Ruud H.G. van Tol, Oct 17 2023
    
  • Python
    def A073642(n):
        a, i = 0, 0
        while n > 0:
            a, n, i = a+(n%2)*i, n//2, i+1
        return a
    print([A073642(n) for n in range(30)]) # A.H.M. Smeets, Aug 17 2019

Formula

a(n) = log_2(A059867(n)).
It seems that for n > 10 a(n) < n/(2*log(n)) and that Sum_{k=1..n} a(k) is asymptotic to C*n*log(n)^2 with 1/2 > C > 0.47.
a(1)=0, a(2n) = a(n) + e1(n), a(2n+1) = a(2n), where e1(n) = A000120(n). - Ralf Stephan, Jun 19 2003
If n = 2^log_2(n) then a(n) = log_2(n); otherwise, a(n) = log_2(n) + a(n-2^log_2(n)), where log_2=A000523. a(2*n+1) = a(2*n), as the least significant bit of n does not contribute to a(n). - Reinhard Zumkeller, Aug 17 2003, edited by A.H.M. Smeets, Aug 17 2019
a(n) = A029931(floor(n/2)). - Franklin T. Adams-Watters, Oct 22 2011
a(n) = Sum_{k=0..A070939(n)-1} k*A030308(n,k). - Reinhard Zumkeller, Jun 01 2013
Conjecture: a(n) = (3*A011371(n) - Sum_{k=1..n} A007814(k)^2)/2 for n > 0. - Velin Yanev, Sep 09 2017
G.f.: (1/(1 - x)) * Sum_{k>=1} k*x^(2^k)/(1 + x^(2^k)). - Ilya Gutkovskiy, Aug 17 2019
From A.H.M. Smeets, Aug 17 2019: (Start)
floor(log_2(n)) <= a(n) <= floor(log_2(n+2)*(log_2(n+2)-1)/2), n > 0.
Lower bound: floor(log_2(n)) = a(n) for n = 2^m or n = 2^m+1, m >= 0.
Upper bound: a(n) = floor(log_2(n+2)*(log_2(n+2)-1)/2) for n = 2^m-2 or n = 2^m-1, m >= 0. (End)
From Aayush Soni, Feb 12 2022: (Start)
For k < 2^n, a((2^n)+k) + a((2^n)-k-1) = n*(n+1)/2.
Proof: Any (n+1)-bit number 111...11_2 can only be split into two numbers 2^n + k and 2^n - k - 1 which never share any bits in common. Since a(111...11_2) = 0+1+2+...+n, this implies the stated formula. (End)

Extensions

a(0)=0 and offset corrected by Philippe Deléham, Apr 20 2009

A161511 Number of 1...0 pairs in the binary representation of 2n.

Original entry on oeis.org

0, 1, 2, 2, 3, 3, 4, 3, 4, 4, 5, 4, 6, 5, 6, 4, 5, 5, 6, 5, 7, 6, 7, 5, 8, 7, 8, 6, 9, 7, 8, 5, 6, 6, 7, 6, 8, 7, 8, 6, 9, 8, 9, 7, 10, 8, 9, 6, 10, 9, 10, 8, 11, 9, 10, 7, 12, 10, 11, 8, 12, 9, 10, 6, 7, 7, 8, 7, 9, 8, 9, 7, 10, 9, 10, 8, 11, 9, 10, 7, 11, 10, 11, 9, 12, 10, 11, 8, 13, 11, 12, 9, 13
Offset: 0

Views

Author

Keywords

Comments

Row (partition) sums of A125106.
a(n) is also the weight (= sum of parts) of the integer partition having viabin number n. The viabin number of an integer partition is defined in the following way. Consider the southeast border of the Ferrers board of the integer partition and consider the binary number obtained by replacing each east step with 1 and each north step, except the last one, with 0. The corresponding decimal form is, by definition, the viabin number of the given integer partition. "Viabin" is coined from "via binary". For example, consider the integer partition [2,2,2,1]. The southeast border of its Ferrers board yields 10100, leading to the viabin number 20. - Emeric Deutsch, Jul 24 2017

Examples

			For n = 5, the binary representation of 2n is 1010; the 1...0 pairs are 10xx, 1xx0, and xx10, so a(5) = 3.
		

Crossrefs

Cf. A000120, A243499 (gives the corresponding products), A227183, A056239, A243503, A006068, A163511.
Sum of prime indices of A005940.
Row sums of A125106.
A reverse version is A359043, row sums of A242628.
A029837 adds up standard compositions, row sums of A066099.
A029931 adds up binary indices, row sums of A048793.

Programs

  • Mathematica
    a[0] = 0; a[n_] := If[EvenQ[n], a[n/2] + DigitCount[n/2, 2, 1], a[(n-1)/2] + 1]; Array[a, 93, 0] (* Jean-François Alcover, Sep 09 2017 *)
  • PARI
    a(n)=local(t,k);t=0;k=1;while(n>0,if(n%2==0,k++,t+=k);n\=2);t
    
  • Python
    def A161511(n):
        a, b = 0, 0
        for i, j in enumerate(bin(n)[:1:-1], 1):
            if int(j):
                a += i-b
                b += 1
        return a # Chai Wah Wu, Jul 26 2023
  • Scheme
    ;; Two variants, the recursive one requiring memoizing definec-macro from Antti Karttunen's IntSeq-library.
    (define (A161511 n) (let loop ((n n) (i 1) (s 0)) (cond ((zero? n) s) ((even? n) (loop (/ n 2) (+ i 1) s)) (else (loop (/ (- n 1) 2) i (+ s i))))))
    (definec (A161511 n) (cond ((zero? n) n) ((even? n) (+ (A000120 n) (A161511 (/ n 2)))) (else (+ 1 (A161511 (/ (- n 1) 2))))))
    ;; Antti Karttunen, Jun 28 2014
    

Formula

a(0) = 0; a(2n) = a(n) + A000120(n); a(2n+1) = a(n) + 1.
From Antti Karttunen, Jun 28 2014: (Start)
Can be also obtained by mapping with an appropriate permutation from the lists of partition sizes computed for other enumerations similar to A125106:
a(n) = A227183(A006068(n)).
a(n) = A056239(A005940(n+1)).
a(n) = A243503(A163511(n)). (End)
a(n) = A029931(n) - binomial(A000120(n),2). - Gus Wiseman, Jan 03 2023
a(n) = a(n - A048896(n-1)) + 1 for n>=1 (see Peter J. Taylor link). - Mikhail Kurkov, Jul 04 2025

A333222 Numbers k such that every restriction of the k-th composition in standard order to a subinterval has a different sum.

Original entry on oeis.org

0, 1, 2, 4, 5, 6, 8, 9, 12, 16, 17, 18, 20, 24, 32, 33, 34, 40, 41, 48, 50, 64, 65, 66, 68, 69, 70, 72, 80, 81, 88, 96, 98, 104, 128, 129, 130, 132, 133, 134, 144, 145, 160, 161, 176, 192, 194, 196, 208, 256, 257, 258, 260, 261, 262, 264, 265, 268, 272, 274
Offset: 1

Views

Author

Gus Wiseman, Mar 17 2020

Keywords

Comments

Also numbers whose binary indices together with 0 define a Golomb ruler.
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.

Examples

			The list of terms together with the corresponding compositions begins:
    0: ()        41: (2,3,1)    130: (6,2)      262: (6,1,2)
    1: (1)       48: (1,5)      132: (5,3)      264: (5,4)
    2: (2)       50: (1,3,2)    133: (5,2,1)    265: (5,3,1)
    4: (3)       64: (7)        134: (5,1,2)    268: (5,1,3)
    5: (2,1)     65: (6,1)      144: (3,5)      272: (4,5)
    6: (1,2)     66: (5,2)      145: (3,4,1)    274: (4,3,2)
    8: (4)       68: (4,3)      160: (2,6)      276: (4,2,3)
    9: (3,1)     69: (4,2,1)    161: (2,5,1)    288: (3,6)
   12: (1,3)     70: (4,1,2)    176: (2,1,5)    289: (3,5,1)
   16: (5)       72: (3,4)      192: (1,7)      290: (3,4,2)
   17: (4,1)     80: (2,5)      194: (1,5,2)    296: (3,2,4)
   18: (3,2)     81: (2,4,1)    196: (1,4,3)    304: (3,1,5)
   20: (2,3)     88: (2,1,4)    208: (1,2,5)    320: (2,7)
   24: (1,4)     96: (1,6)      256: (9)        321: (2,6,1)
   32: (6)       98: (1,4,2)    257: (8,1)      324: (2,4,3)
   33: (5,1)    104: (1,2,4)    258: (7,2)      328: (2,3,4)
   34: (4,2)    128: (8)        260: (6,3)      352: (2,1,6)
   40: (2,4)    129: (7,1)      261: (6,2,1)    384: (1,8)
		

Crossrefs

A subset of A233564.
Also a subset of A333223.
These compositions are counted by A169942 and A325677.
The number of distinct nonzero subsequence-sums is A333224.
The number of distinct subsequence-sums is A333257.
Lengths of optimal Golomb rulers are A003022.
Inequivalent optimal Golomb rulers are counted by A036501.
Complete rulers are A103295, with perfect case A103300.
Knapsack partitions are counted by A108917, with strict case A275972.
Distinct subsequences are counted by A124770 and A124771.
Golomb subsets are counted by A143823.
Heinz numbers of knapsack partitions are A299702.
Knapsack compositions are counted by A325676.
Maximal Golomb rulers are counted by A325683.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Select[Range[0,300],UnsameQ@@ReplaceList[stc[#],{_,s__,_}:>Plus[s]]&]

A359359 Sum of positions of zeros in the binary expansion of n, where positions are read starting with 1 from the left (big-endian).

Original entry on oeis.org

1, 0, 2, 0, 5, 2, 3, 0, 9, 5, 6, 2, 7, 3, 4, 0, 14, 9, 10, 5, 11, 6, 7, 2, 12, 7, 8, 3, 9, 4, 5, 0, 20, 14, 15, 9, 16, 10, 11, 5, 17, 11, 12, 6, 13, 7, 8, 2, 18, 12, 13, 7, 14, 8, 9, 3, 15, 9, 10, 4, 11, 5, 6, 0, 27, 20, 21, 14, 22, 15, 16, 9, 23, 16, 17, 10
Offset: 0

Views

Author

Gus Wiseman, Jan 03 2023

Keywords

Examples

			The binary expansion of 100 is (1,1,0,0,1,0,0), with zeros at positions {3,4,6,7}, so a(100) = 20.
		

Crossrefs

The number of zeros is A023416, partial sums A059015.
For positions of 1's we have A230877, reversed A029931.
The reversed version is A359400.
A003714 lists numbers with no successive binary indices.
A030190 gives binary expansion.
A039004 lists the positions of zeros in A345927.

Programs

  • Mathematica
    Table[Total[Join@@Position[IntegerDigits[n,2],0]],{n,0,100}]

Formula

a(n>0) = binomial(A029837(n)+1,2) - A230877(n).

A372441 Number of binary indices (binary weight) of n minus number of prime indices (bigomega) of n.

Original entry on oeis.org

1, 0, 1, -1, 1, 0, 2, -2, 0, 0, 2, -1, 2, 1, 2, -3, 1, -1, 2, -1, 1, 1, 3, -2, 1, 1, 1, 0, 3, 1, 4, -4, 0, 0, 1, -2, 2, 1, 2, -2, 2, 0, 3, 0, 1, 2, 4, -3, 1, 0, 2, 0, 3, 0, 3, -1, 2, 2, 4, 0, 4, 3, 3, -5, 0, -1, 2, -1, 1, 0, 3, -3, 2, 1, 1, 0, 2, 1, 4, -3, -1
Offset: 1

Views

Author

Gus Wiseman, May 07 2024

Keywords

Comments

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.
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.

Crossrefs

Positions of zeros are A071814.
For sum instead of length we have A372428, zeros A372427.
For minimum instead of length we have A372437, zeros {}.
For maximum instead of length we have A372442, zeros A372436.
Positions of odd terms are A372590, even A372591.
A003963 gives product of prime indices.
A019565 gives Heinz number of binary indices, adjoint A048675.
A029837 gives greatest binary index, least A001511.
A048793 lists binary indices, length A000120, reverse A272020, sum A029931.
A061395 gives greatest prime index, least A055396.
A070939 gives length of binary expansion.
A112798 lists prime indices, length A001222, reverse A296150, sum A056239.

Programs

  • Maple
    f:= proc(n) convert(convert(n,base,2),`+`)-numtheory:-bigomega(n) end proc:
    map(f, [$1..100]); # Robert Israel, May 22 2024
  • Mathematica
    bix[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
    prix[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    Table[Length[bix[n]]-Length[prix[n]],{n,100}]

Formula

a(n) = A000120(n) - A001222(n).

A374519 Numbers k such that the leaders of anti-runs in the k-th composition in standard order (A066099) are identical.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, 17, 18, 20, 21, 22, 24, 25, 27, 28, 29, 30, 31, 32, 33, 34, 36, 37, 38, 40, 41, 42, 44, 45, 48, 49, 50, 51, 52, 54, 55, 56, 57, 59, 60, 61, 62, 63, 64, 65, 66, 68, 69, 70, 72, 73, 76, 77, 80, 81, 82, 84, 85
Offset: 1

Views

Author

Gus Wiseman, Aug 01 2024

Keywords

Comments

The leaders of anti-runs in a sequence are obtained by splitting it into maximal consecutive anti-runs (sequences with no adjacent equal terms) and taking the first term of each.
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.

Examples

			The 346th composition in standard order is (2,2,1,2,2), with anti-runs ((2),(2,1,2),(2)), with leaders (2,2,2), so 346 is in the sequence.
		

Crossrefs

Positions of constant rows in A374515.
Compositions of this type are counted by A374517.
The complement is A374520.
For distinct instead of identical leaders we have A374638, counted by A374518.
Other types of runs (instead of anti-):
- For identical runs we have A272919, counted by A000005.
- For weakly increasing runs we have A374633, counted by A374631.
- For strictly increasing runs we have A374685, counted by A374686.
- For weakly decreasing runs we have A374744, counted by A374742.
- For strictly decreasing runs we have A374759, counted by A374760.
A065120 gives leaders of standard compositions.
A106356 counts compositions by number of maximal anti-runs.
A238279 counts compositions by number of maximal runs.
A238424 counts partitions whose first differences are an anti-run.
All of the following pertain to compositions in standard order:
- Length is A000120.
- Sum is A029837(n+1).
- Parts are listed by A066099.
- Number of adjacent equal pairs is A124762, unequal A333382.
- Anti-runs are ranked by A333489, counted by A003242.
- Run-length transform is A333627, sum A070939.
- Run-compression transform is A373948, sum A373953, excess A373954.
- Ranks of contiguous compositions are A374249, counted by A274174.
Six types of maximal runs:

Programs

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

A333220 The number k such that the k-th composition in standard order consists of the prime indices of n in weakly increasing order.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 8, 7, 10, 12, 16, 14, 32, 24, 20, 15, 64, 26, 128, 28, 40, 48, 256, 30, 36, 96, 42, 56, 512, 52, 1024, 31, 80, 192, 72, 58, 2048, 384, 160, 60, 4096, 104, 8192, 112, 84, 768, 16384, 62, 136, 100, 320, 224, 32768, 106, 144, 120, 640, 1536
Offset: 1

Views

Author

Gus Wiseman, Mar 17 2020

Keywords

Comments

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 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.

Examples

			The sequence of terms together with the corresponding compositions begins:
      0: ()             128: (8)             2048: (12)
      1: (1)             28: (1,1,3)          384: (1,8)
      2: (2)             40: (2,4)            160: (2,6)
      3: (1,1)           48: (1,5)             60: (1,1,1,3)
      4: (3)            256: (9)             4096: (13)
      6: (1,2)           30: (1,1,1,2)        104: (1,2,4)
      8: (4)             36: (3,3)           8192: (14)
      7: (1,1,1)         96: (1,6)            112: (1,1,5)
     10: (2,2)           42: (2,2,2)           84: (2,2,3)
     12: (1,3)           56: (1,1,4)          768: (1,9)
     16: (5)            512: (10)           16384: (15)
     14: (1,1,2)         52: (1,2,3)           62: (1,1,1,1,2)
     32: (6)           1024: (11)             136: (4,4)
     24: (1,4)           31: (1,1,1,1,1)      100: (1,3,3)
     20: (2,3)           80: (2,5)            320: (2,7)
     15: (1,1,1,1)      192: (1,7)            224: (1,1,6)
     64: (7)             72: (3,4)          32768: (16)
     26: (1,2,2)         58: (1,1,2,2)        106: (1,2,2,2)
		

Crossrefs

The version with prime indices taken in weakly decreasing order is A233249.
A partial inverse is A333219.

Programs

  • Mathematica
    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[fbi[Accumulate[Reverse[primeMS[n]]]],{n,100}]

Formula

A000120(a(n)) = A056239(n).

A359755 Positions of first appearances in the sequence of weighted sums of prime indices (A304818).

Original entry on oeis.org

1, 2, 3, 4, 6, 7, 8, 10, 12, 15, 16, 18, 20, 24, 26, 28, 36, 40, 46, 48, 50, 52, 56, 62, 68, 74, 76, 86, 88, 92, 94, 106, 107, 118, 122, 124, 131, 134, 136, 142, 146, 152, 158, 164, 166, 173, 178, 188, 193, 194, 199, 202, 206, 214, 218, 226, 229, 236, 239, 254
Offset: 1

Views

Author

Gus Wiseman, Jan 15 2023

Keywords

Comments

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.
The weighted sum of a sequence (y_1,...,y_k) is Sum_{i=1..k} i*y_i.

Examples

			The terms together with their prime indices begin:
    1: {}
    2: {1}
    3: {2}
    4: {1,1}
    6: {1,2}
    7: {4}
    8: {1,1,1}
   10: {1,3}
   12: {1,1,2}
   15: {2,3}
   16: {1,1,1,1}
   18: {1,2,2}
   20: {1,1,3}
   24: {1,1,1,2}
		

Crossrefs

The version for standard compositions is A089633, zero-based A359756.
Positions of first appearances in A304818, reverse A318283.
The zero-based version is A359675, unsorted A359676.
The reverse zero-based version is A359680, unsorted A359681.
This is the sorted version of A359682, reverse A359679.
The reverse version is A359754.
A053632 counts compositions by weighted sum.
A112798 lists prime indices, length A001222, sum A056239.
A320387 counts multisets by weighted sum, zero-based A359678.
A358136 lists partial sums of prime indices, ranked by A358137, rev A359361.

Programs

  • Mathematica
    nn=1000;
    primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    ots[y_]:=Sum[i*y[[i]],{i,Length[y]}];
    seq=Table[ots[primeMS[n]],{n,1,nn}];
    Select[Range[nn],FreeQ[seq[[Range[#-1]]],seq[[#]]]&]
Previous Showing 71-80 of 314 results. Next