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.

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A062880 Zero together with the numbers which can be written as a sum of distinct odd powers of 2.

Original entry on oeis.org

0, 2, 8, 10, 32, 34, 40, 42, 128, 130, 136, 138, 160, 162, 168, 170, 512, 514, 520, 522, 544, 546, 552, 554, 640, 642, 648, 650, 672, 674, 680, 682, 2048, 2050, 2056, 2058, 2080, 2082, 2088, 2090, 2176, 2178, 2184, 2186, 2208, 2210, 2216, 2218, 2560, 2562
Offset: 0

Views

Author

Antti Karttunen, Jun 26 2001

Keywords

Comments

Binary expansion of n does not contain 1-bits at even positions.
Integers whose base-4 representation consists of only 0's and 2's.
Every nonnegative even number is a unique sum of the form a(k)+2*a(l); moreover, this sequence is unique with such property. - Vladimir Shevelev, Nov 07 2008
Also numbers such that the digital sum base 2 and the digital sum base 4 are in a ratio of 2:4. - Michel Marcus, Sep 23 2013
From Gus Wiseman, Jun 10 2020: (Start)
Numbers k such that the k-th composition in standard order has all even parts. 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. For example, the sequence of all compositions into even parts begins:
0: () 520: (6,4) 2080: (6,6)
2: (2) 522: (6,2,2) 2082: (6,4,2)
8: (4) 544: (4,6) 2088: (6,2,4)
10: (2,2) 546: (4,4,2) 2090: (6,2,2,2)
32: (6) 552: (4,2,4) 2176: (4,8)
34: (4,2) 554: (4,2,2,2) 2178: (4,6,2)
40: (2,4) 640: (2,8) 2184: (4,4,4)
42: (2,2,2) 642: (2,6,2) 2186: (4,4,2,2)
128: (8) 648: (2,4,4) 2208: (4,2,6)
130: (6,2) 650: (2,4,2,2) 2210: (4,2,4,2)
136: (4,4) 672: (2,2,6) 2216: (4,2,2,4)
138: (4,2,2) 674: (2,2,4,2) 2218: (4,2,2,2,2)
160: (2,6) 680: (2,2,2,4) 2560: (2,10)
162: (2,4,2) 682: (2,2,2,2,2) 2562: (2,8,2)
168: (2,2,4) 2048: (12) 2568: (2,6,4)
170: (2,2,2,2) 2050: (10,2) 2570: (2,6,2,2)
512: (10) 2056: (8,4) 2592: (2,4,6)
514: (8,2) 2058: (8,2,2) 2594: (2,4,4,2)
(End)

Crossrefs

Cf. A000695.
Except for first term, n such that A063694(n) = 0. Binary expansion is given in A062033.
Interpreted as Zeckendorf expansion: A062879.
Central diagonal of arrays A163357 and A163359.
Even partitions are counted by A035363.
Numbers with an even number of 1's in binary expansion are A001969.
Numbers whose binary expansion has even length are A053754.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Compositions without even parts are A060142.
- Sum is A070939.
- Product is A124758.
- Strict compositions are A233564.
- Heinz number is A333219.
- Number of distinct parts is A334028.

Programs

  • C
    uint32_t a_next(uint32_t a_n) { return (a_n + 0x55555556) & 0xaaaaaaaa; } /* Falk Hüffner, Jan 22 2022 */
    
  • Haskell
    a062880 n = a062880_list !! n
    a062880_list = filter f [0..] where
       f 0 = True
       f x = (m == 0 || m == 2) && f x'  where (x', m) = divMod x 4
    -- Reinhard Zumkeller, Nov 20 2012
    
  • Maple
    [seq(a(j),j=0..100)]; a := n -> add((floor(n/(2^i)) mod 2)*(2^((2*i)+1)),i=0..floor_log_2(n+1));
  • Mathematica
    b[n_] := BitAnd[n, Sum[2^k, {k, 0, Log[2, n] // Floor, 2}]]; Select[Range[ 0, 10^4], b[#] == 0&] (* Jean-François Alcover, Feb 28 2016 *)
  • Python
    def A062880(n): return int(bin(n)[2:],4)<<1 # Chai Wah Wu, Aug 21 2023

Formula

a(n) = 2 * A000695(n). - Vladimir Shevelev, Nov 07 2008
From Robert Israel, Apr 10 2018: (Start)
a(2*n) = 4*a(n).
a(2*n+1) = 4*a(n)+2.
G.f. g(x) satisfies: g(x) = 4*(1+x)*g(x^2)+2*x/(1-x^2). (End)

A333628 Runs-resistance of the n-th composition in standard order. Number of steps taking run-lengths to reduce the n-th composition in standard order to a singleton.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, Mar 31 2020

Keywords

Comments

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.
For the operation of taking the sequence of run-lengths of a finite sequence, runs-resistance is defined as the number of applications required to reach a singleton.

Examples

			Starting with 13789 and repeatedly applying A333627 gives: 13789 -> 859 -> 110 -> 29 -> 11 -> 6 -> 3 -> 2, corresponding to the compositions: (1,2,2,1,1,2,1,1,2,1) -> (1,2,2,1,2,1,1) -> (1,2,1,1,2) -> (1,1,2,1) -> (2,1,1) -> (1,2) -> (1,1) -> (2), so a(13789) = 7.
		

Crossrefs

Number of times applying A333627 to reach a power of 2, starting with n.
Positions of first appearances are A333629.
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.
- First appearances for specified run-lengths are A333630.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    runsres[q_]:=Length[NestWhileList[Length/@Split[#]&,q,Length[#]>1&]]-1;
    Table[runsres[stc[n]],{n,100}]

A334438 Heinz numbers of all integer partitions sorted first by sum, then by length, and finally reverse-lexicographically.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 7, 10, 9, 12, 16, 11, 14, 15, 20, 18, 24, 32, 13, 22, 21, 25, 28, 30, 27, 40, 36, 48, 64, 17, 26, 33, 35, 44, 42, 50, 45, 56, 60, 54, 80, 72, 96, 128, 19, 34, 39, 55, 49, 52, 66, 70, 63, 75, 88, 84, 100, 90, 81, 112, 120, 108, 160, 144, 192, 256
Offset: 0

Views

Author

Gus Wiseman, May 03 2020

Keywords

Comments

First differs from A185974 shifted left once at a(76) = 99, A185974(75) = 98.
A permutation of the positive integers.
This is the Abramowitz-Stegun ordering of integer partitions (A334433) except that the finer order is reverse-lexicographic instead of lexicographic. The version for reversed partitions is A334435.
The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k). This gives a bijective correspondence between positive integers and integer partitions.
As a triangle with row lengths A000041, the sequence starts {{1},{2},{3,4},{5,6,8},...}, so offset is 0.

Examples

			The sequence of terms together with their prime indices begins:
    1: {}            32: {1,1,1,1,1}       50: {1,3,3}
    2: {1}           13: {6}               45: {2,2,3}
    3: {2}           22: {1,5}             56: {1,1,1,4}
    4: {1,1}         21: {2,4}             60: {1,1,2,3}
    5: {3}           25: {3,3}             54: {1,2,2,2}
    6: {1,2}         28: {1,1,4}           80: {1,1,1,1,3}
    8: {1,1,1}       30: {1,2,3}           72: {1,1,1,2,2}
    7: {4}           27: {2,2,2}           96: {1,1,1,1,1,2}
   10: {1,3}         40: {1,1,1,3}        128: {1,1,1,1,1,1,1}
    9: {2,2}         36: {1,1,2,2}         19: {8}
   12: {1,1,2}       48: {1,1,1,1,2}       34: {1,7}
   16: {1,1,1,1}     64: {1,1,1,1,1,1}     39: {2,6}
   11: {5}           17: {7}               55: {3,5}
   14: {1,4}         26: {1,6}             49: {4,4}
   15: {2,3}         33: {2,5}             52: {1,1,6}
   20: {1,1,3}       35: {3,4}             66: {1,2,5}
   18: {1,2,2}       44: {1,1,5}           70: {1,3,4}
   24: {1,1,1,2}     42: {1,2,4}           63: {2,2,4}
Triangle begins:
   1
   2
   3   4
   5   6   8
   7  10   9  12  16
  11  14  15  20  18  24  32
  13  22  21  25  28  30  27  40  36  48  64
  17  26  33  35  44  42  50  45  56  60  54  80  72  96 128
This corresponds to the following tetrangle:
                  0
                 (1)
               (2)(11)
             (3)(21)(111)
        (4)(31)(22)(211)(1111)
  (5)(41)(32)(311)(221)(2111)(11111)
		

Crossrefs

Row lengths are A000041.
Ignoring length gives A129129.
Compositions under the same order are A296774 (triangle).
The dual version (sum/length/lex) is A334433.
The version for reversed partitions is A334435.
The constructive version is A334439 (triangle).
Lexicographically ordered reversed partitions are A026791.
Reversed partitions in Abramowitz-Stegun (sum/length/lex) order are A036036.
Partitions in increasing-length colexicographic order (sum/length/colex) are A036037.
Reverse-lexicographically ordered partitions are A080577.
Sorting reversed partitions by Heinz number gives A112798.
Graded lexicographically ordered partitions are A193073.
Partitions in colexicographic order (sum/colex) are A211992.
Graded Heinz numbers are given by A215366.
Sorting partitions by Heinz number gives A296150.

Programs

  • Mathematica
    revlensort[f_,c_]:=If[Length[f]!=Length[c],Length[f]
    				

Formula

A001221(a(n)) = A103921(n).
A001222(a(n)) = A036043(n).

A071364 Smallest number with same sequence of exponents in canonical prime factorization as n.

Original entry on oeis.org

1, 2, 2, 4, 2, 6, 2, 8, 4, 6, 2, 12, 2, 6, 6, 16, 2, 18, 2, 12, 6, 6, 2, 24, 4, 6, 8, 12, 2, 30, 2, 32, 6, 6, 6, 36, 2, 6, 6, 24, 2, 30, 2, 12, 12, 6, 2, 48, 4, 18, 6, 12, 2, 54, 6, 24, 6, 6, 2, 60, 2, 6, 12, 64, 6, 30, 2, 12, 6, 30, 2, 72, 2, 6, 18, 12, 6, 30, 2, 48, 16, 6, 2, 60, 6, 6, 6, 24
Offset: 1

Views

Author

Reinhard Zumkeller, May 21 2002

Keywords

Comments

A046523(a(n))=A046523(n); A046523(n)<=a(n)<=n; A001221(a(n))=A001221(n), A001222(a(n))=A001222(n); A020639(a(n))=2, A006530(a(n))=A000040(A001221(n))<=A006530(n); A000005(a(n))=A000005(n);
a(a(n))=a(n); a(n)=2^k iff n=p^k, p prime, k>0 (A000961); if n>1 is not a prime power, then a(n) mod 6 = 0; range of values = A055932, as distinct prime factors of a(n) are consecutive: a(n)=n iff n=A055932(k) for some k;
a(A003586(n))=A003586(n).

Examples

			a(105875) = a(5*5*5*7*11*11) = 2*2*2*3*5*5 = 600.
		

Crossrefs

Cf. A000040.
The range is A055932.
The reversed version is A331580.
Unsorted prime signature is A124010.
Numbers whose prime signature is aperiodic are A329139.

Programs

  • Haskell
    a071364 = product . zipWith (^) a000040_list . a124010_row
    -- Reinhard Zumkeller, Feb 19 2012
    
  • Mathematica
    Table[ e = Last /@ FactorInteger[n]; Product[Prime[i]^e[[i]], {i, Length[e]}], {n, 88}] (* Ray Chandler, Sep 23 2005 *)
  • PARI
    a(n) = f = factor(n); for (i=1, #f~, f[i,1] = prime(i)); factorback(f); \\ Michel Marcus, Jun 13 2014
    
  • Python
    from math import prod
    from sympy import prime, factorint
    def A071364(n): return prod(prime(i+1)**p[1] for i,p in enumerate(sorted(factorint(n).items()))) # Chai Wah Wu, Sep 16 2022

Formula

In prime factorization of n, replace least prime by 2, next least by 3, etc.
a(n) = product(A000040(k)^A124010(k): k=1..A001221(n)). - Reinhard Zumkeller, Apr 27 2013

Extensions

Extended by Ray Chandler, Sep 23 2005

A233249 a(1)=0; for k >= 1, let prime(k) map to 10...0 with k-1 zeros and let prime(k)*prime(m) map to the concatenation in binary of 2^(k-1) and 2^(m-1). For n >= 2, let the prime power factorization of n be mapped to r(n). a(n) is the term in A114994 which is c-equivalent to r(n) (see there our comment).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 8, 7, 10, 9, 16, 11, 32, 17, 18, 15, 64, 21, 128, 19, 34, 33, 256, 23, 36, 65, 42, 35, 512, 37, 1024, 31, 66, 129, 68, 43, 2048, 257, 130, 39, 4096, 69, 8192, 67, 74, 513, 16384, 47, 136, 73, 258, 131, 32768, 85, 132, 71, 514, 1025, 65536, 75
Offset: 1

Views

Author

Vladimir Shevelev, Dec 06 2013

Keywords

Comments

Let (10...0)_i (i>=0) denote 2^i in binary. Under (10...0)_i^k we understand a concatenation of (10...0)_i k times.
If n=Product_{i=1..m} p_i^t_i is the prime power factorization of n, then in the name r(n)=concatenation{i=1..m} ((10...0_(i-1)^t_i).
Numbers q and s are called c-equivalent if their binary expansions contain the same set of parts of the form 10...0. For example, 14=(1)(1)(10)~(10)(1)(1)=11.
Conversely, if n~n_1 such that n_1 is in A114994 and has c-factorization: n_1 = concatenation{i=m,...,0} ((10...0)i^t_i), one can consider "converse" sequence {s(n)}, where s(n) = Product{i=m..0} p_(i+1)^t_i.
For example, for n=22, n_1=21=((10)^2)(1), and s(22)=3^2*2=18.
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 binary expansion of k, prepending 0, taking first differences, and reversing again. Then a(n) is the number k such that the k-th composition in standard order consists of the prime indices of n in weakly decreasing order (the partition with Heinz number n). - Gus Wiseman, Apr 02 2020

Examples

			n=10=2*5 is mapped to (1)(100)~(100)(1). Since 9 is in A114994, then a(10)=9.
From _Gus Wiseman_, Apr 02 2020: (Start)
The sequence together with the corresponding compositions begins:
   0: ()             128: (8)             2048: (12)
   1: (1)             19: (3,1,1)          257: (8,1)
   2: (2)             34: (4,2)            130: (6,2)
   3: (1,1)           33: (5,1)             39: (3,1,1,1)
   4: (3)            256: (9)             4096: (13)
   5: (2,1)           23: (2,1,1,1)         69: (4,2,1)
   8: (4)             36: (3,3)           8192: (14)
   7: (1,1,1)         65: (6,1)             67: (5,1,1)
  10: (2,2)           42: (2,2,2)           74: (3,2,2)
   9: (3,1)           35: (4,1,1)          513: (9,1)
  16: (5)            512: (10)           16384: (15)
  11: (2,1,1)         37: (3,2,1)           47: (2,1,1,1,1)
  32: (6)           1024: (11)             136: (4,4)
  17: (4,1)           31: (1,1,1,1,1)       73: (3,3,1)
  18: (3,2)           66: (5,2)            258: (7,2)
  15: (1,1,1,1)      129: (7,1)            131: (6,1,1)
  64: (7)             68: (4,3)          32768: (16)
  21: (2,2,1)         43: (2,2,1,1)         85: (2,2,2,1)
For example, the Heinz number of (2,2,1) is 18, and the 21st composition in standard order is (2,2,1), so a(18) = 21.
(End)
		

Crossrefs

The sorted version is A114994.
The primorials A002110 map to A246534.
A partial inverse is A333219.
The reversed version is A333220.

Programs

  • Mathematica
    primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    Table[Total[2^Accumulate[primeMS[n]]]/2,{n,100}] (* Gus Wiseman, Apr 02 2020 *)

Formula

A059893(a(n)) = A333220(n). A124767(a(n)) = A001221(n). - Gus Wiseman, Apr 02 2020

Extensions

More terms from Peter J. C. Moses, Dec 07 2013

A334434 Heinz number of the n-th integer partition in graded lexicographic order.

Original entry on oeis.org

1, 2, 4, 3, 8, 6, 5, 16, 12, 9, 10, 7, 32, 24, 18, 20, 15, 14, 11, 64, 48, 36, 27, 40, 30, 25, 28, 21, 22, 13, 128, 96, 72, 54, 80, 60, 45, 50, 56, 42, 35, 44, 33, 26, 17, 256, 192, 144, 108, 81, 160, 120, 90, 100, 75, 112, 84, 63, 70, 49, 88, 66, 55, 52, 39, 34, 19
Offset: 0

Views

Author

Gus Wiseman, May 01 2020

Keywords

Comments

A permutation of the positive integers.
This is the graded reverse of the so-called "Mathematica" order (A080577, A129129).
The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k). This gives a bijective correspondence between positive integers and integer partitions.
As a triangle with row lengths A000041, the sequence starts {{1},{2},{4,3},{8,6,5},...}, so offset is 0.

Examples

			The sequence of terms together with their prime indices begins:
    1: {}              11: {5}                 45: {2,2,3}
    2: {1}             64: {1,1,1,1,1,1}       50: {1,3,3}
    4: {1,1}           48: {1,1,1,1,2}         56: {1,1,1,4}
    3: {2}             36: {1,1,2,2}           42: {1,2,4}
    8: {1,1,1}         27: {2,2,2}             35: {3,4}
    6: {1,2}           40: {1,1,1,3}           44: {1,1,5}
    5: {3}             30: {1,2,3}             33: {2,5}
   16: {1,1,1,1}       25: {3,3}               26: {1,6}
   12: {1,1,2}         28: {1,1,4}             17: {7}
    9: {2,2}           21: {2,4}              256: {1,1,1,1,1,1,1,1}
   10: {1,3}           22: {1,5}              192: {1,1,1,1,1,1,2}
    7: {4}             13: {6}                144: {1,1,1,1,2,2}
   32: {1,1,1,1,1}    128: {1,1,1,1,1,1,1}    108: {1,1,2,2,2}
   24: {1,1,1,2}       96: {1,1,1,1,1,2}       81: {2,2,2,2}
   18: {1,2,2}         72: {1,1,1,2,2}        160: {1,1,1,1,1,3}
   20: {1,1,3}         54: {1,2,2,2}          120: {1,1,1,2,3}
   15: {2,3}           80: {1,1,1,1,3}         90: {1,2,2,3}
   14: {1,4}           60: {1,1,2,3}          100: {1,1,3,3}
Triangle begins:
    1
    2
    4   3
    8   6   5
   16  12   9  10   7
   32  24  18  20  15  14  11
   64  48  36  27  40  30  25  28  21  22  13
  128  96  72  54  80  60  45  50  56  42  35  44  33  26  17
  ...
This corresponds to the tetrangle:
                  0
                 (1)
               (11)(2)
             (111)(21)(3)
        (1111)(211)(22)(31)(4)
  (11111)(2111)(221)(311)(32)(41)(5)
		

Crossrefs

Row lengths are A000041.
The dual version (sum/revlex) is A129129.
The constructive version is A193073.
Compositions under the same order are A228351.
The length-sensitive version is A334433.
The version for reversed (weakly increasing) partitions is A334437.
Lexicographically ordered reversed partitions are A026791.
Reversed partitions in Abramowitz-Stegun order (sum/length/lex) are A036036.
Reverse-lexicographically ordered partitions are A080577.
Sorting reversed partitions by Heinz number gives A112798.
Graded Heinz numbers are A215366.
Sorting partitions by Heinz number gives A296150.
Row sums give A145519.

Programs

  • Maple
    T:= n-> map(p-> mul(ithprime(i), i=p), combinat[partition](n))[]:
    seq(T(n), n=0..8);  # Alois P. Heinz, Jan 26 2025
  • Mathematica
    lexsort[f_,c_]:=OrderedQ[PadRight[{f,c}]];
    Join@@Table[Times@@Prime/@#&/@Sort[IntegerPartitions[n],lexsort],{n,0,8}]
    - or -
    Join@@Table[Times@@Prime/@#&/@Reverse[IntegerPartitions[n]],{n,0,8}]

Formula

A001222(a(n)) appears to be A049085(n).

A334436 Heinz numbers of all reversed integer partitions sorted first by sum and then reverse-lexicographically.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 7, 9, 10, 12, 16, 11, 15, 14, 18, 20, 24, 32, 13, 25, 21, 27, 22, 30, 28, 36, 40, 48, 64, 17, 35, 33, 45, 26, 50, 42, 54, 44, 60, 56, 72, 80, 96, 128, 19, 49, 55, 39, 75, 63, 81, 34, 70, 66, 90, 52, 100, 84, 108, 88, 120, 112, 144, 160, 192, 256
Offset: 0

Views

Author

Gus Wiseman, May 02 2020

Keywords

Comments

First differs from A334435 at a(22) = 27, A334435(22) = 22.
A permutation of the positive integers.
Reversed integer partitions are finite weakly increasing sequences of positive integers. For non-reversed partitions, see A129129 and A228531.
This is the so-called "Mathematica" order (A080577).
The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k). This gives a bijective correspondence between positive integers and integer partitions.

Examples

			The sequence of terms together with their prime indices begins:
    1: {}            32: {1,1,1,1,1}       42: {1,2,4}
    2: {1}           13: {6}               54: {1,2,2,2}
    3: {2}           25: {3,3}             44: {1,1,5}
    4: {1,1}         21: {2,4}             60: {1,1,2,3}
    5: {3}           27: {2,2,2}           56: {1,1,1,4}
    6: {1,2}         22: {1,5}             72: {1,1,1,2,2}
    8: {1,1,1}       30: {1,2,3}           80: {1,1,1,1,3}
    7: {4}           28: {1,1,4}           96: {1,1,1,1,1,2}
    9: {2,2}         36: {1,1,2,2}        128: {1,1,1,1,1,1,1}
   10: {1,3}         40: {1,1,1,3}         19: {8}
   12: {1,1,2}       48: {1,1,1,1,2}       49: {4,4}
   16: {1,1,1,1}     64: {1,1,1,1,1,1}     55: {3,5}
   11: {5}           17: {7}               39: {2,6}
   15: {2,3}         35: {3,4}             75: {2,3,3}
   14: {1,4}         33: {2,5}             63: {2,2,4}
   18: {1,2,2}       45: {2,2,3}           81: {2,2,2,2}
   20: {1,1,3}       26: {1,6}             34: {1,7}
   24: {1,1,1,2}     50: {1,3,3}           70: {1,3,4}
Triangle begins:
   1
   2
   3   4
   5   6   8
   7   9  10  12  16
  11  15  14  18  20  24  32
  13  25  21  27  22  30  28  36  40  48  64
  17  35  33  45  26  50  42  54  44  60  56  72  80  96 128
This corresponds to the following tetrangle:
                  0
                 (1)
               (2)(11)
             (3)(12)(111)
        (4)(22)(13)(112)(1111)
  (5)(23)(14)(122)(113)(1112)(11111)
		

Crossrefs

Row lengths are A000041.
Compositions under the same order are A066099 (triangle).
The version for non-reversed partitions is A129129.
The constructive version is A228531.
The lengths of these partitions are A333486.
The length-sensitive version is A334435.
The dual version (sum/lex) is A334437.
Lexicographically ordered reversed partitions are A026791.
Reversed partitions in Abramowitz-Stegun (sum/length/lex) order are A036036.
Partitions in increasing-length colexicographic order (sum/length/colex) are A036037.
Reverse-lexicographically ordered partitions are A080577.
Sorting reversed partitions by Heinz number gives A112798.
Graded lexicographically ordered partitions are A193073.
Partitions in colexicographic order (sum/colex) are A211992.
Graded Heinz numbers are A215366.
Sorting partitions by Heinz number gives A296150.
Partitions in dual Abramowitz-Stegun (sum/length/revlex) order are A334439.

Programs

  • Mathematica
    lexsort[f_,c_]:=OrderedQ[PadRight[{f,c}]];
    Table[Times@@Prime/@#&/@Reverse[Sort[Sort/@IntegerPartitions[n],lexsort]],{n,0,8}]

Formula

A001222(a(n)) = A333486(n).

A351596 Numbers k such that the k-th composition in standard order has all distinct run-lengths.

Original entry on oeis.org

0, 1, 2, 3, 4, 7, 8, 10, 11, 14, 15, 16, 19, 21, 23, 26, 28, 30, 31, 32, 35, 36, 39, 42, 47, 56, 60, 62, 63, 64, 67, 71, 73, 74, 79, 84, 85, 87, 95, 100, 106, 112, 119, 120, 122, 123, 124, 126, 127, 128, 131, 135, 136, 138, 143, 146, 159, 164, 168, 170, 171
Offset: 1

Views

Author

Gus Wiseman, Feb 24 2022

Keywords

Comments

The n-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 n, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

Examples

			The terms together with their binary expansions and corresponding compositions begin:
   0:      0  ()
   1:      1  (1)
   2:     10  (2)
   3:     11  (1,1)
   4:    100  (3)
   7:    111  (1,1,1)
   8:   1000  (4)
  10:   1010  (2,2)
  11:   1011  (2,1,1)
  14:   1110  (1,1,2)
  15:   1111  (1,1,1,1)
  16:  10000  (5)
  19:  10011  (3,1,1)
  21:  10101  (2,2,1)
  23:  10111  (2,1,1,1)
		

Crossrefs

The version using binary expansions is A044813.
The version for Heinz numbers and prime multiplicities is A130091.
These compositions are counted by A329739, normal A329740.
The version for runs instead of run-lengths is A351290, counted by A351013.
A005811 counts runs in binary expansion, distinct A297770.
A011782 counts integer compositions.
A085207 represents concatenation of standard compositions, reverse A085208.
A333489 ranks anti-runs, complement A348612.
A345167 ranks alternating compositions, counted by A025047.
A351204 counts partitions where every permutation has all distinct runs.
Counting words with all distinct run-lengths:
- A032020 = binary expansions, for runs A351018.
- A351017 = binary words, for runs A351016.
- A351292 = patterns, for runs A351200.
Selected statistics of standard compositions (A066099, A228351):
- Length is A000120.
- Sum is A070939.
- Runs are counted by A124767, distinct A351014.
- Heinz number is A333219.
- Number of distinct parts is A334028.

Programs

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

A351290 Numbers k such that the k-th composition in standard order has all distinct runs.

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, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 47, 48, 50, 51, 52, 55, 56, 57, 58, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 78
Offset: 1

Views

Author

Gus Wiseman, Feb 10 2022

Keywords

Comments

The n-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 n, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

Examples

			The terms together with their binary expansions and corresponding compositions begin:
   0:      0  ()
   1:      1  (1)
   2:     10  (2)
   3:     11  (1,1)
   4:    100  (3)
   5:    101  (2,1)
   6:    110  (1,2)
   7:    111  (1,1,1)
   8:   1000  (4)
   9:   1001  (3,1)
  10:   1010  (2,2)
  11:   1011  (2,1,1)
  12:   1100  (1,3)
  14:   1110  (1,1,2)
  15:   1111  (1,1,1,1)
		

Crossrefs

The version for Heinz numbers and prime multiplicities is A130091.
The version using binary expansions is A175413, complement A351205.
The version for run-lengths instead of runs is A329739.
These compositions are counted by A351013.
The complement is A351291.
A005811 counts runs in binary expansion, distinct A297770.
A011782 counts integer compositions.
A044813 lists numbers whose binary expansion has distinct run-lengths.
A085207 represents concatenation of standard compositions, reverse A085208.
A333489 ranks anti-runs, complement A348612.
A345167 ranks alternating compositions, counted by A025047.
A351204 counts partitions where every permutation has all distinct runs.
Counting words with all distinct runs:
- A351016 = binary words, for run-lengths A351017.
- A351018 = binary expansions, for run-lengths A032020.
- A351200 = patterns, for run-lengths A351292.
- A351202 = permutations of prime factors.
Selected statistics of standard compositions:
- Length is A000120.
- Parts are A066099, reverse A228351.
- Sum is A070939.
- Runs are counted by A124767, distinct A351014.
- Heinz number is A333219.
- Number of distinct parts is A334028.
Selected classes of standard compositions:
- Partitions are A114994, strict A333256.
- Multisets are A225620, strict A333255.
- Strict compositions are A233564.
- Constant compositions are A272919.

Programs

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

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

Views

Author

Gus Wiseman, Mar 17 2020

Keywords

Comments

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.

Examples

			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)
		

Crossrefs

Row lengths are A008480.
Column k = 1 is A233249.
Column k = -1 is A333220.
A related triangle for partitions is A215366.

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[Sort[fbi/@Accumulate/@Permutations[primeMS[n]]],{n,30}]
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