cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 80 results. Next

A285103 Number of odd terms on row n of A053632: a(n) = A000120(A068052(n)).

Original entry on oeis.org

1, 2, 4, 6, 6, 12, 12, 16, 22, 28, 32, 30, 36, 52, 48, 62, 62, 68, 88, 104, 116, 108, 128, 128, 132, 168, 160, 168, 200, 204, 240, 232, 242, 284, 300, 324, 332, 348, 352, 352, 412, 440, 400, 466, 460, 516, 496, 566, 582, 580, 608, 646, 676, 736, 716, 782, 728, 816, 832, 856, 916, 924, 948, 1034, 1008, 1044, 1096, 1154, 1112, 1212, 1204, 1188
Offset: 0

Views

Author

Antti Karttunen, Apr 15 2017

Keywords

Crossrefs

Number of odd term on row n of A053632.

Programs

  • Maple
    b:= proc(n) option remember; `if`(n=0, 1,
          (t-> Bits[Xor](2^n*t, t))(b(n-1)))
        end:
    a:= n-> convert(Bits[Split](b(n)), `+`):
    seq(a(n), n=0..71);  # Alois P. Heinz, Mar 07 2024
  • Mathematica
    b[n_] := b[n] = If[n == 0, 1, With[{t = b[n-1]}, BitXor[2^n*t, t]]];
    a[n_] := DigitCount[b[n], 2, 1];
    Table[a[n], {n, 0, 100}] (* Jean-François Alcover, May 17 2024, after Alois P. Heinz *)
  • Python
    # uses [A000120]
    l=[1]
    for n in range(1, 101):
        x = l[n - 1]
        l.append(x^(2**n*x))
    print([A000120(k) for k in l]) # Indranil Ghosh, Jun 28 2017
  • Scheme
    (define (A285103 n) (A000120 (A068052 n)))
    

Formula

a(n) = A000120(A068052(n)).
a(n) = A001221(A285102(n)) = A001222(A285102(n)).
A285104(n) = 2^n - a(n).
A000124(n) = a(n) + A285105(n).

A285105 Number of even terms on row n of A053632.

Original entry on oeis.org

0, 0, 0, 1, 5, 4, 10, 13, 15, 18, 24, 37, 43, 40, 58, 59, 75, 86, 84, 87, 95, 124, 126, 149, 169, 158, 192, 211, 207, 232, 226, 265, 287, 278, 296, 307, 335, 356, 390, 429, 409, 422, 504, 481, 531, 520, 586, 563, 595, 646, 668, 681, 703, 696, 770, 759, 869, 838, 880, 915, 915, 968, 1006, 983, 1073, 1102, 1116, 1125, 1235, 1204, 1282, 1369, 1349
Offset: 0

Views

Author

Antti Karttunen, Apr 15 2017

Keywords

Crossrefs

Programs

Formula

a(n) = A080791(A068052(n)).
A000124(n) = A285103(n) + a(n).

A070937 Number of times maximal coefficient (A025591) appears in Product_{k<=n} (x^k + 1), i.e., number of times highest value appears in n-th row of A053632 or n-th column of A070936.

Original entry on oeis.org

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

Views

Author

Henry Bottomley, May 12 2002

Keywords

Examples

			a(4)=5 since Product_{k<=4} (x^k + 1) = 1 + x + x^2 + 2x^3 + 2x^4 + 2x^5 + 2x^6 + 2x^7 + x^8 + x^9 + x^10 and 2 appears as a coefficient 5 times.
		

Formula

If n mod 4 = 0 or 3 then a(n) odd, otherwise a(n) even.
For n > 9: a(n) = A014695(n).
From Chai Wah Wu, Apr 10 2021: (Start)
a(n) = a(n-1) - a(n-2) + a(n-3) for n > 12.
G.f.: (2*x^12 - 2*x^11 + 6*x^10 - 4*x^9 + 6*x^8 - 2*x^7 - 2*x^6 + 2*x^5 - 6*x^4 + 2*x^3 - 3*x^2 - x - 1)/((x - 1)*(x^2 + 1)). (End)

A000009 Expansion of Product_{m >= 1} (1 + x^m); number of partitions of n into distinct parts; number of partitions of n into odd parts.

Original entry on oeis.org

1, 1, 1, 2, 2, 3, 4, 5, 6, 8, 10, 12, 15, 18, 22, 27, 32, 38, 46, 54, 64, 76, 89, 104, 122, 142, 165, 192, 222, 256, 296, 340, 390, 448, 512, 585, 668, 760, 864, 982, 1113, 1260, 1426, 1610, 1816, 2048, 2304, 2590, 2910, 3264, 3658, 4097, 4582, 5120, 5718, 6378
Offset: 0

Views

Author

Keywords

Comments

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

Examples

			G.f. = 1 + x + x^2 + 2*x^3 + 2*x^4 + 3*x^5 + 4*x^6 + 5*x^7 + 6*x^8 + 8*x^9 + ...
G.f. = q + q^25 + q^49 + 2*q^73 + 2*q^97 + 3*q^121 + 4*q^145 + 5*q^169 + ...
The partitions of n into distinct parts (see A118457) for small n are:
  1: 1
  2: 2
  3: 3, 21
  4: 4, 31
  5: 5, 41, 32
  6: 6, 51, 42, 321
  7: 7, 61, 52, 43, 421
  8: 8, 71, 62, 53, 521, 431
  ...
From _Reinhard Zumkeller_, Jun 13 2009: (Start)
a(8)=6, A140207(8)=#{5+2+1,4+3+1}=2, A003056(8)=3, A051162(8)=5;
a(9)=8, A140207(9)=#{6+2+1,5+3+1,4+3+2}=3, A003056(9)=3, A051162(9)=6;
a(10)=10, A140207(10)=#{4+3+2+1}=1, A003056(10)=4, A051162(10)=4. (End)
		

References

  • Mohammad K. Azarian, A Generalization of the Climbing Stairs Problem, Mathematics and Computer Education, Vol. 31, No. 1, pp. 24-28, Winter 1997. MathEduc Database (Zentralblatt MATH, 1997c.01891).
  • Mohammad K. Azarian, A Generalization of the Climbing Stairs Problem II, Missouri Journal of Mathematical Sciences, Vol. 16, No. 1, Winter 2004, pp. 12-17. Zentralblatt MATH, Zbl 1071.05501.
  • George E. Andrews, The Theory of Partitions, Cambridge University Press, 1998, p. 19.
  • George E. Andrews, Number Theory, Dover Publications, 1994, Theorem 12-3, pp. 154-5, and (13-1-1) p. 163.
  • Raymond Ayoub, An Introduction to the Analytic Theory of Numbers, Amer. Math. Soc., 1963; see p. 196.
  • T. J. I'a. Bromwich, Introduction to the Theory of Infinite Series, Macmillan, 2nd. ed. 1949, p. 116, Problem 18.
  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 99.
  • William Dunham, The Mathematical Universe, pp. 57-62, J. Wiley, 1994.
  • Leonhard Euler, De partitione numerorum, Novi commentarii academiae scientiarum Petropolitanae 3 (1750/1), 1753, reprinted in: Commentationes Arithmeticae. (Opera Omnia. Series Prima: Opera Mathematica, Volumen Secundum), 1915, Lipsiae et Berolini, 254-294.
  • Ian P. Goulden and David M. Jackson, Combinatorial Enumeration, Wiley, N.Y., 1983, (2.5.1).
  • G. H. Hardy, Ramanujan: twelve lectures on subjects suggested by his life and work, Cambridge, University Press, 1940, p. 86.
  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 277, Theorems 344, 346.
  • Carlos J. Moreno and Samuel S. Wagstaff, Jr., Sums of Squares of Integers, Chapman and Hall, 2006, p. 253.
  • Srinivasa Ramanujan, Collected Papers, Ed. G. H. Hardy et al., Cambridge 1927; Chelsea, NY, 1962. See Table V on page 309.
  • 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).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 288-290.

Crossrefs

Apart from the first term, equals A052839-1. The rows of A053632 converge to this sequence. When reduced modulo 2 equals the absolute values of A010815. The positions of odd terms given by A001318.
a(n) = Sum_{n=1..m} A097306(n, m), row sums of triangle of number of partitions of n into m odd parts.
Cf. A001318, A000041, A000700, A003724, A004111, A007837, A010815, A035294, A068049, A078408, A081360, A088670, A109950, A109968, A132312, A146061, A035363, A010054, A057077, A089806, A091602, A237515, A118457 (the partitions), A118459 (partition lengths), A015723 (total number of parts), A230957 (boustrophedon transform).
Cf. A167377 (complement).
Cf. A067659 (odd number of parts), A067661 (even number of parts).
Number of r-regular partitions for r = 2 through 12: A000009, A000726, A001935, A035959, A219601, A035985, A261775, A104502, A261776, A328545, A328546.

Programs

  • Haskell
    import Data.MemoCombinators (memo2, integral)
    a000009 n = a000009_list !! n
    a000009_list = map (pM 1) [0..] where
       pM = memo2 integral integral p
       p _ 0 = 1
       p k m | m < k     = 0
             | otherwise = pM (k + 1) (m - k) + pM (k + 1) m
    -- Reinhard Zumkeller, Sep 09 2015, Nov 05 2013
    
  • Julia
    # uses A010815
    using Memoize
    @memoize function A000009(n)
        n == 0 && return 1
        s = sum((-1)^k*A000009(n - k^2) for k in 1:isqrt(n))
        A010815(n) - 2*s
    end # Peter Luschny, Sep 09 2021
  • Magma
    Coefficients(&*[1+x^m:m in [1..100]])[1..100] where x is PolynomialRing(Integers()).1; // Sergei Haller (sergei(AT)sergei-haller.de), Dec 21 2006
    
  • Maple
    N := 100; t1 := series(mul(1+x^k,k=1..N),x,N); A000009 := proc(n) coeff(t1,x,n); end;
    spec := [ P, {P=PowerSet(N), N=Sequence(Z,card>=1)} ]: [ seq(combstruct[count](spec, size=n), n=0..58) ];
    spec := [ P, {P=PowerSet(N), N=Sequence(Z,card>=1)} ]: combstruct[allstructs](spec, size=10); # to get the actual partitions for n=10
    A000009 := proc(n)
        local x,m;
        product(1+x^m,m=1..n+1) ;
        expand(%) ;
        coeff(%,x,n) ;
    end proc: # R. J. Mathar, Jun 18 2016
    lim := 99; # Enlarge if more terms are needed.
    simplify(expand(QDifferenceEquations:-QPochhammer(-1, x, lim)/2, x)):
    seq(coeff(%, x, n), n=0..55); # Peter Luschny, Nov 17 2016
    # Alternative:
    a:= proc(n) option remember; `if`(n=0, 1, add(a(n-j)*add(
         `if`(d::odd, d, 0), d=numtheory[divisors](j)), j=1..n)/n)
        end:
    seq(a(n), n=0..55);  # Alois P. Heinz, Jun 24 2025
  • Mathematica
    PartitionsQ[Range[0, 60]] (* Harvey Dale, Jul 27 2009 *)
    a[ n_] := SeriesCoefficient[ Product[ 1 + x^k, {k, n}], {x, 0, n}]; (* Michael Somos, Jul 06 2011 *)
    a[ n_] := SeriesCoefficient[ 1 / Product[ 1 - x^k, {k, 1, n, 2}], {x, 0, n}]; (* Michael Somos, Jul 06 2011 *)
    a[ n_] := With[ {t = Log[q] / (2 Pi I)}, SeriesCoefficient[ q^(-1/24) DedekindEta[2 t] / DedekindEta[ t], {q, 0, n}]]; (* Michael Somos, Jul 06 2011 *)
    a[ n_] := SeriesCoefficient[ 1 / QPochhammer[ x, x^2], {x, 0, n}]; (* Michael Somos, May 24 2013 *)
    a[ n_] := SeriesCoefficient[ Series[ QHypergeometricPFQ[ {q}, {q x}, q, - q x], {q, 0, n}] /. x -> 1, {q, 0, n}]; (* Michael Somos, Mar 04 2014 *)
    a[ n_] := SeriesCoefficient[ QHypergeometricPFQ[{}, {}, q, -1] / 2, {q, 0, n}]; (* Michael Somos, Mar 04 2014 *)
    nmax = 60; CoefficientList[Series[Exp[Sum[(-1)^(k+1)/k*x^k/(1-x^k), {k, 1, nmax}]], {x, 0, nmax}], x] (* Vaclav Kotesovec, Aug 25 2015 *)
    nmax = 100; poly = ConstantArray[0, nmax + 1]; poly[[1]] = 1; poly[[2]] = 1; Do[Do[poly[[j + 1]] += poly[[j - k + 1]], {j, nmax, k, -1}];, {k, 2, nmax}]; poly (* Vaclav Kotesovec, Jan 14 2017 *)
  • Maxima
    num_distinct_partitions(60,list); /* Emanuele Munarini, Feb 24 2014 */
    
  • Maxima
    h(n):=if oddp(n)=true then 1 else 0;
    S(n,m):=if n=0 then 1 else if nVladimir Kruchinin, Sep 07 2014 */
    
  • PARI
    {a(n) = if( n<0, 0, polcoeff( prod( k=1, n, 1 + x^k, 1 + x * O(x^n)), n))}; /* Michael Somos, Nov 17 1999 */
    
  • PARI
    {a(n) = my(A); if( n<0, 0, A = x * O(x^n); polcoeff( eta(x^2 + A) / eta(x + A), n))};
    
  • PARI
    {a(n) = my(c); forpart(p=n, if( n<1 || p[1]<2, c++; for(i=1, #p-1, if( p[i+1] > p[i]+1, c--; break)))); c}; /* Michael Somos, Aug 13 2017 */
    
  • PARI
    lista(nn) = {q='q+O('q^nn); Vec(eta(q^2)/eta(q))} \\ Altug Alkan, Mar 20 2018
    
  • Python
    # uses A010815
    from functools import lru_cache
    from math import isqrt
    @lru_cache(maxsize=None)
    def A000009(n): return 1 if n == 0 else A010815(n)+2*sum((-1)**(k+1)*A000009(n-k**2) for k in range(1,isqrt(n)+1)) # Chai Wah Wu, Sep 08 2021
    
  • Python
    import numpy as np
    n = 1000
    arr = np.zeros(n,dtype=object)
    arr[0] = 1
    for i in range(1,n):
        arr[i:] += arr[:n-i]
    print(arr) # Yigit Oktar, Jul 12 2025
    
  • SageMath
    # uses[EulerTransform from A166861]
    a = BinaryRecurrenceSequence(0, 1)
    b = EulerTransform(a)
    print([b(n) for n in range(56)]) # Peter Luschny, Nov 11 2020
    

Formula

G.f.: Product_{m>=1} (1 + x^m) = 1/Product_{m>=0} (1-x^(2m+1)) = Sum_{k>=0} Product_{i=1..k} x^i/(1-x^i) = Sum_{n>=0} x^(n*(n+1)/2) / Product_{k=1..n} (1-x^k).
G.f.: Sum_{n>=0} x^n*Product_{k=1..n-1} (1+x^k) = 1 + Sum_{n>=1} x^n*Product_{k>=n+1} (1+x^k). - Joerg Arndt, Jan 29 2011
Product_{k>=1} (1+x^(2k)) = Sum_{k>=0} x^(k*(k+1))/Product_{i=1..k} (1-x^(2i)) - Euler (Hardy and Wright, Theorem 346).
Asymptotics: a(n) ~ exp(Pi l_n / sqrt(3)) / ( 4 3^(1/4) l_n^(3/2) ) where l_n = (n-1/24)^(1/2) (Ayoub).
For n > 1, a(n) = (1/n)*Sum_{k=1..n} b(k)*a(n-k), with a(0)=1, b(n) = A000593(n) = sum of odd divisors of n; cf. A000700. - Vladeta Jovovic, Jan 21 2002
a(n) = t(n, 0), t as defined in A079211.
a(n) = Sum_{k=0..n-1} A117195(n,k) = A117192(n) + A117193(n) for n>0. - Reinhard Zumkeller, Mar 03 2006
a(n) = A026837(n) + A026838(n) = A118301(n) + A118302(n); a(A001318(n)) = A051044(n); a(A090864(n)) = A118303(n). - Reinhard Zumkeller, Apr 22 2006
Expansion of 1 / chi(-x) = chi(x) / chi(-x^2) = f(-x) / phi(x) = f(x) / phi(-x^2) = psi(x) / f(-x^2) = f(-x^2) / f(-x) = f(-x^4) / psi(-x) in powers of x where phi(), psi(), chi(), f() are Ramanujan theta functions. - Michael Somos, Mar 12 2011
G.f. is a period 1 Fourier series which satisfies f(-1 / (1152 t)) = 2^(-1/2) / f(t) where q = exp(2 Pi i t). - Michael Somos, Aug 16 2007
Expansion of q^(-1/24) * eta(q^2) / eta(q) in powers of q.
Expansion of q^(-1/24) 2^(-1/2) f2(t) in powers of q = exp(2 Pi i t) where f2() is a Weber function. - Michael Somos, Oct 18 2007
Given g.f. A(x), then B(x) = x * A(x^3)^8 satisfies 0 = f(B(x), B(x^2)) where f(u, v) = v - u^2 + 16*u*v^2 . - Michael Somos, May 31 2005
Given g.f. A(x), then B(x) = x * A(x^8)^3 satisfies 0 = f(B(x), B(x^3)) where f(u, v) = (u^3 - v) * (u + v^3) - 9 * u^3 * v^3. - Michael Somos, Mar 25 2008
From Evangelos Georgiadis, Andrew V. Sutherland, Kiran S. Kedlaya (egeorg(AT)mit.edu), Mar 03 2009: (Start)
a(0)=1; a(n) = 2*(Sum_{k=1..floor(sqrt(n))} (-1)^(k+1) a(n-k^2)) + sigma(n) where sigma(n) = (-1)^j if (n=(j*(3*j+1))/2 OR n=(j*(3*j-1))/2) otherwise sigma(n)=0 (simpler: sigma = A010815). (End)
From Gary W. Adamson, Jun 13 2009: (Start)
The product g.f. = (1/(1-x))*(1/(1-x^3))*(1/(1-x^5))*...; = (1,1,1,...)*
(1,0,0,1,0,0,1,0,0,1,...)*(1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,...) * ...; =
a*b*c*... where a, a*b, a*b*c, ... converge to A000009:
1, 1, 1, 2, 2, 2, 3, 3, 3, 4, ... = a*b
1, 1, 1, 2, 2, 3, 4, 4, 5, 6, ... = a*b*c
1, 1, 1, 2, 2, 3, 4, 5, 6, 7, ... = a*b*c*d
1, 1, 1, 2, 2, 3, 4, 5, 6, 8, ... = a*b*c*d*e
1, 1, 1, 2, 2, 3, 4, 5, 6, 8, ... = a*b*c*d*e*f
... (cf. analogous example in A000041). (End)
a(A004526(n)) = A172033(n). - Reinhard Zumkeller, Jan 23 2010
a(n) = P(n) - P(n-2) - P(n-4) + P(n-10) + P(n-14) + ... + (-1)^m P(n-2p_m) + ..., where P(n) is the partition function (A000041) and p_m = m(3m-1)/2 is the m-th generalized pentagonal number (A001318). - Jerome Malenfant, Feb 16 2011
a(n) = A054242(n,0) = A201377(n,0). - Reinhard Zumkeller, Dec 02 2011
More precise asymptotics: a(n) ~ exp(Pi*sqrt((n-1/24)/3)) / (4*3^(1/4)*(n-1/24)^(3/4)) * (1 + (Pi^2-27)/(24*Pi*sqrt(3*(n-1/24))) + (Pi^4-270*Pi^2-1215)/(3456*Pi^2*(n-1/24))). - Vaclav Kotesovec, Nov 30 2015
a(n) = A067661(n) + A067659(n). Wolfdieter Lang, Jan 18 2016
From Vaclav Kotesovec, May 29 2016: (Start)
a(n) ~ exp(Pi*sqrt(n/3))/(4*3^(1/4)*n^(3/4)) * (1 + (Pi/(48*sqrt(3)) - (3*sqrt(3))/(8*Pi))/sqrt(n) + (Pi^2/13824 - 5/128 - 45/(128*Pi^2))/n).
a(n) ~ exp(Pi*sqrt(n/3) + (Pi/(48*sqrt(3)) - 3*sqrt(3)/(8*Pi))/sqrt(n) - (1/32 + 9/(16*Pi^2))/n) / (4*3^(1/4)*n^(3/4)).
(End)
a(n) = A089806(n)*A010815(floor(n/2)) + a(n-1) + a(n-2) - a(n-5) - a(n-7) + a(n-12) + ... + A057077(m-1)*a(n-A001318(m)) + ..., where n > A001318(m). - Gevorg Hmayakyan, Jul 07 2016
a(n) ~ Pi*BesselI(1, Pi*sqrt((n+1/24)/3)) / sqrt(24*n+1). - Vaclav Kotesovec, Nov 08 2016
a(n) = A000041(n) - A047967(n). - R. J. Mathar, Nov 20 2017
Sum_{n>=1} 1/a(n) = A237515. - Amiram Eldar, Nov 15 2020
From Peter Bala, Jan 15 2021: (Start)
G.f.: (1 + x)*Sum_{n >= 0} x^(n*(n+3)/2)/Product_{k = 1..n} (1 - x^k) =
(1 + x)*(1 + x^2)*Sum_{n >= 0} x^(n*(n+5)/2)/Product_{k = 1..n} (1 - x^k) = (1 + x)*(1 + x^2)*(1 + x^3)*Sum_{n >= 0} x^(n*(n+7)/2)/Product_{k = 1..n} (1 - x^k) = ....
G.f.: (1/2)*Sum_{n >= 0} x^(n*(n-1)/2)/Product_{k = 1..n} (1 - x^k) =
(1/2)*(1/(1 + x))*Sum_{n >= 0} x^((n-1)*(n-2)/2)/Product_{k = 1..n} (1 - x^k) = (1/2)*(1/((1 + x)*(1 + x^2)))*Sum_{n >= 0} x^((n-2)*(n-3)/2)/Product_{k = 1..n} (1 - x^k) = ....
G.f.: Sum_{n >= 0} x^n/Product_{k = 1..n} (1 - x^(2*k)) = (1/(1 - x)) * Sum_{n >= 0} x^(3*n)/Product_{k = 1..n} (1 - x^(2*k)) = (1/((1 - x)*(1 - x^3))) * Sum_{n >= 0} x^(5*n)/Product_{k = 1..n} (1 - x^(2*k)) = (1/((1 - x)*(1 - x^3)*(1 - x^5))) * Sum_{n >= 0} x^(7*n)/Product_{k = 1..n} (1 - x^(2*k)) = .... (End)
From Peter Bala, Feb 02 2021: (Start)
G.f.: A(x) = Sum_{n >= 0} x^(n*(2*n-1))/Product_{k = 1..2*n} (1 - x^k). (Set z = x and q = x^2 in Mc Laughlin et al. (2019 ArXiv version), Section 1.3, Identity 7.)
Similarly, A(x) = Sum_{n >= 0} x^(n*(2*n+1))/Product_{k = 1..2*n+1} (1 - x^k). (End)
a(n) = A001227(n) + A238005(n) + A238006(n). - R. J. Mathar, Sep 08 2021
G.f.: A(x) = exp ( Sum_{n >= 1} x^n/(n*(1 - x^(2*n))) ) = exp ( Sum_{n >= 1} (-1)^(n+1)*x^n/(n*(1 - x^n)) ). - Peter Bala, Dec 23 2021
Sum_{n>=0} a(n)/exp(Pi*n) = exp(Pi/24)/2^(1/8) = A292820. - Simon Plouffe, May 12 2023 [Proof: Sum_{n>=0} a(n)/exp(Pi*n) = phi(exp(-2*Pi)) / phi(exp(-Pi)), where phi(q) is the Euler modular function. We have phi(exp(-2*Pi)) = exp(Pi/12) * Gamma(1/4) / (2 * Pi^(3/4)) and phi(exp(-Pi)) = exp(Pi/24) * Gamma(1/4) / (2^(7/8) * Pi^(3/4)), see formulas (14) and (13) in I. Mező, 2013. - Vaclav Kotesovec, May 12 2023]
a(2*n) = Sum_{j=1..n} p(n+j, 2*j) and a(2*n+1) = Sum_{j=1..n+1} p(n+j,2*j-1), where p(n, s) is the number of partitions of n having exactly s parts. - Gregory L. Simay, Aug 30 2023

A025591 Maximal coefficient of Product_{k<=n} (1 + x^k). Number of solutions to +- 1 +- 2 +- 3 +- ... +- n = 0 or 1.

Original entry on oeis.org

1, 1, 1, 2, 2, 3, 5, 8, 14, 23, 40, 70, 124, 221, 397, 722, 1314, 2410, 4441, 8220, 15272, 28460, 53222, 99820, 187692, 353743, 668273, 1265204, 2399784, 4559828, 8679280, 16547220, 31592878, 60400688, 115633260, 221653776, 425363952, 817175698
Offset: 0

Views

Author

Keywords

Comments

If k is allowed to approach infinity, this gives the partition numbers A000009.
a(n) is the maximal number of subsets of {1,2,...,n} that share the same sum.

Crossrefs

Programs

  • Maple
    b:= proc(n, i) option remember; `if`(n>i*(i+1)/2, 0,
          `if`(i=0, 1, b(n+i, i-1)+b(abs(n-i), i-1)))
        end:
    a:=n-> b(0, n)+b(1, n):
    seq(a(n), n=0..40);  # Alois P. Heinz, Mar 10 2014
  • Mathematica
    f[n_, s_] := f[n, s]=Which[n==0, If[s==0, 1, 0], Abs[s]>(n*(n+1))/2, 0, True, f[n-1, s-n]+f[n-1, s+n]]; Table[Which[Mod[n, 4]==0||Mod[n, 4]==3, f[n, 0], Mod[n, 4]==1||Mod[n, 4]==2, f[n, 1]], {n, 0, 40}]
    (* Second program: *)
    p = 1; Flatten[{1, Table[p = Expand[p*(1 + x^n)]; Max[CoefficientList[p, x]], {n, 1, 50}]}] (* Vaclav Kotesovec, May 04 2018 *)
    b[n_, i_] := b[n, i] = If[n > i(i+1)/2, 0, If[i == 0, 1, b[n+i, i-1] + b[Abs[n-i], i-1]]];
    a[n_] := b[0, n] + b[1, n]; a /@ Range[0, 40] (* Jean-François Alcover, Feb 17 2020, after Alois P. Heinz *)
  • PARI
    a(n)=if(n<0,0,polcoeff(prod(k=1,n,1+x^k),n*(n+1)\4))
    
  • Python
    from collections import Counter
    def A025591(n):
        c = {0:1,1:1}
        for i in range(2,n+1):
            d = Counter(c)
            for k in c:
                d[k+i] += c[k]
            c = d
        return max(c.values()) # Chai Wah Wu, Jan 31 2024

Formula

a(n) = A063865(n) + A063866(n).
a(n) ~ sqrt(6/Pi) * 2^n / n^(3/2) [conjectured by Andrica and Tomescu (2002) and proved by Sullivan (2013)]. - Vaclav Kotesovec, Mar 17 2020
More precise asymptotics: a(n) ~ sqrt(6/Pi) * 2^n / n^(3/2) * (1 - 6/(5*n) + 589/(560*n^2) - 39/(50*n^3) + ...). - Vaclav Kotesovec, Dec 30 2022
a(n) = max_{k>=0} A053632(n,k). - Alois P. Heinz, Jan 20 2023

A063865 Number of solutions to +- 1 +- 2 +- 3 +- ... +- n = 0.

Original entry on oeis.org

1, 0, 0, 2, 2, 0, 0, 8, 14, 0, 0, 70, 124, 0, 0, 722, 1314, 0, 0, 8220, 15272, 0, 0, 99820, 187692, 0, 0, 1265204, 2399784, 0, 0, 16547220, 31592878, 0, 0, 221653776, 425363952, 0, 0, 3025553180, 5830034720, 0, 0, 41931984034, 81072032060, 0, 0
Offset: 0

Views

Author

N. J. A. Sloane, suggested by J. H. Conway, Aug 27 2001

Keywords

Comments

Number of sum partitions of the half of the n-th-triangular number by distinct numbers in the range 1 to n. Example: a(7)=8 since triangular(7)=28 and 14 = 2+3+4+5 = 1+3+4+6 = 1+2+5+6 = 3+5+6 = 7+1+2+4 = 7+3+4 = 7+2+5 = 7+1+6. - Hieronymus Fischer, Oct 20 2010
The asymptotic formula below was stated as a conjecture by Andrica & Tomescu in 2002 and proved by B. D. Sullivan in 2013. See his paper and H.-K. Hwang's review MR 2003j:05005 of the JIS paper. - Jonathan Sondow, Nov 11 2013
a(n) is the number of subsets of {1..n} whose sum is equal to the sum of their complement. See example below. - Gus Wiseman, Jul 04 2019

Examples

			From _Gus Wiseman_, Jul 04 2019: (Start)
For example, the a(0) = 1 through a(8) = 14 subsets (empty columns not shown) are:
  {}  {3}    {1,4}  {1,6,7}    {3,7,8}
      {1,2}  {2,3}  {2,5,7}    {4,6,8}
                    {3,4,7}    {5,6,7}
                    {3,5,6}    {1,2,7,8}
                    {1,2,4,7}  {1,3,6,8}
                    {1,2,5,6}  {1,4,5,8}
                    {1,3,4,6}  {1,4,6,7}
                    {2,3,4,5}  {2,3,5,8}
                               {2,3,6,7}
                               {2,4,5,7}
                               {3,4,5,6}
                               {1,2,3,4,8}
                               {1,2,3,5,7}
                               {1,2,4,5,6}
(End)
		

Crossrefs

"Decimations": A060468 = 2*A060005, A123117 = 2*A104456.
Analogous sequences for sums of squares and cubes are A158092, A158118, see also A019568. - Pietro Majer, Mar 15 2009

Programs

  • Maple
    M:=400; t1:=1; lprint(0,1); for n from 1 to M do t1:=expand(t1*(x^n+1/x^n)); lprint(n, coeff(t1,x,0)); od: # N. J. A. Sloane, Jul 07 2008
  • Mathematica
    f[n_, s_] := f[n, s]=Which[n==0, If[s==0, 1, 0], Abs[s]>(n*(n+1))/2, 0, True, f[ n-1, s-n]+f[n-1, s+n]]; a[n_] := f[n, 0]
    nmax = 50; d = {1}; a1 = {};
    Do[
      i = Ceiling[Length[d]/2];
      AppendTo[a1, If[i > Length[d], 0, d[[i]]]];
      d = PadLeft[d, Length[d] + 2 n] + PadRight[d, Length[d] + 2 n];
      , {n, nmax}];
    a1 (* Ray Chandler, Mar 13 2014 *)
  • PARI
    a(n)=my(x='x); polcoeff(prod(k=1,n,x^k+x^-k)+O(x),0) \\ Charles R Greathouse IV, May 18 2015
    
  • PARI
    a(n)=0^n+floor(prod(k=1,n,2^(n*k)+2^(-n*k)))%(2^n) \\ Tani Akinari, Mar 09 2016

Formula

Asymptotic formula: a(n) ~ sqrt(6/Pi)*n^(-3/2)*2^n for n = 0 or 3 (mod 4) as n approaches infinity.
a(n) = 0 unless n == 0 or 3 (mod 4).
a(n) = constant term in expansion of Product_{ k = 1..n } (x^k + 1/x^k). - N. J. A. Sloane, Jul 07 2008
If n = 0 or 3 (mod 4) then a(n) = coefficient of x^(n(n+1)/4) in Product_{k=1..n} (1+x^k). - D. Andrica and I. Tomescu.
a(n) = 2*A058377(n) for any n > 0. - Rémy Sigrist, Oct 11 2017

Extensions

More terms from Dean Hickerson, Aug 28 2001
Corrected and edited by Steven Finch, Feb 01 2009

A358194 Irregular triangle read by rows where T(n,k) is the number of integer partitions of n with partial sums summing to k, where k ranges from n to n(n+1)/2.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, Dec 31 2022

Keywords

Comments

The partial sums of a sequence (a, b, c, ...) are (a, a+b, a+b+c, ...).

Examples

			Triangle begins:
  1
  1
  1 1
  1 0 1 1
  1 0 1 1 0 1 1
  1 0 0 1 1 0 1 1 0 1 1
  1 0 0 1 1 1 1 0 1 1 0 1 1 0 1 1
  1 0 0 0 1 1 1 0 1 1 1 1 1 0 1 1 0 1 1 0 1 1
  1 0 0 0 1 1 1 1 0 1 1 1 2 1 0 1 1 1 1 1 0 1 1 0 1 1 0 1 1
For example, the T(15,59) = 5 partitions are: (8,2,2,2,1), (7,3,3,1,1), (6,5,2,1,1), (4,3,2,2,2,2), (3,3,3,3,2,1).
		

Crossrefs

Row sums are A000041.
The version for compositions is A053632.
Row lengths are A152947.
The version for reversed partitions is A264034.
A048793 = partial sums of reversed standard compositions, sum A029931.
A358134 = partial sums of standard compositions, sum A359042.
A358136 = partial sums of prime indices, sum A318283.
A359361 = partial sums of reversed prime indices, sum A304818.

Programs

  • Mathematica
    Table[Length[Select[IntegerPartitions[n],Total[Accumulate[#]]==k&]],{n,0,8},{k,n,n*(n+1)/2}]

A008967 Coefficients of Gaussian polynomials q_binomial(n-2, 2). Also triangle of distribution of rank sums: Wilcoxon's statistic. Irregular triangle read by rows.

Original entry on oeis.org

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

Views

Author

Keywords

Comments

Rows are numbers of dominoes with k spots where each half-domino has zero to n spots (in standard domino set: n=6, there are 28 dominoes and row is 1,1,2,2,3,3,4,3,3,2,2,1,1). - Henry Bottomley, Aug 23 2000
These numbers appear in the solution of Cayley's counting problem on covariants as N(p,2,w) = [x^p,q^w] Phi(q,x) with the o.g.f. Phi(q,x) = 1/((1-x)(1-qx)(1-q^2x)) given by Peter Bala in the formula section. See the Hawkins reference, p. 264, were also references are given. - Wolfdieter Lang, Nov 30 2012
The entry a(p,w), p >= 0, w = 0,1,...,2*p, of this irregular triangle is the number of nonnegative solutions of m_0 + m_1 + m_2 = p and 1*m_1 + 2*m_2 = w. See the Hawkins reference p. 264, (4.8). N(p,2,w) there is a(p,w). See also the Cayley reference p. 110, 35. with m = 2, Theta = p and q = w. - Wolfdieter Lang, Dec 01 2012
From Gus Wiseman, Sep 20 2023: (Start)
Also the number of unordered pairs of distinct positive integers up to n with sum k. For example, row n = 9 counts the following pairs:
12 13 14 15 16 17 18 19 29 39 49 59 69 79 89
23 24 25 26 27 28 38 48 58 68 78
34 35 36 37 47 57 67
45 46 56
Allowing repeated parts (x,x) gives A004737.
For strict partitions instead of just pairs we have A053632.
(End)

Examples

			1;
1,1,1;
1,1,2,1,1;
1,1,2,2,2,1,1;
1,1,2,2,3,2,2,1,1;
1,1,2,2,3,3,3,2,2,1,1;
...
Partitions: row p=2 and column w=2 has entry 2 because the 2 solutions of the two equations mentioned in a comment above are: m_0 = 0, m_1 = 2, m_2 = 0 and m_0 = 1, m_1 = 0, m_2 = 1. - _Wolfdieter Lang_, Dec 01 2012
		

References

  • G. E. Andrews, The Theory of Partitions, Addison-Wesley, 1976, p. 242.
  • F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 236.
  • T. Hawkins, Emergence of the Theory of Lie Groups, Springer 2000, ch. 7.4, p. 260-5.

Crossrefs

A version with zeros is A219238.
This is the case of A365541 counting only length-2 subsets.

Programs

  • Maple
    qBinom := proc(n,m,q)
            mul((1-q^(n-i))/(1-q^(i+1)),i=0..m-1) ;
            factor(%) ;
            expand(%) ;
    end proc:
    A008967 := proc(n,k)
            coeftayl( qBinom(n,2,q),q=0,k ) ;
    end proc:
    seq(seq( A008967(n,k),k=0..2*n-4),n=2..10) ; # assumes offset 2. R. J. Mathar, Oct 13 2011
  • Mathematica
    rmax = 11; f[r_] := Product[(x^i - x^(r+1))/(1-x^i), {i, 1, r-2}]/  x^((r-1)*(r-2)/2); row[r_] := CoefficientList[ Series[ f[r], {x, 0, 2rmax}], x]; Flatten[ Table[ row[r], {r, 2, rmax}]] (* Jean-François Alcover, Oct 13 2011, after given formula *)
    T[n_, k_] := SeriesCoefficient[QBinomial[n - 2, 2, q], {q, 0, k}];
    Table[T[n, k], {n, 4, 13}, {k, 0, 2 n - 8}] // Flatten (* Jean-François Alcover, Aug 20 2019 *)
    Table[Length[Select[Subsets[Range[n],{2}],Total[#]==k&]],{n,2,15},{k,3,2n-1}] (* Gus Wiseman, Sep 20 2023 *)
  • SageMath
    print(flatten([q_binomial(n-2, 2).list() for n in (4..13)])) # Peter Luschny, Oct 23 2019

Formula

Let f(r) = Product( (x^i-x^(r+1))/(1-x^i), i = 1..r-2) / x^((r-1)*(r-2)/2); then expanding f(r) in powers of x and taking coefficients gives the successive rows of this triangle (with a different offset).
Expanding (q^n - 1)(q^(n+1) - 1)/((q - 1)(q^2 - 1)) in powers of q and taking coefficients gives the n-th row of the triangle. Ordinary generating function: 1/((1-x)(1-qx)(1-q^2x)) = 1 + x(1 + q + q^2) + x^2(1 + q + 2q^2 + q^3 + q^4) + .... - Peter Bala, Sep 23 2007
For n >= 2, let a(n,i) denote the i-th entry of the (n-1)-st row of this triangle; for every 0 <= i <= n-2, a(n,i) = a(n,2(n-2)-i) = ceiling((i+1)/2). - Christian Barrientos, Aug 08 2019

Extensions

More terms from Christian Barrientos, Aug 08 2019

A320387 Number of partitions of n into distinct parts such that the successive differences of consecutive parts are nonincreasing, and first difference <= first part.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 3, 2, 2, 4, 3, 4, 5, 3, 5, 7, 4, 7, 8, 6, 8, 11, 7, 9, 13, 9, 11, 16, 12, 15, 18, 13, 17, 20, 17, 21, 24, 19, 24, 30, 22, 28, 34, 26, 34, 38, 30, 37, 43, 37, 42, 48, 41, 50, 58, 48, 55, 64, 53, 64, 71, 59, 73, 81, 69, 79, 89, 79, 90, 101, 87, 100, 111
Offset: 0

Views

Author

Seiichi Manyama, Oct 12 2018

Keywords

Comments

Partitions are usually written with parts in descending order, but the conditions are easier to check "visually" if written in ascending order.
Generating function of the "second integrals" of partitions: given a partition (p_1, ..., p_s) written in weakly decreasing order, write the sequence B = (b_1, b_2, ..., b_s) = (p_1, p_1 + p_2, ..., p_1 + ... + p_s). The sequence gives the coefficients of the generating function summing q^(b_1 + ... + b_s) over all partitions of all nonnegative integers. - William J. Keith, Apr 23 2022
From Gus Wiseman, Jan 17 2023: (Start)
Equivalently, a(n) is the number of multisets (weakly increasing sequences of positive integers) with weighted sum n. For example, the Heinz numbers of the a(0) = 1 through a(15) = 7 multisets are:
1 2 3 4 7 6 8 10 15 12 16 18 20 26 24 28
5 11 9 17 19 14 21 22 27 41 30 32
13 23 29 31 33 55 39 34
25 35 37 43 45
49 77 47
65
121
These multisets are counted by A264034. The reverse version is A007294. The zero-based version is A359678.
(End)

Examples

			There are a(29) = 15 such partitions of 29:
  01: [29]
  02: [10, 19]
  03: [11, 18]
  04: [12, 17]
  05: [13, 16]
  06: [14, 15]
  07: [5, 10, 14]
  08: [6, 10, 13]
  09: [6, 11, 12]
  10: [7, 10, 12]
  11: [8, 10, 11]
  12: [3, 6, 9, 11]
  13: [5, 7, 8, 9]
  14: [2, 4, 6, 8, 9]
  15: [3, 5, 6, 7, 8]
There are a(30) = 18 such partitions of 30:
  01: [30]
  02: [10, 20]
  03: [11, 19]
  04: [12, 18]
  05: [13, 17]
  06: [14, 16]
  07: [5, 10, 15]
  08: [6, 10, 14]
  09: [6, 11, 13]
  10: [7, 10, 13]
  11: [7, 11, 12]
  12: [8, 10, 12]
  13: [3, 6, 9, 12]
  14: [9, 10, 11]
  15: [4, 7, 9, 10]
  16: [2, 4, 6, 8, 10]
  17: [6, 7, 8, 9]
  18: [4, 5, 6, 7, 8]
		

Crossrefs

Number of appearances of n > 0 in A304818, reverse A318283.
A053632 counts compositions by weighted sum.
A358194 counts partitions by weighted sum, reverse A264034.
Weighted sum of prime indices: A359497, A359676, A359682, A359754, A359755.

Programs

  • Mathematica
    prix[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    ots[y_]:=Sum[i*y[[i]],{i,Length[y]}];
    Table[Length[Select[Range[2^n],ots[prix[#]]==n&]],{n,10}] (* Gus Wiseman, Jan 17 2023 *)
  • PARI
    seq(n)={Vec(sum(k=1, (sqrtint(8*n+1)+1)\2, my(t=binomial(k,2)); x^t/prod(j=1, k-1, 1 - x^(t-binomial(j,2)) + O(x^(n-t+1)))))} \\ Andrew Howroyd, Jan 22 2023
  • Ruby
    def partition(n, min, max)
      return [[]] if n == 0
      [max, n].min.downto(min).flat_map{|i| partition(n - i, min, i - 1).map{|rest| [i, *rest]}}
    end
    def f(n)
      return 1 if n == 0
      cnt = 0
      partition(n, 1, n).each{|ary|
        ary << 0
        ary0 = (1..ary.size - 1).map{|i| ary[i - 1] - ary[i]}
        cnt += 1 if ary0.sort == ary0
      }
      cnt
    end
    def A320387(n)
      (0..n).map{|i| f(i)}
    end
    p A320387(50)
    

Formula

G.f.: Sum_{k>=1} x^binomial(k,2)/Product_{j=1..k-1} (1 - x^(binomial(k,2)-binomial(j,2))). - Andrew Howroyd, Jan 22 2023

A028362 Total number of self-dual binary codes of length 2n. Totally isotropic spaces of index n in symplectic geometry of dimension 2n.

Original entry on oeis.org

1, 3, 15, 135, 2295, 75735, 4922775, 635037975, 163204759575, 83724041661975, 85817142703524375, 175839325399521444375, 720413716161839357604375, 5902349576513949856852644375, 96709997811181068404530578084375
Offset: 1

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Author

Keywords

Comments

These numbers appear in the second column of A155103. - Mats Granvik, Jan 20 2009
a(n) = n terms in the sequence (1, 2, 4, 8, 16, ...) dot n terms in the sequence (1, 1, 3, 15, 135). Example: a(5) = 2295 = (1, 2, 4, 8, 16) dot (1, 1, 3, 15, 135) = (1 + 2 + 12 + 120 + 2160). - Gary W. Adamson, Aug 02 2010

Examples

			G.f. = x + 3*x^2 + 15*x^3 + 135*x^4 + 2295*x^5 + 75735*x^6 + 4922775*x^7 + ...
		

References

  • F. J. MacWilliams and N. J. A. Sloane, The Theory of Error-Correcting Codes, Elsevier-North Holland, 1978, p. 630.

Crossrefs

Cf. A155103. - Mats Granvik, Jan 20 2009
Cf. A005329, A006088. - Paul D. Hanna, Sep 16 2009

Programs

  • Magma
    [1] cat [&*[ 2^k+1: k in [1..n] ]: n in [1..16]]; // Vincenzo Librandi, Dec 24 2015
    
  • Maple
    seq(mul(1 + 2^j, j = 1..n-1), n = 1..20); # G. C. Greubel, Jun 06 2020
  • Mathematica
    Table[Product[2^i+1,{i,n-1}],{n,15}] (* or *) FoldList[Times,1, 2^Range[15]+1] (* Harvey P. Dale, Nov 21 2011 *)
    Table[QPochhammer[-2, 2, n - 1], {n, 15}] (* Arkadiusz Wesolowski, Oct 29 2012 *)
  • PARI
    {a(n)=polcoeff(sum(m=0,n,2^(m*(m-1)/2)*x^m/prod(k=0,m-1,1-2^k*x+x*O(x^n))),n)} \\ Paul D. Hanna, Sep 16 2009
    
  • PARI
    {a(n) = if( n<1, 0 , prod(k=1, n-1, 2^k + 1))}; /* Michael Somos, Jan 28 2018 */
    
  • PARI
    {a(n) = sum(k=0, n-1, 2^(k*(k+1)/2) * prod(j=1, k, (2^(n-j) - 1) / (2^j - 1)))}; /* Michael Somos, Jan 28 2018 */
    
  • Python
    for n in range(2,40,2):
      product = 1
      for i in range(1,n//2-1 + 1):
        product *= (2**i+1)
      print(product)
    # Nathan J. Russell, Mar 01 2016
    
  • Python
    from math import prod
    def A028362(n): return prod((1<Chai Wah Wu, Jun 20 2022
    
  • Sage
    from ore_algebra import *
    R. = QQ['x']
    A. = OreAlgebra(R, 'Qx', q=2)
    print((Qx - x - 1).to_list([0,1], 10))  # Ralf Stephan, Apr 24 2014
    
  • Sage
    from sage.combinat.q_analogues import q_pochhammer
    [q_pochhammer(n-1,-2,2) for n in (1..20)] # G. C. Greubel, Jun 06 2020
    
  • Scheme
    ;; With memoization-macro definec.
    (define (A028362 n) (A028362off0 (- n 1)))
    (definec (A028362off0 n) (if (zero? n) 1 (+ (A028362off0 (- n 1)) (* (expt 2 n) (A028362off0 (- n 1))))))
    ;; Antti Karttunen, Apr 15 2017

Formula

a(n) = Product_{i=1..n-1} (2^i+1).
Letting a(0)=1, we have a(n) = Sum_{k=0..n-1} 2^k*a(k) for n>0. a(n) is asymptotic to c*sqrt(2)^(n^2-n) where c=2.384231029031371724149899288.... = A079555 = Product_{k>=1} (1 + 1/2^k). - Benoit Cloitre, Jan 25 2003
G.f.: Sum_{n>=1} 2^(n*(n-1)/2) * x^n/(Product_{k=0..n-1} (1-2^k*x)). - Paul D. Hanna, Sep 16 2009
a(n) = 2^(binomial(n,2) - 1)*(-1; 1/2){n}, where (a;q){n} is the q-Pochhammer symbol. - G. C. Greubel, Dec 23 2015
From Antti Karttunen, Apr 15 2017: (Start)
a(n) = A048675(A285101(n-1)).
a(n) = b(n-1), where b(0) = 1, and for n > 0, b(n) = b(n-1) + (2^n)*b(n-1).
a(n) = Sum_{i=1..A000124(n-1)} A053632(n-1,i-1)*(2^(i-1)) [where the indexing of both rows and columns of irregular table A053632(row,col) is considered to start from zero].
(End)
G.f. A(x) satisfies: A(x) = x * (1 + A(2*x)) / (1 - x). - Ilya Gutkovskiy, Jun 06 2020
Conjectural o.g.f. as a continued fraction of Stieltjes type (S-fraction):
1/(1 - 3*x/(1 - 2*x/(1 - 10*x/(1 - 12*x/(1 - 36*x/(1 - 56*x/(1 - 136*x/(1 - 240*x/(1 - ... - 2^(n-1)*(2^n + 1)*x/(1 - 2^n*(2^n - 1)*x/(1 - ... ))))))))))). - Peter Bala, Sep 27 2023
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