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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A008836 Liouville's function lambda(n) = (-1)^k, where k is number of primes dividing n (counted with multiplicity).

Original entry on oeis.org

1, -1, -1, 1, -1, 1, -1, -1, 1, 1, -1, -1, -1, 1, 1, 1, -1, -1, -1, -1, 1, 1, -1, 1, 1, 1, -1, -1, -1, -1, -1, -1, 1, 1, 1, 1, -1, 1, 1, 1, -1, -1, -1, -1, -1, 1, -1, -1, 1, -1, 1, -1, -1, 1, 1, 1, 1, 1, -1, 1, -1, 1, -1, 1, 1, -1, -1, -1, 1, -1, -1, -1, -1, 1, -1, -1, 1, -1, -1, -1, 1, 1, -1, 1, 1, 1, 1, 1, -1, 1, 1, -1, 1, 1, 1, 1, -1, -1, -1, 1, -1
Offset: 1

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Author

Keywords

Comments

Coons and Borwein: "We give a new proof of Fatou's theorem: if an algebraic function has a power series expansion with bounded integer coefficients, then it must be a rational function. This result is applied to show that for any non-trivial completely multiplicative function from N to {-1,1}, the series sum_{n=1..infinity} f(n)z^n is transcendental over {Z}[z]; in particular, sum_{n=1..infinity} lambda(n)z^n is transcendental, where lambda is Liouville's function. The transcendence of sum_{n=1..infinity} mu(n)z^n is also proved." - Jonathan Vos Post, Jun 11 2008
Coons proves that a(n) is not k-automatic for any k > 2. - Jonathan Vos Post, Oct 22 2008
The Riemann hypothesis is equivalent to the statement that for every fixed epsilon > 0, lim_{n -> infinity} (a(1) + a(2) + ... + a(n))/n^(1/2 + epsilon) = 0 (Borwein et al., theorem 1.2). - Arkadiusz Wesolowski, Oct 08 2013

Examples

			a(4) = 1 because since bigomega(4) = 2 (the prime divisor 2 is counted twice), then (-1)^2 = 1.
a(5) = -1 because 5 is prime and therefore bigomega(5) = 1 and (-1)^1 = -1.
		

References

  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 37.
  • P. Borwein, S. Choi, B. Rooney and A. Weirathmueller, The Riemann Hypothesis: A Resource for the Aficionado and Virtuoso Alike, Springer, Berlin, 2008, pp. 1-11.
  • H. Gupta, On a table of values of L(n), Proceedings of the Indian Academy of Sciences. Section A, 12 (1940), 407-409.
  • H. Gupta, A table of values of Liouville's function L(n), Research Bulletin of East Panjab University, No. 3 (Feb. 1950), 45-55.
  • P. Ribenboim, Algebraic Numbers, p. 44.
  • J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 279.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, Exercise 3.3.5 on page 99.
  • J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, p. 112.

Crossrefs

Möbius transform of A010052.
Cf. A182448 (Dgf at s=2), A347328 (Dgf at s=3), A347329 (Dgf at s=4).

Programs

  • Haskell
    a008836 = (1 -) . (* 2) . a066829  -- Reinhard Zumkeller, Nov 19 2011
    
  • Maple
    A008836 := n -> (-1)^numtheory[bigomega](n); # Peter Luschny, Sep 15 2011
    with(numtheory): A008836 := proc(n) local i,it,s; it := ifactors(n): s := (-1)^add(it[2][i][2], i=1..nops(it[2])): RETURN(s) end:
  • Mathematica
    Table[LiouvilleLambda[n], {n, 100}] (* Enrique Pérez Herrero, Dec 28 2009 *)
    Table[If[OddQ[PrimeOmega[n]],-1,1],{n,110}] (* Harvey P. Dale, Sep 10 2014 *)
  • PARI
    {a(n) = if( n<1, 0, n=factor(n); (-1)^sum(i=1, matsize(n)[1], n[i,2]))}; /* Michael Somos, Jan 01 2006 */
    
  • PARI
    a(n)=(-1)^bigomega(n) \\ Charles R Greathouse IV, Jan 09 2013
    
  • Python
    from sympy import factorint
    def A008836(n): return -1 if sum(factorint(n).values()) % 2 else 1 # Chai Wah Wu, May 24 2022

Formula

Dirichlet g.f.: zeta(2s)/zeta(s); Dirichlet inverse of A008966.
Sum_{ d divides n } lambda(d) = 1 if n is a square, otherwise 0.
Completely multiplicative with a(p) = -1, p prime.
a(n) = (-1)^A001222(n) = (-1)^bigomega(n). - Jonathan Vos Post, Apr 16 2006
a(n) = A033999(A001222(n)). - Jaroslav Krizek, Sep 28 2009
Sum_{d|n} a(d) *(A000005(d))^2 = a(n) *Sum{d|n} A000005(d). - Vladimir Shevelev, May 22 2010
a(n) = 1 - 2*A066829(n). - Reinhard Zumkeller, Nov 19 2011
a(n) = i^(tau(n^2)-1) where tau(n) is A000005 and i is the imaginary unit. - Anthony Browne, May 11 2016
a(n) = A106400(A156552(n)). - Antti Karttunen, May 30 2017
Recurrence: a(1)=1, n > 1: a(n) = sign(1/2 - Sum_{dMats Granvik, Oct 11 2017
a(n) = Sum_{ d | n } A008683(d)*A010052(n/d). - Jinyuan Wang, Apr 20 2019
a(1) = 1; a(n) = -Sum_{d|n, d < n} mu(n/d)^2 * a(d). - Ilya Gutkovskiy, Mar 10 2021
a(n) = (-1)^A349905(n). - Antti Karttunen, Apr 26 2022
From Ridouane Oudra, Jun 02 2024: (Start)
a(n) = (-1)^A066829(n);
a(n) = (-1)^A063647(n);
a(n) = A101455(A048691(n));
a(n) = sin(tau(n^2)*Pi/2). (End)

A005043 Riordan numbers: a(n) = (n-1)*(2*a(n-1) + 3*a(n-2))/(n+1).

Original entry on oeis.org

1, 0, 1, 1, 3, 6, 15, 36, 91, 232, 603, 1585, 4213, 11298, 30537, 83097, 227475, 625992, 1730787, 4805595, 13393689, 37458330, 105089229, 295673994, 834086421, 2358641376, 6684761125, 18985057351, 54022715451, 154000562758, 439742222071, 1257643249140
Offset: 0

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Author

Keywords

Comments

Also called Motzkin summands or ring numbers.
The old name was "Motzkin sums", used in certain publications. The sequence has the property that Motzkin(n) = A001006(n) = a(n) + a(n+1), e.g., A001006(4) = 9 = 3 + 6 = a(4) + a(5).
Number of 'Catalan partitions', that is partitions of a set 1,2,3,...,n into parts that are not singletons and whose convex hulls are disjoint when the points are arranged on a circle (so when the parts are all pairs we get Catalan numbers). - Aart Blokhuis (aartb(AT)win.tue.nl), Jul 04 2000
Number of ordered trees with n edges and no vertices of outdegree 1. For n > 1, number of dissections of a convex polygon by nonintersecting diagonals with a total number of n+1 edges. - Emeric Deutsch, Mar 06 2002
Number of Motzkin paths of length n with no horizontal steps at level 0. - Emeric Deutsch, Nov 09 2003
Number of Dyck paths of semilength n with no peaks at odd level. Example: a(4)=3 because we have UUUUDDDD, UUDDUUDD and UUDUDUDD, where U=(1,1), D=(1,-1). Number of Dyck paths of semilength n with no ascents of length 1 (an ascent in a Dyck path is a maximal string of up steps). Example: a(4)=3 because we have UUUUDDDD, UUDDUUDD and UUDUUDDD. - Emeric Deutsch, Dec 05 2003
Arises in Schubert calculus as follows. Let P = complex projective space of dimension n+1. Take n projective subspaces of codimension 3 in P in general position. Then a(n) is the number of lines of P intersecting all these subspaces. - F. Hirzebruch, Feb 09 2004
Difference between central trinomial coefficient and its predecessor. Example: a(6) = 15 = 141 - 126 and (1 + x + x^2)^6 = ... + 126*x^5 + 141*x^6 + ... (Catalan number A000108(n) is the difference between central binomial coefficient and its predecessor.) - David Callan, Feb 07 2004
a(n) = number of 321-avoiding permutations on [n] in which each left-to-right maximum is a descent (i.e., is followed by a smaller number). For example, a(4) counts 4123, 3142, 2143. - David Callan, Jul 20 2005
The Hankel transform of this sequence give A000012 = [1, 1, 1, 1, 1, 1, 1, ...]; example: Det([1, 0, 1, 1; 0, 1, 1, 3; 1, 1, 3, 6; 1, 3, 6, 15]) = 1. - Philippe Deléham, May 28 2005
The number of projective invariants of degree 2 for n labeled points on the projective line. - Benjamin J. Howard (bhoward(AT)ima.umn.edu), Nov 24 2006
Define a random variable X=trA^2, where A is a 2 X 2 unitary symplectic matrix chosen from USp(2) with Haar measure. The n-th central moment of X is E[(X+1)^n] = a(n). - Andrew V. Sutherland, Dec 02 2007
Let V be the adjoint representation of the complex Lie algebra sl(2). The dimension of the invariant subspace of the n-th tensor power of V is a(n). - Samson Black (sblack1(AT)uoregon.edu), Aug 27 2008
Starting with offset 3 = iterates of M * [1,1,1,...], where M = a tridiagonal matrix with [0,1,1,1,...] in the main diagonal and [1,1,1,...] in the super and subdiagonals. - Gary W. Adamson, Jan 08 2009
a(n) has the following standard-Young-tableaux (SYT) interpretation: binomial(n+1,k)*binomial(n-k-1,k-1)/(n+1)=f^(k,k,1^{n-2k}) where f^lambda equals the number of SYT of shape lambda. - Amitai Regev (amotai.regev(AT)weizmann.ac.il), Mar 02 2010
a(n) is also the sum of the numbers of standard Young tableaux of shapes (k,k,1^{n-2k}) for all 1 <= k <= floor(n/2). - Amitai Regev (amotai.regev(AT)weizmann.ac.il), Mar 10 2010
a(n) is the number of derangements of {1,2,...,n} having genus 0. The genus g(p) of a permutation p of {1,2,...,n} is defined by g(p)=(1/2)[n+1-z(p)-z(cp')], where p' is the inverse permutation of p, c = 234...n1 = (1,2,...,n), and z(q) is the number of cycles of the permutation q. Example: a(3)=1 because p=231=(123) is the only derangement of {1,2,3} with genus 0. Indeed, cp'=231*312=123=(1)(2)(3) and so g(p) = (1/2)(3+1-1-3)=0. - Emeric Deutsch, May 29 2010
Apparently: Number of Dyck 2n-paths with all ascents length 2 and no descent length 2. - David Scambler, Apr 17 2012
This is true. Proof: The mapping "insert a peak (UD) after each upstep (U)" is a bijection from all Dyck n-paths to those Dyck (2n)-paths in which each ascent is of length 2. It sends descents of length 1 in the n-path to descents of length 2 in the (2n)-path. But Dyck n-paths with no descents of length 1 are equinumerous with Riordan n-paths (Motzkin n-paths with no flatsteps at ground level) as follows. Given a Dyck n-path with no descents of length 1, split it into consecutive step pairs, then replace UU with U, DD with D, UD with a blue flatstep (F), DU with a red flatstep, and concatenate the new steps to get a colored Motzkin path. Each red F will be (immediately) preceded by a blue F or a D. In the latter case, transfer the red F so that it precedes the matching U of the D. Finally, erase colors to get the required Riordan path. For example, with lowercase f denoting a red flatstep, U^5 D^2 U D^4 U^4 D^3 U D^2 -> (U^2, U^2, UD, DU, D^2, D^2, U^2, U^2 D^2, DU, D^2) -> UUFfDDUUDfD -> UUFFDDUFUDD. - David Callan, Apr 25 2012
From Nolan Wallach, Aug 20 2014: (Start)
Let ch[part1, part2] be the value of the character of the symmetric group on n letters corresponding to the partition part1 of n on the conjucgacy class given by part2. Let A[n] be the set of (n+1) partitions of 2n with parts 1 or 2. Then deleting the first term of the sequence one has a(n) = Sum_{k=1..n+1} binomial(n,k-1)*ch[[n,n], A[n][[k]]])/2^n. This via the Frobenius Character Formula can be interpreted as the dimension of the SL(n,C) invariants in tensor^n (wedge^2 C^n).
Explanation: Let p_j denote sum (x_i)^j the sum in k variables. Then the Frobenius formula says then (p_1)^j_1 (p_2)^j_2 ... (p_r)^j_r is equal to sum(lambda, ch[lambda, 1^j_12^j_2 ... r^j_r] S_lambda) with S_lambda the Schur function corresponding to lambda. This formula implies that the coefficient of S([n,n]) in (((p_1)^1+p_2)/2)^n in its expansion in terms of Schur functions is the right hand side of our formula. If we specialize the number of variables to 2 then S[n,n](x,y)=(xy)^n. Which when restricted to y=x^(-1) is 1. That is it is 1 on SL(2).
On the other hand ((p_1)^2+p_2)/2 is the complete homogeneous symmetric function of degree 2 that is tr(S^2(X)). Thus our formula for a(n) is the same as that of Samson Black above since his V is the same as S^2(C^2) as a representation of SL(2). On the other hand, if we multiply ch(lambda) by sgn you get ch(Transpose(lambda)). So ch([n,n]) becomes ch([2,...,2]) (here there are n 2's). The formula for a(n) is now (1/2^n)*Sum_{j=0..n} ch([2,..,2], 1^(2n-2j) 2^j])*(-1)^j)*binomial(n,j), which calculates the coefficient of S_(2,...,2) in (((p_1)^2-p_2)/2)^n. But ((p_1)^2-p_2)/2 in n variables is the second elementary symmetric function which is the character of wedge^2 C^n and S_(2,...,2) is 1 on SL(n).
(End)
a(n) = number of noncrossing partitions (A000108) of [n] that contain no singletons, also number of nonnesting partitions (A000108) of [n] that contain no singletons. - David Callan, Aug 27 2014
From Tom Copeland, Nov 02 2014: (Start)
Let P(x) = x/(1+x) with comp. inverse Pinv(x) = x/(1-x) = -P[-x], and C(x)= [1-sqrt(1-4x)]/2, an o.g.f. for the shifted Catalan numbers A000108, with inverse Cinv(x) = x * (1-x).
Fin(x) = P[C(x)] = C(x)/[1 + C(x)] is an o.g.f. for the Fine numbers, A000957 with inverse Fin^(-1)(x) = Cinv[Pinv(x)] = Cinv[-P(-x)].
Mot(x) = C[P(x)] = C[-Pinv(-x)] gives an o.g.f. for shifted A005043, the Motzkin or Riordan numbers with comp. inverse Mot^(-1)(x) = Pinv[Cinv(x)] = (x - x^2) / (1 - x + x^2) (cf. A057078).
BTC(x) = C[Pinv(x)] gives A007317, a binomial transform of the Catalan numbers, with BTC^(-1)(x) = P[Cinv(x)].
Fib(x) = -Fin[Cinv(Cinv(-x))] = -P[Cinv(-x)] = x + 2 x^2 + 3 x^3 + 5 x^4 + ... = (x+x^2)/[1-x-x^2] is an o.g.f. for the shifted Fibonacci sequence A000045, so the comp. inverse is Fib^(-1)(x) = -C[Pinv(-x)] = -BTC(-x) and Fib(x) = -BTC^(-1)(-x).
Various relations among the o.g.f.s may be easily constructed, such as Fib[-Mot(-x)] = -P[P(-x)] = x/(1-2*x) a generating fct for 2^n.
Generalizing to P(x,t) = x /(1 + t*x) and Pinv(x,t) = x /(1 - t*x) = -P(-x,t) gives other relations to lattice paths, such as the o.g.f. for A091867, C[P[x,1-t]], and that for A104597, Pinv[Cinv(x),t+1]. (End)
Consistent with David Callan's comment above, A249548, provides a refinement of the Motzkin sums into the individual numbers for the non-crossing partitions he describes. - Tom Copeland, Nov 09 2014
The number of lattice paths from (0,0) to (n,0) that do not cross below the x-axis and use up-step=(1,1) and down-steps=(1,-k) where k is a positive integer. For example, a(4) = 3: [(1,1)(1,1)(1,-1)(1,-1)], [(1,1)(1,-1)(1,1)(1,-1)] and [(1,1)(1,1)(1,1)(1,-3)]. - Nicholas Ham, Aug 19 2015
A series created using 2*(a(n) + a(n+1)) + (a(n+1) + a(n+2)) has Hankel transform of F(2n), offset 3, F being a Fibonacci number, A001906 (Empirical observation). - Tony Foster III, Jul 30 2016
The series a(n) + A001006(n) has Hankel transform F(2n+1), offset n=1, F being the Fibonacci bisection A001519 (empirical observation). - Tony Foster III, Sep 05 2016
The Rubey and Stump reference proves a refinement of a conjecture of René Marczinzik, which they state as: "The number of 2-Gorenstein algebras which are Nakayama algebras with n simple modules and have an oriented line as associated quiver equals the number of Motzkin paths of length n. Moreover, the number of such algebras having the double centraliser property with respect to a minimal faithful projective-injective module equals the number of Riordan paths, that is, Motzkin paths without level-steps at height zero, of length n." - Eric M. Schmidt, Dec 16 2017
A connection to the Thue-Morse sequence: (-1)^a(n) = (-1)^A010060(n) * (-1)^A010060(n+1) = A106400(n) * A106400(n+1). - Vladimir Reshetnikov, Jul 21 2019
Named by Bernhart (1999) after the American mathematician John Riordan (1903-1988). - Amiram Eldar, Apr 15 2021

Examples

			a(5)=6 because the only dissections of a polygon with a total number of 6 edges are: five pentagons with one of the five diagonals and the hexagon with no diagonals.
G.f. = 1 + x^2 + x^3 + 3*x^4 + 6*x^5 + 15*x^6 + 36*x^7 + 91*x^8 + 232*x^9 + ...
From _Gus Wiseman_, Nov 15 2022: (Start)
The a(0) = 1 through a(6) = 15 lone-child-avoiding (no vertices of outdegree 1) ordered rooted trees with n + 1 vertices (ranked by A358376):
  o  .  (oo)  (ooo)  (oooo)   (ooooo)   (oooooo)
                     ((oo)o)  ((oo)oo)  ((oo)ooo)
                     (o(oo))  ((ooo)o)  ((ooo)oo)
                              (o(oo)o)  ((oooo)o)
                              (o(ooo))  (o(oo)oo)
                              (oo(oo))  (o(ooo)o)
                                        (o(oooo))
                                        (oo(oo)o)
                                        (oo(ooo))
                                        (ooo(oo))
                                        (((oo)o)o)
                                        ((o(oo))o)
                                        ((oo)(oo))
                                        (o((oo)o))
                                        (o(o(oo)))
(End)
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Row sums of triangle A020474, first differences of A082395.
First diagonal of triangular array in A059346.
Binomial transform of A126930. - Philippe Deléham, Nov 26 2009
The Hankel transform of a(n+1) is A128834. The Hankel transform of a(n+2) is floor((2*n+4)/3) = A004523(n+2). - Paul Barry, Mar 08 2011
The Kn11 triangle sums of triangle A175136 lead to A005043(n+2), while the Kn12(n) = A005043(n+4)-2^(n+1), Kn13(n) = A005043(n+6)-(n^2+9*n+56)*2^(n-2) and the Kn4(n) = A005043(2*n+2) = A099251(n+1) triangle sums are related to the sequence given above. For the definitions of these triangle sums see A180662. - Johannes W. Meijer, May 06 2011
Cf. A187306 (self-convolution), A348210 (column 1).
Bisections: A099251, A099252.

Programs

  • Haskell
    a005043 n = a005043_list !! n
    a005043_list = 1 : 0 : zipWith div
       (zipWith (*) [1..] (zipWith (+)
           (map (* 2) $ tail a005043_list) (map (* 3) a005043_list))) [3..]
    -- Reinhard Zumkeller, Jan 31 2012
    
  • Maple
    A005043 := proc(n) option remember; if n <= 1 then 1-n else (n-1)*(2*A005043(n-1)+3*A005043(n-2))/(n+1); fi; end;
    Order := 20: solve(series((x-x^2)/(1-x+x^2),x)=y,x); # outputs g.f.
  • Mathematica
    a[0]=1; a[1]=0; a[n_]:= a[n] = (n-1)*(2*a[n-1] + 3*a[n-2])/(n+1); Table[ a[n], {n, 0, 30}] (* Robert G. Wilson v, Jun 14 2005 *)
    Table[(-3)^(1/2)/6 * (-1)^n*(3*Hypergeometric2F1[1/2,n+1,1,4/3]+ Hypergeometric2F1[1/2,n+2,1,4/3]), {n,0,32}] (* cf. Mark van Hoeij in A001006 *) (* Wouter Meeussen, Jan 23 2010 *)
    RecurrenceTable[{a[0]==1,a[1]==0,a[n]==(n-1) (2a[n-1]+3a[n-2])/(n+1)},a,{n,30}] (* Harvey P. Dale, Sep 27 2013 *)
    a[ n_]:= SeriesCoefficient[2/(1+x +Sqrt[1-2x-3x^2]), {x, 0, n}]; (* Michael Somos, Aug 21 2014 *)
    a[ n_]:= If[n<0, 0, 3^(n+3/2) Hypergeometric2F1[3/2, n+2, 2, 4]/I]; (* Michael Somos, Aug 21 2014 *)
    Table[3^(n+3/2) CatalanNumber[n] (4(5+2n)Hypergeometric2F1[3/2, 3/2, 1/2-n, 1/4] -9 Hypergeometric2F1[3/2, 5/2, 1/2 -n, 1/4])/(4^(n+3) (n+1)), {n, 0, 31}] (* Vladimir Reshetnikov, Jul 21 2019 *)
    Table[Sqrt[27]/8 (3/4)^n CatalanNumber[n] Hypergeometric2F1[1/2, 3/2, 1/2 - n, 1/4], {n, 0, 31}] (* Jan Mangaldan, Sep 12 2021 *)
  • Maxima
    a[0]:1$
    a[1]:0$
    a[n]:=(n-1)*(2*a[n-1]+3*a[n-2])/(n+1)$
    makelist(a[n],n,0,12); /* Emanuele Munarini, Mar 02 2011 */
    
  • PARI
    {a(n) = if( n<0, 0, n++; polcoeff( serreverse( (x - x^3) / (1 + x^3) + x * O(x^n)), n))}; /* Michael Somos, May 31 2005 */
    
  • PARI
    my(N=66); Vec(serreverse(x/(1+x*sum(k=1,N,x^k))+O(x^N))) \\ Joerg Arndt, Aug 19 2012
    
  • Python
    from functools import cache
    @cache
    def A005043(n: int) -> int:
        if n <= 1: return 1 - n
        return (n - 1) * (2 * A005043(n - 1) + 3 * A005043(n - 2)) // (n + 1)
    print([A005043(n) for n in range(32)]) # Peter Luschny, Nov 20 2022
  • Sage
    A005043 = lambda n: (-1)^n*jacobi_P(n,1,-n-3/2,-7)/(n+1)
    [simplify(A005043(n)) for n in (0..29)]
    # Peter Luschny, Sep 23 2014
    
  • Sage
    def ms():
        a, b, c, d, n = 0, 1, 1, -1, 1
        yield 1
        while True:
            yield -b + (-1)^n*d
            n += 1
            a, b = b, (3*(n-1)*n*a+(2*n-1)*n*b)/((n+1)*(n-1))
            c, d = d, (3*(n-1)*c-(2*n-1)*d)/n
    A005043 = ms()
    print([next(A005043) for  in range(32)]) # _Peter Luschny, May 16 2016
    

Formula

a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*A000108(k). a(n) = (1/(n+1)) * Sum_{k=0..ceiling(n/2)} binomial(n+1, k)*binomial(n-k-1, k-1), for n > 1. - Len Smiley. [Comment from Amitai Regev (amitai.regev(AT)weizmann.ac.il), Mar 02 2010: the latter sum should be over the range k=1..floor(n/2).]
G.f.: (1 + x - sqrt(1-2*x-3*x^2))/(2*x*(1+x)).
G.f.: 2/(1+x+sqrt(1-2*x-3*x^2)). - Paul Peart (ppeart(AT)fac.howard.edu), May 27 2000
a(n+1) + (-1)^n = a(0)*a(n) + a(1)*a(n-1) + ... + a(n)*a(0). - Bernhart
a(n) = (1/(n+1)) * Sum_{i} (-1)^i*binomial(n+1, i)*binomial(2*n-2*i, n-i). - Bernhart
G.f. A(x) satisfies A = 1/(1+x) + x*A^2.
E.g.f.: exp(x)*(BesselI(0, 2*x) - BesselI(1, 2*x)). - Vladeta Jovovic, Apr 28 2003
a(n) = A001006(n-1) - a(n-1).
a(n+1) = Sum_{k=0..n} (-1)^k*A026300(n, k), where A026300 is the Motzkin triangle.
a(n) = Sum_{k=0..n} (-1)^k*binomial(n, k)*binomial(k, floor(k/2)). - Paul Barry, Jan 27 2005
a(n) = Sum_{k>=0} A086810(n-k, k). - Philippe Deléham, May 30 2005
a(n+2) = Sum_{k>=0} A064189(n-k, k). - Philippe Deléham, May 31 2005
Moment representation: a(n) = (1/(2*Pi))*Int(x^n*sqrt((1+x)(3-x))/(1+x),x,-1,3). - Paul Barry, Jul 09 2006
Inverse binomial transform of A000108 (Catalan numbers). - Philippe Deléham, Oct 20 2006
a(n) = (2/Pi)* Integral_{x=0..Pi} (4*cos(x)^2-1)^n*sin(x)^2 dx. - Andrew V. Sutherland, Dec 02 2007
G.f.: 1/(1-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-... (continued fraction). - Paul Barry, Jan 22 2009
G.f.: 1/(1+x-x/(1-x/(1+x-x/(1-x/(1+x-x/(1-... (continued fraction). - Paul Barry, May 16 2009
G.f.: 1/(1-x^2/(1-x/(1-x/(1-x^2/(1-x/(1-x/(1-x^2/(1-x/(1-... (continued fraction). - Paul Barry, Mar 02 2010
a(n) = -(-1)^n * hypergeom([1/2, n+2],[2],4/3) / sqrt(-3). - Mark van Hoeij, Jul 02 2010
a(n) = (-1)^n*hypergeometric([-n,1/2],[2],4). - Peter Luschny, Aug 15 2012
Let A(x) be the g.f., then x*A(x) is the reversion of x/(1 + x^2*Sum_{k>=0} x^k); see A215340 for the correspondence to Dyck paths without length-1 ascents. - Joerg Arndt, Aug 19 2012 and Apr 16 2013
a(n) ~ 3^(n+3/2)/(8*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 02 2012
G.f.: 2/(1+x+1/G(0)), where G(k) = 1 + x*(2+3*x)*(4*k+1)/( 4*k+2 - x*(2+3*x)*(4*k+2)*(4*k+3)/(x*(2+3*x)*(4*k+3) + 4*(k+1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 05 2013
D-finite (an alternative): (n+1)*a(n) = 3*(n-2)*a(n-3) + (5*n-7)*a(n-2) + (n-2)*a(n-1), n >= 3. - Fung Lam, Mar 22 2014
Asymptotics: a(n) = (3^(n+2)/(sqrt(3*n*Pi)*(8*n)))*(1-21/(16*n) + O(1/n^2)) (with contribution by Vaclav Kotesovec). - Fung Lam, Mar 22 2014
a(n) = T(2*n-1,n)/n, where T(n,k) = triangle of A180177. - Vladimir Kruchinin, Sep 23 2014
a(n) = (-1)^n*JacobiP(n,1,-n-3/2,-7)/(n+1). - Peter Luschny, Sep 23 2014
a(n) = Sum_{k=0..n} C(n,k)*(C(k,n-k)-C(k,n-k-1)). - Peter Luschny, Oct 01 2014
Conjecture: a(n) = A002426(n) - A005717(n), n > 0. - Mikhail Kurkov, Feb 24 2019 [The conjecture is true. - Amiram Eldar, May 17 2024]
a(n) = A309303(n) + A309303(n+1). - Vladimir Reshetnikov, Jul 22 2019
From Peter Bala, Feb 11 2022: (Start)
a(n) = A005773(n+1) - 2*A005717(n) for n >= 1.
Conjectures: for n >= 1, n divides a(2*n+1) and 2*n-1 divides a(2*n). (End)

Extensions

Thanks to Laura L. M. Yang (yanglm(AT)hotmail.com) for a correction, Aug 29 2004
Name changed to Riordan numbers following a suggestion from Ira M. Gessel. - N. J. A. Sloane, Jul 24 2020

A018819 Binary partition function: number of partitions of n into powers of 2.

Original entry on oeis.org

1, 1, 2, 2, 4, 4, 6, 6, 10, 10, 14, 14, 20, 20, 26, 26, 36, 36, 46, 46, 60, 60, 74, 74, 94, 94, 114, 114, 140, 140, 166, 166, 202, 202, 238, 238, 284, 284, 330, 330, 390, 390, 450, 450, 524, 524, 598, 598, 692, 692, 786, 786, 900, 900, 1014, 1014, 1154, 1154, 1294, 1294
Offset: 0

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Author

Keywords

Comments

First differences of A000123; also A000123 with terms repeated. See the relevant proof that follows the first formula below.
Among these partitions there is exactly one partition with all distinct terms, as every number can be expressed as the sum of the distinct powers of 2.
Euler transform of A036987 with offset 1.
a(n) is the number of "non-squashing" partitions of n, that is, partitions n = 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. - N. J. A. Sloane, Nov 30 2003
Normally the OEIS does not include sequences like this where every term is repeated, but an exception was made for this one because of its importance. The unrepeated sequence A000123 is the main entry.
Number of different partial sums from 1 + [1, *2] + [1, *2] + ..., where [1, *2] means we can either add 1 or multiply by 2. E.g., a(6) = 6 because we have 6 = 1 + 1 + 1 + 1 + 1 + 1 = (1+1) * 2 + 1 + 1 = 1 * 2 * 2 + 1 + 1 = (1+1+1) * 2 = 1 * 2 + 1 + 1 + 1 + 1 = (1*2+1) * 2 where the connection is defined via expanding each bracket; e.g., this is 6 = 1 + 1 + 1 + 1 + 1 + 1 = 2 + 2 + 1 + 1 = 4 + 1 + 1 = 2 + 2 + 2 = 2 + 1 + 1 + 1 + 1 = 4 + 2. - Jon Perry, Jan 01 2004
Number of partitions p of n such that the number of compositions generated by p is odd. For proof see the Alekseyev and Adams-Watters link. - Vladeta Jovovic, Aug 06 2007
Differs from A008645 first at a(64). - R. J. Mathar, May 28 2008
Appears to be row sums of A155077. - Mats Granvik, Jan 19 2009
Number of partitions (p_1, p_2, ..., p_k) of n, with p_1 >= p_2 >= ... >= p_k, such that for each i, p_i >= p_{i+1} + ... + p_k. - John MCKAY (mckay(AT)encs.concordia.ca), Mar 06 2009 (these are the "non-squashing" partitions as nonincreasing lists).
Equals rightmost diagonal of triangle of A168261. Starting with offset 1 = eigensequence of triangle A115361 and row sums of triangle A168261. - Gary W. Adamson, Nov 21 2009
Equals convolution square root of A171238: (1, 2, 5, 8, 16, 24, 40, 56, 88, ...). - Gary W. Adamson, Dec 05 2009
Let B = the n-th convolution power of the sequence and C = the aerated variant of B. It appears that B/C = the binomial sequence beginning (1, n, ...). Example: Third convolution power of the sequence is (1, 3, 9, 19, 42, 78, 146, ...), with C = (1, 0, 3, 0, 9, 0, 19, ...). Then B/C = (1, 3, 6, 10, 15, 21, ...). - Gary W. Adamson, Aug 15 2016
From Gary W. Adamson, Sep 08 2016: (Start)
The limit of the matrix power M^k as n-->inf results in a single column vector equal to the sequence, where M is the following production matrix:
1, 0, 0, 0, 0, ...
1, 0, 0, 0, 0, ...
1, 1, 0, 0, 0, ...
1, 1, 0, 0, 0, ...
1, 1, 1, 0, 0, ...
1, 1, 1, 0, 0, ...
1, 1, 1, 1, 0, ...
1, 1, 1, 1, 0, ...
1, 1, 1, 1, 1, ...
... (End)
a(n) is the number of "non-borrowing" partitions of n, meaning binary subtraction of a smaller part from a larger part will never require place-value borrowing. - David V. Feldman, Jan 29 2020
From Gus Wiseman, May 25 2024: (Start)
Also the number of multisets of positive integers whose binary rank is n, where the binary rank of a multiset m is given by Sum_i 2^(m_i-1). For example, the a(1) = 1 through a(8) = 10 multisets are:
{1} {2} {12} {3} {13} {23} {123} {4}
{11} {111} {22} {122} {113} {1113} {33}
{112} {1112} {222} {1222} {223}
{1111} {11111} {1122} {11122} {1123}
{11112} {111112} {2222}
{111111} {1111111} {11113}
{11222}
{111122}
{1111112}
{11111111}
(End)

Examples

			G.f. = 1 + x + 2*x^2 + 2*x^3 + 4*x^4 + 4*x^5 + 6*x^6 + 6*x^7 + 10*x^8 + ...
a(4) = 4: the partitions are 4, 2 + 2, 2 + 1 + 1, 1 + 1 + 1 + 1.
a(7) = 6: the partitions are 4 + 2 + 1, 4 + 1 + 1 + 1, 2 + 2 + 2 + 1, 2 + 2 + 1 + 1 + 1, 2 + 1 + 1 + 1 + 1 + 1, 1 + 1 + 1 + 1 + 1 + 1 + 1.
From _Joerg Arndt_, Dec 17 2012: (Start)
The a(10) = 14 binary partitions of 10 are (in lexicographic order)
[ 1]  [ 1 1 1 1 1 1 1 1 1 1 ]
[ 2]  [ 2 1 1 1 1 1 1 1 1 ]
[ 3]  [ 2 2 1 1 1 1 1 1 ]
[ 4]  [ 2 2 2 1 1 1 1 ]
[ 5]  [ 2 2 2 2 1 1 ]
[ 6]  [ 2 2 2 2 2 ]
[ 7]  [ 4 1 1 1 1 1 1 ]
[ 8]  [ 4 2 1 1 1 1 ]
[ 9]  [ 4 2 2 1 1 ]
[10]  [ 4 2 2 2 ]
[11]  [ 4 4 1 1 ]
[12]  [ 4 4 2 ]
[13]  [ 8 1 1 ]
[14]  [ 8 2 ]
The a(11) = 14 binary partitions of 11 are obtained by appending 1 to each partition in the list.
The a(10) = 14 non-squashing partitions of 10 are (in lexicographic order)
[ 1]  [ 6 3 1 1 ]
[ 2]  [ 6 3 2 ]
[ 3]  [ 6 4 1 ]
[ 4]  [ 6 5 ]
[ 5]  [ 7 2 1 1 ]
[ 6]  [ 7 2 2 ]
[ 7]  [ 7 3 1 ]
[ 8]  [ 7 4 ]
[ 9]  [ 8 2 1 ]
[10]  [ 8 3 ]
[11]  [ 9 1 1 ]
[12]  [ 9 2 ]
[13]  [ 10 1 ]
[14]  [ 11 ]
The a(11) = 14 non-squashing partitions of 11 are obtained by adding 1 to the first part in each partition in the list.
(End)
From _David V. Feldman_, Jan 29 2020: (Start)
The a(10) = 14 non-borrowing partitions of 10 are (in lexicographic order)
[ 1] [1 1 1 1 1 1 1 1 1 1]
[ 2] [2 2 2 2 2]
[ 3] [3 1 1 1 1 1 1 1]
[ 4] [3 3 1 1 1 1]
[ 5] [3 3 2 2]
[ 6] [3 3 3 1]
[ 7] [5 1 1 1 1 1]
[ 8] [5 5]
[ 9] [6 2 2]
[10] [6 4]
[11] [7 1 1 1]
[12] [7 3]
[13] [9 1]
[14] [10]
The a(11) = 14 non-borrowing partitions of 11 are obtained either by adding 1 to the first even part in each partition (if any) or else appending a 1 after the last part.
(End)
For example, the five partitions of 4, written in nonincreasing order, are [1, 1, 1, 1], [2, 1, 1], [2, 2], [3, 1], [4]. The last four satisfy the condition, and a(4) = 4. The Maple program below verifies this for small values of n.
		

Crossrefs

A000123 is the main entry for the binary partition function and gives many more properties and references.
Cf. A115625 (labeled binary partitions), A115626 (labeled non-squashing partitions).
Convolution inverse of A106400.
Multiplicity of n in A048675, for distinct prime indices A087207.
Row lengths of A277905.
A118462 lists binary ranks of strict integer partitions, row sums A372888.
A372890 adds up binary ranks of integer partitions.

Programs

  • Haskell
    a018819 n = a018819_list !! n
    a018819_list = 1 : f (tail a008619_list) where
       f (x:xs) = (sum $ take x a018819_list) : f xs
    -- Reinhard Zumkeller, Jan 28 2012
    
  • Haskell
    import Data.List (intersperse)
    a018819 = (a018819_list !!)
    a018819_list = 1 : 1 : (<*>) (zipWith (+)) (intersperse 0) (tail a018819_list)
    -- Johan Wiltink, Nov 08 2018
    
  • Maple
    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
    max = 59; a[0] = a[1] = 1; a[n_?OddQ] := a[n] = a[n-1]; a[n_?EvenQ] := a[n] = a[n-1] + a[n/2]; Table[a[n], {n, 0, max}]
    (* or *) CoefficientList[Series[1/Product[(1-x^(2^j)), {j, 0, Log[2, max] // Ceiling}], {x, 0, max}], x] (* Jean-François Alcover, May 17 2011, updated Feb 17 2014 *)
    a[ n_] := If[n<1, Boole[n==0], a[n] = a[n-1] + If[EvenQ@n, a[Quotient[n,2]], 0]]; (* Michael Somos, May 04 2022 *)
    Table[Count[IntegerPartitions[n],?(AllTrue[Log2[#],IntegerQ]&)],{n,0,60}] (* _Harvey P. Dale, Jun 20 2024 *)
  • PARI
    { n=15; v=vector(n); for (i=1,n,v[i]=vector(2^(i-1))); v[1][1]=1; for (i=2,n, k=length(v[i-1]); for (j=1,k, v[i][j]=v[i-1][j]+1; v[i][j+k]=v[i-1][j]*2)); c=vector(n); for (i=1,n, for (j=1,2^(i-1), if (v[i][j]<=n, c[v[i][j]]++))); c } /* Jon Perry */
    
  • 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)); polcoeff(A, n))}; /* Michael Somos, Aug 25 2003 */
    
  • PARI
    {a(n) = if( n<1, n==0, if( n%2, a(n-1), a(n/2)+a(n-1)))}; /* Michael Somos, Aug 25 2003 */
    
  • Python
    from functools import lru_cache
    @lru_cache(maxsize=None)
    def A018819(n): return 1 if n == 0 else A018819(n-1) + (0 if n % 2 else A018819(n//2)) # Chai Wah Wu, Jan 18 2022

Formula

a(2m+1) = a(2m), a(2m) = a(2m-1) + a(m). Proof: If n is odd there is a part of size 1; removing it gives a partition of n - 1. If n is even either there is a part of size 1, whose removal gives a partition of n - 1, or else all parts have even sizes and dividing each part by 2 gives a partition of n/2.
G.f.: 1 / Product_{j>=0} (1-x^(2^j)).
a(n) = (1/n)*Sum_{k = 1..n} A038712(k)*a(n-k), n > 1, a(0) = 1. - Vladeta Jovovic, Aug 22 2002
a(2*n) = a(2*n + 1) = A000123(n). - Michael Somos, Aug 25 2003
a(n) = 1 if n = 0, Sum_{j = 0..floor(n/2)} a(j) if n > 0. - David W. Wilson, Aug 16 2007
G.f. A(x) satisfies A(x^2) = (1-x) * A(x). - Michael Somos, Aug 25 2003
G.f. A(x) satisfies 0 = f(A(x), A(x^2), A(x^4)) where f(u, v, w) = u^2*w - 2*u*v^2 + v^3. - Michael Somos, Apr 10 2005
G.f. A(x) satisfies 0 = f(A(x), A(x^2), A(x^3), A(x^6)) where f(u1, u2, u3, u6) = u6 * u1^3 - 3*u3*u2*u1^2 + 3*u3*u2^2*u1 - u3*u2^3. - Michael Somos, Oct 15 2006
G.f.: 1/( Sum_{n >= 0} x^evil(n) - x^odious(n) ), where evil(n) = A001969(n) and odious(n) = A000069(n). - Paul D. Hanna, Jan 23 2012
Let A(x) by the g.f. and B(x) = A(x^k), then 0 = B*((1-A)^k - (-A)^k) + (-A)^k, see fxtbook link. - Joerg Arndt, Dec 17 2012
G.f.: Product_{n>=0} (1+x^(2^n))^(n+1), see the fxtbook link. - Joerg Arndt, Feb 28 2014
G.f.: 1 + Sum_{i>=0} x^(2^i) / Product_{j=0..i} (1 - x^(2^j)). - Ilya Gutkovskiy, May 07 2017

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

A124754 Alternating sum of compositions in standard order.

Original entry on oeis.org

0, 1, 2, 0, 3, 1, -1, 1, 4, 2, 0, 2, -2, 0, 2, 0, 5, 3, 1, 3, -1, 1, 3, 1, -3, -1, 1, -1, 3, 1, -1, 1, 6, 4, 2, 4, 0, 2, 4, 2, -2, 0, 2, 0, 4, 2, 0, 2, -4, -2, 0, -2, 2, 0, -2, 0, 4, 2, 0, 2, -2, 0, 2, 0, 7, 5, 3, 5, 1, 3, 5, 3, -1, 1, 3, 1, 5, 3, 1, 3, -3, -1, 1, -1, 3, 1, -1, 1, 5, 3, 1, 3, -1, 1, 3, 1, -5, -3, -1, -3, 1, -1, -3, -1, 3
Offset: 0

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Author

Keywords

Comments

The standard order of compositions is given by A066099.
The sum of row n is 2^{n-1} for n>0.

Examples

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

Crossrefs

Cf. A066099, A070939, A124756, A011782 (row lengths), A106400.

Formula

For a composition b(1),...,b(k), a(n) = Sum_{i=1}^k (-1)^{i-1} b(i).
a(2^k) = k+1. If n = 2^e_1 + 2^e_2 + k, 0 <= k < 2^e_2 < 2^e_1, then a(n) = (e_1 - e_2) - a(2^e_2 + k).
a(0) = 0; for n>0, a(n) = a(floor(n/2)) - A106400(n).

A010059 Another version of the Thue-Morse sequence: let A_k denote the first 2^k terms; then A_0 = 1 and for k >= 0, A_{k+1} = A_k B_k, where B_k is obtained from A_k by interchanging 0's and 1's.

Original entry on oeis.org

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

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Keywords

Comments

Characteristic function of A001969 (evil numbers). - Ralf Stephan, Jun 20 2003
From Gary W. Adamson, Aug 24 2008: (Start)
Parity of A143579 (Odious numbers (A000069) interleaved with Evil numbers (A001969)).
Two conjectures: If n is even, the ratio of 1's to 0's = 1:1.
There are no three adjacent terms of the same parity. (End)
Conjecture (verified for the first 280000 entries): this is the characteristic function of A001969. - R. J. Mathar, Sep 05 2008
From Michel Dekking, Jan 05 2021: (Start)
Proof of these three conjectures: the first two follow directly from the third, because the sequence A010059 is the binary complement of the Thue-Morse sequence A010060.
For the third conjecture: the odious and evil numbers occur as quadruples EOOE and OEEO, simply by their definition. To obtain the mod 2 version of the interleave of the odious and evil numbers we therefore have to apply a transformation
EOOE -> OEOE, OEEO -> OEOE to these quadruples.
But this changes the parities from the corresponding 4n, 4n+1, 4n+2, 4n+3 quadruples from 0101 to 1001 in the first case, and from 0101 to 0110 in the second case. Since the quadruples EOOE and OEEO occur in a Thue Morse pattern, then also the quadruples 1001 and 0110 occur in a Thue Morse pattern, finishing the proof.
(End)

Examples

			The evolution starting at 1 is:
.1
.1, 0
.1, 0, 0, 1,
.1, 0, 0, 1, 0, 1, 1, 0
.1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1
.1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0
...........
		

References

  • W. H. Gottschalk and G. A. Hedlund, Topological Dynamics. American Mathematical Society, Colloquium Publications, Vol. 36, Providence, RI, 1955, p. 105.
  • M. Lothaire, Combinatorics on Words. Addison-Wesley, Reading, MA, 1983, p. 23.
  • A. Salomaa, Jewels of Formal Language Theory. Computer Science Press, Rockville, MD, 1981, p. 6.

Crossrefs

Cf. A001285 (1, 2 version), A010060 (0, 1 version), A106400 (+1, -1 version), A059448 (with reversed subsections).
Cf. also A000069, A026147, A159481.

Programs

  • Haskell
    a010059 = (1 -) . a010060  -- Reinhard Zumkeller, Feb 04 2013
    
  • Maple
    A010059 := n->1-A010060(n);
    map(t->49-t,convert(StringTools[ThueMorse](1000),bytes)); # Robert Israel, Feb 02 2016
    # second Maple program:
    a := n -> ifelse(type(add(convert(n, base, 2)), even), 1, 0):
    seq(a(n), n = 0 .. 104); # Peter Luschny, Mar 11 2024
  • Mathematica
    Mod[ CoefficientList[Series[(1 + Sqrt[(1 - 3x)/(1 + x)])/(2(1 + x)), {x, 0, 111}], x], 2] (* Stephan Wolfram *)
    CoefficientList[ Series[1/(1 - x) + Product[1 - x^2^k, {k, 0, 10}], {x, 0, 111}]/2, x] (* Robert G. Wilson v, Jul 16 2004 *)
    Nest[ Flatten[ # /. {0 -> {0, 1}, 1 -> {1, 0}}] &, {1}, 7] (* Robert G. Wilson v Sep 26 2006 *)
    od = Select[ Range[0, 129], OddQ@ DigitCount[ #, 2, 1] &]; ev = Select[ Range[0, 129], EvenQ@ DigitCount[ #, 2, 1] &]; Mod[ Flatten@ Transpose[{od, ev}], 2] (* Robert G. Wilson v, Apr 14 2009 *)
    Nest[ Join[ #, Mod[2# + 1, 3]] &, {1}, 7] (* Robert G. Wilson v, Jul 27 2014 *)
    {{1}}~Join~SubstitutionSystem[{0 -> {0, 1}, 1 -> {1, 0}}, {0}, 6] // Flatten (* Michael De Vlieger, Aug 15 2016, Version 10.2 *)
    1 - ThueMorse[Range[0, 100]] (* Paolo Xausa, Oct 25 2024 *)
  • PARI
    a(n)=!(hammingweight(n)%2); \\ Charles R Greathouse IV, Mar 29 2013
    
  • Python
    def A010059(n): return n.bit_count()&1^1 # Chai Wah Wu, Mar 01 2023
  • R
    maxrow <- 8 # by choice
    b01 <- 0
    for(m in 0:maxrow) for(k in 0:(2^m-1)){
    b01[2^(m+1)+    k] <-   b01[2^m+k]
    b01[2^(m+1)+2^m+k] <- 1-b01[2^m+k]
    }
    (b01 <- c(1,b01))
    # Yosu Yurramendi, Apr 10 2017
    

Formula

G.f.: (1/2) * (1/(1-x) + Product_{k>=0} (1 - x^2^k)). - Ralf Stephan, Jun 20 2003
a(n) = A143579(n) mod 2. - Gary W. Adamson, Aug 24 2008
a(n) + A010060(n) = 1 for all n.
a(n) = A159481(n+1) - A159481(n). - Reinhard Zumkeller, Apr 16 2009
a(n) + A026147(n-1) = 2n for n >= 1. - Clark Kimberling, Oct 06 2014
a(n) = A000069(n+1) (mod 2). - John M. Campbell, Jun 30 2016
a(n) = A059448(A054429(n)) = (A106400(n)+1)/2 = (1+A008836(A005940(1+n)))/2. - Antti Karttunen, May 30 2017
If A(n)=(a(0),a(2),...,a(2^n-1)), then A(n+1)=(A(n),1-A(n)). - Arie Bos, Jul 27 2022
a(n) = (1 + (-1)^A000120(n))/2. - Lorenzo Sauras Altuzarra, Mar 10 2024

A299757 Weight of the strict integer partition with FDH number n.

Original entry on oeis.org

0, 1, 2, 3, 4, 3, 5, 4, 6, 5, 7, 5, 8, 6, 6, 9, 10, 7, 11, 7, 7, 8, 12, 6, 13, 9, 8, 8, 14, 7, 15, 10, 9, 11, 9, 9, 16, 12, 10, 8, 17, 8, 18, 10, 10, 13, 19, 11, 20, 14, 12, 11, 21, 9, 11, 9, 13, 15, 22, 9, 23, 16, 11, 12, 12, 10, 24, 13, 14, 10, 25, 10, 26, 17
Offset: 1

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Author

Gus Wiseman, Feb 18 2018

Keywords

Comments

Let f(n) = A050376(n) be the n-th Fermi-Dirac prime. Every positive integer n has a unique factorization of the form n = f(s_1)*...*f(s_k) where the s_i are strictly increasing positive integers. This determines a unique strict integer partition (s_k...s_1) whose FDH number is then defined to be n.
In analogy with the Heinz number correspondence between integer partitions and positive integers (see A056239), FDH numbers give a correspondence between strict integer partitions and positive integers.

Examples

			Sequence of strict integer partitions begins: () (1) (2) (3) (4) (2,1) (5) (3,1) (6) (4,1) (7) (3,2) (8) (5,1) (4,2) (9).
		

Crossrefs

Programs

  • Mathematica
    FDfactor[n_]:=If[n===1,{},Sort[Join@@Cases[FactorInteger[n],{p_,k_}:>Power[p,Cases[Position[IntegerDigits[k,2]//Reverse,1],{m_}->2^(m-1)]]]]];
    nn=200;FDprimeList=Array[FDfactor,nn,1,Union];
    FDrules=MapIndexed[(#1->#2[[1]])&,FDprimeList];
    Table[Total[FDfactor[n]/.FDrules],{n,nn}]

A059448 The parity of the number of zero digits when n is written in binary.

Original entry on oeis.org

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

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Author

Henry Bottomley, Feb 02 2001

Keywords

Comments

Old name was: "If A_k are the terms from 2^(k-1) through to 2^k-1, then A_(k+1) is B_k A_k where B_k is A_k with 0's and 1's swapped, starting from a(1)=0; also parity of number of zero digits when n is written in binary. a(0) not given as it could be 1 or 0 depending on the definition or formula used." - Michel Dekking, Sep 11 2020
The sequence (when prefixed by 0) is overlap-free [Allouche and Shallit].
From Vladimir Shevelev, May 23 2017: (Start)
Theorem: The sequence is cubefree.
Here we show only that the sequence contains no three consecutive equal terms. Indeed, using the recursions below, we have
a(4*n)=a(n), a(4*n+1)=1-a(n), a(4*n+2)=1-a(n), a(4*n+3)=a(n), n >= 1, and our statement easily follows. In general, the Theorem could be proved either directly (cf. A269027) or using the remark below from Jeffrey Shallit and the well-known fact [first proved not later than 1912 by Axel Thue (private communication from Jean-Paul Allouche)] that the Thue-Morse sequence is cubefree.
Note that, by the formulas modulo 4, the sequence is constructed over four terms {a(4*n),a(4*n+1),a(4*n+2),a(4*n+3)} which, starting with a(4), are either {0,1,1,0} or {1,0,0,1}, the first elements of which form {a(n)}. (End)

References

  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 26, Problem 23.

Crossrefs

Characteristic function of A059009.
Cf. A298952 (complement), A242179 (values +-1).

Programs

  • Haskell
    a059448 = (`mod` 2) . a023416  -- Reinhard Zumkeller, Mar 01 2012
    
  • Maple
    s1:=[];
    for n from 1 to 200 do
    t1:=convert(n,base,2); t2:=subs(1=NULL,t1); s1:=[op(s1),nops(t2) mod 2]; od:
    s1;
  • Mathematica
    Table[Boole[OddQ[Count[IntegerDigits[n, 2], 0]]], {n, 1, 105}] (* Jean-François Alcover, Apr 05 2013 *)
  • PARI
    a(n)=(#binary(n)-hammingweight(n))%2;
    vector(99,n,a(n)) /* Joerg Arndt, Sep 11 2020 */
    
  • Python
    def A059448(n): return (n.bit_length()^n.bit_count())&1 # Chai Wah Wu, Jul 26 2023

Formula

a(2n) = 1 - a(n); a(2n+1) = a(n) = 1 - a(2n). If 2^k <= n < 2^(k+1) then a(n) = 1 - a(n-2^(k-1)). a(n) = A023416(n) mod 2 = A059009(n) - 2n = 2n + 1 - A059010(n) = |A010060(n) - A030300(n-1)|.
Let b(1)=1 and b(n) = b(n-ceiling(n/2)) - b(n-floor(n/2)); then for n >= 1, a(n) = (1/2)*(1-b(2n+1)). - Benoit Cloitre, Apr 26 2005
Alternatively, if x is the sequence, then x = 010 mu^2(x), where mu is the Thue-Morse morphism sending 0 to 01 and 1 to 10. - Jeffrey Shallit, Jun 06 2016
a(n) = A010059(A054429(n)) = (1+A008836(A163511(n)))/2. - Antti Karttunen, May 30 2017
Alternatively, if x is the sequence, then x = 0 tau(x), where tau is the "twisted" Thue-Morse morphism sending 0 to 10 and 1 to 01. Note that tau^2 = mu^2, giving x = 010 mu^2(x). - Michel Dekking, Sep 30 2020

Extensions

Name changed by Michel Dekking, Sep 11 2020

A019568 a(n) = smallest k >= 1 such that {1^n, 2^n, 3^n, ..., k^n} can be partitioned into two sets with equal sum.

Original entry on oeis.org

2, 3, 7, 12, 16, 24, 31, 39, 47, 44, 60, 71, 79, 79, 87
Offset: 0

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Author

Keywords

Comments

a(n) is least integer k such that at least one signed sum of the first k n-th powers equals zero.
a(n) < 2^(n+1). The partition of the set {k: 0 <= k < 2^(n+1)} into two sets A,B according to the parity of the number of 1s in the binary expansion of k, has the property that Sum_{k in A} p(k) = Sum_{k in B} p(k) for any polynomial p of degree <= n. Equivalently, if e(k) is the Thue-Morse sequence A106400, then Sum_{0 <= k < 2^m} e(k)p(k) = 0 for any polynomial p with deg(p) < m. - Pietro Majer, Mar 14 2009

Examples

			For n=1 and 2 we have: 1+2-3 = 0 (so a(1)=3), 1+4-9+16-25-36+49 = 0 (so a(2)=7).
The sum of the ninth powers of 3 5 9 10 14 19 20 21 25 26 28 31 35 36 37 38 40 41 42 is half the sum of the ninth powers of 1..44, so a(9)=44. - _Don Reble_, Oct 21 2005
Example: the signs (+--+-++--++-+--+) in (+0)-1-8+27-64+125+216-...+3375=0 are those of the expansion of Q(x):=(1-x)(1-x^2)(1-x^4)(1-x^8) = +1-x-x^2+x^3-..+x^15. Since (1-x)^4 divides Q(x), if S is the shift operator on sequences, the operator Q(S) has the fourth discrete difference (I-S)^4 as factor, hence annihilates the sequence of cubes. - _Pietro Majer_, Mar 14 2009
		

References

  • Posting to sci.math Nov 11 1996 by fredh(AT)ix.netcom.com (Fred W. Helenius).

Crossrefs

Cf. A240070 (partitioned into 3 sets).

Programs

  • Mathematica
    Table[k = 1; found = False; While[s = Range[k]^n; sm = Total[s]; If[EvenQ[sm], sm = sm/2; found = MemberQ[Total /@ Subsets[s], sm]]; ! found, k++]; k, {n, 0, 4}] (* T. D. Noe, Apr 01 2014 *)

Formula

a(n) == 0 or 3 (mod 4) for n >= 1 - David W. Wilson, Oct 20 2005

Extensions

More terms from Don Reble, Oct 21 2005
Definition simplified by Pietro Majer, Mar 15 2009
a(13)-a(14) from Sean A. Irvine, Mar 27 2019

A215016 Decimal expansion of the product of 1 - 1/2^2^n over all n >= 0.

Original entry on oeis.org

3, 5, 0, 1, 8, 3, 8, 6, 5, 4, 3, 9, 5, 6, 9, 6, 0, 8, 8, 6, 6, 5, 5, 4, 5, 2, 6, 9, 6, 6, 1, 7, 8, 8, 6, 7, 6, 4, 2, 0, 8, 6, 5, 0, 2, 1, 7, 6, 9, 2, 1, 7, 6, 9, 7, 0, 6, 4, 8, 2, 3, 3, 8, 6, 0, 4, 8, 2, 5, 6, 3, 0, 5, 3, 6, 8, 6, 9, 6, 4, 4, 1
Offset: 0

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Author

Keywords

Comments

Can be used to efficiently compute A014571: A014571 = 1/2 - (1/4) * A215016.

Examples

			0.35018386543956960886655452696617886764208650217692176970648233860482563...
		

Crossrefs

Programs

  • Mathematica
    RealDigits[NProduct[1 - 1/2^2^n, {n, 0, Infinity}, WorkingPrecision -> 120]][[1]] (* Alonso del Arte, Jul 31 2012 *)
  • PARI
    prodinf(n=0,1-1.>>2^n)

Formula

Equals Sum_{n>=0} A106400(n)/2^n. - Robert FERREOL, Jan 10 2022
From Amiram Eldar, Feb 19 2024: (Start)
Equals Product_{n>=0} (1 - 1/A001146(n)).
Equals 2/A258716.
Equals 1/(3/2 + A258714). (End)
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