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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A001405 a(n) = binomial(n, floor(n/2)).

Original entry on oeis.org

1, 1, 2, 3, 6, 10, 20, 35, 70, 126, 252, 462, 924, 1716, 3432, 6435, 12870, 24310, 48620, 92378, 184756, 352716, 705432, 1352078, 2704156, 5200300, 10400600, 20058300, 40116600, 77558760, 155117520, 300540195, 601080390, 1166803110
Offset: 0

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Author

Keywords

Comments

Sperner's theorem says that this is the maximal number of subsets of an n-set such that no one contains another.
When computed from index -1, [seq(binomial(n,floor(n/2)), n = -1..30)]; -> [1,1,1,2,3,6,10,20,35,70,126,...] and convolved with aerated Catalan numbers [seq(((n+1) mod 2)*binomial(n,n/2)/((n/2)+1), n = 0..30)]; -> [1,0,1,0,2,0,5,0,14,0,42,0,132,0,...] shifts left by one: [1,1,2,3,6,10,20,35,70,126,252,...] and if again convolved with aerated Catalan numbers, gives A037952 apart from the initial term. - Antti Karttunen, Jun 05 2001 [This is correct because the g.f.'s satisfy (1+x*g001405(x))*g126120(x) = g001405(x) and g001405(x)*g126120(x) = g037952(x)/x. - R. J. Mathar, Sep 23 2021]
Number of ordered trees with n+1 edges, having nonroot nodes of outdegree 0 or 2. - Emeric Deutsch, Aug 02 2002
Gives for n >= 1 the maximum absolute column sum norm of the inverse of the Vandermonde matrix (a_ij) i=0..n-1, j=0..n-1 with a_00=1 and a_ij=i^j for (i,j) != (0,0). - Torsten Muetze, Feb 06 2004
Image of Catalan numbers A000108 under the Riordan array (1/(1-2x),-x/(1-2x)) or A065109. - Paul Barry, Jan 27 2005
Number of left factors of Dyck paths, consisting of n steps. Example: a(4)=6 because we have UDUD, UDUU, UUDD, UUDU, UUUD and UUUU, where U=(1,1) and D=(1,-1). - Emeric Deutsch, Apr 23 2005
Number of dispersed Dyck paths of length n; they are defined as concatenations of Dyck paths and (1,0)-steps on the x-axis; equivalently, Motzkin paths with no (1,0)-steps at positive height. Example: a(4)=6 because we have HHHH, HHUD, HUDH, UDHH, UDUD, and UUDD, where U=(1,1), H=(1,0), and D=(1,-1). - Emeric Deutsch, Jun 04 2011
a(n) is odd iff n=2^k-1. - Jon Perry, May 05 2005
An inverse Chebyshev transform of binomial(1,n)=(1,1,0,0,0,...) where g(x)->(1/sqrt(1-4*x^2))*g(x*c(x^2)), with c(x) the g.f. of A000108. - Paul Barry, May 13 2005
In a random walk on the number line, starting at 0 and with 0 absorbing after the first step, number of ways of ending up at a positive integer after n steps. - Joshua Zucker, Jul 31 2005
Maximum number of sums of the form Sum_{i=1..n} e(i)*a(i) that are congruent to 0 mod q, where e_i=0 or 1 and gcd(a_i,q)=1, provided that q > ceiling(n/2). - Ralf Stephan, Apr 27 2003
Also the number of standard tableaux of height <= 2. - Mike Zabrocki, Mar 24 2007
Hankel transform of this sequence forms A000012 = [1,1,1,1,1,1,1,...]. - Philippe Deléham, Oct 24 2007
A001263 * [1, -2, 3, -4, 5, ...] = [1, -1, -2, 3, 6, -10, -20, 35, 70, -126, ...]. - Gary W. Adamson, Jan 02 2008
Equals right border of triangle A153585. - Gary W. Adamson, Dec 28 2008
Second binomial transform of A168491. - Philippe Deléham, Nov 27 2009
a(n) is also the number of distinct strings of length n, each of which is a prefix of a string of balanced parentheses; see example. - Lee A. Newberg, Apr 26 2010
Number of symmetric balanced strings of n pairs of parentheses; see example. - Joerg Arndt, Jul 25 2011
a(n) is the number of permutation patterns modulo 2. - Olivier Gérard, Feb 25 2011
For n >= 2, a(n-1) is the number of incongruent two-color bracelets of 2*n-1 beads, n of which are black (A007123), having a diameter of symmetry. - Vladimir Shevelev, May 03 2011
The number of permutations of n elements where p(k-2) < p(k) for all k. - Joerg Arndt, Jul 23 2011
Also size of the equivalence class of S_{n+1} containing the identity permutation under transformations of positionally adjacent elements of the form abc <--> cba where a < b < c, cf. A210668. - Tom Roby, May 15 2012
a(n) is the number of symmetric Dyck paths of length 2n. - Matt Watson, Sep 26 2012
a(n) is divisible by A000108(floor(n/2)) = abs(A129996(n-2)). - Paul Curtz, Oct 23 2012
a(n) is the number of permutations of length n avoiding both 213 and 231 in the classical sense which are breadth-first search reading words of increasing unary-binary trees. For more details, see the entry for permutations avoiding 231 at A245898. - Manda Riehl, Aug 05 2014
Number of symmetric standard Young tableaux of shape (n,n). - Ran Pan, Apr 10 2015
From Luciano Ancora, May 09 2015: (Start)
Also "stepped path" in the array formed by partial sums of the all 1's sequence (or a Pascal's triangle displayed as a square). Example:
[1], [1], 1, 1, 1, 1, 1, ... A000012
1, [2], [3], 4, 5, 6, 7, ...
1, 3, [6], [10], 15, 21, 28, ...
1, 4, 10, [20], [35], 56, 84, ...
1, 5, 15, 35, [70], [126], 210, ...
Sequences in second formula are the mixed diagonals shown in this array. (End)
a(n) = A265848(n,n). - Reinhard Zumkeller, Dec 24 2015
The constant Sum_{n >= 0} a(n)/n! is 1 + A130820. - Peter Bala, Jul 02 2016
Number of meanders (walks starting at the origin and ending at any altitude >= 0 that may touch but never go below the x-axis) with n steps from {-1,1}. - David Nguyen, Dec 20 2016
a(n) is also the number of paths of n steps (either up or down by 1) that end at the maximal value achieved along the path. - Winston Luo, Jun 01 2017
Number of binary n-tuples such that the number of 1's in the even positions is the same as the number of 1's in the odd positions. - Juan A. Olmos, Dec 21 2017
Equivalently, a(n) is the number of subsets of {1,...,n} containing as many even numbers as odd numbers. - Gus Wiseman, Mar 17 2018
a(n) is the number of Dyck paths with semilength = n+1, returns to the x-axis = floor((n+3)/2) and up movements in odd positions = floor((n+3)/2). Example: a(4)=6, U=up movement in odd position, u=up movement in even position, d=down movement, -=return to x-axis: Uududd-Ud-Ud-, Ud-Uudd-Uudd-, Uudd-Uudd-Ud-, Ud-Ud-Uududd-, Uudd-Ud-Uudd-, Ud-Uududd-Ud-. - Roger Ford, Dec 29 2017
Let C_n(R, H) denote the transition matrix from the ribbon basis to the homogeneous basis of the graded component of the algebra of noncommutative symmetric functions of order n. Letting I(2^(n-1)) denote the identity matrix of order 2^(n-1), it has been conjectured that the dimension of the kernel of C_n(R, H) - I(2^(n-1)) is always equal to a(n-1). - John M. Campbell, Mar 30 2018
The number of U-equivalence classes of Łukasiewicz paths. Łukasiewicz paths are U-equivalent iff the positions of pattern U are identical in these paths. - Sergey Kirgizov, Apr 08 2018
All binary self-dual codes of length 2n, for n > 0, must contain at least a(n) codewords of weight n. More to the point, there will always be at least one, perhaps unique, binary self-dual code of length 2n that will contain exactly a(n) codewords that have a hamming weight equal to half the length of the code (n). This code can be constructed by direct summing the unique binary self-dual code of length 2 (up to permutation equivalence) to itself n times. A permutation equivalent code can be constructed by augmenting two identity matrices of length n together. - Nathan J. Russell, Nov 25 2018
Closed under addition. - Torlach Rush, Apr 18 2019
The sequence starting (1, 2, 3, 6, ...) is the invert transform of A097331: (1, 1, 0, 1, 0, 2, 0, 5, 0, 14, 0, 42, ...). - Gary W. Adamson, Feb 22 2020
From Gary W. Adamson, Feb 24 2020: (Start)
The sequence is the culminating limit of an infinite set of sequences with convergents of 2*cos(Pi/N), N = (3, 5, 7, 9, ...).
The first few such sequences are:
N = 3: (1, 1, 1, 1, 1, 1, 1, 1, ...)
N = 5: (1, 1, 2, 3, 5, 8, 13, 21, ...) = A000045
N = 7: (1, 1, 2, 3, 6, 10, 19, 33, ...) = A028495, a(n)/a(n-1) tends to 1.801937...
N = 9 (1, 1, 2, 3, 6, 10, 20, 35, ...) = A061551, a(n)/a(n_1) tends to 1.879385...
...
In the limit one gets the current sequence with ratio 2. (End)
a(n) is also the number of monotone lattice paths from (0,0) to (floor(n/2),ceiling(n/2)). These are the number of Grand Dyck paths when n is even. - Nachum Dershowitz, Aug 12 2020
The maximum number of preimages that a permutation of length n+1 can have under the consecutive-132-avoiding stack-sorting map. - Colin Defant, Aug 28 2020
Counts faro permutations of length n. Faro permutations are permutations avoiding the three consecutive patterns 231, 321 and 312. They are obtained by a perfect faro shuffle of two nondecreasing words of lengths differing by at most one. - Sergey Kirgizov, Jan 12 2021
Per "Sperner's Theorem", the largest possible familes of finite sets none of which contain any other sets in the family. - Renzo Benedetti, May 26 2021
a(n-1) are the incomplete, primitive Dyck paths of n steps without a first return: paths of U and D steps starting at the origin, never touching the horizontal axis later on, and ending above the horizontal axis. n=1: {U}, n=2: {UU}, n=3: {UUU, UUD}, n=4: {UUUU, UUUD, UUDU}, n=5: {UUUUU, UUUUD, UUUDD, UUDUU, UUUDU, UUDUD}. For comparison: A037952 counts incomplete Dyck paths with n steps with any number of intermediate returns to the horizontal axis, ending above the horizontal axis. - R. J. Mathar, Sep 24 2021
a(n) is the number of noncrossing partitions of [n] whose nontrivial blocks are of type {a,b}, with a <= n/2, b > n/2. - Francesca Aicardi, May 29 2022
Maximal coefficient of (1+x)^n. - Vaclav Kotesovec, Dec 30 2022
Sums of lower-left-to-upper-right diagonals of the Catalan Triangle A001263. - Howard A. Landman, Sep 16 2024

Examples

			For n = 4, the a(4) = 6 distinct strings of length 4, each of which is a prefix of a string of balanced parentheses, are ((((, (((), (()(, ()((, ()(), and (()). - _Lee A. Newberg_, Apr 26 2010
There are a(5)=10 symmetric balanced strings of 5 pairs of parentheses:
[ 1] ((((()))))
[ 2] (((()())))
[ 3] ((()()()))
[ 4] ((())(()))
[ 5] (()()()())
[ 6] (()(())())
[ 7] (())()(())
[ 8] ()()()()()
[ 9] ()((()))()
[10] ()(()())() - _Joerg Arndt_, Jul 25 2011
G.f. = 1 + x + 2*x^2 + 3*x^3 + 6*x^4 + 10*x^5 + 20*x^6 + 35*x^7 + 70*x^8 + ...
The a(4)=6 binary 4-tuples such that the number of 1's in the even positions is the same as the number of 1's in the odd positions are 0000, 1100, 1001, 0110, 0011, 1111. - _Juan A. Olmos_, Dec 21 2017
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.
  • M. Aigner and G. M. Ziegler, Proofs from The Book, Springer-Verlag, Berlin, 1999; see p. 135.
  • K. Engel, Sperner Theory, Camb. Univ. Press, 1997; Theorem 1.1.1.
  • P. Frankl, Extremal sets systems, Chap. 24 of R. L. Graham et al., eds, Handbook of Combinatorics, North-Holland.
  • J. C. P. Miller, editor, Table of Binomial Coefficients. Royal Society Mathematical Tables, Vol. 3, Cambridge Univ. Press, 1954.
  • 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).
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 7.16(b), p. 452.

Crossrefs

Row sums of Catalan triangle A053121 and of symmetric Dyck paths A088855.
Enumerates the structures encoded by A061854 and A061855.
First differences are in A037952.
Apparently a(n) = lim_{k->infinity} A094718(k, n).
Partial sums are in A036256. Column k=2 of A182172. Column k=1 of A335570.
Bisections: A000984 (even part), A001700 (odd part).
Cf. A097331.
Cf. A107373, A340567, A340568, A340569 (popularity of certain patterns in faro permutations).

Programs

  • GAP
    List([0..40],n->Binomial(n,Int(n/2))); # Muniru A Asiru, Apr 08 2018
    
  • Haskell
    a001405 n = a007318_row n !! (n `div` 2) -- Reinhard Zumkeller, Nov 09 2011
    
  • Magma
    [Binomial(n, Floor(n/2)): n in [0..40]]; // Vincenzo Librandi, Nov 16 2014
    
  • Maple
    A001405 := n->binomial(n, floor(n/2)): seq(A001405(n), n=0..33);
  • Mathematica
    Table[Binomial[n, Floor[n/2]], {n, 0, 40}] (* Stefan Steinerberger, Apr 08 2006 *)
    Table[DifferenceRoot[Function[{a,n},{-4 n a[n]-2 a[1+n]+(2+n) a[2+n] == 0,a[1] == 1,a[2] == 1}]][n], {n, 30}] (* Luciano Ancora, Jul 08 2015 *)
    Array[Binomial[#,Floor[#/2]]&,40,0] (* Harvey P. Dale, Mar 05 2018 *)
  • Maxima
    A001405(n):=binomial(n,floor(n/2))$
    makelist(A001405(n),n,0,30); /* Martin Ettl, Nov 01 2012 */
    
  • PARI
    a(n) = binomial(n, n\2);
    
  • PARI
    first(n) = x='x+O('x^n); Vec((-1+2*x+sqrt(1-4*x^2))/(2*x-4*x^2)) \\ Iain Fox, Dec 20 2017 (edited by Iain Fox, May 07 2018)
    
  • Python
    from math import comb
    def A001405(n): return comb(n,n//2) # Chai Wah Wu, Jun 07 2022

Formula

a(n) = max_{k=0..n} binomial(n, k).
a(2*n) = A000984(n), a(2*n+1) = A001700(n).
By symmetry, a(n) = binomial(n, ceiling(n/2)). - Labos Elemer, Mar 20 2003
P-recursive with recurrence: a(0) = 1, a(1) = 1, and for n >= 2, (n+1)*a(n) = 2*a(n-1) + 4*(n-1)*a(n-2). - Peter Bala, Feb 28 2011
G.f.: (1+x*c(x^2))/sqrt(1-4*x^2) = 1/(1 - x - x^2*c(x^2)); where c(x) = g.f. for Catalan numbers A000108.
G.f.: (-1 + 2*x + sqrt(1-4*x^2))/(2*x - 4*x^2). - Lee A. Newberg, Apr 26 2010
G.f.: 1/(1 - x - x^2/(1 - x^2/(1 - x^2/(1 - x^2/(1 - ... (continued fraction). - Paul Barry, Aug 12 2009
a(0) = 1; a(2*m+2) = 2*a(2*m+1); a(2*m+1) = Sum_{k = 0..2*m} (-1)^k*a(k)*a(2*m-k). - Len Smiley, Dec 09 2001
G.f.: (sqrt((1+2*x)/(1-2*x)) - 1)/(2*x). - Vladeta Jovovic, Apr 28 2003
The o.g.f. A(x) satisfies A(x) + x*A^2(x) = 1/(1-2*x). - Peter Bala, Feb 28 2011
E.g.f.: BesselI(0, 2*x) + BesselI(1, 2*x). - Vladeta Jovovic, Apr 28 2003
a(0) = 1; a(2*m+2) = 2*a(2*m+1); a(2*m+1) = 2*a(2*m) - c(m), where c(m)=A000108(m) are the Catalan numbers. - Christopher Hanusa (chanusa(AT)washington.edu), Nov 25 2003
a(n) = Sum_{k=0..n} (-1)^k*2^(n-k)*binomial(n, k)*A000108(k). - Paul Barry, Jan 27 2005
a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)*binomial(1, n-2*k). - Paul Barry, May 13 2005
From Paul Barry, Nov 02 2004: (Start)
a(n) = Sum_{k=0..floor((n+1)/2)} (binomial(n+1, k)*(cos((n-2*k+1)*Pi/2) + sin((n-2*k+1)*Pi/2))).
a(n) = Sum_{k=0..n+1}, (binomial(n+1, (n-k+1)/2)*(1-(-1)^(n-k))*(cos(k*Pi/2) + sin(k*Pi))/2). (End)
a(n) = Sum_{k=floor(n/2)..n} (binomial(n,n-k) - binomial(n,n-k-1)). - Paul Barry, Sep 06 2007
Inverse binomial transform of A005773 starting (1, 2, 5, 13, 35, 96, ...) and double inverse binomial transform of A001700. Row sums of triangle A132815. - Gary W. Adamson, Aug 31 2007
a(n) = Sum_{k=0..n} A120730(n,k). - Philippe Deléham, Oct 16 2008
a(n) = Sum_{k = 0..floor(n/2)} (binomial(n,k) - binomial(n,k-1)). - Nishant Doshi (doshinikki2004(AT)gmail.com), Apr 06 2009
Sum_{n>=0} a(n)/10^(n+1) = 0.1123724... = (sqrt(3)-sqrt(2))/(2*sqrt(2)); Sum_{n>=0} a(n)/100^(n+1) = 0.0101020306102035... = (sqrt(51)-sqrt(49))/(2*sqrt(49)). - Mark Dols, Jul 15 2010
Conjectured: a(n) = 2^n*2F1(1/2,-n;2;2), useful for number of paths in 1-d for which the coordinate is never negative. - Benjamin Phillabaum, Feb 20 2011
a(2*m+1) = (2*m+1)*a(2*m)/(m+1), e.g., a(7) = (7/4)*a(6) = (7/4)*20 = 35. - Jon Perry, Jan 20 2011
From Peter Bala, Feb 28 2011: (Start)
Let F(x) be the logarithmic derivative of the o.g.f. A(x). Then 1+x*F(x) is the o.g.f. for A027306.
Let G(x) be the logarithmic derivative of 1+x*A(x). Then x*G(x) is the o.g.f. for A058622. (End)
Let M = an infinite tridiagonal matrix with 1's in the super and subdiagonals and [1,0,0,0,...] in the main diagonal; and V = the vector [1,0,0,0,...]. a(n) = M^n*V, leftmost term. - Gary W. Adamson, Jun 13 2011
Let M = an infinite tridiagonal matrix with 1's in the super and subdiagonals and [1,0,0,0,...] in the main diagonal. a(n) = M^n_{1,1}. - Corrected by Gary W. Adamson, Jan 30 2012
a(n) = A007318(n, floor(n/2)). - Reinhard Zumkeller, Nov 09 2011
a(n+1) = Sum_{k=0..n} a(n-k)*A097331(k) = a(n) + Sum_{k=0..(n-1)/2} A000108(k)*a(n-2*k-1). - Philippe Deléham, Nov 27 2011
a(n) = A214282(n) - A214283(n), for n > 0. - Reinhard Zumkeller, Jul 14 2012
a(n) = Sum_{k=0..n} A168511(n,k)*(-1)^(n-k). - Philippe Deléham, Mar 19 2013
a(n+2*p-2) = Sum_{k=0..floor(n/2)} A009766(n-k+p-1, k+p-1) + binomial(n+2*p-2, p-2), for p >= 1. - Johannes W. Meijer, Aug 02 2013
O.g.f.: (1-x*c(x^2))/(1-2*x), with the o.g.f. c(x) of Catalan numbers A000108. See the rewritten formula given by Lee A. Newberg above. This is the o.g.f. for the row sums the Riordan triangle A053121. - Wolfdieter Lang, Sep 22 2013
a(n) ~ 2^n / sqrt(Pi * n/2). - Charles R Greathouse IV, Oct 23 2015
a(n) = 2^n*hypergeom([1/2,-n], [2], 2). - Vladimir Reshetnikov, Nov 02 2015
a(2*k) = Sum_{i=0..k} binomial(k, i)*binomial(k, i), a(2*k+1) = Sum_{i=0..k} binomial(k+1, i)*binomial(k, i). - Juan A. Olmos, Dec 21 2017
a(0) = 1, a(n) = 2 * a(n-1) for even n, a(n) = (2*n/(n+1)) * a(n-1) for odd n. - James East, Sep 25 2019
a(n) = A037952(n) + A000108(n/2) where A(.)=0 for non-integer argument. - R. J. Mathar, Sep 23 2021
From Amiram Eldar, Mar 10 2022: (Start)
Sum_{n>=0} 1/a(n) = 2*Pi/(3*sqrt(3)) + 2.
Sum_{n>=0} (-1)^n/a(n) = 2/3 - 2*Pi/(9*sqrt(3)). (End)
For k>2, Sum_{n>=0} a(n)/k^n = (sqrt((k+2)/(k-2)) - 1)*k/2. - Vaclav Kotesovec, May 13 2022
From Peter Bala, Mar 24 2023: (Start)
a(n) = Sum_{k = 0..n+1} (-1)^(k+binomial(n+2,2)) * k/(n+1) * binomial(n+1,k)^2.
(n + 1)*(2*n - 1)*a(n) = (-1)^(n+1)*2*a(n-1) + 4*(n - 1)*(2*n + 1)*a(n-2) with a(0) = a(1) = 1. (End)
a(n) = Integral_{x=-2..2} x^n*W(x)*dx, n>=0, where W(x) = sqrt((2+x)/(2-x))/(2*Pi) is a positive function on x=(-2,2) and is singular at x = 2. Therefore a(n) is a positive definite sequence. - Karol A. Penson, May 12 2025

A325698 Numbers with as many even as odd prime indices, counted with multiplicity.

Original entry on oeis.org

1, 6, 14, 15, 26, 33, 35, 36, 38, 51, 58, 65, 69, 74, 77, 84, 86, 90, 93, 95, 106, 119, 122, 123, 141, 142, 143, 145, 156, 158, 161, 177, 178, 185, 196, 198, 201, 202, 209, 210, 214, 215, 216, 217, 219, 221, 225, 226, 228, 249, 262, 265, 278, 287, 291, 299
Offset: 1

Views

Author

Gus Wiseman, May 17 2019

Keywords

Comments

These are Heinz numbers of the integer partitions counted by A045931.
A prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798.
The integers in the multiplicative subgroup of positive rational numbers generated by the products of two consecutive primes (A006094). The sequence is closed under multiplication, prime shift (A003961), and - where the result is an integer - under division. Using these closures, all the terms can be derived from the presence of 6. For example, A003961(6) = 15, A003961(15) = 35, 6 * 35 = 210, 210/15 = 14. Closed also under A297845, since A297845 can be defined using squaring, prime shift and multiplication. - Peter Munn, Oct 05 2020

Examples

			The sequence of terms together with their prime indices begins:
    1: {}
    6: {1,2}
   14: {1,4}
   15: {2,3}
   26: {1,6}
   33: {2,5}
   35: {3,4}
   36: {1,1,2,2}
   38: {1,8}
   51: {2,7}
   58: {1,10}
   65: {3,6}
   69: {2,9}
   74: {1,12}
   77: {4,5}
   84: {1,1,2,4}
   86: {1,14}
   90: {1,2,2,3}
   93: {2,11}
   95: {3,8}
		

Crossrefs

Positions of 0's in A195017.
A257992(n) = A257991(n).
Closed under: A003961, A003991, A297845.
Subsequence of A028260, A332820.

Programs

  • Mathematica
    Select[Range[100],Total[Cases[If[#==1,{},FactorInteger[#]],{p_,k_}:>k*(-1)^PrimePi[p]]]==0&]
  • PARI
    is(n) = {my(v = vector(2), f = factor(n));for(i = 1, #f~,v[1 + primepi(f[i, 1])%2]+=f[i, 2]);v[1] == v[2]} \\ David A. Corneth, Oct 06 2020
    
  • Python
    from sympy import factorint, primepi
    def ok(n):
        v = [0, 0]
        for p, e in factorint(n).items(): v[primepi(p)%2] += e
        return v[0] == v[1]
    print([k for k in range(300) if ok(k)]) # Michael S. Branicky, Apr 16 2022 after David A. Corneth

A045931 Number of partitions of n with equal number of even and odd parts.

Original entry on oeis.org

1, 0, 0, 1, 0, 2, 1, 3, 2, 5, 5, 7, 9, 11, 16, 18, 25, 28, 41, 44, 62, 70, 94, 107, 140, 163, 207, 245, 302, 361, 440, 527, 632, 763, 904, 1090, 1285, 1544, 1812, 2173, 2539, 3031, 3538, 4202, 4896, 5793, 6736, 7934, 9221, 10811, 12549, 14661, 16994, 19780
Offset: 0

Views

Author

Keywords

Comments

The trivariate g.f. with x marking weight (i.e., sum of the parts), t marking number of odd parts and s marking number of even parts, is 1/product((1-tx^(2j-1))(1-sx^(2j)), j=1..infinity). - Emeric Deutsch, Mar 30 2006

Examples

			a(9) = 5 because we have [8,1], [7,2], [6,3], [5,4] and [2,2,2,1,1,1].
From _Gus Wiseman_, Jan 23 2022: (Start)
The a(0) = 1 through a(12) = 9 partitions (A = 10, empty columns indicated by dots):
  ()  .  .  21   .  32   2211   43   3221   54       3322   65       4332
                    41          52   4211   63       4321   74       4431
                                61          72       4411   83       5322
                                            81       5221   92       5421
                                            222111   6211   A1       6321
                                                            322211   6411
                                                            422111   7221
                                                                     8211
                                                                     22221111
(End)
		

Crossrefs

The version for subsets of {1..n} is A001405.
Dominated by A027187 (partitions of even length).
More odd/even parts: A108950/A108949.
More or same number of odd/even parts: A130780/A171966.
The strict case is A239241.
This is column k = 0 of the triangle A240009.
Counting only distinct parts gives A241638, ranked by A325700.
A half-conjugate version is A277579.
These partitions are ranked by A325698.
A000041 counts integer partitions, strict A000009.
A047993 counts balanced partitions, ranked by A106529.
A257991/A257992 count odd/even parts by Heinz number.

Programs

  • Maple
    g:=1/product((1-t*x^(2*j-1))*(1-s*x^(2*j)),j=1..30): gser:=simplify(series(g,x=0,56)): P[0]:=1: for n from 1 to 53 do P[n]:=subs(s=1/t,coeff(gser,x^n)) od: seq(coeff(t*P[n],t),n=0..53); # Emeric Deutsch, Mar 30 2006
  • Mathematica
    p[n_] := p[n] = Select[IntegerPartitions[n], Count[#, ?OddQ] == Count[#, ?EvenQ] &]; t = Table[p[n], {n, 0, 10}] (* partitions of n with # odd parts = # even parts *)
    TableForm[t] (* partitions, vertical format *)
    Table[Length[p[n]], {n, 0, 30}] (* A045931 *)
    (* Peter J. C. Moses, Mar 10 2014 *)
    nmax = 100; CoefficientList[Series[Sum[x^(3*k) / Product[(1 - x^(2*j))^2, {j, 1, k}], {k, 0, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Jun 15 2025 *)

Formula

G.f.: Sum_{k>=0} x^(3*k)/Product_{i=1..k} (1-x^(2*i))^2. - Vladeta Jovovic, Aug 18 2007
a(n) = A000041(n)-A171967(n) = A130780(n)-A108950(n) = A171966(n)-A108949(n). - Reinhard Zumkeller, Jan 21 2010
a(n) = A000041(n) - A108950(n) - A108949(n) = A130780(n) + A171966(n) - A000041(n). - Gus Wiseman, Jan 23 2022
a(n) ~ Pi * exp(Pi*sqrt(2*n/3)) / (48*n^(3/2)). - Vaclav Kotesovec, Jun 15 2025

A195017 If n = Product_{k >= 1} (p_k)^(c_k) where p_k is k-th prime and c_k >= 0 then a(n) = Sum_{k >= 1} c_k*((-1)^(k-1)).

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Feb 06 2012

Keywords

Comments

Let p(n,x) be the completely additive polynomial-valued function such that p(1,x) = 0 and p(prime(n),x) = x^(n-1), like is defined in A206284 (although here we are not limited to just irreducible polynomials). Then a(n) is the value of the polynomial encoded in such a manner by n, when it is evaluated at x=-1. - The original definition rewritten and clarified by Antti Karttunen, Oct 03 2018
Positions of 0 give the values of n for which the polynomial p(n,x) is divisible by x+1. For related sequences, see the Mathematica section.
Also the number of odd prime indices of n minus the number of even prime indices of n (both counted with multiplicity), where a prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798. - Gus Wiseman, Oct 24 2023

Examples

			The sequence can be read from a list of the polynomials:
  p(n,x)      with x = -1, gives a(n)
------------------------------------------
  p(1,x) = 0           0
  p(2,x) = 1x^0        1
  p(3,x) = x          -1
  p(4,x) = 2x^0        2
  p(5,x) = x^2         1
  p(6,x) = 1+x         0
  p(7,x) = x^3        -1
  p(8,x) = 3x^0        3
  p(9,x) = 2x         -2
  p(10,x) = x^2 + 1    2.
(The list runs through all the polynomials whose coefficients are nonnegative integers.)
		

Crossrefs

For other evaluation functions of such encoded polynomials, see A001222, A048675, A056239, A090880, A248663.
Zeros are A325698, distinct A325700.
For sum instead of count we have A366749 = A366531 - A366528.
A000009 counts partitions into odd parts, ranked by A066208.
A035363 counts partitions into even parts, ranked by A066207.
A112798 lists prime indices, reverse A296150, sum A056239.
A257991 counts odd prime indices, even A257992.
A300061 lists numbers with even sum of prime indices, odd A300063.

Programs

  • Mathematica
    b[n_] := Table[x^k, {k, 0, n}];
    f[n_] := f[n] = FactorInteger[n]; z = 200;
    t[n_, m_, k_] := If[PrimeQ[f[n][[m, 1]]] && f[n][[m, 1]]
    == Prime[k], f[n][[m, 2]], 0];
    u = Table[Apply[Plus,
        Table[Table[t[n, m, k], {k, 1, PrimePi[n]}], {m, 1,
          Length[f[n]]}]], {n, 1, z}];
    p[n_, x_] := u[[n]].b[-1 + Length[u[[n]]]]
    Table[p[n, x] /. x -> 0, {n, 1, z/2}]   (* A007814 *)
    Table[p[2 n, x] /. x -> 0, {n, 1, z/2}] (* A001511 *)
    Table[p[n, x] /. x -> 1, {n, 1, z}]     (* A001222 *)
    Table[p[n, x] /. x -> 2, {n, 1, z}]     (* A048675 *)
    Table[p[n, x] /. x -> 3, {n, 1, z}]     (* A090880 *)
    Table[p[n, x] /. x -> -1, {n, 1, z}]    (* A195017 *)
    z = 100; Sum[-(-1)^k IntegerExponent[Range[z], Prime[k]], {k, 1, PrimePi[z]}] (* Friedjof Tellkamp, Aug 05 2024 *)
  • PARI
    A195017(n) = { my(f); if(1==n, 0, f=factor(n); sum(i=1, #f~, f[i,2] * (-1)^(1+primepi(f[i,1])))); } \\ Antti Karttunen, Oct 03 2018

Formula

Totally additive with a(p^e) = e * (-1)^(1+PrimePi(p)), where PrimePi(n) = A000720(n). - Antti Karttunen, Oct 03 2018
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{p prime} = (-1)^(primepi(p)+1)/(p-1) = Sum_{k>=1} (-1)^(k+1)/A006093(k) = A078437 + Sum_{k>=1} (-1)^(k+1)/A036689(k) = 0.6339266524059... . - Amiram Eldar, Sep 29 2023
a(n) = A257991(n) - A257992(n). - Gus Wiseman, Oct 24 2023
a(n) = -Sum_{k=1..pi(n)} (-1)^k * valuation(n, prime(k)). - Friedjof Tellkamp, Aug 05 2024

Extensions

More terms, name changed and example-section edited by Antti Karttunen, Oct 03 2018

A063886 Number of n-step walks on a line starting from the origin but not returning to it.

Original entry on oeis.org

1, 2, 2, 4, 6, 12, 20, 40, 70, 140, 252, 504, 924, 1848, 3432, 6864, 12870, 25740, 48620, 97240, 184756, 369512, 705432, 1410864, 2704156, 5408312, 10400600, 20801200, 40116600, 80233200, 155117520, 310235040, 601080390, 1202160780, 2333606220, 4667212440
Offset: 0

Views

Author

Henry Bottomley, Aug 28 2001

Keywords

Comments

A Chebyshev transform of A007877(n+1). The g.f. is transformed to (1+x)/((1-x)(1+x^2)) under the mapping G(x)->(1/(1+x^2))G(1/(1+x^2)). - Paul Barry, Oct 12 2004
a(n-1) = 2*C(n-2, floor((n-2)/2)) is also the number of bit strings of length n in which the number of 00 substrings is equal to the number of 11 substrings. For example, when n = 4 we have 4 such bit strings: 0011, 0101, 1010, and 1100. - Angel Plaza, Apr 23 2009
Hankel transform is A120617. - Paul Barry, Aug 10 2009
The Hankel transform of a(n) is (-2)^C(n+1,2). The Hankel transform of (-1)^C(n+1,2)*a(n) is (-1)^C(n+1,2)*A164584(n). - Paul Barry, Aug 17 2009
For n > 1, a(n) is also the number of n-step walks starting from the origin and returning to it exactly once. - Geoffrey Critzer, Jan 24 2010
-a(n) is the Z-sequence for the Riordan array A130777. (See the W. Lang link under A006232 for A- and Z-sequences for Riordan matrices). - Wolfdieter Lang, Jul 12 2011
Number of subsets of {1,...,n} in which the even elements appear as often at even positions as at odd positions. - Gus Wiseman, Mar 17 2018

Examples

			a(4) = 6 because there are six length four walks that do not return to the origin: {-1, -2, -3, -4}, {-1, -2, -3, -2}, {-1, -2, -1, -2}, {1, 2, 1, 2}, {1, 2, 3, 2}, {1, 2, 3, 4}. There are also six such walks that return exactly one time: {-1, -2, -1, 0}, {-1, 0, -1, -2}, {-1, 0, 1, 2}, {1, 0, -1, -2}, {1, 0, 1, 2}, {1, 2, 1, 0}. - _Geoffrey Critzer_, Jan 24 2010
The a(5) = 12 subsets in which the even elements appear as often at even positions as at odd positions: {}, {1}, {3}, {5}, {1,3}, {1,5}, {2,4}, {3,5}, {1,2,4}, {1,3,5}, {2,4,5}, {1,2,4,5}. - _Gus Wiseman_, Mar 17 2018
		

Crossrefs

Programs

  • Magma
    [1] cat [2*Binomial(n-1, Floor((n-1)/2)): n in [1..40]]; // G. C. Greubel, Jun 07 2023
    
  • Maple
    seq(seq(binomial(2*j,j)*i, i=1..2),j=0..16); # Zerinvary Lajos, Apr 28 2007
    # second Maple program:
    a:= proc(n) option remember; `if`(n<2, n+1,
           4*a(n-2) +2*(a(n-1) -4*a(n-2))/n)
        end:
    seq(a(n), n=0..40);  # Alois P. Heinz, Feb 10 2014
    # third program:
    A063886 := series(BesselI(0, 2*x)*(1 + x*2 + x*Pi*StruveL(1, 2*x)) - Pi*x*BesselI(1, 2*x)*StruveL(0, 2*x), x = 0, 34): seq(n!*coeff(A063886, x, n), n = 0 .. 33); # Mélika Tebni, Jun 17 2024
  • Mathematica
    Table[Length[Select[Map[Accumulate, Strings[{-1, 1}, n]], Count[ #, 0] == 0 &]], {n, 0, 20}] (* Geoffrey Critzer, Jan 24 2010 *)
    CoefficientList[Series[Sqrt[(1+2x)/(1-2x)],{x,0,40}],x] (* Harvey P. Dale, Apr 28 2016 *)
  • PARI
    a(n)=(n==0)+2*binomial(n-1,(n-1)\2)
    
  • PARI
    a(n) = 2^n*prod(k=0,n-1,(k/n+1/n)^((-1)^k)); \\ Michel Marcus, Dec 03 2013
    
  • Python
    from math import ceil
    from sympy import binomial
    def a(n):
        if n==0: return 1
        return 2*binomial(n-1,(n-1)//2)
    print([a(n) for n in range(18)])
    # David Nacin, Feb 29 2012
    
  • SageMath
    [2*binomial(n-1, (n-1)//2) + int(n==0) for n in range(41)] # G. C. Greubel, Jun 07 2023

Formula

G.f.: sqrt((1+2*x)/(1-2*x)).
a(n+1) = 2*C(n, floor(n/2)) = 2*A001405(n); a(2n) = C(2n, n) = A000984(n) = 4*a(2n-2)-|A002420(n)| = 4*a(2n-2)-2*A000108(n-1) = 2*A001700(n-1); a(2n+1) = 2*a(2n) = A028329(n).
2*a(n) = A047073(n+1).
a(n) = Sum_{k=0..n} abs(A106180(n,k)). - Philippe Deléham, Oct 06 2006
a(n) = Sum_{k=0..n} (k+1)binomial(n, (n-k)/2) ( 1-cos((k+1)*Pi/2) (1+(-1)^(n-k))/(n+k+2) ). - Paul Barry, Oct 12 2004
G.f.: 1/(1-2*x/(1+x/(1+x/(1-x/(1-x/(1+x/(1+x/(1-x/(1-x/(1+ ... (continued fraction). - Paul Barry, Aug 10 2009
G.f.: 1 + 2*x/(G(0)-x+x^2) where G(k)= 1 - 2*x^2 - x^4/G(k+1); (continued fraction, 1-step). - Sergei N. Gladkovskii, Aug 10 2012
D-finite with recurrence: n*a(n) = 2*a(n-1) + 4*(n-2)*a(n-2). - R. J. Mathar, Dec 03 2012
From Sergei N. Gladkovskii, Jul 26 2013: (Start)
G.f.: 1/G(0), where G(k) = 1 - 2*x/(1 + 2*x/(1 + 1/G(k+1) )); (continued fraction).
G.f.: G(0), where G(k) = 1 + 2*x/(1 - 2*x/(1 + 1/G(k+1) )); (continued fraction).
G.f.: W(0)/2*(1+2*x), where W(k) = 1 + 1/(1 - 2*x/(2*x + (k+1)/(x*(2*k+1))/W(k+1) )), abs(x) < 1/2; (continued fraction). (End)
a(n) = 2^n*Product_{k=0..n-1} (k/n + 1/n)^((-1)^k). - Peter Luschny, Dec 02 2013
G.f.: G(0), where G(k) = 1 + 2*x*(4*k+1)/((2*k+1)*(1+2*x) - (2*k+1)*(4*k+3)*x*(1+2*x)/((4*k+3)*x + (k+1)*(1+2*x)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 19 2014
From Peter Bala, Mar 29 2024: (Start)
a(n) = 2^n * Sum_{k = 0..n} (-1)^(n+k)*binomial(1/2, k)*binomial(- 1/2, n-k) = 2^n * A000246(n)/n!.
a(n) = (1/2^n) * binomial(2*n, n) * hypergeom([-1/2, -n], [1/2 - n], -1). (End)
E.g.f.: BesselI(0, 2*x)*(1 + x*(2 + Pi)*StruveL(1, 2*x)) - Pi*x*BesselI(1, 2*x)*StruveL(0, 2*x). - Stefano Spezia, May 11 2024
a(n) = A089849(n) + A138364(n). - Mélika Tebni, Jun 17 2024
From Amiram Eldar, Aug 15 2025: (Start)
Sum_{n>=0} 1/a(n) = Pi/(3*sqrt(3)) + 2.
Sum_{n>=0} (-1)^n/a(n) = 2/3 + Pi/(9*sqrt(3)). (End)

A325700 Numbers with as many distinct even as distinct odd prime indices.

Original entry on oeis.org

1, 6, 12, 14, 15, 18, 24, 26, 28, 33, 35, 36, 38, 45, 48, 51, 52, 54, 56, 58, 65, 69, 72, 74, 75, 76, 77, 86, 93, 95, 96, 98, 99, 104, 106, 108, 112, 116, 119, 122, 123, 135, 141, 142, 143, 144, 145, 148, 152, 153, 158, 161, 162, 172, 175, 177, 178, 185, 192
Offset: 1

Views

Author

Gus Wiseman, May 17 2019

Keywords

Comments

These are the Heinz numbers of the integer partitions counted by A241638.
A prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798.

Examples

			The sequence of terms together with their prime indices begins:
    1: {}
    6: {1,2}
   12: {1,1,2}
   14: {1,4}
   15: {2,3}
   18: {1,2,2}
   24: {1,1,1,2}
   26: {1,6}
   28: {1,1,4}
   33: {2,5}
   35: {3,4}
   36: {1,1,2,2}
   38: {1,8}
   45: {2,2,3}
   48: {1,1,1,1,2}
   51: {2,7}
   52: {1,1,6}
   54: {1,2,2,2}
   56: {1,1,1,4}
   58: {1,10}
		

Crossrefs

Programs

  • Mathematica
    Select[Range[100],0==Total[(-1)^PrimePi/@First/@If[#==1,{},FactorInteger[#]]]&]

A349157 Heinz numbers of integer partitions where the number of even parts is equal to the number of odd conjugate parts.

Original entry on oeis.org

1, 4, 6, 15, 16, 21, 24, 25, 35, 60, 64, 77, 84, 90, 91, 96, 100, 121, 126, 140, 143, 150, 210, 221, 240, 247, 256, 289, 297, 308, 323, 336, 351, 360, 364, 375, 384, 400, 437, 462, 484, 490, 495, 504, 525, 529, 546, 551, 560, 572, 585, 600, 625, 667, 686, 726
Offset: 1

Views

Author

Gus Wiseman, Jan 21 2022

Keywords

Comments

The Heinz number of a partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k), so these are numbers with the same number of even prime indices as odd conjugate prime indices.
These are also partitions for which the number of even parts is equal to the positive alternating sum of the parts.

Examples

			The terms and their prime indices begin:
    1: ()
    4: (1,1)
    6: (2,1)
   15: (3,2)
   16: (1,1,1,1)
   21: (4,2)
   24: (2,1,1,1)
   25: (3,3)
   35: (4,3)
   60: (3,2,1,1)
   64: (1,1,1,1,1,1)
   77: (5,4)
   84: (4,2,1,1)
   90: (3,2,2,1)
   91: (6,4)
   96: (2,1,1,1,1,1)
		

Crossrefs

A subset of A028260 (even bigomega), counted by A027187.
These partitions are counted by A277579.
This is the half-conjugate version of A325698, counted by A045931.
A000041 counts partitions, strict A000009.
A047993 counts balanced partitions, ranked by A106529.
A056239 adds up prime indices, row sums of A112798, counted by A001222.
A100824 counts partitions with at most one odd part, ranked by A349150.
A108950/A108949 count partitions with more odd/even parts.
A122111 represents conjugation using Heinz numbers.
A130780/A171966 count partitions with more or equal odd/even parts.
A257991/A257992 count odd/even prime indices.
A316524 gives the alternating sum of prime indices (reverse: A344616).

Programs

  • Mathematica
    primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    conj[y_]:=If[Length[y]==0,y,Table[Length[Select[y,#>=k&]],{k,1,Max[y]}]];
    Select[Range[100],Count[primeMS[#],?EvenQ]==Count[conj[primeMS[#]],?OddQ]&]

Formula

A257992(a(n)) = A257991(A122111(a(n))).

A350847 Number of even parts in the conjugate of the integer partition with Heinz number n.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, Mar 14 2022

Keywords

Comments

The Heinz number of a partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k), so a(n) counts even prime indices of n.

Crossrefs

Positions of first appearances are A001248.
The triangular version is A116482.
Positions of zeros are A346635.
Subtracting from the number of odd conjugate parts gives A350941.
Subtracting from the number of odd parts gives A350942.
Subtracting from the number of even parts gives A350950.
There are four statistics:
- A257991 = # of odd parts, conjugate A344616.
- A257992 = # of even parts, conjugate A350847 (this sequence).
There are six possible pairings of statistics:
- A325698: # of even parts = # of odd parts, counted by A045931.
- A349157: # of even parts = # of odd conjugate parts, counted by A277579.
- A350848: # of even conj parts = # of odd conj parts, counted by A045931.
- A350943: # of even conjugate parts = # of odd parts, counted by A277579.
- A350944: # of odd parts = # of odd conjugate parts, counted by A277103.
- A350945: # of even parts = # of even conjugate parts, counted by A350948.
There are three possible double-pairings of statistics:
- A350946, counted by A351977.
- A350949, counted by A351976.
- A351980, counted by A351981.
The case of all four statistics equal is A350947, counted by A351978.
A056239 adds up prime indices, counted by A001222, row sums of A112798.
A122111 represents partition conjugation using Heinz numbers.

Programs

  • Mathematica
    primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    conj[y_]:=If[Length[y]==0,y,Table[Length[Select[y,#>=k&]],{k,1,Max[y]}]];
    Table[Count[conj[primeMS[n]],_?EvenQ],{n,100}]

Formula

a(n) = A344616(n) - A350941(n).
a(n) = A257992(A122111(n)).

A350848 Heinz numbers of integer partitions for which the number of even conjugate parts is equal to the number of odd conjugate parts.

Original entry on oeis.org

1, 6, 18, 21, 24, 54, 65, 70, 72, 84, 96, 133, 147, 162, 182, 189, 210, 216, 260, 280, 288, 319, 336, 384, 418, 429, 481, 486, 490, 525, 532, 546, 585, 588, 630, 648, 728, 731, 741, 754, 756, 840, 845, 864, 1007, 1029, 1040, 1120, 1152, 1197, 1254, 1258, 1276
Offset: 1

Views

Author

Gus Wiseman, Jan 27 2022

Keywords

Comments

The Heinz number of a partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k). This gives a bijective correspondence between positive integers and integer partitions.

Examples

			The terms together with their prime indices begin:
   1: ()
   6: (2,1)
  18: (2,2,1)
  21: (4,2)
  24: (2,1,1,1)
  54: (2,2,2,1)
  65: (6,3)
  70: (4,3,1)
  72: (2,2,1,1,1)
  84: (4,2,1,1)
  96: (2,1,1,1,1,1)
		

Crossrefs

These partitions are counted by A045931.
The conjugate strict version is counted by A239241.
The conjugate version is A325698.
These are the positions of 0's in A350941.
Adding the conjugate condition gives A350946, all four equal A350947.
A257991 counts odd parts, conjugate A344616.
A257992 counts even parts, conjugate A350847.
A325698: # of even parts = # of odd parts.
A349157: # of even parts = # of odd conjugate parts, counted by A277579.
A350848: # even conjugate parts = # odd conjugate parts, counted by A045931.
A350943: # of even conjugate parts = # of odd parts, counted by A277579.
A350944: # of odd parts = # of odd conjugate parts, counted by A277103.
A350945: # of even parts = # of even conjugate parts, counted by A350948.
A000041 = integer partitions, strict A000009.
A056239 adds up prime indices, counted by A001222, row sums of A112798.
A316524 = alternating sum of prime indices, reverse A344616.

Programs

  • Mathematica
    primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    conj[y_]:=If[Length[y]==0,y,Table[Length[Select[y,#>=k&]],{k,1,Max[y]}]];
    Select[Range[100],Count[conj[primeMS[#]],?EvenQ]==Count[conj[primeMS[#]],?OddQ]&]

Formula

A344616(a(n)) = A350847(a(n)).
A257991(A122111(a(n))) = A257992(A122111(a(n))).

A350948 Number of integer partitions of n with as many even parts as even conjugate parts.

Original entry on oeis.org

1, 1, 0, 3, 1, 5, 3, 7, 6, 10, 10, 18, 19, 27, 31, 40, 47, 65, 75, 98, 115, 142, 170, 217, 257, 316, 376, 458, 544, 671, 792, 952, 1129, 1351, 1598, 1919, 2259, 2681, 3155, 3739, 4384, 5181, 6064, 7129, 8331, 9764, 11380, 13308, 15477, 18047, 20944
Offset: 0

Views

Author

Gus Wiseman, Mar 14 2022

Keywords

Examples

			The a(0) = 1 through a(8) = 6 partitions (empty column indicated by dot):
  ()  (1)  .  (3)    (22)  (5)      (42)    (7)        (62)
              (21)         (41)     (321)   (61)       (332)
              (111)        (311)    (2211)  (511)      (521)
                           (2111)           (4111)     (4211)
                           (11111)          (31111)    (32111)
                                            (211111)   (221111)
                                            (1111111)
For example, both (3,2,1,1,1) and its conjugate (5,2,1) have exactly 1 even part, so are counted under a(8).
		

Crossrefs

Comparing even to odd parts gives A045931, ranked by A325698.
The odd version is A277103, even rank case A345196, ranked by A350944.
Comparing even to odd conjugate parts gives A277579, ranked by A349157.
Comparing product of parts to product of conjugate parts gives A325039.
These partitions are ranked by A350945, the zeros of A350950.
A000041 counts integer partitions, strict A000009.
A103919 counts partitions by sum and alternating sum, reverse A344612.
A116482 counts partitions by number of even (or even conjugate) parts.
A122111 represents partition conjugation using Heinz numbers.
A257991 counts odd parts, conjugate A344616.
A257992 counts even parts, conjugate A350847.
A351976: # even = # even conj, # odd = # odd conj, ranked by A350949.
A351977: # even = # odd, # even conj = # odd conj, ranked by A350946.
A351978: # even = # odd = # even conj = # odd conj, ranked by A350947.
A351981: # even = # odd conj, # odd = # even conj, ranked by A351980.

Programs

  • Mathematica
    conj[y_]:=If[Length[y]==0,y,Table[Length[Select[y,#>=k&]],{k,1,Max[y]}]];
    Table[Length[Select[IntegerPartitions[n],Count[#,?EvenQ]==Count[conj[#],?EvenQ]&]],{n,0,30}]
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