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 1-10 of 35 results. Next

A040039 First differences of A033485; also A033485 with terms repeated.

Original entry on oeis.org

1, 1, 2, 2, 3, 3, 5, 5, 7, 7, 10, 10, 13, 13, 18, 18, 23, 23, 30, 30, 37, 37, 47, 47, 57, 57, 70, 70, 83, 83, 101, 101, 119, 119, 142, 142, 165, 165, 195, 195, 225, 225, 262, 262, 299, 299, 346, 346, 393, 393, 450, 450, 507, 507, 577, 577, 647, 647, 730, 730, 813, 813, 914, 914, 1015, 1015, 1134, 1134, 1253, 1253, 1395, 1395
Offset: 0

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Author

Keywords

Comments

Apparently a(n) = number of partitions (p_1, p_2, ..., p_k) of n+1, with p_1 >= p_2 >= ... >= p_k, such that for each i, p_i > p_{i+1}+...+p_k. - John McKay (mac(AT)mathstat.concordia.ca), Mar 06 2009
Comment from John McKay confirmed in paper by Bessenrodt, Olsson, and Sellers. Such partitions are called "strongly decreasing" partitions in the paper, see the function s(n) therein.
Also the number of unlabeled binary rooted trees with 2*n + 3 nodes in which the two branches directly under any given non-leaf node are either equal or at least one of them is a leaf. - Gus Wiseman, Oct 08 2018
From Gus Wiseman, Apr 06 2021: (Start)
This sequence counts both of the following essentially equivalent things:
1. Sets of distinct positive integers with maximum n + 1 in which all adjacent elements have quotients < 1/2. For example, the a(0) = 1 through a(8) = 7 subsets are:
{1} {2} {3} {4} {5} {6} {7} {8} {9}
{1,3} {1,4} {1,5} {1,6} {1,7} {1,8} {1,9}
{2,5} {2,6} {2,7} {2,8} {2,9}
{3,7} {3,8} {3,9}
{1,3,7} {1,3,8} {4,9}
{1,3,9}
{1,4,9}
2. Sets of distinct positive integers with maximum n + 1 whose first differences are term-wise greater than their decapitation (remove the maximum). For example, the set q = {1,4,9} has first differences (3,5), which are greater than (1,4), so q is counted under a(8). On the other hand, r = {1,5,9} has first differences (4,4), which are not greater than (1,5), so r is not counted under a(8).
Also the number of partitions of n + 1 into powers of 2 covering an initial interval of powers of 2. For example, the a(0) = 1 through a(8) = 7 partitions are:
1 11 21 211 221 2211 421 4211 4221
111 1111 2111 21111 2221 22211 22221
11111 111111 22111 221111 42111
211111 2111111 222111
1111111 11111111 2211111
21111111
111111111
(End)

Examples

			From _Joerg Arndt_, Dec 17 2012: (Start)
The a(19-1)=30 strongly decreasing partitions of 19 are (in lexicographic order)
[ 1]    [ 10 5 3 1 ]
[ 2]    [ 10 5 4 ]
[ 3]    [ 10 6 2 1 ]
[ 4]    [ 10 6 3 ]
[ 5]    [ 10 7 2 ]
[ 6]    [ 10 8 1 ]
[ 7]    [ 10 9 ]
[ 8]    [ 11 5 2 1 ]
[ 9]    [ 11 5 3 ]
[10]    [ 11 6 2 ]
[11]    [ 11 7 1 ]
[12]    [ 11 8 ]
[13]    [ 12 4 2 1 ]
[14]    [ 12 4 3 ]
[15]    [ 12 5 2 ]
[16]    [ 12 6 1 ]
[17]    [ 12 7 ]
[18]    [ 13 4 2 ]
[19]    [ 13 5 1 ]
[20]    [ 13 6 ]
[21]    [ 14 3 2 ]
[22]    [ 14 4 1 ]
[23]    [ 14 5 ]
[24]    [ 15 3 1 ]
[25]    [ 15 4 ]
[26]    [ 16 2 1 ]
[27]    [ 16 3 ]
[28]    [ 17 2 ]
[29]    [ 18 1 ]
[30]    [ 19 ]
The a(20-1)=30 strongly decreasing partitions of 20 are obtained by adding 1 to the first part in each partition in the list.
(End)
From _Gus Wiseman_, Oct 08 2018: (Start)
The a(-1) = 1 through a(4) = 3 semichiral binary rooted trees:
  o  (oo)  (o(oo))  ((oo)(oo))  (o((oo)(oo)))  ((o(oo))(o(oo)))
                    (o(o(oo)))  (o(o(o(oo))))  (o(o((oo)(oo))))
                                               (o(o(o(o(oo)))))
(End)
		

Crossrefs

Cf. A000123.
The equal case is A001511.
The reflected version is A045690.
The unequal (anti-run) version is A045691.
A000929 counts partitions with all adjacent parts x >= 2y.
A002843 counts compositions with all adjacent parts x <= 2y.
A018819 counts partitions into powers of 2.
A154402 counts partitions with all adjacent parts x = 2y.
A274199 counts compositions with all adjacent parts x < 2y.
A342094 counts partitions with all adjacent parts x <= 2y (strict: A342095).
A342096 counts partitions without adjacent x >= 2y (strict: A342097).
A342098 counts partitions with all adjacent parts x > 2y.
A342337 counts partitions with all adjacent parts x = y or x = 2y.

Programs

  • Maple
    # For example, the five partitions of 4, written in nonincreasing order, are
    # [1,1,1,1], [2,1,1], [2,2], [3,1], [4].
    # Only the last two satisfy the condition, and a(3)=2.
    # The Maple program below verifies this for small values of n.
    with(combinat); N:=8; a:=array(1..N); c:=array(1..N);
    for n from 1 to N do p:=partition(n); np:=nops(p); t:=0;
    for s to np do r:=p[s]; r:=sort(r,`>`); nr:=nops(r); j:=1;
    while jsum(r[k],k=j+1..nr) do j:=j+1;od; # gives A040039
    #while j= sum(r[k],k=j+1..nr) do j:=j+1;od; # gives A018819
    if j=nr then t:=t+1;fi od; a[n]:=t; od;
    # John McKay
  • Mathematica
    T[n_, m_] := T[n, m] = Sum[Binomial[n-2k-1, n-2k-m] Sum[Binomial[m, i] T[k, i], {i, 1, k}], {k, 0, (n-m)/2}] + Binomial[n-1, n-m];
    a[n_] := T[n+1, 1];
    Table[a[n], {n, 0, 80}] (* Jean-François Alcover, Jul 27 2018, after Vladimir Kruchinin *)
    Table[Length[Select[Subsets[Range[n]],MemberQ[#,n]&&And@@Table[#[[i-1]]/#[[i]]<1/2,{i,2,Length[#]}]&]],{n,15}] (* Gus Wiseman, Apr 06 2021 *)
  • Maxima
    T(n,m):=sum(binomial(n-2*k-1,n-2*k-m)*sum(binomial(m,i)*T(k,i),i,1,k),k,0,(n-m)/2)+binomial(n-1,n-m);
    makelist(T(n+1,1),n,0,40); /* Vladimir Kruchinin, Mar 19 2015 */
    
  • PARI
    /* compute as "A033485 with terms repeated" */
    b(n) = if(n<2, 1, b(floor(n/2))+b(n-1));  /* A033485 */
    a(n) = b(n\2+1); /* note different offsets */
    for(n=0,99, print1(a(n),", ")); /* Joerg Arndt, Jan 21 2011 */
    
  • Python
    from itertools import islice
    from collections import deque
    def A040039_gen(): # generator of terms
        aqueue, f, b, a = deque([2]), True, 1, 2
        yield from (1, 1, 2, 2)
        while True:
            a += b
            yield from (a, a)
            aqueue.append(a)
            if f: b = aqueue.popleft()
            f = not f
    A040039_list = list(islice(A040039_gen(),40)) # Chai Wah Wu, Jun 07 2022

Formula

Let T(x) be the g.f, then T(x) = 1 + x/(1-x)*T(x^2) = 1 + x/(1-x) * ( 1 + x^2/(1-x^2) * ( 1 + x^4/(1-x^4) * ( 1 + x^8/(1-x^8) *(...) ))). [Joerg Arndt, May 11 2010]
From Joerg Arndt, Oct 02 2013: (Start)
G.f.: sum(k>=1, x^(2^k-1) / prod(j=0..k-1, 1-x^(2^k) ) ) [Bessenrodt/Olsson/Sellers].
G.f.: 1/(2*x^2) * ( 1/prod(k>=0, 1 - x^(2^k) ) - (1 + x) ).
a(n) = 1/2 * A018819(n+2).
(End)
a(n) = T(n+1,1), where T(n,m)=sum(k..0,(n-m)/2, binomial(n-2*k-1,n-2*k-m)*sum(i=1..k, binomial(m,i)*T(k,i)))+binomial(n-1,n-m). - Vladimir Kruchinin, Mar 19 2015
Using offset 1: a(1) = 1; a(n even) = a(n-1); a(n odd) = a(n-1) + a((n-1)/2). - Gus Wiseman, Oct 08 2018

A178855 Partial sums of A033485.

Original entry on oeis.org

1, 3, 6, 11, 18, 28, 41, 59, 82, 112, 149, 196, 253, 323, 406, 507, 626, 768, 933, 1128, 1353, 1615, 1914, 2260, 2653, 3103, 3610, 4187, 4834, 5564, 6377, 7291, 8306, 9440, 10693, 12088, 13625, 15327, 17194, 19256, 21513, 23995, 26702, 29671, 32902, 36432
Offset: 1

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Author

Philippe Deléham, Jun 19 2010

Keywords

Crossrefs

Programs

  • Maple
    b:= proc(n) option remember;
          `if`(n<2, n, b(n-1)+b(iquo(n, 2)))
        end:
    a:= n-> (b(2*n+1)-1)/2:
    seq(a(n), n=1..60);  # Alois P. Heinz, Feb 17 2022
  • Python
    from itertools import islice
    from collections import deque
    def A178855_gen(): # generator of terms
        aqueue, f, b, a = deque([2]), True, 1, 2
        while True:
            a += b
            aqueue.append(a)
            if f:
                yield (a-1)//2
                b = aqueue.popleft()
            f = not f
    A178885_list = list(islice(A178855_gen(),40)) # Chai Wah Wu, Jun 08 2022

Formula

a(n) = (A033485(2n+1)-1)/2.

Extensions

a(40) corrected by Georg Fischer, Aug 28 2020

A094451 a(n) = A033485(n) modulo 3.

Original entry on oeis.org

1, 2, 0, 2, 1, 1, 1, 0, 2, 0, 1, 2, 0, 1, 2, 2, 2, 1, 0, 0, 0, 1, 2, 1, 0, 0, 0, 1, 2, 1, 0, 2, 1, 0, 2, 0, 1, 1, 1, 1, 1, 1, 1, 2, 0, 2, 1, 2, 0, 0, 0, 0, 0, 0, 0, 1, 2, 1, 0, 1, 2, 2, 2, 1, 0, 1, 2, 2, 2, 1, 0, 0, 0, 1, 2, 0, 1, 2
Offset: 1

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Author

Philippe Deléham, Jun 04 2004

Keywords

Comments

Partials sums modulo 3 of the sequence : 1, a(1), a(1), a(2), a(2), a(3), a(3), a(4), a(4), a(5), ...

Crossrefs

Cf. A033485.

Programs

  • Python
    from itertools import islice
    from collections import deque
    def A094451_gen(): # generator of terms
        aqueue, f, b, a = deque([2]), True, 1, 2
        yield from (1, 2)
        while True:
            a = (a+b) % 3
            yield a
            aqueue.append(a)
            if f: b = aqueue.popleft()
            f = not f
    A094451_list = list(islice(A094451_gen(),40)) # Chai Wah Wu, Jun 08 2022

A179045 Triangle T(n,k), 1<=k<=n, read by rows, related to A033485.

Original entry on oeis.org

1, 2, 1, 3, 1, 1, 5, 2, 1, 1, 7, 3, 1, 1, 1, 10, 4, 2, 1, 1, 1, 13, 5, 3, 1, 1, 1, 1, 18, 7, 4, 2, 1, 1, 1, 1, 23, 9, 5, 3, 1, 1, 1, 1, 1, 30, 12, 6, 4, 2, 1, 1, 1, 1, 1, 37, 15, 7, 5, 3, 1, 1, 1, 1, 1, 1, 47, 19, 9, 6, 4, 2, 1, 1, 1, 1, 1, 1, 57, 23, 11, 7, 5, 3, 1, 1, 1, 1, 1, 1, 1, 70, 28, 14, 8, 6, 4
Offset: 1

Views

Author

Philippe Deléham, Jun 26 2010

Keywords

Comments

Column k=1 : A033485 ; column k=2 : A062188 ; row sums : A102378.

Examples

			Triangle begins : 1 ; 2,1 ; 3,1,1 ; 5,2,1,1 ; 7,3,1,1,1 ; ...
		

Crossrefs

Formula

T(n,k)=T(n-1,k)+T([n/2],k), T(n,n)=1, T(n,k)=0 if k>n.

A000123 Number of binary partitions: number of partitions of 2n into powers of 2.

Original entry on oeis.org

1, 2, 4, 6, 10, 14, 20, 26, 36, 46, 60, 74, 94, 114, 140, 166, 202, 238, 284, 330, 390, 450, 524, 598, 692, 786, 900, 1014, 1154, 1294, 1460, 1626, 1828, 2030, 2268, 2506, 2790, 3074, 3404, 3734, 4124, 4514, 4964, 5414, 5938, 6462, 7060, 7658, 8350, 9042, 9828
Offset: 0

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Author

Keywords

Comments

Also, a(n) = number of "non-squashing" partitions of 2n (or 2n+1), that is, partitions 2n = p_1 + p_2 + ... + p_k with 1 <= p_1 <= p_2 <= ... <= p_k and p_1 + p_2 + ... + p_i <= p_{i+1} for all 1 <= i < k [Hirschhorn and Sellers].
Row sums of A101566. - Paul Barry, Jan 03 2005
Equals infinite convolution product of [1,2,2,2,2,2,2,2,2] aerated A000079 - 1 times, i.e., [1,2,2,2,2,2,2,2,2] * [1,0,2,0,2,0,2,0,2] * [1,0,0,0,2,0,0,0,2]. - Mats Granvik and Gary W. Adamson, Aug 04 2009
Which can be further decomposed to the infinite convolution product of finally supported sequences, namely [1,1] aerated A000079 - 1 times with multiplicity A000027 + 1 times, i.e., [1,1] * [1,1] * [1,0,1] * [1,0,1] * [1,0,1] * ... (next terms are [1,0,0,0,1] 4 times, etc.). - Eitan Y. Levine, Jun 18 2023
Given A018819 = A000123 with repeats, polcoeff (1, 1, 2, 2, 4, 4, ...) * (1, 1, 1, ...) = (1, 2, 4, 6, 10, ...) = (1, 0, 2, 0, 4, 0, 6, ...) * (1, 2, 2, 2, ...). - Gary W. Adamson, Dec 16 2009
Let M = an infinite lower triangular matrix with (1, 2, 2, 2, ...) in every column shifted down twice. A000123 = lim_{n->infinity} M^n, the left-shifted vector considered as a sequence. Replacing (1, 2, 2, 2, ...) with (1, 3, 3, 3, ...) and following the same procedure, we obtain A171370: (1, 3, 6, 12, 18, 30, 42, 66, 84, 120, ...). - Gary W. Adamson, Dec 06 2009
First differences of the sequence are (1, 2, 2, 4, 4, 6, 6, 10, ...), A018819, i.e., the sequence itself with each term duplicated except for the first one (unless a 0 is prefixed before taking the first differences), as shown by the formula a(n) - a(n-1) = a(floor(n/2)), valid for all n including n = 0 if we let a(-1) = 0. - M. F. Hasler, Feb 19 2019
Sum over k <= n of number of partitions of k into powers of 2, A018819. - Peter Munn, Feb 21 2020

Examples

			For non-squashing partitions and binary partitions see the example in A018819.
For n=3, the a(3)=6 admitted partitions of 2n=6 are 1+1+1+1+1+1, 1+1+1+1+2, 1+1+2+2, 2+2+2, 1+1+4 and 2+4. - _R. J. Mathar_, Aug 11 2021
		

References

  • G. E. Andrews, The Theory of Partitions, Addison-Wesley, 1976.
  • R. F. Churchhouse, Binary partitions, pp. 397-400 of A. O. L. Atkin and B. J. Birch, editors, Computers in Number Theory. Academic Press, NY, 1971.
  • N. G. de Bruijn, On Mahler's partition problem, Indagationes Mathematicae, vol. X (1948), 210-220.
  • G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
  • H. Gupta, A simple proof of the Churchhouse conjecture concerning binary partitions, Indian J. Pure Appl. Math. 3 (1972), 791-794.
  • H. Gupta, A direct proof of the Churchhouse conjecture concerning binary partitions, Indian J. Math. 18 (1976), 1-5.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A000041, A002033, A002487, A002577, A005704-A005706, A023359, A040039, A100529. Partial sums and bisection of A018819.
A column of A072170. Row sums of A089177. Twice A033485.
Cf. A145515. - Alois P. Heinz, Apr 16 2009
Cf. A171370. - Gary W. Adamson, Dec 06 2009

Programs

  • Haskell
    import Data.List (transpose)
    a000123 n = a000123_list !! n
    a000123_list = 1 : zipWith (+)
       a000123_list (tail $ concat $ transpose [a000123_list, a000123_list])
    -- Reinhard Zumkeller, Nov 15 2012, Aug 01 2011
    
  • Magma
    [1] cat [n eq 1 select n+1 else Self(n-1) + Self(n div 2): n in [1..70]]; // Vincenzo Librandi, Dec 17 2016
    
  • Maple
    A000123 := proc(n) option remember; if n=0 then 1 else A000123(n-1)+A000123(floor(n/2)); fi; end; [ seq(A000123(i),i=0..50) ];
    # second Maple program: more efficient for large n; try: a( 10^25 );
    g:= proc(b, n) option remember; `if`(b<0, 0, `if`(b=0 or
          n=0, 1, `if`(b>=n, add((-1)^(t+1)*binomial(n+1, t)
          *g(b-t, n), t=1..n+1), g(b-1, n)+g(2*b, n-1))))
        end:
    a:= n-> (t-> g(n/2^(t-1), t))(max(ilog2(2*n), 1)):
    seq(a(n), n=0..60); # Alois P. Heinz, Apr 16 2009, revised Apr 14 2016
  • Mathematica
    a[0] = 1; a[n_] := a[n] = a[Floor[n/2]] + a[n-1]; Array[a,49,0] (* Jean-François Alcover, Apr 11 2011, after M. F. Hasler *)
    Fold[Append[#1, Total[Take[Flatten[Transpose[{#1, #1}]], #2]]] &, {1}, Range[2, 49]] (* Birkas Gyorgy, Apr 18 2011 *)
  • PARI
    {a(n) = my(A, m); if( n<1, n==0, m=1; A = 1 + O(x); while(m<=n, m*=2; A = subst(A, x, x^2) * (1+x) / (1-x)); polcoeff(A, n))}; /* Michael Somos, Aug 25 2003 */
    
  • PARI
    {a(n) = if( n<1, n==0, a(n\2) + a(n-1))}; /* Michael Somos, Aug 25 2003 */
    
  • PARI
    A123=[];A000123(n)={ n<3 && return(2^n); if( n<=#A123, A123[n] && return(A123[n]); A123[n-1] && return( A123[n] = A123[n-1]+A000123(n\2) ), n>2*#A123 && A123=concat(A123,vector((n-#A123)\2))); A123[if(n>#A123,1,n)]=2*sum(k=1,n\2-1,A000123(k),1)+(n%2+1)*A000123(n\2)} \\ Stores results in global vector A123 dynamically resized to at most 3n/4 when size is less than n/2. Gives a(n*10^6) in ~ n sec. - M. F. Hasler, Apr 30 2009
    
  • PARI
    {a(n)=polcoeff(exp(sum(m=1,n,2^valuation(2*m,2)*x^m/m)+x*O(x^n)),n)} \\ Paul D. Hanna, Oct 30 2012
    
  • Python
    from functools import lru_cache
    @lru_cache(maxsize=None)
    def A000123(n): return 1 if n == 0 else A000123(n-1) + A000123(n//2) # Chai Wah Wu, Jan 18 2022

Formula

a(n) = A018819(2*n).
a(n) = a(n-1) + a(floor(n/2)). For proof see A018819.
2 * a(n) = a(n+1) + a(n-1) if n is even. - Michael Somos, Jan 07 2011
G.f.: (1-x)^(-1) Product_{n>=0} (1 - x^(2^n))^(-1).
a(n) = Sum_{i=0..n} a(floor(i/2)) [O'Shea].
a(n) = (1/n)*Sum_{k=1..n} (A038712(k)+1)*a(n-k), n > 1, a(0)=1. - Vladeta Jovovic, Aug 22 2002
Conjecture: Limit_{n ->infinity} (log(n)*a(2n))/(n*a(n)) = c = 1.63... - Benoit Cloitre, Jan 26 2003 [The constant c is equal to 2*log(2) = 1.38629436... =A016627. - Vaclav Kotesovec, Aug 07 2019]
G.f. A(x) satisfies A(x^2) = ((1-x)/(1+x)) * A(x). - Michael Somos, Aug 25 2003
G.f.: Product_{k>=0} (1+x^(2^k))/(1-x^(2^k)) = (Product_{k>=0} (1+x^(2^k))^(k+1) )/(1-x) = Product_{k>=0} (1+x^(2^k))^(k+2). - Joerg Arndt, Apr 24 2005
From Philippe Flajolet, Sep 06 2008: (Start)
The asymptotic rate of growth is known precisely - see De Bruijn's paper. With p(n) the number of partitions of n into powers of two, the asymptotic formula of de Bruijn is: log(p(2*n)) = 1/(2*L2)*(log(n/log(n)))^2 + (1/2 + 1/L2 + LL2/L2)*log(n) - (1 + LL2/L2)*log(log(n)) + Phi(log(n/log(n))/L2), where L2=log(2), LL2=log(log(2)) and Phi(x) is a certain periodic function with period 1 and a tiny amplitude.
Numerically, Phi(x) appears to have a mean value around 0.66. An expansion up to O(1) term had been obtained earlier by Kurt Mahler. (End)
G.f.: exp( Sum_{n>=1} 2^A001511(n) * x^n/n ), where 2^A001511(n) is the highest power of 2 that divides 2*n. - Paul D. Hanna, Oct 30 2012
(n/2)*a(n) = Sum_{k = 0..n-1} (n-k)/A000265(n-k)*a(k). - Peter Bala, Mar 03 2019
Conjectures from Mikhail Kurkov, May 04 2025: (Start)
Sum_{k=0..n} a(2^m*k)*A106400(n-k) = A125790(m,2*n) for m >= 0, n >= 0.
Sum_{k=0..n} a(2^m*(2*k+1))*A106400(n-k) = A125790(m+1,2*n+1) for m >= 0, n >= 0.
More generally, if we define b(n,m,p,q) = Sum_{k=0..n} a(2^m*(2*p*k+2*q+1))*A106400(n-k) for m >= 0, p > 0, q >= 0, n >= 0, then it also looks like that we have b(n,m,p,q) = Sum_{k=0..m+1} A078121(m+1,k)*b(n,k,p/2,(q-1)/2), b(n,m,p,q) = Sum_{k=0..m+1} A078121(m+1,k)*b(n,k,p/2,q/2)*(-1)^(m+k+1) for m >= 0, p > 0, q >= 0, n >= 0. (End)
Conjecture: Sum_{i>=0} a(2^m*i + k)*x^i = f(k,x) / Product_{q>=0} (1 - x^(2^q)) for m > 0, 2^(m-1) <= k < 2^m where f(k,x) is g.f. for k-th row of A381810. - Mikhail Kurkov, May 17 2025

Extensions

More terms from Robin Trew (trew(AT)hcs.harvard.edu)
Values up to a(10^4) checked with given PARI code by M. F. Hasler, Apr 30 2009

A036554 Numbers whose binary representation ends in an odd number of zeros.

Original entry on oeis.org

2, 6, 8, 10, 14, 18, 22, 24, 26, 30, 32, 34, 38, 40, 42, 46, 50, 54, 56, 58, 62, 66, 70, 72, 74, 78, 82, 86, 88, 90, 94, 96, 98, 102, 104, 106, 110, 114, 118, 120, 122, 126, 128, 130, 134, 136, 138, 142, 146, 150, 152, 154, 158, 160, 162, 166, 168, 170, 174
Offset: 1

Views

Author

Keywords

Comments

Fraenkel (2010) called these the "dopey" numbers.
Also n such that A035263(n)=0 or A050292(n) = A050292(n-1).
Indices of even numbers in A033485. - Philippe Deléham, Mar 16 2004
a(n) is an odious number (see A000069) for n odd; a(n) is an evil number (see A001969) for n even. - Philippe Deléham, Mar 16 2004
Indices of even numbers in A007913, in A001511. - Philippe Deléham, Mar 27 2004
This sequence consists of the increasing values of n such that A097357(n) is even. - Creighton Dement, Aug 14 2004
Numbers with an odd number of 2's in their prime factorization (e.g., 8 = 2*2*2). - Mark Dow, Sep 04 2007
Equals the set of natural numbers not in A003159 or A141290. - Gary W. Adamson, Jun 22 2008
Represents the set of CCW n-th moves in the standard Tower of Hanoi game; and terms in even rows of a [1, 3, 5, 7, 9, ...] * [1, 2, 4, 8, 16, ...] multiplication table. Refer to the example. - Gary W. Adamson, Mar 20 2010
Refer to the comments in A003159 relating to A000041 and A174065. - Gary W. Adamson, Mar 21 2010
If the upper s-Wythoff sequence of s is s, then s=A036554. (See A184117 for the definition of lower and upper s-Wythoff sequences.) Starting with any nondecreasing sequence s of positive integers, A036554 is the limit when the upper s-Wythoff operation is iterated. For example, starting with s=(1,4,9,16,...) = (n^2), we obtain lower and upper s-Wythoff sequences
a=(1,3,4,5,6,8,9,10,11,12,14,...) = A184427;
b=(2,7,12,21,31,44,58,74,...) = A184428.
Then putting s=a and repeating the operation gives
b'=(2,6,8,10,13,17,20,...), which has the same first four terms as A036554. - Clark Kimberling, Jan 14 2011
Or numbers having infinitary divisor 2, or the same, having factor 2 in Fermi-Dirac representation as a product of distinct terms of A050376. - Vladimir Shevelev, Mar 18 2013
Thus, numbers not in A300841 or in A302792. Equally, sequence 2*A300841(n) sorted into ascending order. - Antti Karttunen, Apr 23 2018

Examples

			From _Gary W. Adamson_, Mar 20 2010: (Start)
Equals terms in even numbered rows in the following multiplication table:
(rows are labeled 1,2,3,... as with the Towers of Hanoi disks)
   1,  3,  5,  7,  9, 11, ...
   2,  6, 10, 14, 18, 22, ...
   4, 12, 20, 28, 36, 44, ...
   8, 24, 40, 56, 72, 88, ...
   ...
As shown, 2, 6, 8, 10, 14, ...; are in even numbered rows, given the row with (1, 3, 5, 7, ...) = row 1.
The term "5" is in an odd row, so the 5th Towers of Hanoi move is CW, moving disc #1 (in the first row).
a(3) = 8 in row 4, indicating that the 8th Tower of Hanoi move is CCW, moving disc #4.
A036554 bisects the positive nonzero natural numbers into those in the A036554 set comprising 1/3 of the total numbers, given sufficiently large n.
This corresponds to 1/3 of the TOH moves being CCW and 2/3 CW. Row 1 of the multiplication table = 1/2 of the natural numbers, row 2 = 1/4, row 3 = 1/8 and so on, or 1 = (1/2 + 1/4 + 1/8 + 1/16 + ...). Taking the odd-indexed terms of this series given offset 1, we obtain 2/3 = 1/2 + 1/8 + 1/32 + ..., while sum of the even-indexed terms is 1/3. (End)
		

Crossrefs

Indices of odd numbers in A007814. Subsequence of A036552. Complement of A003159. Also double of A003159.
Cf. A000041, A003157, A003158, A005408, A052330, A072939, A079523, A096268 (characteristic function, when interpreted with offset 1), A141290, A174065, A300841.

Programs

  • Haskell
    a036554 = (+ 1) . a079523  -- Reinhard Zumkeller, Mar 01 2012
    
  • Magma
    [2*m:m in [1..100] | Valuation(m,2) mod 2 eq 0]; // Marius A. Burtea, Aug 29 2019
    
  • Mathematica
    Select[Range[200],OddQ[IntegerExponent[#,2]]&] (* Harvey P. Dale, Oct 19 2011 *)
  • PARI
    is(n)=valuation(n,2)%2 \\ Charles R Greathouse IV, Nov 20 2012
    
  • Python
    def ok(n):
      c = 0
      while n%2 == 0: n //= 2; c += 1
      return c%2 == 1
    print([m for m in range(1, 175) if ok(m)]) # Michael S. Branicky, Feb 06 2021
    
  • Python
    from itertools import count, islice
    def A036554_gen(startvalue=1): return filter(lambda n:(~n & n-1).bit_length()&1,count(max(startvalue,1))) # generator of terms >= startvalue
    A036554_list = list(islice(A036554_gen(),30)) # Chai Wah Wu, Jul 05 2022
    
  • Python
    is_A036554 = lambda n: A001511(n)&1==0 # M. F. Hasler, Nov 26 2024
    
  • Python
    def A036554(n):
        def bisection(f,kmin=0,kmax=1):
            while f(kmax) > kmax: kmax <<= 1
            kmin = kmax >> 1
            while kmax-kmin > 1:
                kmid = kmax+kmin>>1
                if f(kmid) <= kmid:
                    kmax = kmid
                else:
                    kmin = kmid
            return kmax
        def f(x):
            c, s = n+x, bin(x)[2:]
            l = len(s)
            for i in range(l&1,l,2):
                c -= int(s[i])+int('0'+s[:i],2)
            return c
        return bisection(f,n,n) # Chai Wah Wu, Jan 29 2025

Formula

a(n) = A079523(n)+1 = A072939(n)-1.
a(n) = A003156(n) + n = A003157(n) - n = A003158(n) - n + 1. - Philippe Deléham, Apr 10 2004
Values of k such that A091297(k) = 2. - Philippe Deléham, Feb 25 2004
a(n) ~ 3n. - Charles R Greathouse IV, Nov 20 2012 [In fact, a(n) = 3n + O(log n). - Charles R Greathouse IV, Nov 27 2024]
a(n) = 2*A003159(n). - Clark Kimberling, Sep 30 2014
{a(n)} = A052330({A005408(n)}), where {a(n)} denotes the set of integers in the sequence. - Peter Munn, Aug 26 2019

Extensions

Incorrect equation removed from formula by Peter Munn, Dec 04 2020

A035263 Trajectory of 1 under the morphism 0 -> 11, 1 -> 10; parity of 2-adic valuation of 2n: a(n) = A000035(A001511(n)).

Original entry on oeis.org

1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1
Offset: 1

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Author

Keywords

Comments

First Feigenbaum symbolic (or period-doubling) sequence, corresponding to the accumulation point of the 2^{k} cycles through successive bifurcations.
To construct the sequence: start with 1 and concatenate: 1,1, then change the last term (1->0; 0->1) gives: 1,0. Concatenate those 2 terms: 1,0,1,0, change the last term: 1,0,1,1. Concatenate those 4 terms: 1,0,1,1,1,0,1,1 change the last term: 1,0,1,1,1,0,1,0, etc. - Benoit Cloitre, Dec 17 2002
Let T denote the present sequence. Here is another way to construct T. Start with the sequence S = 1,0,1,,1,0,1,,1,0,1,,1,0,1,,... and fill in the successive holes with the successive terms of the sequence T (from paper by Allouche et al.). - Emeric Deutsch, Jan 08 2003 [Note that if we fill in the holes with the terms of S itself, we get A141260. - N. J. A. Sloane, Jan 14 2009]
From N. J. A. Sloane, Feb 27 2009: (Start)
In more detail: define S to be 1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1,0,1___...
If we fill the holes with S we get A141260:
1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0,
........1.........0.........1.........1.........0.......1.........1.........0...
- the result is
1..0..1.1.1..0..1.0.1..0..1.1.1..0..1.1.1..0..1.0.1.... = A141260.
But instead, if we define T recursively by filling the holes in S with the terms of T itself, we get A035263:
1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0,
........1.........0.........1.........1.........1.......0.........1.........0...
- the result is
1..0..1.1.1..0..1.0.1..0..1.1.1..0..1.1.1..0..1.1.1.0.1.0.1..0..1.1.1..0..1.0.1.. = A035263. (End)
Characteristic function of A003159, i.e., A035263(n)=1 if n is in A003159 and A035263(n)=0 otherwise (from paper by Allouche et al.). - Emeric Deutsch, Jan 15 2003
This is the sequence of R (=1), L (=0) moves in the Towers of Hanoi puzzle: R, L, R, R, R, L, R, L, R, L, R, R, R, ... - Gary W. Adamson, Sep 21 2003
Manfred Schroeder, p. 279 states, "... the kneading sequences for unimodal maps in the binary notation, 0, 1, 0, 1, 1, 1, 0, 1..., are obtained from the Morse-Thue sequence by taking sums mod 2 of adjacent elements." On p. 278, in the chapter "Self-Similarity in the Logistic Parabola", he writes, "Is there a closer connection between the Morse-Thue sequence and the symbolic dynamics of the superstable orbits? There is indeed. To see this, let us replace R by 1 and C and L by 0." - Gary W. Adamson, Sep 21 2003
Partial sums modulo 2 of the sequence 1, a(1), a(1), a(2), a(2), a(3), a(3), a(4), a(4), a(5), a(5), a(6), a(6), ... . - Philippe Deléham, Jan 02 2004
Parity of A007913, A065882 and A065883. - Philippe Deléham, Mar 28 2004
The length of n-th run of 1's in this sequence is A080426(n). - Philippe Deléham, Apr 19 2004
Also parity of A005043, A005773, A026378, A104455, A117641. - Philippe Deléham, Apr 28 2007
Equals parity of the Towers of Hanoi, or ruler sequence (A001511), where the Towers of Hanoi sequence (1, 2, 1, 3, 1, 2, 1, 4, ...) denotes the disc moved, labeled (1, 2, 3, ...) starting from the top; and the parity of (1, 2, 1, 3, ...) denotes the direction of the move, CW or CCW. The frequency of CW moves converges to 2/3. - Gary W. Adamson, May 11 2007
A conjectured identity relating to the partition sequence, A000041: p(x) = A(x) * A(x^2) when A(x) = the Euler transform of A035263 = polcoeff A174065: (1 + x + x^2 + 2x^3 + 3x^4 + 4x^5 + ...). - Gary W. Adamson, Mar 21 2010
a(n) is 1 if the number of trailing zeros in the binary representation of n is even. - Ralf Stephan, Aug 22 2013
From Gary W. Adamson, Mar 25 2015: (Start)
A conjectured identity relating to the partition sequence, A000041 as polcoeff p(x); A003159, and its characteristic function A035263: (1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, ...); and A036554 indicating n-th terms with zeros in A035263: (2, 6, 8, 10, 14, 18, 22, ...).
The conjecture states that p(x) = A(x) = A(x^2) when A(x) = polcoeffA174065 = the Euler transform of A035263 = 1/(1-x)*(1-x^3)*(1-x^4)*(1-x^5)*... = (1 + x + x^2 + 2x^3 + 3x^4 + 4x^5 + ...) and the aerated variant = the Euler transform of the complement of A035263: 1/(1-x^2)*(1-x^6)*(1-x^8)*... = (1 + x^2 + x^4 + 2x^6 + 3x^8 + 4x^10 + ...).
(End)
The conjecture above was proved by Jean-Paul Allouche on Dec 21 2013.
Regarded as a column vector, this sequence is the product of A047999 (Sierpinski's gasket) regarded as an infinite lower triangular matrix and A036497 (the Fredholm-Rueppel sequence) where the 1's have alternating signs, 1, -1, 0, 1, 0, 0, 0, -1, .... - Gary W. Adamson, Jun 02 2021
The numbers of 1's through n (A050292) can be determined by starting with the binary (say for 19 = 1 0 0 1 1) and writing: next term is twice current term if 0, otherwise twice plus 1. The result is 1, 2, 4, 9, 19. Take the difference row, = 1, 1, 2, 5, 10; and add the odd-indexed terms from the right: 5, 4, 3, 2, 1 = 10 + 2 + 1 = 13. The algorithm is the basis for determining the disc configurations in the tower of Hanoi game, as shown in the Jul 24 2021 comment of A060572. - Gary W. Adamson, Jul 28 2021

References

  • Karamanos, Kostas. "From symbolic dynamics to a digital approach." International Journal of Bifurcation and Chaos 11.06 (2001): 1683-1694. (Full version. See p. 1685)
  • Karamanos, K. (2000). From symbolic dynamics to a digital approach: chaos and transcendence. In Michel Planat (Ed.), Noise, Oscillators and Algebraic Randomness (Lecture Notes in Physics, pp. 357-371). Springer, Berlin, Heidelberg. (Short version. See p. 359)
  • Manfred R. Schroeder, "Fractals, Chaos, Power Laws", W. H. Freeman, 1991
  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 892, column 2, Note on p. 84, part (a).

Crossrefs

Parity of A001511. Anti-parity of A007814.
Absolute values of first differences of A010060. Apart from signs, same as A029883. Essentially the same as A056832.
Swapping 0 and 1 gives A096268.
Cf. A033485, A050292 (partial sums), A089608, A088172, A019300, A039982, A073675, A121701, A141260, A000041, A174065, A220466, A154269 (Mobius transform).
Limit of A317957(n) for large n.

Programs

  • Haskell
    import Data.Bits (xor)
    a035263 n = a035263_list !! (n-1)
    a035263_list = zipWith xor a010060_list $ tail a010060_list
    -- Reinhard Zumkeller, Mar 01 2012
    
  • Maple
    nmax:=105: for p from 0 to ceil(simplify(log[2](nmax))) do for n from 1 to ceil(nmax/(p+2)) do a((2*n-1)*2^p) := (p+1) mod 2 od: od: seq(a(n), n=1..nmax); # Johannes W. Meijer, Feb 07 2013
    A035263 := n -> 1 - padic[ordp](n, 2) mod 2:
    seq(A035263(n), n=1..105); # Peter Luschny, Oct 02 2018
  • Mathematica
    a[n_] := a[n] = If[ EvenQ[n], 1 - a[n/2], 1]; Table[ a[n], {n, 1, 105}] (* Or *)
    Rest[ CoefficientList[ Series[ Sum[ x^(2^k)/(1 + (-1)^k*x^(2^k)), {k, 0, 20}], {x, 0, 105}], x]]
    f[1] := True; f[x_] := Xor[f[x - 1], f[Floor[x/2]]]; a[x_] := Boole[f[x]] (* Ben Branman, Oct 04 2010 *)
    a[n_] := If[n == 0, 0, 1 - Mod[ IntegerExponent[n, 2], 2]]; (* Jean-François Alcover, Jul 19 2013, after Michael Somos *)
    Nest[ Flatten[# /. {0 -> {1, 1}, 1 -> {1, 0}}] &, {0}, 7] (* Robert G. Wilson v, Jul 23 2014 *)
    SubstitutionSystem[{0->{1,1},1->{1,0}},1,{7}][[1]] (* Harvey P. Dale, Jun 06 2022 *)
  • PARI
    {a(n) = if( n==0, 0, 1 - valuation(n, 2)%2)}; /* Michael Somos, Sep 04 2006 */
    
  • PARI
    {a(n) = if( n==0, 0, n = abs(n); subst( Pol(binary(n)) - Pol(binary(n-1)), x, 1)%2)}; /* Michael Somos, Sep 04 2006 */
    
  • PARI
    {a(n) = if( n==0, 0, n = abs(n); direuler(p=2, n, 1 / (1 - X^((p<3) + 1)))[n])}; /* Michael Somos, Sep 04 2006 */
    
  • Python
    def A035263(n): return (n&-n).bit_length()&1 # Chai Wah Wu, Jan 09 2023
  • Scheme
    (define (A035263 n) (let loop ((n n) (i 1)) (cond ((odd? n) (modulo i 2)) (else (loop (/ n 2) (+ 1 i)))))) ;; (Use mod instead of modulo in R6RS) Antti Karttunen, Sep 11 2017
    

Formula

Absolute values of first differences (A029883) of Thue-Morse sequence (A001285 or A010060). Self-similar under 10->1 and 11->0.
Series expansion: (1/x) * Sum_{i>=0} (-1)^(i+1)*x^(2^i)/(x^(2^i)-1). - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 17 2003
a(n) = Sum_{k>=0} (-1)^k*(floor((n+1)/2^k)-floor(n/2^k)). - Benoit Cloitre, Jun 03 2003
Another g.f.: Sum_{k>=0} x^(2^k)/(1+(-1)^k*x^(2^k)). - Ralf Stephan, Jun 13 2003
a(2*n) = 1-a(n), a(2*n+1) = 1. - Ralf Stephan, Jun 13 2003
a(n) = parity of A033485(n). - Philippe Deléham, Aug 13 2003
Equals A088172 mod 2, where A088172 = 1, 2, 3, 7, 13, 26, 53, 106, 211, 422, 845, ... (first differences of A019300). - Gary W. Adamson, Sep 21 2003
a(n) = a(n-1) - (-1)^n*a(floor(n/2)). - Benoit Cloitre, Dec 02 2003
a(1) = 1 and a(n) = abs(a(n-1) - a(floor(n/2))). - Benoit Cloitre, Dec 02 2003
a(n) = 1 - A096268(n+1); A050292 gives partial sums. - Reinhard Zumkeller, Aug 16 2006
Multiplicative with a(2^k) = 1 - (k mod 2), a(p^k) = 1, p > 2. Dirichlet g.f.: Product_{n = 4 or an odd prime} (1/(1-1/n^s)). - Christian G. Bower, May 18 2005
a(-n) = a(n). a(0)=0. - Michael Somos, Sep 04 2006
Dirichlet g.f.: zeta(s)*2^s/(2^s+1). - Ralf Stephan, Jun 17 2007
a(n+1) = a(n) XOR a(ceiling(n/2)), a(1) = 1. - Reinhard Zumkeller, Jun 11 2009
Let D(x) be the generating function, then D(x) + D(x^2) == x/(1-x). - Joerg Arndt, May 11 2010
a(n) = A010060(n) XOR A010060(n+1); a(A079523(n)) = 0; a(A121539(n)) = 1. - Reinhard Zumkeller, Mar 01 2012
a((2*n-1)*2^p) = (p+1) mod 2, p >= 0 and n >= 1. - Johannes W. Meijer, Feb 07 2013
a(n) = A000035(A001511(n)). - Omar E. Pol, Oct 29 2013
a(n) = 2-A056832(n) = (5-A089608(n))/4. - Antti Karttunen, Sep 11 2017, after Benoit Cloitre
For n >= 0, a(n+1) = M(2n) mod 2 where M(n) is the Motzkin number A001006 (see Deutsch and Sagan 2006 link). - David Callan, Oct 02 2018
a(n) = A038712(n) mod 3. - Kevin Ryde, Jul 11 2019
Given any n in the form (k * 2^m, k odd), extract k and m. Categorize the results into two outcomes of (k, m, even or odd). If (k, m) is (odd, even) substitute 1. If (odd, odd), denote the result 0. Example: 5 = (5 * 2^0), (odd, even, = 1). (6 = 3 * 2^1), (odd, odd, = 0). - Gary W. Adamson, Jun 23 2021

Extensions

Alternative description added to the name by Antti Karttunen, Sep 11 2017

A001401 Number of partitions of n into at most 5 parts.

Original entry on oeis.org

1, 1, 2, 3, 5, 7, 10, 13, 18, 23, 30, 37, 47, 57, 70, 84, 101, 119, 141, 164, 192, 221, 255, 291, 333, 377, 427, 480, 540, 603, 674, 748, 831, 918, 1014, 1115, 1226, 1342, 1469, 1602, 1747, 1898, 2062, 2233, 2418, 2611, 2818, 3034, 3266, 3507, 3765, 4033, 4319
Offset: 0

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Author

Keywords

Comments

a(n) = T_{r}(n) for r large, where T_{r}(n) = number of outcomes in which r indistinguishable dice yield a sum r+n-1.
a(n) = coefficient of q^n in the expansion of (m choose 5)_q as m goes to infinity. - Y. Kelly Itakura (yitkr(AT)mta.ca), Aug 21 2002
For n > 4: also number of partitions of n into parts <= 5: a(n) = A026820(n,5). - Reinhard Zumkeller, Jan 21 2010
Number of different distributions of n+15 identical balls in 5 boxes as x,y,z,p,q where 0 < x < y < z < p < q. - Ece Uslu and Esin Becenen, Jan 11 2016 [i.e., a(n) is the number of partitions of n+15 into 5 distinct parts. - R. J. Mathar, Feb 28 2021]
Tengely and Ulas prove that a(n) is a square only for n=1 and 2027. - Michel Marcus, Feb 11 2021

Examples

			(5 choose 5)_q = 1;
(6 choose 5)_q = q^5 + q^4 + q^3 + q^2 + q + 1;
(7 choose 5)_q = q^10 + q^9 + 2*q^8 + 2*q^7 + 3*q^6 + 3*q^5 + 3*q^4 + 2*q^3 + 2*q^2 + q + 1;
(8 choose 5)_q = q^15 + q^14 + 2*q^13 + 3*q^12 + 4*q^11 + 5*q^10 + 6*q^9 + 6*q^8 + 6*q^7 + 6*q^6 + 5*q^5 + 4*q^4 + 3*q^3 + 2*q^2 + q + 1;
so the coefficient of q^0 converges to 1, q^1 to 1, q^2 to 2 and so on.
a(3) = 3, i.e., {1,2,3,4,8}, {1,2,3,5,7}, {1,2,4,5,6}. Number of different distributions of 18 identical balls in 5 boxes as x,y,z,p,q where 0 < x < y < z < p < q. - _Ece Uslu_, Esin Becenen, Jan 11 2016
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 115, row m=5 of Q(m,n) table.
  • H. Gupta et al., Tables of Partitions. Royal Society Mathematical Tables, Vol. 4, Cambridge Univ. Press, 1958, p. 2.
  • D. E. Knuth, The Art of Computer Programming, vol. 4, fascicle 3, Generating All Combinations and Partitions, Section 7.2.1.4., p. 56, exercise 31.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

a(n) = A008284(n+5, 5), n >= 0.
Cf. A008619, A001400, A001399, A008667 (first differences), A008804.
First differences of A002622.

Programs

  • Maple
    with(combstruct):ZL6:=[S,{S=Set(Cycle(Z,card<6))}, unlabeled]:seq(count(ZL6,size=n),n=0..52); # Zerinvary Lajos, Sep 24 2007
    a:= n-> (Matrix(15, (i,j)-> if (i=j-1) then 1 elif j=1 then [1, 1, 0, 0, -1, -1, -1, 1, 1, 1, 0, 0, -1, -1, 1][i] else 0 fi)^n)[1,1]: seq(a(n), n=0..60); # Alois P. Heinz, Jul 31 2008
    B:=[S,{S = Set(Sequence(Z,1 <= card),card <=5)},unlabelled]: seq(combstruct[count](B, size=n), n=0..52); # Zerinvary Lajos, Mar 21 2009
  • Mathematica
    CoefficientList[ Series[ 1/((1 - x)*(1 - x^2)*(1 - x^3)*(1 - x^4)*(1 - x^5)), {x, 0, 60} ], x ]
    a[n_] := IntegerPartitions[n, 5] // Length; Table[a[n], {n, 0, 52}] (* Jean-François Alcover, Jul 13 2012 *)
    LinearRecurrence[{1,1,0,0,-1,-1,-1,1,1,1,0,0,-1,-1,1},{1,1,2,3,5,7,10,13,18,23,30,37,47,57,70},60] (* Harvey P. Dale, Jan 05 2019 *)
  • PARI
    a(n)=#partitions(n,,5) \\ Charles R Greathouse IV, Sep 15 2014
    
  • PARI
    a(n) = (n^4 + 30*n^3 + 310*n^2 + 1320*n - 90*n*(n%2) + 2880)\2880 \\ Hoang Xuan Thanh, Aug 12 2025

Formula

G.f.: 1/((1-x)*(1-x^2)*(1-x^3)*(1-x^4)*(1-x^5)).
a(n) = 1 + (a(n-2) + a(n-3) + a(n-4)) - (a(n-6) + (2*a(n-7)) + a(n-8)) + (a(n-10) + a(n-11) + a(n-12)) - a(n-14). - Norman J. Meluch (norm(AT)iss.gm.com), Mar 09 2000
Let a1(n) = Sum_{i=0..floor(n/3)} (1 + ceiling((n-3*i-1)/2)), a2(n) = Sum_{i=0..floor(n/4)} (1 + ceiling((n-4*i-1)/2) + a1(n-4*i-3)), then a(n) = Sum_{i=0..floor(n/5)} (1 + ceiling((n-5*i-1)/2) + a1(n-5*i-3) + a2(n-5*i-4)). - Jon Perry, Jun 27 2003
(n choose 5)_q=(q^n-1)*(q^(n-1)-1)*(q^(n-2)-1)*(q^(n-3)-1)*(q^(n-4)-1)/((q^5-1)*(q^4-1)*(q^3-1)*(q^2-1)*(q-1)).
a(n) = round(((n+5)^4 + 10*((n+5)^3 + (n+5)^2) - 75*(n+5) - 45*(n+5)*(-1)^(n+5))/2880). - Washington Bomfim, Jul 03 2012
a(n) = a(n-1) + a(n-2) - a(n-5) - a(n-6) - a(n-7) + a(n-8) + a(n-9) + a(n-10) - a(n-13) - a(n-14) + a(n+15). - David Neil McGrath, Sep 13 2014
a(n+5) = a(n) + A001400(n) = A001400(n)+A026811(n). - Ece Uslu, Esin Becenen, Jan 11 2016
From Vladimír Modrák, Jul 13 2022: (Start)
a(n) = Sum_{k=0..floor(n/5)} Sum_{j=0..floor(n/4)} Sum_{i=0..floor(n/3)} ceiling((max(0, n + 1 - 3*i - 4*j - 5*k))/2).
a(n) = Sum_{j=0..floor(n/5)} Sum_{i=0..floor(n/4)} floor(((max(0, n + 3 - 4*i - 5*j))^2+4)/12). (End)
a(2n) = a(2n-1) + a(n) - a(n-8) = a(n) + Sum_{k=0..n-1} A008804(k). - David García Herrero, Aug 26 2024
a(n) = floor((n^4 + 30*n^3 + 310*n^2 + 1275*n + 45*n*(-1)^n+2880)/2880). - Hoang Xuan Thanh, Aug 12 2025

Extensions

Additional comments from Michael Somos and Branislav Kisacanin (branislav.kisacanin(AT)delphiauto.com)

A007413 A squarefree (or Thue-Morse) ternary sequence: closed under 1->123, 2->13, 3->2. Start with 1.

Original entry on oeis.org

1, 2, 3, 1, 3, 2, 1, 2, 3, 2, 1, 3, 1, 2, 3, 1, 3, 2, 1, 3, 1, 2, 3, 2, 1, 2, 3, 1, 3, 2, 1, 2, 3, 2, 1, 3, 1, 2, 3, 2, 1, 2, 3, 1, 3, 2, 1, 3, 1, 2, 3, 1, 3, 2, 1, 2, 3, 2, 1, 3, 1, 2, 3, 1, 3, 2, 1, 3, 1, 2, 3, 2, 1, 2, 3, 1, 3, 2, 1, 3, 1, 2, 3, 1, 3, 2, 1, 2, 3, 2, 1, 3, 1, 2, 3, 2, 1, 2, 3, 1, 3, 2, 1, 2, 3
Offset: 1

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Comments

a(n)=2 if and only if n-1 is in A079523. - Benoit Cloitre, Mar 10 2003
Partial sums modulo 4 of the sequence 1, a(1), a(1), a(2), a(2), a(3), a(3), a(4), a(4), a(5), a(5), a(6), a(6), ... - Philippe Deléham, Mar 04 2004
To construct the sequence: start with 1 and concatenate 4 -1 = 3: 1, 3, then change the last term (2 -> 1, 3 ->2 ) gives 1, 2. Concatenate 1, 2 with 4 -1 = 3, 4 - 2 = 2: 1, 2, 3, 2 and change the last term: 1, 2, 3, 1. Concatenate 1, 2, 3, 1 with 4 - 1 = 3, 4 - 2 = 2, 4 - 3 = 1, 4 - 1 = 3: 1, 2, 3, 1, 3, 2, 1, 3 and change the last term: 1, 2, 3, 1, 3, 2, 1, 2 etc. - Philippe Deléham, Mar 04 2004
To construct the sequence: start with the Thue-Morse sequence A010060 = 0, 1, 1, 0, 1, 0, 0, 1, ... Then change 0 -> 1, 2, 3, and 1 -> 3, 2, 1, gives: 1, 2, 3, , 3, 2, 1, ,3, 2, 1, , 1, 2, 3, , 3, 2, 1, , ... and fill in the successive holes with the successive terms of the sequence itself. - _Philippe Deléham, Mar 04 2004
To construct the sequence: to insert the number 2 between the A003156(k)-th term and the (1 + A003156(k))-th term of the sequence 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, ... - Philippe Deléham, Mar 04 2004
Conjecture. The sequence is formed by the numbers of 1's between every pair of consecutive 2's in A076826. - Vladimir Shevelev, May 31 2009

Examples

			Here are the first 5 stages in the construction of this sequence, together with Mma code, taken from Keranen's article. His alphabet is a,b,c rather than 1,2,3.
productions = {"a" -> "abc ", "b" -> "ac ", "c" -> "b ", " " -> ""};
NestList[g, "a", 5] // TableForm
a
abc
abc ac b
abc ac b abc b ac
abc ac b abc b ac abc ac b ac abc b
abc ac b abc b ac abc ac b ac abc b abc ac b abc b ac abc b abc ac b ac
		

References

  • Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.
  • J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 18.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • A. Thue, Über unendliche Zeichenreihen, Norske Vid. Selsk. Skr. I. Mat. Nat. Kl. Christiania, No. 7 (1906), 1-22.

Crossrefs

First differences of A000069.
Equals A036580(n-1) + 1.

Programs

  • Mathematica
    Nest[ Flatten[ # /. {1 -> {1, 2, 3}, 2 -> {1, 3}, 3 -> {2}}] &, {1}, 7] (* Robert G. Wilson v, May 07 2005 *)
    2 - Differences[ThueMorse[Range[0, 100]]] (* Paolo Xausa, Oct 25 2024 *)
  • PARI
    {a(n) = if( n<1 || valuation(n, 2)%2, 2, 2 + (-1)^subst( Pol(binary(n)), x,1))};
    
  • Python
    def A007413(n): return 2-(n.bit_count()&1)+((n-1).bit_count()&1) # Chai Wah Wu, Mar 03 2023

Formula

a(n) modulo 2 = A035263(n). a(A036554(n)) = 2. a(A003159(n)) = 1 if n odd. a(A003159(n)) = 3 if n even. a(n) = A033485(n) mod 4. a(n) = 4 - A036585(n-1). - Philippe Deléham, Mar 04 2004
a(n) = 2 - A029883(n) = 3 - A036577(n). - Philippe Deléham, Mar 20 2004
For n>=1, we have: 1) a(A108269(n))=A010684(n-1); 2) a(A079523(n))=A010684(n-1); 3) a(A081706(2n))=A010684(n). - Vladimir Shevelev, Jun 22 2009

A022328 Exponent of 2 (value of i) in n-th number of form 2^i*3^j, i >= 0, j >= 0 (see A003586).

Original entry on oeis.org

0, 1, 0, 2, 1, 3, 0, 2, 4, 1, 3, 0, 5, 2, 4, 1, 6, 3, 0, 5, 2, 7, 4, 1, 6, 3, 0, 8, 5, 2, 7, 4, 1, 9, 6, 3, 0, 8, 5, 2, 10, 7, 4, 1, 9, 6, 3, 11, 0, 8, 5, 2, 10, 7, 4, 12, 1, 9, 6, 3, 11, 0, 8, 5, 13, 2, 10, 7, 4, 12, 1, 9, 6, 14, 3, 11, 0, 8, 5, 13, 2, 10, 7, 15, 4, 12, 1, 9, 6, 14, 3, 11, 0, 8, 16, 5, 13, 2
Offset: 1

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From Clark Kimberling, Mar 18 2015 and May 21 2015: (Start)
This is the signature sequence of log(3)/log(2) and is a fractal sequence; e.g., if the first occurrence of each n is removed, the resulting sequence is the original sequence.
Moreover, if the sequence is partitioned into segments starting with 0 as follows:
0,1
0,2,1,3
0,2,4,1,3
0,5,2,4,1,6,3,
and so on, then deleting the greatest number in each segment leaves
0
0,2,1
0,2,1,3
0,5,2,4,1,3,
and so on, which, concatenated to (0,0,2,1,0,2,1,3,0,5,2,4,1,3,...), is another fractal sequence, in today's usual meaning of that term. When introduced in 1995, one of the defining properties of a fractal sequence was, essentially, that before each n appears, every k < n must have already appeared; this requirement ensures that the sequence yields a dispersion; e.g., A114577 yields A114537. However, the usual meaning of "fractal sequence" nowadays is simply "a sequence that contains itself as a proper subsequence". It is proposed here that the original version be renamed "strongly fractal". Thus, the operations called upper trimming and lower trimming (e.g., A084531, A167237), when applied to strongly fractal sequences, yield strongly fractal sequences. The operation introduced here, which can be called "segment-upper trimming", carries fractal sequences to fractal sequences, but not strongly fractal to strongly fractal.
Associated with the signature sequence S of each positive irrational number is an interspersion (or equivalently, a dispersion), in which row n >= 0 consists of the positions of n in S. The interspersion associated with the signature sequence of log(3)/log(2) is A255975.
(End)
Comment from Allan C. Wechsler, May 26 2024 (Start):
More generally, the "signature sequence" of an irrational number H can be defined as follows. Consider all the numbers of the form a + bH, where a and b are positive integers, and sort them into increasing order (there are no cluster points or other obstacles). The sequence of a-values is then the *signature sequence of H.
If the coefficients a and b are allowed to be 0, you get the same sequence but with all the entries decremented by 1.
(End)
a(n) = A069352(n) - A022329(n). - Reinhard Zumkeller, May 16 2015

Crossrefs

Programs

  • Haskell
    import Data.Set (singleton, deleteFindMin, insert)
    a022328 n = a022328_list !! (n-1)
    (a022328_list, a022329_list) = unzip $ f $ singleton (1, (0, 0)) where
       f s = (i, j) :
             f (insert (2 * y, (i + 1, j)) $ insert (3 * y, (i, j + 1)) s')
             where ((y, (i, j)), s') = deleteFindMin s
    -- Reinhard Zumkeller, Nov 19 2015, May 16 2015
    
  • Mathematica
    t = Sort[Flatten[Table[2^i 3^j, {i, 0, 200}, {j, 0, 200}]]];
    Table[IntegerExponent[t[[n]], 2], {n, 1, 200}]  (* A022338 *)
    (* Clark Kimberling, Mar 18 2015 *)
  • Python
    from sympy import integer_log
    def A022328(n):
        def bisection(f,kmin=0,kmax=1):
            while f(kmax) > kmax: kmax <<= 1
            while kmax-kmin > 1:
                kmid = kmax+kmin>>1
                if f(kmid) <= kmid:
                    kmax = kmid
                else:
                    kmin = kmid
            return kmax
        def f(x): return n+x-sum((x//3**i).bit_length() for i in range(integer_log(x,3)[0]+1))
        return (~(m:=bisection(f,n,n))&m-1).bit_length() # Chai Wah Wu, Sep 15 2024

Formula

a(n) = A191475(n) - 1. - Franklin T. Adams-Watters, Mar 19 2009 [Corrected by N. J. A. Sloane, May 26 2024]
A003586(n) = 2^a(n)*3^A022329(n). - N. J. A. Sloane, Mar 19 2009
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