cp's OEIS Frontend

Partitions into distinct parts are sometimes called "strict partitions".

Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).

The result that number of partitions of n into distinct parts = number of partitions of n into odd parts is due to Euler.

Bijection: given n = L1* 1 + L2*3 + L3*5 + L7*7 + ..., a partition into odd parts, write each Li in binary, Li = 2^a1 + 2^a2 + 2^a3 + ... where the aj's are all different, then expand n = (2^a1 * 1 + ...)*1 + ... by removing the brackets and we get a partition into distinct parts. For the reverse operation, just keep splitting any even number into halves until no evens remain.

Euler transform of period 2 sequence [1,0,1,0,...]. - Michael Somos, Dec 16 2002

Number of different partial sums 1+[1,2]+[1,3]+[1,4]+..., where [1,x] indicates a choice. E.g., a(6)=4, as we can write 1+1+1+1+1+1, 1+2+3, 1+2+1+1+1, 1+1+3+1. - Jon Perry, Dec 31 2003

a(n) is the sum of the number of partitions of x_j into at most j parts, where j is the index for the j-th triangular number and n-T(j)=x_j. For example; a(12)=partitions into <= 4 parts of 12-T(4)=2 + partitions into <= 3 parts of 12-T(3)=6 + partitions into <= 2 parts of 12-T(2)=9 + partitions into 1 part of 12-T(1)=11 = (2)(11) + (6)(51)(42)(411)(33)(321)(222) + (9)(81)(72)(63)(54)+(11) = 2+7+5+1 = 15. - Jon Perry, Jan 13 2004

Number of partitions of n where if k is the largest part, all parts 1..k are present. - Jon Perry, Sep 21 2005

Jack Grahl and Franklin T. Adams-Watters prove this claim of Jon Perry's by observing that the Ferrers dual of a "gapless" partition is guaranteed to have distinct parts; since the Ferrers dual is an involution, this establishes a bijection between the two sets of partitions. - Allan C. Wechsler, Sep 28 2021

The number of connected threshold graphs having n edges. - Michael D. Barrus (mbarrus2(AT)uiuc.edu), Jul 12 2007

Starting with offset 1 = row sums of triangle A146061 and the INVERT transform of A000700 starting: (1, 0, 1, -1, 1, -1, 1, -2, 2, -2, 2, -3, 3, -3, 4, -5, ...). - Gary W. Adamson, Oct 26 2008

Number of partitions of n in which the largest part occurs an odd number of times and all other parts occur an even number of times. (Such partitions are the duals of the partitions with odd parts.) - David Wasserman, Mar 04 2009

Equals A035363 convolved with A010054. The convolution square of A000009 = A022567 = A000041 convolved with A010054. A000041 = A000009 convolved with A035363. - Gary W. Adamson, Jun 11 2009

Considering all partitions of n into distinct parts: there are A140207(n) partitions of maximal size which is A003056(n), and A051162(n) is the greatest number occurring in these partitions. - Reinhard Zumkeller, Jun 13 2009

Equals left border of triangle A091602 starting with offset 1. - Gary W. Adamson, Mar 13 2010

Number of symmetric unimodal compositions of n+1 where the maximal part appears once. Also number of symmetric unimodal compositions of n where the maximal part appears an odd number of times. - Joerg Arndt, Jun 11 2013

Because for these partitions the exponents of the parts 1, 2, ... are either 0 or 1 (j^0 meaning that part j is absent) one could call these partitions also 'fermionic partitions'. The parts are the levels, that is the positive integers, and the occupation number is either 0 or 1 (like Pauli's exclusion principle). The 'fermionic states' are denoted by these partitions of n. - Wolfdieter Lang, May 14 2014

The set of partitions containing only odd parts forms a monoid under the product described in comments to A047993. - Richard Locke Peterson, Aug 16 2018

Ewell (1973) gives a number of recurrences. - N. J. A. Sloane, Jan 14 2020

a(n) equals the number of permutations p of the set {1,2,...,n+1}, written in one line notation as p = p_1p_2...p_(n+1), satisfying p_(i+1) - p_i <= 1 for 1 <= i <= n, (i.e., those permutations that, when read from left to right, never increase by more than 1) whose major index maj(p) := Sum_{p_i > p_(i+1)} i equals n. For example, of the 16 permutations on 5 letters satisfying p_(i+1) - p_i <= 1, 1 <= i <= 4, there are exactly two permutations whose major index is 4, namely, 5 3 4 1 2 and 2 3 4 5 1. Hence a(4) = 2. See the Bala link in A007318 for a proof. - Peter Bala, Mar 30 2022

Conjecture: Each positive integer n can be written as a_1 + ... + a_k, where a_1,...,a_k are strict partition numbers (i.e., terms of the current sequence) with no one dividing another. This has been verified for n = 1..1350. - Zhi-Wei Sun, Apr 14 2023

Conjecture: For each integer n > 7, a(n) divides none of p(n), p(n) - 1 and p(n) + 1, where p(n) is the number of partitions of n given by A000041. This has been verified for n up to 10^5. - Zhi-Wei Sun, May 20 2023 [Verified for n <= 2*10^6. - Vaclav Kotesovec, May 23 2023]

The g.f. Product_{k >= 0} 1 + x^k = Product_{k >= 0} 1 - x^k + 2*x^k == Product_{k >= 0} 1 - x^k == Sum_{k in Z} (-1)^k*x^(k*(3*k-1)/2) (mod 2) by Euler's pentagonal number theorem. It follows that a(n) is odd iff n = k*(3*k - 1)/2 for some integer k, i.e., iff n is a generalized pentagonal number A001318. - Peter Bala, Jan 07 2025

When convolved with the partition numbers A000041 gives 1, 0, 0, 0, 0, ...

Also, number of different partitions of n into parts of -1 different kinds (based upon formal analogy). - Michele Dondi (blazar(AT)lcm.mi.infn.it), Jun 29 2004

The comment that "when convolved with the partition numbers gives [1, 0, 0, 0, ...]" is equivalent to row sums of triangle A145975 = [1, 0, 0, 0, ...]; where A145975 is a partition number convolution triangle. - Gary W. Adamson, Oct 25 2008

When convolved with n-th partial sums of A000041 = the binomial sequence starting (1, n, ...). Example: A010815 convolved with A014160 (partial sum operation applied thrice to the partition numbers) = (1, 3, 6, 10, ...). - Gary W. Adamson, Nov 11 2008

(A000012^(-n) * A000041) convolved with A010815 = n-th row of the inverse of Pascal's triangle, (as a vector, followed by zeros); where A000012^(-1) = the pairwise difference operator. Example: (A000012^(-4) * A000041) convolved with A010815 = (1, -4, 6, -4, 1, 0, 0, 0, ...). - Gary W. Adamson, Nov 11 2008

Also sum of [product of (1-2/(hook lengths)^2)] over all partitions of n. - Wouter Meeussen, Sep 16 2010

Cayley (1895) begins article 387 with "Write for shortness sqrt(2k'K / pi) / [1-q^{2m-1}]^2 = G, ..." which is a convoluted way of writing G = [1-q^{2m}] = (1-q^2)(1-q^4)... - Michael Somos, Aug 01 2011

This is an example of the quintuple product identity in the form f(a*b^4, a^2/b) - (a/b) * f(a^4*b, b^2/a) = f(-a*b, -a^2*b^2) * f(-a/b, -b^2) / f(a, b) where a = x^3, b = x. - Michael Somos, Jan 21 2012

Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).

Number 1 of the 14 primitive eta-products which are holomorphic modular forms of weight 1/2 listed by D. Zagier on page 30 of "The 1-2-3 of Modular Forms". - Michael Somos, May 04 2016

Might be called the "Lichtenberg sequence" after Georg Christoph Lichtenberg, who discussed it in 1769 in connection with the Chinese Rings puzzle (baguenaudier). - Andreas M. Hinz, Feb 15 2017

Number of steps to change from a binary string of n 0's to n 1's using a Gray code. - Jon Stadler (jstadler(AT)coastal.edu)

Popular puzzles such as Spin-Out and The Brain Puzzler are based on the Gray binary system and require a(n) steps to complete for some number n.

Conjecture: {a(n)} also gives all j for which A048702(j) = A000217(j); i.e., if we take the a(n)-th triangular number (a(n)^2 + a(n))/2 and multiply it by 3, we get a(n)-th even-length binary palindrome A048701(a(n)) concatenated from a(n) and its reverse. E.g., a(4) = 10, which is 1010 in binary; the tenth triangular number is 55, and 55*3 = 165 = 10100101 in binary. - Antti Karttunen, circa 1999. (This has been now proved by Paul K. Stockmeyer in his arXiv:1608.08245 paper.) - Antti Karttunen, Aug 31 2016

Number of ways to tie a tie of n or fewer half turns, excluding mirror images. Also number of walks of length n or less on a triangular lattice with the following restrictions; given l, r and c as the lattice axes. 1. All steps are taken in the positive axis direction. 2. No two consecutive steps are taken on the same axis. 3. All walks begin with l. 4. All walks end with rlc or lrc. - Bill Blewett, Dec 21 2000

a(n) is the minimal number of vertices to be selected in a vertex-cover of the balanced tree B_n. - Sen-peng Eu, Jun 15 2002

A087117(a(n)) = A038374(a(n)) = 1 for n > 1; see also A090050. - Reinhard Zumkeller, Nov 20 2003

Intersection of A003754 and A003714; complement of A107907. - Reinhard Zumkeller, May 28 2005

Equivalently, numbers m whose binary representation is effectively, for some number k, both the lazy Fibonacci and the Zeckendorf representation of k (in which case k = A022290(m)). - Peter Munn, Oct 06 2022

a(n+1) gives row sums of Riordan array (1/(1-x), x(1+2x)). - Paul Barry, Jul 18 2005

Total number of initial 01's in all binary words of length n+1. Example: a(3) = 5 because the binary words of length 4 that start with 01 are (01)00, (01)(01), (01)10 and (01)11 and the total number of initial 01's is 5 (shown between parentheses). a(n) = Sum_{k >= 0} k*A119440(n+1, k). - Emeric Deutsch, May 19 2006

In Norway we call the 10-ring puzzle "strikketoy" or "knitwear" (see link). It takes 682 moves to free the two parts. - Hans Isdahl, Jan 07 2008

Equals A002450 and A020988 interlaced. - Zak Seidov, Feb 10 2008

For n > 1, let B_n = the complete binary tree with vertex set V where |V| = 2^n - 1. If VC is a minimum-size vertex cover of B_n, Sen-Peng Eu points out that a(n) = |VC|. It also follows that if IS = V\VC, a(n+1) = |IS|. - K.V.Iyer, Apr 13 2009

Starting with 1 and convolved with [1, 2, 2, 2, ...] = A000295. - Gary W. Adamson, Jun 02 2009

a(n) written in base 2 is sequence A056830(n). - Jaroslav Krizek, Aug 05 2009

This is the sequence A(0, 1; 1, 2; 1) of the family of sequences [a,b:c,d:k] considered by G. Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below. - Wolfdieter Lang, Oct 18 2010

From Vladimir Shevelev, Jan 30 2012, Feb 13 2012: (Start)

1) Denote by {n, k} the number of permutations of 1, ..., n with the up-down index k (for definition, see comment in A203827). Then max_k{n, k} = {n, a(n)} = A000111(n).

2) a(n) is the minimal number > a(n-1) with the Hamming distance d_H(a(n-1), a(n)) = n. Thus this sequence is the Hamming analog of triangular numbers 0, 1, 3, 6, 10, ... (End)

From Hieronymus Fischer, Nov 22 2012: (Start)

Represented in binary form each term after the second one contains every previous term as a substring.

The terms a(2) = 2 and a(3) = 5 are the only primes. Proof: For even n we get a(n) = 2*(2^(2*n) - 1)/3, which shows that a(n) is even, too, and obviously a(n) > 2 for all even n > 2. For odd n we have a(n) = (2^(n+1) - 1)/3 = (2^((n+1)/2) - 1) * (2^((n+1)/2) + 1)/3. Evidently, at least one of these factors is divisible by 3, both are greater than 6, provided n > 3. Hence it follows that a(n) is composite for all odd n > 3.

Represented as a binary number, a(n+1) has exactly n prime substrings. Proof: Evidently, a(1) = 1_2 has zero and a(2) = 10_2 has 1 prime substring. Let n > 1. Written in binary, a(n+1) is 101010101...01 (if n + 1 is odd) and is 101010101...10 (if n + 1 is even) with n + 1 digits. Only the 2- and 3-digits substrings 10_2 (=2) and 101_2 (=5) are prime substrings. All the other substrings are nonprime since every substring is a previous term and all terms unequal to 2 and 5 are nonprime. For even n + 1, the number of prime substrings equal to 2 = 10_2 is (n+1)/2, and the number of prime substrings equal to 5 = 101_2 is (n-1)/2, makes a sum of n. For odd n + 1 we get n/2 for both, the number of 2's and 5's prime substrings, in any case, the sum is n. (End)

Number of different 3-colorings for the vertices of all triangulated planar polygons on a base with n+2 vertices if the colors of the two base vertices are fixed. - Patrick Labarque, Feb 09 2013

A090079(a(n)) = a(n) and A090079(m) <> a(n) for m < a(n). - Reinhard Zumkeller, Feb 16 2013

a(n) is the number of length n binary words containing at least one 1 and ending in an even number (possibly zero) of 0's. a(3) = 5 because we have: 001, 011, 100, 101, 111. - Geoffrey Critzer, Dec 15 2013

a(n) is the number of permutations of length n+1 having exactly one descent such that the first element of the permutation is an even number. - Ran Pan, Apr 18 2015

a(n) is the sequence of the last row of the Hadamard matrix H(2^n) obtained via Sylvester's construction: H(2) = [1,1;1,-1], H(2^n) = H(2^(n-1))*H(2), where * is the Kronecker product. - William P. Orrick, Jun 28 2015

Conjectured record values of A264784: a(n) = A264784(A155051(n-1)). - Reinhard Zumkeller, Dec 04 2015. (This is proved by Paul K. Stockmeyer in his arXiv:1608.08245 paper.) - Antti Karttunen, Aug 31 2016

Decimal representation of the x-axis, from the origin to the right edge, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 131", based on the 5-celled von Neumann neighborhood. See A279053 for references and links. - Robert Price, Dec 05 2016

For n > 4, a(n-2) is the second-largest number in row n of A127824. - Dmitry Kamenetsky, Feb 11 2017

Conjecture: a(n+1) is the number of compositions of n with two kinds of parts, n and n', where the order of the 1 and 1' does not matter. For n=2, a(3) = 5 compositions, enumerated as follows: 2; 2'; 1,1; 1',1 = 1',1; 1',1'. - Gregory L. Simay, Sep 02 2017

Conjecture proved by recognizing the appropriate g.f. is x/(1 - x)(1 - x)(1 - 2*x^2 - 2x^3 - ...) = x/(1 - 2*x - x^2 + 2x^3). - Gregory L. Simay, Sep 10 2017

a(n) = 2^(n-1) + 2^(n-3) + 2^(n-5) + ... a(2*k -1) = A002450(k) is the sum of the powers of 4. a(2*k) = 2*A002450(k). - Gregory L. Simay, Sep 27 2017

a(2*n) = n times the string [10] in binary representation, a(2*n+1) = n times the string [10] followed with [1] in binary representation. Example: a(7) = 85 = (1010101) in binary, a(8) = 170 = (10101010) in binary. - Ctibor O. Zizka, Nov 06 2018

Except for 0, these are the positive integers whose binary expansion has cuts-resistance 1. For the operation of shortening all runs by 1, cuts-resistance is the number of applications required to reach an empty word. Cuts-resistance 2 is A329862. - Gus Wiseman, Nov 27 2019

From Markus Sigg, Sep 14 2020: (Start)

Let s(k) be the length of the Collatz orbit of k, e.g. s(1) = 1, s(2) = 2, s(3) = 5. Then s(a(n)) = n+3 for n >= 3. Proof by induction: s(a(3)) = s(5) = 6 = 3+3. For odd n >= 5 we have s(a(n)) = s(4*a(n-2)+1) = s(12*a(n-2)+4)+1 = s(3*a(n-2)+1)+3 = s(a(n-2))+2 = (n-2)+3+2 = n+3, and for even n >= 4 this gives s(a(n)) = s(2*a(n-1)) = s(a(n-1))+1 = (n-1)+3+1 = n+3.

Conjecture: For n >= 3, a(n) is the second largest natural number whose Collatz orbit has length n+3. (End)

From Gary W. Adamson, May 14 2021: (Start)

With offset 1 the sequence equals the numbers of 1's from n = 1 to 3, 3 to 7, 7 to 15, ...; of A035263; as shown below:

..1 3 7 15...

..1 0 1 1 1 0 1 0 1 0 1 1 1 0 1...

..1.....2...........5......................10...; a(n) = Sum_(k=1..2n-1)A035263(k)

.....1...........2.......................5...; as to zeros.

..1's in the Tower of Hanoi game represent CW moves On disks in the pattern:

..0, 1, 2, 0, 1, 2, ... whereas even numbered disks move in the pattern:

..0, 2, 1, 0, 2, 1, ... (End)

Except for 0, numbers that are repunits in Gray-code representation (A014550). - Amiram Eldar, May 21 2021

From Gus Wiseman, Apr 20 2023: (Start)

Also the number of nonempty subsets of {1..n} with integer median, where the median of a multiset is the middle part in the odd-length case, and the average of the two middle parts in the even-length case. For example, the a(1) = 1 through a(4) = 10 subsets are:

{1} {1} {1} {1}

{2} {2} {2}

{3} {3}

{1,3} {4}

{1,2,3} {1,3}

{2,4}

{1,2,3}

{1,2,4}

{1,3,4}

{2,3,4}

The complement is counted by A005578.

For mean instead of median we have A051293, counting empty sets A327475.

For normal multisets we have A056450, strongly normal A361202.

For partitions we have A325347, strict A359907, complement A307683.

(End)

In binary form, because the sequence 10101_2... keeps repeating, one can represent a(n) = floor(0.(10)2 * 2^n). Because 0.(10)_2 = 0.(10)_2 * 4 - 2, then 0.(10)_2 = 2/3. Substituting this value in the first expression, we obtain a(n) = floor(2^(n+1)/3). This also agrees with the formula of Paul Barry, Oct 08 2005. - _Giovanni Ciriani, Mar 31 2025

Number of partitions of n in which greatest part is odd.

Number of partitions of n+1 into an even number of parts, the least being 1. Example: a(5)=4 because we have [5,1], [3,1,1,1], [2,1,1] and [1,1,1,1,1,1].

Also number of partitions of n+1 such that the largest part is even and occurs only once. Example: a(5)=4 because we have [6], [4,2], [4,1,1] and [2,1,1,1,1]. - Emeric Deutsch, Apr 05 2006

Also the number of partitions of n such that the number of odd parts and the number of even parts have opposite parities. Example: a(8)=10 is a count of these partitions: 8, 611, 521, 431, 422, 41111, 332, 32111, 22211, 2111111. - Clark Kimberling, Feb 01 2014, corrected Jan 06 2021

In Chaves 2011 see page 38 equation (3.20). - Michael Somos, Dec 28 2014

Suppose that c(0) = 1, that c(1), c(2), ... are indeterminates, that d(0) = 1, and that d(n) = -c(n) - c(n-1)*d(1) - ... - c(0)*d(n-1). When d(n) is expanded as a polynomial in c(1), c(2),..,c(n), the terms are of the form H*c(i_1)*c(i_2)*...*c(i_k). Let P(n) = [c(i_1), c(i_2), ..., c(i_k)], a partition of n. Then H is negative if P has an odd number of parts, and H is positive if P has an even number of parts. That is, d(n) has A027193(n) negative coefficients, A027187(n) positive coefficients, and A000041 terms. The maximal coefficient in d(n), in absolute value, is A102462(n). - Clark Kimberling, Dec 15 2016

The median of a multiset is either the middle part (for odd length), or the average of the two middle parts (for even length).

The median of a multiset is either the middle part (for odd length), or the average of the two middle parts (for even length).

The reverse-alternating sum of a partition (y_1,...,y_k) is Sum_i (-1)^(k-i) y_i.

Also the number of reversed integer partitions of n with alternating sum >= 0.

A formula for the reverse-alternating sum of a partition is: (-1)^(k-1) times the number of odd parts in the conjugate partition, where k is the number of parts. So a(n) is the number of integer partitions of n whose conjugate parts are all even or whose length is odd. By conjugation, this is also the number of integer partitions of n whose parts are all even or whose greatest part is odd.

All integer partitions have alternating sum >= 0, so the non-reversed version is A000041.

Is this sequence weakly increasing? In particular, is A344611(n) <= A160786(n)?

The median of a multiset is either the middle part (for odd length), or the average of the two middle parts (for even length).

Ramanujan theta functions: phi(q) (A000122), chi(q) (A000700).

Also number of partitions of n into exactly 4 parts.

Also the number of weighted cubic graphs on 4 nodes (=the tetrahedron) with weight n. - R. J. Mathar, Nov 03 2018

From Gus Wiseman, Jun 27 2021: (Start)

Also the number of strict integer partitions of 2n with alternating sum 4, or (by conjugation) partitions of 2n covering an initial interval of positive integers with exactly 4 odd parts. The strict partitions with alternating sum 4 are:

(4) (5,1) (6,2) (7,3) (8,4) (9,5) (10,6)

(5,2,1) (5,3,2) (5,4,3) (6,5,3) (7,6,3)

(6,3,1) (6,4,2) (7,5,2) (8,6,2)

(7,4,1) (8,5,1) (9,6,1)

(6,3,2,1) (6,4,3,1) (6,5,4,1)

(7,4,2,1) (7,4,3,2)

(7,5,3,1)

(8,5,2,1)

(6,4,3,2,1)

(End)