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.

Showing 1-8 of 8 results.

A260743 Sequence A261220 shown in factorial base: a(n) = A007623(A261220(n)).

Original entry on oeis.org

0, 1, 10, 20, 100, 101, 200, 220, 300, 310, 1000, 1001, 1010, 1020, 2000, 2001, 2200, 2300, 3000, 3020, 3100, 3300, 4000, 4010, 4100, 4200, 10000, 10001, 10010, 10020, 10100, 10101, 10200, 10220, 10300, 10310, 20000, 20001, 20010, 20020, 22000, 22001, 23000, 23020, 24000, 24010, 30000, 30001, 30200, 30300, 31000, 31001, 33000, 33300, 34000, 34200, 40000, 40020, 40100, 40300, 41000, 41020, 42000, 42300
Offset: 0

Views

Author

Antti Karttunen, Aug 26 2015

Keywords

Crossrefs

Subsequence: A014417.

Programs

Formula

a(n) = A007623(A261220(n)).

A060125 Self-inverse infinite permutation which shows the position of the inverse of each finite permutation in A060117 (or A060118) in the same sequence; or equally, the cross-indexing between A060117 and A060118.

Original entry on oeis.org

0, 1, 2, 5, 4, 3, 6, 7, 14, 23, 22, 15, 12, 19, 8, 11, 16, 21, 18, 13, 20, 17, 10, 9, 24, 25, 26, 29, 28, 27, 54, 55, 86, 119, 118, 87, 84, 115, 56, 59, 88, 117, 114, 85, 116, 89, 58, 57, 48, 49, 74, 101, 100, 75, 30, 31, 38, 47, 46, 39, 60, 67, 80, 107, 112, 93, 66, 61, 92
Offset: 0

Views

Author

Antti Karttunen, Mar 02 2001

Keywords

Comments

PermRank3Aux is a slight modification of rank2 algorithm presented in Myrvold-Ruskey article.

Crossrefs

Cf. A261220 (fixed points).
Cf. A056019 (compare the scatter plots).

Programs

  • Maple
    with(group); permul := (a,b) -> mulperms(b,a); swap := (p,i,j) -> convert(permul(convert(p,'disjcyc'),[[i,j]]),'permlist',nops(p));
    PermRank3Aux := proc(n, p, q) if(1 = n) then RETURN(0); else RETURN((n-p[n])*((n-1)!) + PermRank3Aux(n-1,swap(p,n,q[n]),swap(q,n,p[n]))); fi; end;
    PermRank3R := p -> PermRank3Aux(nops(p),p,convert(invperm(convert(p,'disjcyc')),'permlist',nops(p)));
    PermRank3L := p -> PermRank3Aux(nops(p),convert(invperm(convert(p,'disjcyc')),'permlist',nops(p)),p);
    # a(n) = PermRank3L(PermUnrank3R(n)) or PermRank3R(PermUnrank3L(n)) or PermRank3L(convert(invperm(convert(PermUnrank3L(j), 'disjcyc')), 'permlist', nops(PermUnrank3L(j))))

A060131 a(n) = A072411(A275725(n)); order of each permutation listed in tables A060117 and A060118, i.e., the least common multiple of the cycle sizes.

Original entry on oeis.org

1, 2, 2, 3, 2, 3, 2, 2, 3, 4, 3, 4, 2, 3, 3, 4, 2, 4, 2, 3, 2, 4, 3, 4, 2, 2, 2, 6, 2, 6, 3, 6, 4, 5, 4, 5, 3, 4, 4, 5, 6, 5, 3, 4, 6, 5, 4, 5, 2, 2, 3, 4, 3, 4, 3, 6, 4, 5, 4, 5, 2, 6, 4, 5, 6, 5, 2, 6, 6, 5, 4, 5, 2, 3, 3, 4, 2, 4, 2, 6, 4, 5, 6, 5, 3, 4, 4, 5, 6, 5, 2, 4, 6, 5, 6, 5, 2, 3, 2, 4, 3, 4, 2, 6, 6, 5, 4, 5, 2, 4, 6, 5, 6, 5, 3, 4, 6, 5, 4, 5, 2
Offset: 0

Views

Author

Antti Karttunen, Mar 05 2001

Keywords

Crossrefs

Cf. A261220 (gives the positions of 1 and 2's).
Cf. A275813 (indices of odd terms), A275814 (indices of even terms).

Programs

Formula

From Antti Karttunen, Aug 09 2016: (Start)
a(n) = A072411(A275725(n)).
a(n) = A055092(A060120(n)).
(End)

A275804 Numbers with at most one nonzero digit on each digit slope of the factorial base representation of n.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 7, 8, 9, 10, 12, 13, 16, 18, 20, 24, 25, 26, 27, 28, 30, 31, 32, 33, 34, 36, 37, 40, 42, 44, 48, 49, 50, 51, 52, 60, 61, 64, 66, 68, 72, 73, 76, 78, 79, 82, 90, 96, 98, 102, 104, 108, 120, 121, 122, 123, 124, 126, 127, 128, 129, 130, 132, 133, 136, 138, 140, 144, 145, 146, 147, 148, 150, 151, 152, 153, 154, 156, 157, 160
Offset: 0

Views

Author

Antti Karttunen, Aug 10 2016

Keywords

Comments

Indexing starts from zero, because a(0) = 0 is a special case in this sequence.
Numbers n for which A275947(n) = 0 or equally, for which A275811(n) <= 1.
Numbers n for which A008683(A275734(n)) <> 0, that is, indices of squarefree terms in A275734.
Numbers n for which A060130(n) = A060502(n).
Numbers with at most one nonzero digit on each digit slope of the factorial base representation of n (see A275811 and A060502 for the definition of slopes in this context). More exactly: numbers n in whose factorial base representation (A007623) there does not exist a pair of digit positions i_1 and i_2 with nonzero digits d_1 and d_2, such that (i_1 - d_1) = (i_2 - d_2).

Crossrefs

Complement: A275805.
Indices of zeros in A275947 and A275962.
Intersection with A276005 gives A261220.
Cf. A059590 (a subsequence).

Programs

  • Python
    from operator import mul
    from sympy import prime, factorial as f
    from sympy.ntheory.factor_ import core
    def a007623(n, p=2): return n if n

    0 else '0' for i in x)[::-1] return 0 if n==1 else sum([int(y[i])*f(i + 1) for i in range(len(y))]) def a(n): return 1 if n==0 else a275732(n)*a(a257684(n)) def ok(n): return 1 if n==0 else core(a(n))==a(n) print([n for n in range(201) if ok(n)]) # Indranil Ghosh, Jun 19 2017

A276005 Numbers with hit-free factorial base representations; positions of zeros in A276004 & A276007.

Original entry on oeis.org

0, 1, 2, 4, 5, 6, 7, 12, 14, 16, 18, 19, 20, 22, 23, 24, 25, 26, 28, 29, 48, 49, 54, 55, 60, 66, 67, 72, 74, 76, 78, 84, 86, 88, 90, 92, 94, 96, 97, 98, 100, 101, 102, 103, 108, 110, 112, 114, 115, 116, 118, 119, 120, 121, 122, 124, 125, 126, 127, 132, 134, 136, 138, 139, 140, 142, 143, 240, 241, 242, 244, 245, 264, 265, 266, 268, 269, 288, 289, 312, 314, 316
Offset: 0

Views

Author

Antti Karttunen, Aug 17 2016

Keywords

Comments

We say there is a "hit" in factorial base representation (A007623) of n when there is any such pair of nonzero digits d_i and d_j in positions i > j so that (i - d_i) = j. Here the rightmost (least significant digit) occurs at position 1. This sequence gives all "hit-free" numbers, meaning that for every nonzero digit d_i (in position i) in their factorial base representation the digit at the position (i - d_i) is 0.
Also numbers n for which A060502(n) = A060128(n), in other words, the numbers n for which the number of slopes in their factorial base representation (A007623) is equal to the number of non-singleton cycles of the permutation listed as n-th permutation in the list A060117 (or A060118).
This can be viewed as a factorial base analog of base-2 related A003714.

Examples

			n=14 (factorial base "210") is included because 2 occurs in position 3 and 1 occurs in position 2, thus as (3-2) = 1 <> 2, 2 does not "hit" digit 1.
n=15 ("211") is NOT included because 2 occurring in position 3 hits the rightmost 1 in position 1 (as 3-2 = 1), and moreover, also the middle 1 hits the rightmost 1 as 2-1 = 1.
		

Crossrefs

Complement: A276006.
Cf. A060112 (a subsequence).
Intersection with A275804 gives A261220.
Cf. also A003714, A060117 and A060118.

Formula

Other identities. For all n >= 1:
a(A000110(n)) = n! = A000142(n). [To be proved.]

A261219 Main diagonal of A261216: a(n) = A261216(n,n).

Original entry on oeis.org

0, 0, 0, 5, 0, 3, 0, 0, 14, 16, 22, 20, 0, 19, 8, 20, 0, 7, 0, 13, 0, 7, 10, 16, 0, 0, 0, 5, 0, 3, 54, 54, 60, 65, 66, 69, 84, 90, 78, 95, 84, 81, 114, 108, 114, 107, 102, 111, 0, 0, 74, 76, 100, 98, 30, 30, 78, 83, 102, 105, 0, 19, 26, 45, 100, 119, 0, 13, 74, 87, 28, 41, 0, 97, 50, 98, 0, 49, 0, 97, 26, 117, 22, 47, 36, 108, 60, 113, 36, 63, 0, 25, 50, 33, 10, 59, 0, 73, 0, 49, 52
Offset: 0

Views

Author

Antti Karttunen, Aug 26 2015

Keywords

Comments

Equally: main diagonal of A261217.
For permutation p, which has rank n in permutation list A060117, a(n) gives the rank of the "square" of that permutation (obtained by composing it with itself as: q(i) = p(p(i))) in the same list. Equally, if permutation p has rank n in the order used in list A060118, a(n) gives the rank of the p*p in that same list. Thus zeros (which mark the identity permutation, with rank 0 in both orders) occur at positions where the permutations of A060117 (equally: of A060118) are involutions, listed by A261220.

Crossrefs

Main diagonal of A261216 and A261217.
Cf. A261220 (the positions of zeros).
Cf. also A261099, A089841.
Related permutations: A060119, A060126.

Formula

a(n) = A261216(n,n) = A261217(n,n).
By conjugating a similar sequence:
a(n) = A060126(A261099(A060119(n))).

A275803 a(n) = A051903(A275725(n)); maximal cycle sizes of finite permutations listed in the order A060117 / A060118.

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Aug 10 2016

Keywords

Examples

			For n=27, which in factorial base (A007623) is "1011" and encodes (in A060118-order) permutation "23154" with one 3-cycle and one 2-cycle, the longest cycle has three elements, thus a(27) = 3.
		

Crossrefs

Cf. A261220 (gives the positions of 1 and 2's).
Differs from A060131 for the first time at n=27, where a(27) = 3, while A060131(27) = 6.

Programs

Formula

a(n) = A051903(A275725(n)).

A014489 Positions of involutions (permutations whose square is the identity) in reverse colexicographic order (A055089/A195663).

Original entry on oeis.org

0, 1, 2, 5, 6, 7, 14, 16, 21, 23, 24, 25, 26, 29, 54, 55, 60, 67, 80, 82, 86, 94, 105, 107, 111, 119, 120, 121, 122, 125, 126, 127, 134, 136, 141, 143, 264, 265, 266, 269, 288, 289, 314, 316, 339, 341, 390, 391, 396, 403, 414, 415, 444, 450, 469
Offset: 0

Views

Author

Keywords

Crossrefs

Positions of zeros in A261099.
From a(1)=1 onward also positions of 2's in A055092.
Subsequences: A060112, A064640.
Cf. also A261220.

Programs

  • Maple
    N:= 100: # to get a(0) to a(N)
    M:= 0: A[0]:= 0: count:= 0:
    for m from 2 while count < N do
      P:= remove(t -> t[1]=1, combinat:-permute(m));
      P:= map(t -> ListTools:-Reverse(subs([seq(i=m+1-i,i=1..m)],t)),P);
      R:= select(t -> max(map(nops,convert(P[t],disjcyc))) = 2, [$1..nops(P)]);
      for r in R do
         count:= count+1;
         A[count]:= r+M;
         if count = N then break fi;
      od:
      M:= M + nops(P);
    od:
    seq(A[i],i=0..count); # Robert Israel, Oct 28 2015

Extensions

Name changed by Antti Karttunen, Aug 30 2015
Showing 1-8 of 8 results.