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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A005836 Numbers whose base-3 representation contains no 2.

Original entry on oeis.org

0, 1, 3, 4, 9, 10, 12, 13, 27, 28, 30, 31, 36, 37, 39, 40, 81, 82, 84, 85, 90, 91, 93, 94, 108, 109, 111, 112, 117, 118, 120, 121, 243, 244, 246, 247, 252, 253, 255, 256, 270, 271, 273, 274, 279, 280, 282, 283, 324, 325, 327, 328, 333, 334, 336, 337, 351, 352
Offset: 1

Views

Author

N. J. A. Sloane, Jeffrey Shallit

Keywords

Comments

3 does not divide binomial(2s, s) if and only if s is a member of this sequence, where binomial(2s, s) = A000984(s) are the central binomial coefficients.
This is the lexicographically earliest increasing sequence of nonnegative numbers that contains no arithmetic progression of length 3. - Robert Craigen (craigenr(AT)cc.umanitoba.ca), Jan 29 2001
In the notation of A185256 this is the Stanley Sequence S(0,1). - N. J. A. Sloane, Mar 19 2010
Complement of A074940. - Reinhard Zumkeller, Mar 23 2003
Sums of distinct powers of 3. - Ralf Stephan, Apr 27 2003
Numbers n such that central trinomial coefficient A002426(n) == 1 (mod 3). - Emeric Deutsch and Bruce E. Sagan, Dec 04 2003
A039966(a(n)+1) = 1; A104406(n) = number of terms <= n.
Subsequence of A125292; A125291(a(n)) = 1 for n>1. - Reinhard Zumkeller, Nov 26 2006
Also final value of n - 1 written in base 2 and then read in base 3 and with finally the result translated in base 10. - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 23 2007
a(n) modulo 2 is the Thue-Morse sequence A010060. - Dennis Tseng, Jul 16 2009
Also numbers such that the balanced ternary representation is the same as the base 3 representation. - Alonso del Arte, Feb 25 2011
Fixed point of the morphism: 0 -> 01; 1 -> 34; 2 -> 67; ...; n -> (3n)(3n+1), starting from a(1) = 0. - Philippe Deléham, Oct 22 2011
It appears that this sequence lists the values of n which satisfy the condition sum(binomial(n, k)^(2*j), k = 0..n) mod 3 <> 0, for any j, with offset 0. See Maple code. - Gary Detlefs, Nov 28 2011
Also, it follows from the above comment by Philippe Lallouet that the sequence must be generated by the rules: a(1) = 0, and if m is in the sequence then so are 3*m and 3*m + 1. - L. Edson Jeffery, Nov 20 2015
Add 1 to each term and we get A003278. - N. J. A. Sloane, Dec 01 2019

Examples

			12 is a term because 12 = 110_3.
This sequence regarded as a triangle with rows of lengths 1, 1, 2, 4, 8, 16, ...:
   0
   1
   3,  4
   9, 10, 12, 13
  27, 28, 30, 31, 36, 37, 39, 40
  81, 82, 84, 85, 90, 91, 93, 94, 108, 109, 111, 112, 117, 118, 120, 121
... - _Philippe Deléham_, Jun 06 2015

References

  • Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section E10, pp. 317-323.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Cf. A039966 (characteristic function).
Cf. A002426, A004793, A005823, A007088, A007089, A032924, A033042-A033052, A054591, A055246, A062548, A065361, A074940, A081601, A081603, A081611, A083096, A089118, A121153, A170943, A185256.
For generating functions Product_{k>=0} (1+a*x^(b^k)) for the following values of (a,b) see: (1,2) A000012 and A000027, (1,3) A039966 and A005836, (1,4) A151666 and A000695, (1,5) A151667 and A033042, (2,2) A001316, (2,3) A151668, (2,4) A151669, (2,5) A151670, (3,2) A048883, (3,3) A117940, (3,4) A151665, (3,5) A151671, (4,2) A102376, (4,3) A151672, (4,4) A151673, (4,5) A151674.
Row 3 of array A104257.
Summary of increasing sequences avoiding arithmetic progressions of specified lengths (the second of each pair is obtained by adding 1 to the first):
3-term AP: A005836 (>=0), A003278 (>0);
4-term AP: A005839 (>=0), A005837 (>0);
5-term AP: A020654 (>=0), A020655 (>0);
6-term AP: A020656 (>=0), A005838 (>0);
7-term AP: A020657 (>=0), A020658 (>0);
8-term AP: A020659 (>=0), A020660 (>0);
9-term AP: A020661 (>=0), A020662 (>0);
10-term AP: A020663 (>=0), A020664 (>0).
See also A000452.

Programs

  • Haskell
    a005836 n = a005836_list !! (n-1)
a005836_list = filter ((== 1) . a039966) [0..]
-- Reinhard Zumkeller, Jun 09 2012, Sep 29 2011
  • Julia
    function a(n)
    m, r, b = n, 0, 1
    while m > 0
        m, q = divrem(m, 2)
        r += b * q
        b *= 3
    end
r end; [a(n) for n in 0:57] |> println # Peter Luschny, Jan 03 2021
  • Maple
    t := (j, n) -> add(binomial(n,k)^j, k=0..n):
for i from 1 to 400 do
    if(t(4,i) mod 3 <>0) then print(i) fi
od; # Gary Detlefs, Nov 28 2011
# alternative Maple program:
a:= proc(n) option remember: local k, m:
if n=1 then 0 elif n=2 then 1 elif n>2 then k:=floor(log[2](n-1)): m:=n-2^k: procname(m)+3^k: fi: end proc:
seq(a(n), n=1.. 20); # Paul Weisenhorn, Mar 22 2020
# third Maple program:
a:= n-> `if`(n=1, 0, irem(n-1, 2, 'q')+3*a(q+1)):
seq(a(n), n=1..100);  # Alois P. Heinz, Jan 26 2022
  • Mathematica
    Table[FromDigits[IntegerDigits[k, 2], 3], {k, 60}]
Select[Range[0, 400], DigitCount[#, 3, 2] == 0 &] (* Harvey P. Dale, Jan 04 2012 *)
Join[{0}, Accumulate[Table[(3^IntegerExponent[n, 2] + 1)/2, {n, 57}]]] (* IWABUCHI Yu(u)ki, Aug 01 2012 *)
FromDigits[#,3]&/@Tuples[{0,1},7] (* Harvey P. Dale, May 10 2019 *)
  • PARI
    A=vector(100);for(n=2,#A,A[n]=if(n%2,3*A[n\2+1],A[n-1]+1));A \\ Charles R Greathouse IV, Jul 24 2012
  • PARI
    is(n)=while(n,if(n%3>1,return(0));n\=3);1 \\ Charles R Greathouse IV, Mar 07 2013
  • PARI
    a(n) = fromdigits(binary(n-1),3);  \\ Gheorghe Coserea, Jun 15 2018
  • Python
    def A005836(n):
    return int(format(n-1,'b'),3) # Chai Wah Wu, Jan 04 2015

Formula

a(n) = A005823(n)/2 = A003278(n)-1 = A033159(n)-2 = A033162(n)-3.
Numbers n such that the coefficient of x^n is > 0 in prod (k >= 0, 1 + x^(3^k)). - Benoit Cloitre, Jul 29 2003
a(n+1) = Sum_{k=0..m} b(k)* 3^k and n = Sum( b(k)* 2^k ).
a(2n+1) = 3a(n+1), a(2n+2) = a(2n+1) + 1, a(0) = 0.
a(n+1) = 3*a(floor(n/2)) + n - 2*floor(n/2). - Ralf Stephan, Apr 27 2003
G.f.: (x/(1-x)) * Sum_{k>=0} 3^k*x^2^k/(1+x^2^k). - Ralf Stephan, Apr 27 2003
a(n) = Sum_{k = 1..n-1} (1 + 3^A007814(k)) / 2. - Philippe Deléham, Jul 09 2005
From Reinhard Zumkeller, Mar 02 2008: (Start)
A081603(a(n)) = 0.
If the offset were changed to zero, then: a(0) = 0, a(n+1) = f(a(n)+1, a(n)+1) where f(x, y) = if x < 3 and x <> 2 then y else if x mod 3 = 2 then f(y+1, y+1) else f(floor(x/3), y). (End)
With offset a(0) = 0: a(n) = Sum_{k>=0} A030308(n,k)*3^k. - Philippe Deléham, Oct 15 2011
a(2^n) = A003462(n). - Philippe Deléham, Jun 06 2015
We have liminf_{n->infinity} a(n)/n^(log(3)/log(2)) = 1/2 and limsup_{n->infinity} a(n)/n^(log(3)/log(2)) = 1. - Gheorghe Coserea, Sep 13 2015
a(2^k+m) = a(m) + 3^k with 1 <= m <= 2^k and 1 <= k, a(1)=0, a(2)=1. - Paul Weisenhorn, Mar 22 2020
Sum_{n>=2} 1/a(n) = 2.682853110966175430853916904584699374821677091415714815171756609672281184705... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Feb 12 2022
A065361(a(n)) = n-1. - Rémy Sigrist, Feb 06 2023
a(n) ≍ n^k, where k = log 3/log 2 = 1.5849625007. (I believe the constant varies from 1/2 to 1.) - Charles R Greathouse IV, Mar 29 2024

Extensions

Offset corrected by N. J. A. Sloane, Mar 02 2008
Edited by the Associate Editors of the OEIS, Apr 07 2009

A001316 Gould's sequence: a(n) = Sum_{k=0..n} (binomial(n,k) mod 2); number of odd entries in row n of Pascal's triangle (A007318); a(n) = 2^A000120(n).

Original entry on oeis.org

1, 2, 2, 4, 2, 4, 4, 8, 2, 4, 4, 8, 4, 8, 8, 16, 2, 4, 4, 8, 4, 8, 8, 16, 4, 8, 8, 16, 8, 16, 16, 32, 2, 4, 4, 8, 4, 8, 8, 16, 4, 8, 8, 16, 8, 16, 16, 32, 4, 8, 8, 16, 8, 16, 16, 32, 8, 16, 16, 32, 16, 32, 32, 64, 2, 4, 4, 8, 4, 8, 8, 16, 4, 8, 8, 16, 8, 16, 16, 32, 4, 8, 8, 16, 8, 16, 16, 32
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Also called Dress's sequence.
This sequence might be better called Glaisher's sequence, since James Glaisher showed that odd binomial coefficients are counted by 2^A000120(n) in 1899. - Eric Rowland, Mar 17 2017 [However, the name "Gould's sequence" is deeply entrenched in the literature. - N. J. A. Sloane, Mar 17 2017] [Named after the American mathematician Henry Wadsworth Gould (b. 1928). - Amiram Eldar, Jun 19 2021]
All terms are powers of 2. The first occurrence of 2^k is at n = 2^k - 1; e.g., the first occurrence of 16 is at n = 15. - Robert G. Wilson v, Dec 06 2000
a(n) is the highest power of 2 dividing binomial(2n,n) = A000984(n). - Benoit Cloitre, Jan 23 2002
Also number of 1's in n-th row of triangle in A070886. - Hans Havermann, May 26 2002. Equivalently, number of live cells in generation n of a one-dimensional cellular automaton, Rule 90, starting with a single live cell. - Ben Branman, Feb 28 2009. Ditto for Rule 18. - N. J. A. Sloane, Aug 09 2014. This is also the odd-rule cellular automaton defined by OddRule 003 (see Ekhad-Sloane-Zeilberger "Odd-Rule Cellular Automata on the Square Grid" link). - N. J. A. Sloane, Feb 25 2015
Also number of numbers k, 0<=k<=n, such that (k OR n) = n (bitwise logical OR): a(n) = #{k : T(n,k)=n, 0<=k<=n}, where T is defined as in A080098. - Reinhard Zumkeller, Jan 28 2003
To construct the sequence, start with 1 and use the rule: If k >= 0 and a(0),a(1),...,a(2^k-1) are the first 2^k terms, then the next 2^k terms are 2*a(0),2*a(1),...,2*a(2^k-1). - Benoit Cloitre, Jan 30 2003
Also, numerator((2^k)/k!). - Mohammed Bouayoun (mohammed.bouayoun(AT)sanef.com), Mar 03 2004
The odd entries in Pascal's triangle form the Sierpiński Gasket (a fractal). - Amarnath Murthy, Nov 20 2004
Row sums of Sierpiński's Gasket A047999. - Johannes W. Meijer, Jun 05 2011
Fixed point of the morphism "1" -> "1,2", "2" -> "2,4", "4" -> "4,8", ..., "2^k" -> "2^k,2^(k+1)", ... starting with a(0) = 1; 1 -> 12 -> 1224 -> = 12242448 -> 122424482448488(16) -> ... . - Philippe Deléham, Jun 18 2005
a(n) = number of 1's of stage n of the one-dimensional cellular automaton with Rule 90. - Andras Erszegi (erszegi.andras(AT)chello.hu), Apr 01 2006
a(33)..a(63) = A117973(1)..A117973(31). - Stephen Crowley, Mar 21 2007
Or the number of solutions of the equation: A000120(x) + A000120(n-x) = A000120(n). - Vladimir Shevelev, Jul 19 2009
For positive n, a(n) equals the denominator of the permanent of the n X n matrix consisting entirely of (1/2)'s. - John M. Campbell, May 26 2011
Companions to A001316 are A048896, A105321, A117973, A151930 and A191488. They all have the same structure. We observe that for all these sequences a((2*n+1)*2^p-1) = C(p)*A001316(n), p >= 0. If C(p) = 2^p then a(n) = A001316(n), if C(p) = 1 then a(n) = A048896(n), if C(p) = 2^p+2 then a(n) = A105321(n+1), if C(p) = 2^(p+1) then a(n) = A117973(n), if C(p) = 2^p-2 then a(n) = (-1)*A151930(n) and if C(p) = 2^(p+1)+2 then a(n) = A191488(n). Furthermore for all a(2^p - 1) = C(p). - Johannes W. Meijer, Jun 05 2011
a(n) = number of zeros in n-th row of A219463 = number of ones in n-th row of A047999. - Reinhard Zumkeller, Nov 30 2012
This is the Run Length Transform of S(n) = {1,2,4,8,16,...} (cf. A000079). The Run Length Transform of a sequence {S(n), n>=0} is defined to be the sequence {T(n), n>=0} given by T(n) = Product_i S(i), where i runs through the lengths of runs of 1's in the binary expansion of n. E.g., 19 is 10011 in binary, which has two runs of 1's, of lengths 1 and 2. So T(19) = S(1)*S(2). T(0)=1 (the empty product). - N. J. A. Sloane, Sep 05 2014
A105321(n+1) = a(n+1) + a(n). - Reinhard Zumkeller, Nov 14 2014
a(n) = A261363(n,n) = number of distinct terms in row n of A261363 = number of odd terms in row n+1 of A261363. - Reinhard Zumkeller, Aug 16 2015
From Gary W. Adamson, Aug 26 2016: (Start)
A production matrix for the sequence is lim_{k->infinity} M^k, the left-shifted vector of M:
1, 0, 0, 0, 0, ...
2, 0, 0, 0, 0, ...
0, 1, 0, 0, 0, ...
0, 2, 0, 0, 0, ...
0, 0, 1, 0, 0, ...
0, 0, 2, 0, 0, ...
0, 0, 0, 1, 0, ...
...
The result is equivalent to the g.f. of Apr 06 2003: Product_{k>=0} (1 + 2*z^(2^k)). (End)
Number of binary palindromes of length n for which the first floor(n/2) symbols are themselves a palindrome (Ji and Wilf 2008). - Jeffrey Shallit, Jun 15 2017

Examples

			Has a natural structure as a triangle:
  1,
  2,
  2,4,
  2,4,4,8,
  2,4,4,8,4,8,8,16,
  2,4,4,8,4,8,8,16,4,8,8,16,8,16,16,32,
  2,4,4,8,4,8,8,16,4,8,8,16,8,16,16,32,4,8,8,16,8,16,16,32,8,16,16,32,16,32,32,64,
  ...
The rows converge to A117973.
From _Omar E. Pol_, Jun 07 2009: (Start)
Also, triangle begins:
   1;
   2,2;
   4,2,4,4;
   8,2,4,4,8,4,8,8;
  16,2,4,4,8,4,8,8,16,4,8,8,16,8,16,16;
  32,2,4,4,8,4,8,8,16,4,8,8,16,8,16,16,32,4,8,8,16,8,16,16,32,8,16,16,32,16,32,32;
  64,2,4,4,8,4,8,8,16,4,8,8,16,8,16,16,32,4,8,8,16,8,16,16,32,8,16,16,32,16,32,...
(End)
G.f. = 1 + 2*x + 2*x^2 + 4*x^3 + 2*x^4 + 4*x^5 + 4*x^6 + 8*x^7 + 2*x^8 + ... - _Michael Somos_, Aug 26 2015

References

  • Arthur T. Benjamin and Jennifer J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A., 2003, p. 75ff.
  • Steven R. Finch, Mathematical Constants, Cambridge, 2003, pp. 145-151.
  • James W. L. Glaisher, On the residue of a binomial-theorem coefficient with respect to a prime modulus, Quarterly Journal of Pure and Applied Mathematics, Vol. 30 (1899), pp. 150-156.
  • H. W. Gould, Exponential Binomial Coefficient Series. Tech. Rep. 4, Math. Dept., West Virginia Univ., Morgantown, WV, Sep 1961.
  • Olivier Martin, Andrew M. Odlyzko, and Stephen Wolfram, Algebraic properties of cellular automata, Comm. Math. Physics, Vol. 93 (1984), pp. 219-258. Reprinted in Theory and Applications of Cellular Automata, S Wolfram, Ed., World Scientific, 1986, pp. 51-90 and in Cellular Automata and Complexity: Collected Papers of Stephen Wolfram, Addison-Wesley, 1994, pp. 71-113
  • Manfred R. Schroeder, Fractals, Chaos, Power Laws, W. H. Freeman, NY, 1991, page 383.
  • 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).
  • Andrew Wuensche, Exploring Discrete Dynamics, Luniver Press, 2011. See Fig. 2.3.

Links

Crossrefs

Equals left border of triangle A166548. - Gary W. Adamson, Oct 16 2009
For generating functions Product_{k>=0} (1+a*x^(b^k)) for the following values of (a,b) see: (1,2) A000012 and A000027, (1,3) A039966 and A005836, (1,4) A151666 and A000695, (1,5) A151667 and A033042, (2,2) A001316, (2,3) A151668, (2,4) A151669, (2,5) A151670, (3,2) A048883, (3,3) A117940, (3,4) A151665, (3,5) A151671, (4,2) A102376, (4,3) A151672, (4,4) A151673, (4,5) A151674.
For partial sums see A006046. For first differences see A151930.
This is the numerator of 2^n/n!, while A049606 gives the denominator.
Cf. A051638, A048967, A007318, A094959, A048896, A117973, A008977, A139541, A048883, A102376, A038573, A159913, A000079, A166548, A006047, A089898, A105321, A061142.
Cf. A047999, A261363, A261366.
If we subtract 1 from the terms we get a pair of essentially identical sequences, A038573 and A159913.
A163000 and A163577 count binomial coefficients with 2-adic valuation 1 and 2. A275012 gives a measure of complexity of these sequences. - Eric Rowland, Mar 15 2017
Cf. A286575 (run-length transform), A368655 (binomial transform), also A037445.

Programs

  • Haskell
    import Data.List (transpose)
a001316 = sum . a047999_row  -- Reinhard Zumkeller, Nov 24 2012
a001316_list = 1 : zs where
   zs = 2 : (concat $ transpose [zs, map (* 2) zs])
-- Reinhard Zumkeller, Aug 27 2014, Sep 16 2011
(Sage, Python)
from functools import cache
@cache
def A001316(n):
    if n <= 1: return n+1
    return A001316(n//2) << n%2
print([A001316(n) for n in range(88)])  # Peter Luschny, Nov 19 2012
  • Maple
    A001316 := proc(n) local k; add(binomial(n,k) mod 2, k=0..n); end;
S:=[1]; S:=[op(S),op(2*s)]; # repeat ad infinitum!
a := n -> 2^add(i,i=convert(n,base,2)); # Peter Luschny, Mar 11 2009
  • Mathematica
    Table[ Sum[ Mod[ Binomial[n, k], 2], {k, 0, n} ], {n, 0, 100} ]
Nest[ Join[#, 2#] &, {1}, 7] (* Robert G. Wilson v, Jan 24 2006 and modified Jul 27 2014 *)
Map[Function[Apply[Plus,Flatten[ #1]]], CellularAutomaton[90,{{1},0},100]] (* Produces counts of ON cells. N. J. A. Sloane, Aug 10 2009 *)
ArrayPlot[CellularAutomaton[90, {{1}, 0}, 20]] (* Illustration of first 20 generations. - N. J. A. Sloane, Aug 14 2014 *)
Table[2^(RealDigits[n - 1, 2][[1]] // Total), {n, 1, 100}] (* Gabriel C. Benamy, Dec 08 2009 *)
CoefficientList[Series[Exp[2*x], {x, 0, 100}], x] // Numerator (* Jean-François Alcover, Oct 25 2013 *)
Count[#,?OddQ]&/@Table[Binomial[n,k],{n,0,90},{k,0,n}] (* _Harvey P. Dale, Sep 22 2015 *)
2^DigitSum[Range[0, 100], 2] (* Paolo Xausa, Jul 31 2025 *)
  • PARI
    {a(n) = if( n<0, 0, numerator(2^n / n!))};
  • PARI
    A001316(n)=1<M. F. Hasler, May 03 2009
  • PARI
    a(n)=2^hammingweight(n) \\ Charles R Greathouse IV, Jan 04 2013
  • Python
    def A001316(n):
    return 2**bin(n)[2:].count("1") # Indranil Ghosh, Feb 06 2017
  • Python
    def A001316(n): return 1<Karl-Heinz Hofmann, Aug 01 2025
  • Python
    import numpy # (version >= 2.0.0)
n_up_to = 2**22
A000079 = 1 << numpy.arange(n_up_to.bit_length())
A001316 = A000079[numpy.bitwise_count(numpy.arange(n_up_to))]
print(A001316[0:100]) # Karl-Heinz Hofmann, Aug 01 2025
  • Scheme
    (define (A001316 n) (let loop ((n n) (z 1)) (cond ((zero? n) z) ((even? n) (loop (/ n 2) z)) (else (loop (/ (- n 1) 2) (* z 2)))))) ;; Antti Karttunen, May 29 2017

Formula

a(n) = 2^A000120(n).
a(0) = 1; for n > 0, write n = 2^i + j where 0 <= j < 2^i; then a(n) = 2*a(j).
a(n) = 2*a(n-1)/A006519(n) = A000079(n)*A049606(n)/A000142(n).
a(n) = A038573(n) + 1.
G.f.: Product_{k>=0} (1+2*z^(2^k)). - Ralf Stephan, Apr 06 2003
a(n) = Sum_{i=0..2*n} (binomial(2*n, i) mod 2)*(-1)^i. - Benoit Cloitre, Nov 16 2003
a(n) mod 3 = A001285(n). - Benoit Cloitre, May 09 2004
a(n) = 2^n - 2*Sum_{k=0..n} floor(binomial(n, k)/2). - Paul Barry, Dec 24 2004
a(n) = Product_{k=0..log_2(n)} 2^b(n, k), b(n, k) = coefficient of 2^k in binary expansion of n. - Paul D. Hanna
Sum_{k=0..n-1} a(k) = A006046(n).
a(n) = n/2 + 1/2 + (1/2)*Sum_{k=0..n} (-(-1)^binomial(n,k)). - Stephen Crowley, Mar 21 2007
G.f. for a(n)/A156769(n): (1/2)*z^(1/2)*sinh(2*z^(1/2)). - Johannes W. Meijer, Feb 20 2009
Equals infinite convolution product of [1,2,0,0,0,0,0,0,0] aerated (A000079 - 1) times, i.e., [1,2,0,0,0,0,0,0,0] * [1,0,2,0,0,0,0,0,0] * [1,0,0,0,2,0,0,0,0]. - Mats Granvik, Gary W. Adamson, Oct 02 2009
a(n) = f(n, 1) with f(x, y) = if x = 0 then y otherwise f(floor(x/2), y*(1 + x mod 2)). - Reinhard Zumkeller, Nov 21 2009
a(n) = 2^(number of 1's in binary form of (n-1)). - Gabriel C. Benamy, Dec 08 2009
a((2*n+1)*2^p-1) = (2^p)*a(n), p >= 0. - Johannes W. Meijer, Jun 05 2011
a(n) = A000120(A001317(n)). - Reinhard Zumkeller, Nov 24 2012
a(n) = A226078(n,1). - Reinhard Zumkeller, May 25 2013
a(n) = lcm(n!, 2^n) / n!. - Daniel Suteu, Apr 28 2017
a(n) = A061142(A005940(1+n)). - Antti Karttunen, May 29 2017
a(0) = 1, a(2*n) = a(n), a(2*n+1) = 2*a(n). - Daniele Parisse, Feb 15 2024
a(n*m) <= a(n)^A000120(m). - Joe Amos, Mar 27 2025

Extensions

Additional comments from Henry Bottomley, Mar 12 2001
Further comments from N. J. A. Sloane, May 30 2009

A001850 Central Delannoy numbers: a(n) = Sum_{k=0..n} C(n,k)*C(n+k,k).

Original entry on oeis.org

1, 3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719, 251595969, 1409933619, 7923848253, 44642381823, 252055236609, 1425834724419, 8079317057869, 45849429914943, 260543813797441, 1482376214227923, 8443414161166173, 48141245001931263
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Number of paths from (0,0) to (n,n) in an n X n grid using only steps north, northeast and east (i.e., steps (1,0), (1,1), and (0,1)).
Also the number of ways of aligning two sequences (e.g., of nucleotides or amino acids) of length n, with at most 2*n gaps (-) inserted, so that while unnecessary gappings: - -a a- - are forbidden, both b- and -b are allowed. (If only other of the latter is allowed, then the sequence A000984 gives the number of alignments.) There is an easy bijection from grid walks given by Dickau to such set of alignments (e.g., the straight diagonal corresponds to the perfect alignment with no gaps). - Antti Karttunen, Oct 10 2001
Also main diagonal of array A008288 defined by m(i,1) = m(1,j) = 1, m(i,j) = m(i-1,j-1) + m(i-1,j) + m(i,j-1). - Benoit Cloitre, May 03 2002
So, as a special case of Dmitry Zaitsev's Dec 10 2015 comment on A008288, a(n) is the number of points in Z^n that are L1 (Manhattan) distance <= n from any given point. These terms occur in the crystal ball sequences: a(n) here is the n-th term in the sequence for the n-dimensional cubic lattice. See A008288 for a list of crystal ball sequences (rows or columns of A008288). - Shel Kaphan, Dec 26 2022
a(n) is the number of n-matchings of a comb-like graph with 2*n teeth. Example: a(2) = 13 because the graph consisting of a horizontal path ABCD and the teeth Aa, Bb, Cc, Dd has 13 2-matchings: any of the six possible pairs of teeth and {Aa, BC}, {Aa, CD}, {Bb, CD}, {Cc, AB}, {Dd, AB}, {Dd, BC}, {AB, CD}. - Emeric Deutsch, Jul 02 2002
Number of ordered trees with 2*n+1 edges, having root of odd degree, nonroot nodes of outdegree at most 2 and branches of odd length. - Emeric Deutsch, Aug 02 2002
The sum of the first n coefficients of ((1 - x) / (1 - 2*x))^n is a(n-1). - Michael Somos, Sep 28 2003
Row sums of A063007 and A105870. - Paul Barry, Apr 23 2005
The Hankel transform (see A001906 for definition) of this sequence is A036442: 1, 4, 32, 512, 16384, ... . - Philippe Deléham, Jul 03 2005
Also number of paths from (0,0) to (n,0) using only steps U = (1,1), H = (1,0) and D =(1,-1), U can have 2 colors and H can have 3 colors. - N-E. Fahssi, Jan 27 2008
Equals row sums of triangle A152250 and INVERT transform of A109980: (1, 2, 8, 36, 172, 852, ...). - Gary W. Adamson, Nov 30 2008
Number of overpartitions in the n X n box (treat a walk of the type in the first comment as an overpartition, by interpreting a NE step as N, E with the part thus created being overlined). - William J. Keith, May 19 2017
Diagonal of rational functions 1/(1 - x - y - x*y), 1/(1 - x - y*z - x*y*z). - Gheorghe Coserea, Jul 03 2018
Dimensions of endomorphism algebras End(R^{(n)}) in the Delannoy category attached to the oligomorphic group of order preserving self-bijections of the real line. - Noah Snyder, Mar 22 2023
a(n) is the number of ways to tile a strip of length n with white squares, black squares, and red dominos, where we must have an equal number of white and black squares. - Greg Dresden and Leo Zhang, Jul 11 2025

Examples

			G.f. = 1 + 3*x + 13*x^2 + 63*x^3 + 321*x^4 + 1683*x^5 + 8989*x^6 + ...

References

  • Frits Beukers, Arithmetic properties of Picard-Fuchs equations, Séminaire de Théorie des nombres de Paris, 1982-83, Birkhäuser Boston, Inc.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 593.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 81.
  • L. Moser and W. Zayachkowski, Lattice paths with diagonal steps, Scripta Math., 26 (1961), 223-229.
  • 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, Wadsworth, Vol. 2, 1999; see Example 6.3.8 and Problem 6.49.
  • D. B. West, Combinatorial Mathematics, Cambridge, 2021, p. 28.

Links

Crossrefs

Cf. A008288, bisection of A026003, A027618, A047665, A052141, A084773, A152250, A109980, A000129, A078057, A241023, A243949.
Main diagonal of A064861.
Column k=2 of A262809 and A263159.

Programs

  • Maple
    seq(add(multinomial(n+k,n-k,k,k),k=0..n),n=0..20); # Zerinvary Lajos, Oct 18 2006
seq(orthopoly[P](n,3), n=0..100); # Robert Israel, Nov 03 2015
  • Mathematica
    f[n_] := Sum[ Binomial[n, k] Binomial[n + k, k], {k, 0, n}]; Array[f, 21, 0] (* Or *)
a[0] = 1; a[1] = 3; a[n_] := a[n] = (3(2 n - 1)a[n - 1] - (n - 1)a[n - 2])/n; Array[a, 21, 0] (* Or *)
CoefficientList[ Series[1/Sqrt[1 - 6x + x^2], {x, 0, 20}], x] (* Robert G. Wilson v *)
Table[LegendreP[n, 3], {n, 0, 22}] (* Jean-François Alcover, Jul 16 2012, from first formula *)
a[n_] := Hypergeometric2F1[-n, n+1, 1, -1]; Table[a[n], {n, 0, 22}] (* Jean-François Alcover, Feb 26 2013 *)
a[ n_] := With[ {m = If[n < 0, -1 - n, n]}, SeriesCoefficient[ (1 - 6 x + x^2)^(-1/2), {x, 0, m}]]; (* Michael Somos, Jun 10 2015 *)
  • Maxima
    a(n):=coeff(expand((1+3*x+2*x^2)^n),x,n);
makelist(a(n),n,0,12); /* Emanuele Munarini, Mar 02 2011 */
  • PARI
    {a(n) = if( n<0, n = -1 - n); polcoeff( 1 / sqrt(1 - 6*x + x^2 + x * O(x^n)), n)}; /* Michael Somos, Sep 23 2006 */
  • PARI
    {a(n) = if( n<0, n = -1 - n); subst( pollegendre(n), x, 3)}; /* Michael Somos, Sep 23 2006 */
  • PARI
    {a(n) = if( n<0, n = -1 - n); n++; subst( Pol(((1 - x) / (1 - 2*x) + O(x^n))^n), x, 1);} /* Michael Somos, Sep 23 2006 */
  • PARI
    a(n)=if(n<0, 0, polcoeff((1+3*x+2*x^2)^n, n)) \\ Paul Barry, Aug 22 2007
  • PARI
    /* same as in A092566 but use */
steps=[[1,0], [0,1], [1,1]]; /* Joerg Arndt, Jun 30 2011 */
  • PARI
    a(n)=sum(k=0,n,binomial(n,k)*binomial(n+k,k)); \\ Joerg Arndt, May 11 2013
  • PARI
    my(x='x+O('x^30)); Vec(1/sqrt(1 - 6*x + x^2)) \\ Altug Alkan, Oct 17 2015
  • Python
    # from Nick Hobson.
def f(a, b):
    if a == 0 or b == 0:
        return 1
    return f(a, b - 1) + f(a - 1, b) + f(a - 1, b - 1)
[f(n, n) for n in range(7)]
  • Python
    from gmpy2 import divexact
A001850 = [1, 3]
for n in range(2,10**3):
    A001850.append(divexact(A001850[-1]*(6*n-3)-(n-1)*A001850[-2],n))
# Chai Wah Wu, Sep 01 2014
  • Sage
    a = lambda n: hypergeometric([-n, -n], [1], 2)
[simplify(a(n)) for n in range(23)] # Peter Luschny, Nov 19 2014

Formula

a(n) = P_n(3), where P_n is n-th Legendre polynomial.
G.f.: 1 / sqrt(1 - 6*x + x^2).
a(n) = a(n-1) + 2*A002002(n) = Sum_{j} A063007(n, j). - Henry Bottomley, Jul 02 2001
Dominant term in asymptotic expansion is binomial(2*n, n)/2^(1/4)*((sqrt(2) + 1)/2)^(2*n + 1)*(1 + c_1/n + c_2/n^2 + ...). - Michael David Hirschhorn
a(n) = Sum_{i=0..n} (A000079(i)*A008459(n, i)) = Sum_{i=0..n} (2^i * C(n, i)^2). - Antti Karttunen, Oct 10 2001
a(n) = Sum_{k=0..n} C(n+k, n-k)*C(2*k, k). - Benoit Cloitre, Feb 13 2003
a(n) = Sum_{k=0..n} C(n, k)^2 * 2^k. - Michael Somos, Oct 08 2003
a(n - 1) = coefficient of x^n in A120588(x)^n if n>=0. - Michael Somos, Apr 11 2012
G.f. of a(n-1) = 1 / (1 - x / (1 - 2*x / (1 - 2*x / (1 - x / (1 - 2*x / (1 - x / ...)))))). - Michael Somos, May 11 2012
INVERT transform is A109980. BINOMIAL transform is A080609. BINOMIAL transform of A006139. PSUM transform is A089165. PSUMSIGN transform is A026933. First backward difference is A110170. - Michael Somos, May 11 2012
E.g.f.: exp(3*x)*BesselI(0, 2*sqrt(2)*x). - Vladeta Jovovic, Mar 21 2004
a(n) = Sum_{k=0..n} C(2*n-k, n)*C(n, k). - Paul Barry, Apr 23 2005
a(n) = Sum_{k>=n} binomial(k, n)^2/2^(k+1). - Vladeta Jovovic, Aug 25 2006
a(n) = a(-1 - n) for all n in Z. - Michael Somos, Sep 23 2006
D-finite with recurrence: a(-1) = a(0) = 1; n*a(n) = 3*(2*n-1)*a(n-1) - (n-1)*a(n-2). Eq (4) in T. D. Noe's article in JIS 9 (2006) #06.2.7.
Define general Delannoy numbers by (i,j > 0): d(i,0) = d(0,j) = 1 =: d(0,0) and d(i,j) = d(i-1,j-1) + d(i-2,j-1) + d(i-1,j). Then a(k) = Sum_{j >= 0} d(k,j)^2 + d(k-1,j)^2 = A026933(n)+A026933(n-1). This is a special case of the following formula for general Delannoy numbers: d(k,j) = Sum_{i >= 0, p=0..n} d(p, i) * d(n-p, j-i) + d(p-1, i) * d(n-p-1, j-i-1). - Peter E John, Oct 19 2006
Coefficient of x^n in (1 + 3*x + 2*x^2)^n. - N-E. Fahssi, Jan 11 2008
a(n) = A008288(A046092(n)). - Philippe Deléham, Apr 08 2009
G.f.: 1/(1 - x - 2*x/(1 - x - x/(1 - x - x/(1 - x - x/(1 - ... (continued fraction). - Paul Barry, May 28 2009
G.f.: d/dx log(1/(1 - x*A001003(x))). - Vladimir Kruchinin, Apr 19 2011
G.f.: 1/(2*Q(0) + x - 1) where Q(k) = 1 + k*(1-x) - x - x*(k + 1)*(k + 2)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Mar 14 2013
a(n) = Sum_{k=0..n} C(n,k) * C(n+k,k). - Joerg Arndt, May 11 2013
G.f.: G(0), where G(k) = 1 + x*(6 - x)*(4*k + 1)/(4*k + 2 - 2*x*(6-x)*(2*k + 1)*(4*k + 3)/(x*(6 - x)*(4*k + 3) + 4*(k + 1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 22 2013
G.f.: 2/G(0), where G(k) = 1 + 1/(1 - x*(6 - x)*(2*k - 1)/(x*(6 - x)*(2*k - 1) + 2*(k + 1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 16 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(6 - x)*(2*k + 1)/(x*(6 - x)*(2*k + 1) + 2*(k + 1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jul 17 2013
a(n)^2 = Sum_{k=0..n} 2^k * C(2*k, k)^2 * C(n+k, n-k) = A243949(n). - Paul D. Hanna, Aug 17 2014
a(n) = hypergeom([-n, -n], [1], 2). - Peter Luschny, Nov 19 2014
a(n) = Sum_{k=0..n/2} C(n-k,k) * 3^(n-2*k) * 2^k * C(n,k). - Vladimir Kruchinin, Jun 29 2015
a(n) = A049600(n, n-1).
a(n) = Sum_{0 <= j, k <= n} (-1)^(n+j)*C(n,k)*C(n,j)*C(n+k,k)*C(n+k+j,k+j). Cf. A126086 and A274668. - Peter Bala, Jan 15 2020
a(n) ~ c * (3 + 2*sqrt(2))^n / sqrt(n), where c = 1/sqrt(4*Pi*(3*sqrt(2)-4)) = 0.572681... (Banderier and Schwer, 2005). - Amiram Eldar, Jun 07 2020
a(n+1) = 3*a(n) + 2*Sum_{l=1..n} A006318(l)*a(n-l). [Eq. (1.16) in Qi-Shi-Guo (2016)]
a(n) ~ (1 + sqrt(2))^(2*n+1) / (2^(5/4) * sqrt(Pi*n)). - Vaclav Kotesovec, Jan 09 2023
a(n-1) + a(n) = A241023(n) for n >= 1. - Peter Bala, Sep 18 2024
a(n) = Sum_{k=0..n} C(n+k, 2*k) * C(2*k, k). - Greg Dresden and Leo Zhang, Jul 11 2025

Extensions

New name and reference Sep 15 1995
Formula and more references from Don Knuth, May 15 1996

A088218 Total number of leaves in all rooted ordered trees with n edges.

Original entry on oeis.org

1, 1, 3, 10, 35, 126, 462, 1716, 6435, 24310, 92378, 352716, 1352078, 5200300, 20058300, 77558760, 300540195, 1166803110, 4537567650, 17672631900, 68923264410, 269128937220, 1052049481860, 4116715363800, 16123801841550, 63205303218876, 247959266474052
Offset: 0

Views

Author

Michael Somos, Sep 24 2003

Keywords

Comments

Essentially the same as A001700, which has more information.
Note that the unique rooted tree with no edges has no leaves, so a(0)=1 is by convention. - Michael Somos, Jul 30 2011
Number of ordered partitions of n into n parts, allowing zeros (cf. A097070) is binomial(2*n-1,n) = a(n) = essentially A001700. - Vladeta Jovovic, Sep 15 2004
Hankel transform is A000027; example: Det([1,1,3,10;1,3,10,35;3,10,35,126; 10,35,126,462]) = 4. - Philippe Deléham, Apr 13 2007
a(n) is the number of functions f:[n]->[n] such that for all x,y in [n] if xA045992(n). - Geoffrey Critzer, Apr 02 2009
Hankel transform of the aeration of this sequence is A000027 doubled: 1,1,2,2,3,3,... - Paul Barry, Sep 26 2009
The Fi1 and Fi2 triangle sums of A039599 are given by the terms of this sequence. For the definitions of these triangle sums see A180662. - Johannes W. Meijer, Apr 20 2011
Alternating row sums of Riordan triangle A094527. See the Philippe Deléham formula. - Wolfdieter Lang, Nov 22 2012
(-2)*a(n) is the Z-sequence for the Riordan triangle A110162. For the notion of Z- and A-sequences for Riordan arrays see the W. Lang link under A006232 with details and references. - Wolfdieter Lang, Nov 22 2012
From Gus Wiseman, Jun 27 2021: (Start)
Also the number of integer compositions of 2n with alternating (or reverse-alternating) sum 0 (ranked by A344619). This is equivalent to Ran Pan's comment at A001700. For example, the a(0) = 1 through a(3) = 10 compositions are:
() (11) (22) (33)
(121) (132)
(1111) (231)
(1122)
(1221)
(2112)
(2211)
(11121)
(12111)
(111111)
For n > 0, a(n) is also the number of integer compositions of 2n with alternating sum 2.
(End)
Number of terms in the expansion of (x_1+x_2+...+x_n)^n. - César Eliud Lozada, Jan 08 2022

Examples

			G.f. = 1 + x + 3*x^2 + 10*x^3 + 35*x^4 + 126*x^5 + 462*x^6 + 1716*x^7 + ...
The five rooted ordered trees with 3 edges have 10 leaves.
..x........................
..o..x.x..x......x.........
..o...o...o.x..x.o..x.x.x..
..r...r....r....r.....r....

References

  • L. W. Shapiro and C. J. Wang, Generating identities via 2 X 2 matrices, Congressus Numerantium, 205 (2010), 33-46.

Links

Crossrefs

Same as A001700 modulo initial term and offset.
First differences are A024718.
Main diagonal of A071919 and of A305161.
A signed version is A110556.
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
A003242 counts anti-run compositions.
A025047 counts wiggly compositions (ascend: A025048, descend: A025049).
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A106356 counts compositions by number of maximal anti-runs.
A124754 gives the alternating sum of standard compositions.
A345197 counts compositions by sum, length, and alternating sum.
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0: counted by A088218 (this sequence), ranked by A344619/A344619.
- k = 1: counted by A000984, ranked by A345909/A345911.
- k = -1: counted by A001791, ranked by A345910/A345912.
- k = 2: counted by A088218 (this sequence), ranked by A345925/A345922.
- k = -2: counted by A002054, ranked by A345924/A345923.
- k >= 0: counted by A116406, ranked by A345913/A345914.
- k <= 0: counted by A058622(n-1), ranked by A345915/A345916.
- k > 0: counted by A027306, ranked by A345917/A345918.
- k < 0: counted by A294175, ranked by A345919/A345920.
- k != 0: counted by A058622, ranked by A345921/A345921.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd: counted by A000302, ranked by A053738/A053738.
Cf. A000027, A000070, A000097, A000108, A001622, A006232, A008965, A039599, A045992, A058696, A094527, A097070, A110162, A110555, A180662, A238279, A239830, A325534, A325535, A333213, A344607, A344611, A344617.

Programs

  • Magma
    [Binomial(2*n-1, n): n in [0..30]]; // Vincenzo Librandi, Aug 07 2014
  • Maple
    seq(binomial(2*n-1, n),n=0..24); # Peter Luschny, Sep 22 2014
  • Mathematica
    a[ n_] := SeriesCoefficient[(1 - x)^-n, {x, 0, n}];
c = (1 - (1 - 4 x)^(1/2))/(2 x);CoefficientList[Series[1/(1-(c-1)),{x,0,20}],x] (* Geoffrey Critzer, Dec 02 2010 *)
Table[Binomial[2 n - 1, n], {n, 0, 20}] (* Vincenzo Librandi, Aug 07 2014 *)
a[ n_] := If[ n < 0, 0, With[ {m = 2 n}, m! SeriesCoefficient[ (1 + BesselI[0, 2 x]) / 2, {x, 0, m}]]]; (* Michael Somos, Nov 22 2014 *)
  • PARI
    {a(n) = sum( i=0, n, binomial(n+i-2,i))};
  • PARI
    {a(n) = if( n<0, 0, polcoeff( (1 + 1 / sqrt(1 - 4*x + x * O(x^n))) / 2, n))};
  • PARI
    {a(n) = if( n<0, 0, polcoeff( 1 / (1 - x + x * O(x^n))^n, n))};
  • PARI
    {a(n) = if( n<0, 0, binomial( 2*n - 1, n))};
  • PARI
    {a(n) = if( n<1, n==0, polcoeff( subst((1 - x) / (1 - 2*x), x, serreverse( x - x^2 + x * O(x^n))), n))};
  • Sage
    def A088218(n):
    return rising_factorial(n,n)/falling_factorial(n,n)
[A088218(n) for n in (0..24)]  # Peter Luschny, Nov 21 2012

Formula

G.f.: (1 + 1 / sqrt(1 - 4*x)) / 2.
a(n) = binomial(2*n - 1, n).
a(n) = (n+1)*A000108(n)/2, n>=1. - B. Dubalski (dubalski(AT)atr.bydgoszcz.pl), Feb 05 2002 (in A060150)
a(n) = (0^n + C(2n, n))/2. - Paul Barry, May 21 2004
a(n) is the coefficient of x^n in 1 / (1 - x)^n and also the sum of the first n coefficients of 1 / (1 - x)^n. Given B(x) with the property that the coefficient of x^n in B(x)^n equals the sum of the first n coefficients of B(x)^n, then B(x) = B(0) / (1 - x).
G.f.: 1 / (2 - C(x)) = (1 - x*C(x))/sqrt(1-4*x) where C(x) is g.f. for Catalan numbers A000108. Second equation added by Wolfdieter Lang, Nov 22 2012.
From Paul Barry, Nov 02 2004: (Start)
a(n) = Sum_{k=0..n} binomial(2*n, k)*cos((n-k)*Pi);
a(n) = Sum_{k=0..n} binomial(n, (n-k)/2)*(1+(-1)^(n-k))*cos(k*Pi/2)/2 (with interpolated zeros);
a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)*cos((n-2*k)*Pi/2) (with interpolated zeros); (End)
a(n) = A110556(n)*(-1)^n, central terms in triangle A110555. - Reinhard Zumkeller, Jul 27 2005
a(n) = Sum_{0<=k<=n} A094527(n,k)*(-1)^k. - Philippe Deléham, Mar 14 2007
From Paul Barry, Mar 29 2010: (Start)
G.f.: 1/(1-x/(1-2x/(1-(1/2)x/(1-(3/2)x/(1-(2/3)x/(1-(4/3)x/(1-(3/4)x/(1-(5/4)x/(1-... (continued fraction);
E.g.f.: (of aerated sequence) (1 + Bessel_I(0, 2*x))/2. (End)
a(n + 1) = A001700(n). a(n) = A024718(n) - A024718(n - 1).
E.g.f.: E(x) = 1+x/(G(0)-2*x) ; G(k) = (k+1)^2+2*x*(2*k+1)-2*x*(2*k+3)*((k+1)^2)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec 21 2011
a(n) = Sum_{k=0..n}(-1)^k*binomial(2*n,n+k). - Mircea Merca, Jan 28 2012
a(n) = rf(n,n)/ff(n,n), where rf is the rising factorial and ff the falling factorial. - Peter Luschny, Nov 21 2012
D-finite with recurrence: n*a(n) +2*(-2*n+1)*a(n-1) = 0. - R. J. Mathar, Dec 04 2012
a(n) = hypergeom([1-n,-n],[1],1). - Peter Luschny, Sep 22 2014
G.f.: 1 + x/W(0), where W(k) = 4*k+1 - (4*k+3)*x/(1 - (4*k+1)*x/(4*k+3 - (4*k+5)*x/(1 - (4*k+3)*x/W(k+1) ))) ; (continued fraction). - Sergei N. Gladkovskii, Nov 13 2014
a(n) = A000984(n) + A001791(n). - Gus Wiseman, Jun 28 2021
E.g.f.: (1 + exp(2*x) * BesselI(0,2*x)) / 2. - Ilya Gutkovskiy, Nov 03 2021
From Amiram Eldar, Mar 12 2023: (Start)
Sum_{n>=0} 1/a(n) = 5/3 + 4*Pi/(9*sqrt(3)).
Sum_{n>=0} (-1)^n/a(n) = 3/5 - 8*log(phi)/(5*sqrt(5)), where phi is the golden ratio (A001622). (End)
a(n) ~ 2^(2*n-1)/sqrt(n*Pi). - Stefano Spezia, Apr 17 2024

A097805 Number of compositions of n with k parts, T(n, k) = binomial(n-1, k-1) for n, k >= 1 and T(n, 0) = 0^n, triangle read by rows for n >= 0 and 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 3, 1, 0, 1, 4, 6, 4, 1, 0, 1, 5, 10, 10, 5, 1, 0, 1, 6, 15, 20, 15, 6, 1, 0, 1, 7, 21, 35, 35, 21, 7, 1, 0, 1, 8, 28, 56, 70, 56, 28, 8, 1, 0, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1, 0, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1, 0, 1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1
Offset: 0

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Author

Paul Barry, Aug 25 2004

Keywords

Comments

Previous name was: Riordan array (1, 1/(1-x)) read by rows.
Note this Riordan array would be denoted (1, x/(1-x)) by some authors.
Columns have g.f. (x/(1-x))^k. Reverse of A071919. Row sums are A011782. Antidiagonal sums are Fibonacci(n-1). Inverse as Riordan array is (1, 1/(1+x)). A097805=B*A059260*B^(-1), where B is the binomial matrix.
(0,1)-Pascal triangle. - Philippe Deléham, Nov 21 2006
(n+1) * each term of row n generates triangle A127952: (1; 0, 2; 0, 3, 3; 0, 4, 8, 4; ...). - Gary W. Adamson, Feb 09 2007
Triangle T(n,k), 0<=k<=n, read by rows, given by [0,1,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 12 2008
From Paul Weisenhorn, Feb 09 2011: (Start)
Triangle read by rows: T(r,c) is the number of unordered partitions of n=r*(r+1)/2+c into (r+1) parts < (r+1) and at most pairs of equal parts and parts in neighboring pairs have difference 2.
Triangle read by rows: T(r,c) is the number of unordered partitions of the number n=r*(r+1)/2+(c-1) into r parts < (r+1) and at most pairs of equal parts and parts in neighboring pairs have difference 2. (End)
Triangle read by rows: T(r,c) is the number of ordered partitions (compositions) of r into c parts. - Juergen Will, Jan 04 2016
From Tom Copeland, Oct 25 2012: (Start)
Given a basis composed of a sequence of polynomials p_n(x) characterized by ladder (creation / annihilation, or raising / lowering) operators defined by R p_n(x) = p_(n+1)(x) and L p_n(x) = n p_(n-1)(x) with p_0(x)=1, giving the number operator # p_n(x) = RL p_n(x) = n p_n(x), the lower triangular padded Pascal matrix Pd (A097805) serves as a matrix representation of the operator exp(R^2*L) = exp(R#) =
1) exp(x^2D) for the set x^n and
2) D^(-1) exp(t*x)D for the set x^n/n! (see A218234).
(End)
From James East, Apr 11 2014: (Start)
Square array a(m,n) with m,n=0,1,2,... read by off-diagonals.
a(m,n) gives the number of order-preserving functions f:{1,...,m}->{1,...,n}. Order-preserving means that x
a(n,n)=A088218(n) is the size of the semigroup O_n of all order-preserving transformations of {1,...,n}.
Read as a triangle, this sequence may be obtained by augmenting Pascal's triangle by appending the column 1,0,0,0,... on the left.
(End)
A formula based on the partitions of n with largest part k is given as a Sage program below. The 'conjugate' formula leads to A048004. - Peter Luschny, Jul 13 2015
From Wolfdieter Lang, Feb 17 2017: (Start)
The transposed of this lower triangular Riordan matrix of the associated type T provides the transition matrix between the monomial basis {x^n}, n >= 0, and the basis {y^n}, n >= 0, with y = x/(1-x): x^0 = 1 = y^0, x^n = Sum_{m >= n} Ttrans(n,m) y^m, for n >= 1, with Ttrans(n,m) = binomial(m-1,n-1).
Therefore, if a transformation with this Riordan matrix from a sequence {a} to the sequence {b} is given by b(n) = Sum_{m=0..n} T(n, m)*a(m), with T(n, m) = binomial(n-1, m-1), for n >= 1, then Sum_{n >= 0} a(n)*x^n = Sum_{n >= 0} b(n)*y^n, with y = x/(1-x) and vice versa. This is a modified binomial transformation; the usual one belongs to the Pascal Riordan matrix A007318. (End)
From Gus Wiseman, Jan 23 2022: (Start)
Also the number of compositions of n with alternating sum k, with k ranging from -n to n in steps of 2. For example, row n = 6 counts the following compositions (empty column indicated by dot):
. (15) (24) (33) (42) (51) (6)
(141) (132) (123) (114)
(1113) (231) (222) (213)
(1212) (1122) (321) (312)
(1311) (1221) (1131) (411)
(2112) (2121)
(2211) (3111)
(11121) (11112)
(12111) (11211)
(111111) (21111)
The reverse-alternating version is the same. Counting compositions by all three parameters (sum, length, alternating sum) gives A345197. Compositions of 2n with alternating sum 2k with k ranging from -n + 1 to n are A034871. (End)
Also the convolution triangle of A000012. - Peter Luschny, Oct 07 2022
From Sergey Kitaev, Nov 18 2023: (Start)
Number of permutations of length n avoiding simultaneously the patterns 123 and 132 with k right-to-left maxima. A right-to-left maximum in a permutation a(1)a(2)...a(n) is position i such that a(j) < a(i) for all i < j.
Number of permutations of length n avoiding simultaneously the patterns 231 and 312 with k right-to-left minima (resp., left-to-right maxima). A right-to-left minimum (resp., left-to-right maximum) in a permutation a(1)a(2)...a(n) is position i such that a(j) > a(i) for all j > i (resp., a(j) < a(i) for all j < i).
Number of permutations of length n avoiding simultaneously the patterns 213 and 312 with k right-to-left maxima (resp., left-to-right maxima).
Number of permutations of length n avoiding simultaneously the patterns 213 and 231 with k right-to-left maxima (resp., right-to-left minima). (End)

Examples

			G.f. = 1 + x * (x + x^3 * (1 + x) + x^6 * (1 + x)^2 + x^10 * (1 + x)^3 + ...). - _Michael Somos_, Aug 20 2006
The triangle T(n, k) begins:
n\k  0 1 2  3  4   5   6  7  8 9 10 ...
0:   1
1:   0 1
2:   0 1 1
3:   0 1 2  1
4:   0 1 3  3  1
5:   0 1 4  6  4   1
6:   0 1 5 10 10   5   1
7:   0 1 6 15 20  15   6  1
8:   0 1 7 21 35  35  21  7  1
9:   0 1 8 28 56  70  56 28  8 1
10:  0 1 9 36 84 126 126 84 36 9  1
... reformatted _Wolfdieter Lang_, Jul 31 2017
From _Paul Weisenhorn_, Feb 09 2011: (Start)
T(r=5,c=3) = binomial(4,2) = 6 unordered partitions of the number n = r*(r+1)/2+c = 18 with (r+1)=6 summands: (5+5+4+2+1+1), (5+5+3+3+1+1), (5+4+4+3+1+1), (5+5+3+2+2+1), (5+4+4+2+2+1), (5+4+3+3+2+1).
T(r=5,c=3) = binomial(4,2) = 6 unordered partitions of the number n = r*(r+1)/2+(c-1) = 17 with r=5 summands: (5+5+4+2+1), (5+5+3+3+1), (5+5+3+2+2), (5+4+4+3+1), (5+4+4+2+2), (5+4+3+3+2).  (End)
From _James East_, Apr 11 2014: (Start)
a(0,0)=1 since there is a unique (order-preserving) function {}->{}.
a(m,0)=0 for m>0 since there is no function from a nonempty set to the empty set.
a(3,2)=4 because there are four order-preserving functions {1,2,3}->{1,2}: these are [1,1,1], [2,2,2], [1,1,2], [1,2,2]. Here f=[a,b,c] denotes the function defined by f(1)=a, f(2)=b, f(3)=c.
a(2,3)=6 because there are six order-preserving functions {1,2}->{1,2,3}: these are [1,1], [1,2], [1,3], [2,2], [2,3], [3,3].
(End)

References

  • D. E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, Part 1, Section 7.2.1.3, 2011.

Links

Crossrefs

Case m=0 of the polynomials defined in A278073.
Cf. A000012 (diagonal), A011782 (row sums), A088218 (central terms).
Cf. A007318, A048004, A127952, A118800, A200139, A130595, A167374.
The terms just left of center in odd-indexed rows are A001791, even A002054.
The odd-indexed rows are A034871.
Row sums without the center are A058622.
The unordered version is A072233, without zeros A008284.
Right half without center has row sums A027306(n-1).
Right half with center has row sums A116406(n).
Left half without center has row sums A294175(n-1).
Left half with center has row sums A058622(n-1).
A025047 counts alternating compositions.
A098124 counts balanced compositions, unordered A047993.
A106356 counts compositions by number of maximal anti-runs.
A344651 counts partitions by sum and alternating sum.
A345197 counts compositions by sum, length, and alternating sum.
Cf. A000346, A000984, A001700, A008549, A008965, A114121, A124754, A238279, A345907, A345908, A346632.

Programs

  • Maple
    b:= proc(n, i, p) option remember; `if`(n=0, p!, `if`(i<1, 0,
      expand(add(b(n-i*j, i-1, p+j)/j!*x^j, j=0..n/i))))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n$2, 0)):
seq(T(n), n=0..20);  # Alois P. Heinz, May 25 2014
# Alternatively:
T := proc(k,n) option remember;
if k=n then 1 elif k=0 then 0 else
add(T(k-1,n-i), i=1..n-k+1) fi end:
A097805 := (n,k) -> T(k,n):
for n from 0 to 12 do seq(A097805(n,k), k=0..n) od; # Peter Luschny, Mar 12 2016
# Uses function PMatrix from A357368.
PMatrix(10, n -> 1);  # Peter Luschny, Oct 07 2022
  • Mathematica
    T[0, 0] = 1; T[n_, k_] := Binomial[n-1, k-1]; Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Sep 03 2014, after Paul Weisenhorn *)
Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],Length[#]==k&]],{n,0,10},{k,0,n}] (* Gus Wiseman, Jan 23 2022 *)
  • PARI
    {a(n) = my(m); if( n<2, n==0, n--; m = (sqrtint(8*n + 1) - 1)\2; binomial(m-1, n - m*(m + 1)/2))}; /* Michael Somos, Aug 20 2006 */
  • PARI
    T(n,k) = if (k==0, 0^n, binomial(n-1, k-1)); \\ Michel Marcus, May 06 2022
  • PARI
    row(n) = vector(n+1, k, k--; if (k==0, 0^n, binomial(n-1, k-1))); \\ Michel Marcus, May 06 2022
  • Python
    from math import comb
def T(n, k): return comb(n-1, k-1) if k != 0 else k**n  # Peter Luschny, May 06 2022
  • Sage
    # Illustrates a basic partition formula, is not efficient as a program for large n.
def A097805_row(n):
    r = []
    for k in (0..n):
        s = 0
        for q in Partitions(n, max_part=k, inner=[k]):
            s += mul(binomial(q[j],q[j+1]) for j in range(len(q)-1))
        r.append(s)
    return r
[A097805_row(n) for n in (0..9)] # Peter Luschny, Jul 13 2015

Formula

Number triangle T(n, k) defined by T(n,k) = Sum_{j=0..n} binomial(n, j)*if(k<=j, (-1)^(j-k), 0).
T(r,c) = binomial(r-1,c-1), 0 <= c <= r. - Paul Weisenhorn, Feb 09 2011
G.f.: (-1+x)/(-1+x+x*y). - R. J. Mathar, Aug 11 2015
a(0,0) = 1, a(n,k) = binomial(n-1,n-k) = binomial(n-1,k-1) Juergen Will, Jan 04 2016
G.f.: (x^1 + x^2 + x^3 + ...)^k = (x/(1-x))^k. - Juergen Will, Jan 04 2016
From Tom Copeland, Nov 15 2016: (Start)
E.g.f.: 1 + x*[e^((x+1)t)-1]/(x+1).
This padded Pascal matrix with the odd columns negated is NpdP = M*S = S^(-1)*M^(-1) = S^(-1)*M, where M(n,k) = (-1)^n A130595(n,k), the inverse Pascal matrix with the odd rows negated, S is the summation matrix A000012, the lower triangular matrix with all elements unity, and S^(-1) = A167374, a finite difference matrix. NpdP is self-inverse, i.e., (M*S)^2 = the identity matrix, and has the e.g.f. 1 - x*[e^((1-x)t)-1]/(1-x).
M = NpdP*S^(-1) follows from the well-known recursion property of the Pascal matrix, implying NpdP = M*S.
The self-inverse property of -NpdP is implied by the self-inverse relation of its embedded signed Pascal submatrix M (cf. A130595). Also see A118800 for another proof.
Let P^(-1) be A130595, the inverse Pascal matrix. Then T = A200139*P^(-1) and T^(-1) = padded P^(-1) = P*A097808*P^(-1). (End)
The center (n>0) is T(2n+1,n+1) = A000984(n) = 2*A001700(n-1) = 2*A088218(n) = A126869(2n) = 2*A138364(2n-1). - Gus Wiseman, Jan 25 2022

Extensions

Corrected by Philippe Deléham, Oct 05 2005
New name using classical terminology by Peter Luschny, Feb 05 2019

A001791 a(n) = binomial coefficient C(2n, n-1).

Original entry on oeis.org

0, 1, 4, 15, 56, 210, 792, 3003, 11440, 43758, 167960, 646646, 2496144, 9657700, 37442160, 145422675, 565722720, 2203961430, 8597496600, 33578000610, 131282408400, 513791607420, 2012616400080, 7890371113950, 30957699535776, 121548660036300, 477551179875952
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Number of peaks at even level in all Dyck paths of semilength n+1. Example: a(2)=4 because UDUDUD, UDUU*DD, UU*DDUD, UU*DU*DD, UUUDDD, where U=(1,1), D=(1,-1) and the peaks at even level are shown by *. - Emeric Deutsch, Dec 05 2003
Also number of long ascents (i.e., ascents of length at least two) in all Dyck paths of semilength n+1. Example: a(2)=4 because in the five Dyck paths of semilength 3, namely UDUDUD, UD(UU)DD, (UU)DDUD, (UU)DUDD and (UUU)DDD, we have four long ascents (shown between parentheses). Here U=(1,1) and D=(1,-1). Also number of branch nodes (i.e., vertices of outdegree at least two) in all ordered trees with n+1 edges. - Emeric Deutsch, Feb 22 2004
Number of lattice paths from (0,0) to (n,n) with steps E=(1,0) and N=(0,1) which touch or cross the line x-y=1. Example: For n=2 these are the paths EENN, ENEN, ENNE and NEEN. - Herbert Kociemba, May 23 2004
Narayana transform (A001263) of [1, 3, 5, 7, 9, ...] = (1, 4, 15, 56, 210, ...). Row sums of triangles A136534 and A136536. - Gary W. Adamson, Jan 04 2008
Starting with offset 1 = the Catalan sequence starting (1, 2, 5, 14, ...) convolved with A000984: (1, 2, 6, 20, ...). - Gary W. Adamson, May 17 2009
Also number of peaks plus number of valleys in all Dyck n-paths. - David Scambler, Oct 08 2012
Apparently counts UDDUD in all Dyck paths of semilength n+2. - David Scambler, Apr 22 2013
Apparently the number of peaks strictly left of the midpoint in all Dyck paths of semilength n+1. - David Scambler, Apr 30 2013
For n>0, a(n) is the number of compositions of n into at most n parts if zeros are allowed as parts (so-called "weak" compositions). - L. Edson Jeffery, Jul 24 2014
Number of paths in the half-plane x >= 0, from (0,0) to (2n,2), and consisting of steps U=(1,1) and D=(1,-1). For example, for n=2, we have the 4 paths: UUUD, UUDU, UDUU, DUUU. - José Luis Ramírez Ramírez, Apr 19 2015
For n>1, 1/a(n) is the probability that when a stick is broken up at n points independently and uniformly chosen at random along its length any triple of pieces of the n+1 pieces can form a triangle. The corresponding probability for the existence of at least one triple is A339392(n)/A339393(n). - Amiram Eldar, Dec 04 2020
a(n) is the number of lattice paths of 2n steps taken from the step set {U=(1,1), D=(1,-1)} that start at the origin, never go below the x-axis, and end strictly above the x-axis; more succinctly, proper left factors of Dyck paths. For example, a(2)=4 counts UUUU, UUUD, UUDU, UDUU. - David Callan and Emeric Deutsch, Jan 25 2021
From Gus Wiseman, Jul 21 2021: (Start)
Also the number of integer compositions of 2n+1 with alternating sum -1, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. For example, the a(1) = 1 through a(3) = 15 compositions are:
(1,2) (2,3) (3,4)
(1,3,1) (1,4,2)
(1,1,1,2) (2,4,1)
(1,2,1,1) (1,1,2,3)
(1,2,2,2)
(1,3,2,1)
(2,1,1,3)
(2,2,1,2)
(2,3,1,1)
(1,1,1,3,1)
(1,2,1,2,1)
(1,3,1,1,1)
(1,1,1,1,1,2)
(1,1,1,2,1,1)
(1,2,1,1,1,1)
The following relate to these compositions.
- The unordered version is A000070.
- Allowing any negative alternating sum gives A000346.
- The opposite (positive 1) version is A000984.
- The version for reverse-alternating sum is also A001791 (this sequence).
- Taking alternating sum -2 instead of -1 gives A002054.
- The shifted version for alternating sum 0 is counted by A088218 and ranked by A344619.
- Ranked by A345910 (reverse: A345912).
Equivalently, a(n) counts binary numbers with 2n+1 digits and one more 0 than 1's. For example, the a(2) = 4 binary numbers are: 10001, 10010, 10100, 11000.
(End)
The diagonal of a square n X n matrix where cells of the first row are the nonnegative integers and cells of subsequent rows are sums of cells of the previous row up to and including n. - Torlach Rush, Apr 24 2024
For n>=1, a(n) is the independence number of the odd graph O_{n+1}. - Miquel A. Fiol, Jul 07 2024

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.
  • Cornelius Lanczos, Applied Analysis, Prentice-Hall, Englewood Cliffs, NJ, 1956, p. 517.
  • R. C. Mullin, E. Nemeth and P. J. Schellenberg, The enumeration of almost cubic maps, pp. 281-295 in Proceedings of the Louisiana Conference on Combinatorics, Graph Theory and Computer Science. Vol. 1, edited R. C. Mullin et al., 1970.
  • 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).

Links

Crossrefs

Diagonal 3 of triangle A100257.
First differences are in A076540.
Cf. A000108, A000984, A002378, A002390, A025566, A132812.
A345197 counts compositions by length and alternating sum.
Cf. A000070, A000302, A000346, A002054, A008549, A032443, A088218, A097805, A163493, A202736, A345910.

Programs

  • GAP
    List([0..30],n->Binomial(2*n,n-1)); # Muniru A Asiru, Aug 09 2018
  • Magma
    [Binomial(2*n, n-1): n in [0..30]]; // Vincenzo Librandi, Apr 20 2015
  • Mathematica
    Table[Binomial[2n,n-1],{n,0,30}] (* Harvey P. Dale, Jul 12 2012 *)
CoefficientList[ Series[(1 - 2x - Sqrt[1 - 4x])/(2x*Sqrt[1 - 4x]), {x, 0, 26}], x] (* Robert G. Wilson v, Aug 10 2018 *)
  • Maxima
    A001791(n):=binomial(2*n,n-1)$
makelist(A001791(n),n,0,30); /* Martin Ettl, Nov 05 2012 */
  • PARI
    a(n)=if(n<1,0,(2*n)!/(n+1)!/(n-1)!)

Formula

a(n) = n*A000108(n).
G.f.: x*(d/dx)c(x) where c(x) = Catalan g.f. - Wolfdieter Lang
Convolution of A001700 (central binomial of odd order) and A000108 (Catalan): a(n+1) = Sum_{k=0..n} C(k)*binomial(2*(n-k)+1, n-k), C(k): Catalan. - Wolfdieter Lang
E.g.f.: exp(2x) * I_1(2x), where I_1 is Bessel function. - Michael Somos, Sep 08 2002
a(n) = Sum_{k=0..n} C(n, k)*C(n, k+1). - Paul Barry, May 15 2003
a(n) = Sum_{i=1..n} binomial(i+n-1, n).
G.f.: (1-2x-sqrt(1-4x))/(2x*sqrt(1-4x)). - Emeric Deutsch, Dec 05 2003
a(n) = A092956/(n!). - Amarnath Murthy, Jun 16 2004
a(n) = binomial(2n,n) - A000108(n). - Paul Barry, Apr 21 2005
a(n) = (1/(2*Pi))*Integral_{x=0..4} (x^n*(x-2)/sqrt(x(4-x))) is the moment sequence representation. - Paul Barry, Jan 11 2007
Row sums of triangle A132812 starting (1, 4, 15, 56, 210, ...). - Gary W. Adamson, Sep 01 2007
Starting (1, 4, 15, 56, 210, ...) gives the binomial transform of A025566 starting (1, 3, 8, 22, 61, 171, ...). - Gary W. Adamson, Sep 01 2007
For n >= 1, a(2^n) = 2^(n+1)*A001795(2^(n-1)). - Vladimir Shevelev, Sep 05 2010
D-finite with recurrence: (n-1)*(n+1)*a(n) = 2*n*(2n-1)*a(n-1). - R. J. Mathar, Dec 17 2011
From Sergei N. Gladkovskii, Jul 07 2012: (Start)
G.f.: -1/(2*x) - G(0) where G(k) = 1 - 1/(2*x - 8*x^3*(2*k+1)/(4*x^2*(2*k+1)- (k+1)/G(k+1))); (continued fraction, 3rd kind, 3-step);
E.g.f.: BesselI(1,2*x)*exp(2*x) = x*G(0) where G(k) = 1 + 2*x*(4*k+3)/((2*k+1)*(2*k+3) - x*(2*k+1)*(2*k+3)*(4*k+5)/(x*(4*k+5) + 2*(k+1)*(k+2)/G(k+1))); (continued fraction, 3rd kind, 3-step).
(End)
G.f.: c(x)^3/(2-c(x)) where c(x) is the g.f. for A000108. - Cheyne Homberger, May 05 2014
G.f.: z*C(z)^2/(1-2*z*C(z)), where C(z) is the g.f. of Catalan numbers. - José Luis Ramírez Ramírez, Apr 19 2015
G.f.: x*2F1(3/2,2;3;4x). - R. J. Mathar, Aug 09 2015
a(n) = Sum_{i=1..n} binomial(2*i-2,i-1)*binomial(2*(n-i+1),n-i+2)/(n-i+1). - Vladimir Kruchinin, Sep 07 2015
L.g.f.: 1/(1 - x/(1 - x/(1 - x/(1 - x/(1 - x/(1 - ...)))))) = Sum_{n>=1} a(n)*x^n/n. - Ilya Gutkovskiy, May 10 2017
Sum_{n>=1} 1/a(n) = 1/3 + 5*Pi/(9*sqrt(3)). - Amiram Eldar, Dec 04 2020
Sum_{n>=1} (-1)^(n+1)/a(n) = 1/5 + 14*sqrt(5)*log(phi)/25, where log(phi) = A002390. - Amiram Eldar, Feb 20 2021
a(n) = Product_{i=1..(n - 1)} (((4*i + 6)*i + 2)/((i + 2)*i)), for n>=1. - Neven Sajko, Oct 10 2021
a(n) = 2^(2*n)*gamma(n + 1/2)/(sqrt(Pi)*gamma(n)*(n+1)) for n > 0, and a(0) = lim_{n->0} a(n). - Karol A. Penson, Apr 24 2025

A002457 a(n) = (2n+1)!/n!^2.

Original entry on oeis.org

1, 6, 30, 140, 630, 2772, 12012, 51480, 218790, 923780, 3879876, 16224936, 67603900, 280816200, 1163381400, 4808643120, 19835652870, 81676217700, 335780006100, 1378465288200, 5651707681620, 23145088600920, 94684453367400, 386971244197200, 1580132580471900
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Expected number of matches remaining in Banach's modified matchbox problem (counted when last match is drawn from one of the two boxes), multiplied by 4^(n-1). - Michael Steyer, Apr 13 2001
Hankel transform is (-1)^n*A014480(n). - Paul Barry, Apr 26 2009
Convolved with A000108: (1, 1, 1, 5, 14, 42, ...) = A000531: (1, 7, 38, 187, 874, ...). - Gary W. Adamson, May 14 2009
Convolution of A000302 and A000984. - Philippe Deléham, May 18 2009
1/a(n) is the integral of (x(1-x))^n on interval [0,1]. Apparently John Wallis computed these integrals for n=0,1,2,3,.... A004731, shifted left by one, gives numerators/denominators of related integrals (1-x^2)^n on interval [0,1]. - Marc van Leeuwen, Apr 14 2010
Extend the triangular peaks of Dyck paths of semilength n down to the baseline forming (possibly) larger and overlapping triangles. a(n) = sum of areas of these triangles. Also a(n) = triangular(n) * Catalan(n). - David Scambler, Nov 25 2010
Let H be the n X n Hilbert matrix H(i,j) = 1/(i+j-1) for 1 <= i,j <= n. Let B be the inverse matrix of H. The sum of the elements in row n of B equals a(n-1). - T. D. Noe, May 01 2011
Apparently the number of peaks in all symmetric Dyck paths with semilength 2n+1. - David Scambler, Apr 29 2013
Denominator of central elements of Leibniz's Harmonic Triangle A003506.
Central terms of triangle A116666. - Reinhard Zumkeller, Nov 02 2013
Number of distinct strings of length 2n+1 using n letters A, n letters B, and 1 letter C. - Hans Havermann, May 06 2014
Number of edges in the Hasse diagram of the poset of partitions in the n X n box ordered by containment (from Havermann's comment above, C represents the square added in the edge). - William J. Keith, Aug 18 2015
Let V(n, r) denote the volume of an n-dimensional sphere with radius r then V(n, 1/2^n) = V(n-1, 1/2^n) / a((n-1)/2) for all odd n. - Peter Luschny, Oct 12 2015
a(n) is the result of processing the n+1 row of Pascal's triangle A007318 with the method of A067056. Example: Let n=3. Given the 4th row of Pascal's triangle 1,4,6,4,1, we get 1*(4+6+4+1) + (1+4)*(6+4+1) + (1+4+6)*(4+1) + (1+4+6+4)*1 = 15+55+55+15 = 140 = a(3). - J. M. Bergot, May 26 2017
a(n) is the number of (n+1) X 2 Young tableaux with a two horizontal walls between the first and second column. If there is a wall between two cells, the entries may be decreasing; see [Banderier, Wallner 2021] and A000984 for one horizontal wall. - Michael Wallner, Jan 31 2022
a(n) is the number of facets of the symmetric edge polytope of the cycle graph on 2n+1 vertices. - Mariel Supina, May 12 2022
Diagonal of the rational function 1 / (1 - x - y)^2. - Ilya Gutkovskiy, Apr 23 2025

Examples

			G.f. = 1 + 6*x + 30*x^2 + 140*x^3 + 630*x^4 + 2772*x^5 + 12012*x^6 + 51480*x^7 + ...

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 159.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 83, Problem 25; p. 168, #30.
  • W. Feller, An Introduction to Probability Theory and Its Applications, Vol. I.
  • C. Jordan, Calculus of Finite Differences. Röttig and Romwalter, Budapest, 1939; Chelsea, NY, 1965, p. 449.
  • M. Klamkin, ed., Problems in Applied Mathematics: Selections from SIAM Review, SIAM, 1990; see pp. 127-129.
  • C. Lanczos, Applied Analysis. Prentice-Hall, Englewood Cliffs, NJ, 1956, p. 514.
  • A. P. Prudnikov, Yu. A. Brychkov and O.I. Marichev, "Integrals and Series", Volume 1: "Elementary Functions", Chapter 4: "Finite Sums", New York, Gordon and Breach Science Publishers, 1986-1992.
  • J. Ser, Les Calculs Formels des Séries de Factorielles. Gauthier-Villars, Paris, 1933, p. 92.
  • 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).
  • J. Wallis, Operum Mathematicorum, pars altera, Oxford, 1656, pp 31,34 [Marc van Leeuwen, Apr 14 2010]

Links

Crossrefs

Cf. A000531 (Banach's original match problem).
Cf. A033876, A000984, A001803, A132818, A046521 (second column).
Cf. A000108, A000217.
A diagonal of A331430.
The rightmost diagonal of the triangle A331431.

Programs

Formula

G.f.: (1-4x)^(-3/2) = 1F0(3/2;;4x).
a(n-1) = binomial(2*n, n)*n/2 = binomial(2*n-1, n)*n.
a(n-1) = 4^(n-1)*Sum_{i=0..n-1} binomial(n-1+i, i)*(n-i)/2^(n-1+i).
a(n) ~ 2*Pi^(-1/2)*n^(1/2)*2^(2*n)*{1 + 3/8*n^-1 + ...}. - Joe Keane (jgk(AT)jgk.org), Nov 21 2001
(2*n+2)!/(2*n!*(n+1)!) = (n+n+1)!/(n!*n!) = 1/beta(n+1, n+1) in A061928.
Sum_{i=0..n} i * binomial(n, i)^2 = n*binomial(2*n, n)/2. - Yong Kong (ykong(AT)curagen.com), Dec 26 2000
a(n) ~ 2*Pi^(-1/2)*n^(1/2)*2^(2*n). - Joe Keane (jgk(AT)jgk.org), Jun 07 2002
a(n) = 1/Integral_{x=0..1} x^n (1-x)^n dx. - Fred W. Helenius (fredh(AT)ix.netcom.com), Jun 10 2003
E.g.f.: exp(2*x)*((1+4*x)*BesselI(0, 2*x) + 4*x*BesselI(1, 2*x)). - Vladeta Jovovic, Sep 22 2003
a(n) = Sum_{i+j+k=n} binomial(2i, i)*binomial(2j, j)*binomial(2k, k). - Benoit Cloitre, Nov 09 2003
a(n) = (2*n+1)*A000984(n) = A005408(n)*A000984(n). - Zerinvary Lajos, Dec 12 2010
a(n-1) = Sum_{k=0..n} A039599(n,k)*A000217(k), for n >= 1. - Philippe Deléham, Jun 10 2007
Sum of (n+1)-th row terms of triangle A132818. - Gary W. Adamson, Sep 02 2007
Sum_{n>=0} 1/a(n) = 2*Pi/3^(3/2). - Jaume Oliver Lafont, Mar 07 2009
a(n) = Sum_{k=0..n} binomial(2k,k)*4^(n-k). - Paul Barry, Apr 26 2009
a(n) = A000217(n) * A000108(n). - David Scambler, Nov 25 2010
a(n) = f(n, n-3) where f is given in A034261.
a(n) = A005430(n+1)/2 = A002011(n)/4.
a(n) = binomial(2n+2, 2) * binomial(2n, n) / binomial(n+1, 1), a(n) = binomial(n+1, 1) * binomial(2n+2, n+1) / binomial(2, 1) = binomial(2n+2, n+1) * (n+1)/2. - Rui Duarte, Oct 08 2011
G.f.: (G(0) - 1)/(4*x) where G(k) = 1 + 2*x*((2*k + 3)*G(k+1) - 1)/(k + 1). - Sergei N. Gladkovskii, Dec 03 2011 [Edited by Michael Somos, Dec 06 2013]
G.f.: 1 - 6*x/(G(0)+6*x) where G(k) = 1 + (4*x+1)*k - 6*x - (k+1)*(4*k-2)/G(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Aug 13 2012
G.f.: Q(0), where Q(k) = 1 + 4*(2*k + 1)*x*(2*k + 2 + Q(k+1))/(k+1). - Sergei N. Gladkovskii, May 10 2013 [Edited by Michael Somos, Dec 06 2013]
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - 4*x*(2*k+3)/(4*x*(2*k+3) + 2*(k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 06 2013
a(n) = 2^(4n)/Sum_{k=0..n} (-1)^k*C(2n+1,n-k)/(2k+1). - Mircea Merca, Nov 12 2013
a(n) = (2*n)!*[x^(2*n)] HeunC(0,0,-2,-1/4,7/4,4*x^2) where [x^n] f(x) is the coefficient of x^n in f(x) and HeunC is the Heun confluent function. - Peter Luschny, Nov 22 2013
0 = a(n) * (16*a(n+1) - 2*a(n+2)) + a(n+1) * (a(n+2) - 6*a(n+1)) for all n in Z. - Michael Somos, Dec 06 2013
a(n) = 4^n*binomial(n+1/2, 1/2). - Peter Luschny, Apr 24 2014
a(n) = 4^n*hypergeom([-2*n,-2*n-1,1/2],[-2*n-2,1],2)*(n+1)*(2*n+1). - Peter Luschny, Sep 22 2014
a(n) = 4^n*hypergeom([-n,-1/2],[1],1). - Peter Luschny, May 19 2015
a(n) = 2*4^n*Gamma(3/2+n)/(sqrt(Pi)*Gamma(1+n)). - Peter Luschny, Dec 14 2015
Sum_{n >= 0} 2^(n+1)/a(n) = Pi, related to Newton/Euler's Pi convergence transformation series. - Tony Foster III, Jul 28 2016. See the Weisstein Pi link, eq. (23). - Wolfdieter Lang, Aug 26 2016
Boas-Buck recurrence: a(n) = (6/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, and a(0) = 1. Proof from a(n) = A046521(n+1,1). See comment in A046521. - Wolfdieter Lang, Aug 10 2017
a(n) = (1/3)*Sum_{i = 0..n+1} C(n+1,i)*C(n+1,2*n+1-i)*C(3*n+2-i,n+1) = (1/3)*Sum_{i = 0..2*n+1} (-1)^(i+1)*C(2*n+1,i)*C(n+i+1,i)^2. - Peter Bala, Feb 07 2018
a(n) = (2*n+1)*binomial(2*n, n). - Kolosov Petro, Apr 16 2018
a(n) = (-4)^n*binomial(-3/2, n). - Peter Luschny, Oct 23 2018
a(n) = 1 / Sum_{s=0..n} (-1)^s * binomial(n, s) / (n+s+1). - Kolosov Petro, Jan 22 2019
a(n) = Sum_{k = 0..n} (2*k + 1)*binomial(2*n + 1, n - k). - Peter Bala, Feb 25 2019
4^n/a(n) = Integral_{x=0..1} (1 - x^2)^n. - Michael Somos, Jun 13 2019
D-finite with recurrence: 0 = a(n)*(6 + 4*n) - a(n+1)*(n + 1) for all n in Z. - Michael Somos, Jun 13 2019
Sum_{n>=0} (-1)^n/a(n) = 4*arcsinh(1/2)/sqrt(5). - Amiram Eldar, Sep 10 2020
From Jianing Song, Apr 10 2022: (Start)
G.f. for {1/a(n)}: 4*arcsin(sqrt(x)/2) / sqrt(x*(4-x)).
E.g.f. for {1/a(n)}: exp(x/4)*sqrt(Pi/x)*erf(sqrt(x)/2). (End)
G.f. for {1/a(n)}: 4*arctan(sqrt(x/(4-x))) / sqrt(x*(4-x)). - Michael Somos, Jun 17 2023
a(n) = Sum_{k = 0..n} (-1)^(n+k) * (n + 2*k + 1)*binomial(n+k, k). This is the particular case m = 1 of the identity Sum_{k = 0..m*n} (-1)^k * (n + 2*k + 1) * binomial(n+k, k) = (-1)^(m*n) * (m*n + 1) * binomial((m+1)*n+1, n). Cf. A090816 and A306290. - Peter Bala, Nov 02 2024
a(n) = (1/Pi)*(2*n + 1)*(2^(2*n + 1))*Integral_{x=0..oo} 1/(x^2 + 1)^(n + 1) dx. - Velin Yanev, Jan 28 2025

A056040 Swinging factorial, a(n) = 2^(n-(n mod 2))*Product_{k=1..n} k^((-1)^(k+1)).

Original entry on oeis.org

1, 1, 2, 6, 6, 30, 20, 140, 70, 630, 252, 2772, 924, 12012, 3432, 51480, 12870, 218790, 48620, 923780, 184756, 3879876, 705432, 16224936, 2704156, 67603900, 10400600, 280816200, 40116600, 1163381400, 155117520, 4808643120, 601080390, 19835652870, 2333606220
Offset: 0

Views

Author

Labos Elemer, Jul 25 2000

Keywords

Comments

a(n) is the number of 'swinging orbitals' which are enumerated by the trinomial n over [floor(n/2), n mod 2, floor(n/2)].
Similar to but different from A001405(n) = binomial(n, floor(n/2)), a(n) = lcm(A001405(n-1), A001405(n)) (for n>0).
A055773(n) divides a(n), A001316(floor(n/2)) divides a(n).
Exactly p consecutive multiples of p follow the least positive multiple of p if p is an odd prime. Compare with the similar property of A100071. - Peter Luschny, Aug 27 2012
a(n) is the number of vertices of the polytope resulting from the intersection of an n-hypercube with the hyperplane perpendicular to and bisecting one of its long diagonals. - Didier Guillet, Jun 11 2018 [Edited by Peter Munn, Dec 06 2022]

Examples

			a(10) = 10!/5!^2 = trinomial(10,[5,0,5]);
a(11) = 11!/5!^2 = trinomial(11,[5,1,5]).

Links

Crossrefs

Bisections are A000984 and A002457.
Cf. A000142, A001405, A000188, A055772, A056042, A211226, A162246, A189231, A212303.

Programs

  • Magma
    [(Factorial(n)/(Factorial(Floor(n/2)))^2): n in [0..40]]; // Vincenzo Librandi, Sep 11 2011
  • Maple
    SeriesCoeff := proc(s,n) series(s(w,n),w,n+2);
convert(%,polynom); coeff(%,w,n) end;
a1 := proc(n) local k;
2^(n-(n mod 2))*mul(k^((-1)^(k+1)),k=1..n) end:
a2 := proc(n) option remember;
`if`(n=0,1,n^irem(n,2)*(4/n)^irem(n+1,2)*a2(n-1)) end;
a3 := n -> n!/iquo(n,2)!^2;
g4 := z -> BesselI(0,2*z)*(1+z);
a4 := n -> n!*SeriesCoeff(g4,n);
g5 := z -> (1+z/(1-4*z^2))/sqrt(1-4*z^2);
a5 := n -> SeriesCoeff(g5,n);
g6 := (z,n) -> (1+z^2)^n+n*z*(1+z^2)^(n-1);
a6 := n -> SeriesCoeff(g6,n);
a7 := n -> combinat[multinomial](n,floor(n/2),n mod 2,floor(n/2));
h := n -> binomial(n,floor(n/2)); # A001405
a8 := n -> ilcm(h(n-1),h(n));
F := [a1, a2, a3, a4, a5, a6, a7, a8];
for a in F do seq(a(i), i=0..32) od;
  • Mathematica
    f[n_] := 2^(n - Mod[n, 2])*Product[k^((-1)^(k + 1)), {k, n}]; Array[f, 33, 0] (* Robert G. Wilson v, Aug 02 2010 *)
f[n_] := If[OddQ@n, n*Binomial[n - 1, (n - 1)/2], Binomial[n, n/2]]; Array[f, 33, 0] (* Robert G. Wilson v, Aug 10 2010 *)
sf[n_] := With[{f = Floor[n/2]}, Pochhammer[f+1, n-f]/f!]; (* or, twice faster: *) sf[n_] := n!/Quotient[n, 2]!^2; Table[sf[n], {n, 0, 32}] (* Jean-François Alcover, Jul 26 2013, updated Feb 11 2015 *)
  • PARI
    a(n)=n!/(n\2)!^2 \\ Charles R Greathouse IV, May 02 2011
  • Sage
    def A056040():
    r, n = 1, 0
    while True:
        yield r
        n += 1
        r *= 4/n if is_even(n) else n
a = A056040(); [next(a) for i in range(36)]  # Peter Luschny, Oct 24 2013

Formula

a(n) = n!/floor(n/2)!^2. [Essentially the original name.]
a(0) = 1, a(n) = n^(n mod 2)*(4/n)^(n+1 mod 2)*a(n-1) for n>=1.
E.g.f.: (1+x)*BesselI(0, 2*x). - Vladeta Jovovic, Jan 19 2004
O.g.f.: a(n) = SeriesCoeff_{n}((1+z/(1-4*z^2))/sqrt(1-4*z^2)).
P.g.f.: a(n) = PolyCoeff_{n}((1+z^2)^n+n*z*(1+z^2)^(n-1)).
a(2n+1) = A046212(2n+1) = A100071(2n+1). - M. F. Hasler, Jan 25 2012
a(2*n) = binomial(2*n,n); a(2*n+1) = (2*n+1)*binomial(2*n,n). Central terms of triangle A211226. - Peter Bala, Apr 10 2012
D-finite with recurrence: n*a(n) + (n-2)*a(n-1) + 4*(-2*n+3)*a(n-2) + 4*(-n+1)*a(n-3) + 16*(n-3)*a(n-4) = 0. - Alexander R. Povolotsky, Aug 17 2012
Sum_{n>=0} 1/a(n) = 4/3 + 8*Pi/(9*sqrt(3)). - Alexander R. Povolotsky, Aug 18 2012
E.g.f.: U(0) where U(k)= 1 + x/(1 - x/(x + (k+1)*(k+1)/U(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Oct 19 2012
Central column of the coefficients of the swinging polynomials A162246. - Peter Luschny, Oct 22 2013
a(n) = Sum_{k=0..n} A189231(n, 2*k). (Cf. A212303 for the odd case.) - Peter Luschny, Oct 30 2013
a(n) = hypergeometric([-n,-n-1,1/2],[-n-2,1],2)*2^(n-1)*(n+2). - Peter Luschny, Sep 22 2014
a(n) = 4^floor(n/2)*hypergeometric([-floor(n/2), (-1)^n/2], [1], 1). - Peter Luschny, May 19 2015
Sum_{n>=0} (-1)^n/a(n) = 4/3 - 4*Pi/(9*sqrt(3)). - Amiram Eldar, Mar 10 2022

Extensions

Extended and edited by Peter Luschny, Jun 28 2009

A000172 The Franel number a(n) = Sum_{k = 0..n} binomial(n,k)^3.

Original entry on oeis.org

1, 2, 10, 56, 346, 2252, 15184, 104960, 739162, 5280932, 38165260, 278415920, 2046924400, 15148345760, 112738423360, 843126957056, 6332299624282, 47737325577620, 361077477684436, 2739270870994736, 20836827035351596, 158883473753259752, 1214171997616258240
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Cusick gives a general method of deriving recurrences for the r-th order Franel numbers (this is the sequence of third-order Franel numbers), with floor((r+3)/2) terms.
This is the Taylor expansion of a special point on a curve described by Beauville. - Matthijs Coster, Apr 28 2004
An identity of V. Strehl states that a(n) = Sum_{k = 0..n} C(n,k)^2 * binomial(2*k,n). Zhi-Wei Sun conjectured that for every n = 2,3,... the polynomial f_n(x) = Sum_{k = 0..n} binomial(n,k)^2 * binomial(2*k,n) * x^(n-k) is irreducible over the field of rational numbers. - Zhi-Wei Sun, Mar 21 2013
Conjecture: a(n) == 2 (mod n^3) iff n is prime. - Gary Detlefs, Mar 22 2013
a(p) == 2 (mod p^3) for any prime p since p | C(p,k) for all k = 1,...,p-1. - Zhi-Wei Sun, Aug 14 2013
a(n) is the maximal number of totally mixed Nash equilibria in games of 3 players, each with n+1 pure options. - Raimundas Vidunas, Jan 22 2014
This is one of the Apéry-like sequences - see Cross-references. - Hugo Pfoertner, Aug 06 2017
Diagonal of rational functions 1/(1 - x*y - y*z - x*z - 2*x*y*z), 1/(1 - x - y - z + 4*x*y*z), 1/(1 + y + z + x*y + y*z + x*z + 2*x*y*z), 1/(1 + x + y + z + 2*(x*y + y*z + x*z) + 4*x*y*z). - Gheorghe Coserea, Jul 04 2018
a(n) is the constant term in the expansion of ((1 + x) * (1 + y) + (1 + 1/x) * (1 + 1/y))^n. - Seiichi Manyama, Oct 27 2019
Diagonal of rational function 1 / ((1-x)*(1-y)*(1-z) - x*y*z). - Seiichi Manyama, Jul 11 2020
Named after the Swiss mathematician Jérôme Franel (1859-1939). - Amiram Eldar, Jun 15 2021
It appears that a(n) is equal to the coefficient of (x*y*z)^n in the expansion of (1 + x + y - z)^n * (1 + x - y + z)^n * (1 - x + y + z)^n. Cf. A036917. - Peter Bala, Sep 20 2021

Examples

			O.g.f.: A(x) = 1 + 2*x + 10*x^2 + 56*x^3 + 346*x^4 + 2252*x^5 + ...
O.g.f.: A(x) = 1/(1-2*x) + 3!*x^2/(1-2*x)^4 + (6!/2!^3)*x^4/(1-2*x)^7 + (9!/3!^3)*x^6/(1-2*x)^10 + (12!/4!^3)*x^8/(1-2*x)^13 + ... - _Paul D. Hanna_, Oct 30 2010
Let g.f. A(x) = Sum_{n >= 0} a(n)*x^n/n!^3, then
A(x) = 1 + 2*x + 10*x^2/2!^3 + 56*x^3/3!^3 + 346*x^4/4!^3 + ... where
A(x) = [1 + x + x^2/2!^3 + x^3/3!^3 + x^4/4!^3 + ...]^2. - _Paul D. Hanna_

References

  • Matthijs Coster, Over 6 families van krommen [On 6 families of curves], Master's Thesis (unpublished), Aug 26 1983.
  • Jérôme Franel, On a question of Laisant, Intermédiaire des Mathématiciens, vol 1 1894 pp 45-47
  • H. W. Gould, Combinatorial Identities, Morgantown, 1972, (X.14), p. 56.
  • Murray Klamkin, ed., Problems in Applied Mathematics: Selections from SIAM Review, SIAM, 1990; see pp. 148-149.
  • John Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 193.
  • 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).

Links

Crossrefs

Cf. A002893, A052144, A005260, A096191, A033581, A189791. Second row of array A094424.
Cf. A181543, A006480, A141057, A000578, A007318.
The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)
For primes that do not divide the terms of the sequences A000172, A005258, A002893, A081085, A006077, A093388, A125143, A229111, A002895, A290575, A290576, A005259 see A260793, A291275-A291284 and A133370 respectively.
Sum_{k = 0..n} C(n,k)^m for m = 1..12: A000079, A000984, A000172, A005260, A005261, A069865, A182421, A182422, A182446, A182447, A342294, A342295.
Column k=3 of A372307.

Programs

  • Haskell
    a000172 = sum . map a000578 . a007318_row
-- Reinhard Zumkeller, Jan 06 2013
  • Maple
    A000172 := proc(n)
    add(binomial(n,k)^3,k=0..n) ;
end proc:
seq(A000172(n),n=0..10) ; # R. J. Mathar, Jul 26 2014
A000172_list := proc(len) series(hypergeom([], [1, 1], x)^2, x, len);
seq((n!)^3*coeff(%, x, n), n=0..len-1) end:
A000172_list(21); # Peter Luschny, May 31 2017
  • Mathematica
    Table[Sum[Binomial[n,k]^3,{k,0,n}],{n,0,30}] (* Harvey P. Dale, Aug 24 2011 *)
Table[ HypergeometricPFQ[{-n, -n, -n}, {1, 1}, -1], {n, 0, 20}]  (* Jean-François Alcover, Jul 16 2012, after symbolic sum *)
a[n_] := Sum[ Binomial[2k, n]*Binomial[2k, k]*Binomial[2(n-k), n-k], {k, 0, n}]/2^n; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Mar 20 2013, after Zhi-Wei Sun *)
a[ n_] := SeriesCoefficient[ Hypergeometric2F1[ 1/3, 2/3, 1, 27 x^2 / (1 - 2 x)^3] / (1 - 2 x), {x, 0, n}]; (* Michael Somos, Jul 16 2014 *)
  • PARI
    {a(n)=polcoeff(sum(m=0,n,(3*m)!/m!^3*x^(2*m)/(1-2*x+x*O(x^n))^(3*m+1)),n)} \\ Paul D. Hanna, Oct 30 2010
  • PARI
    {a(n)=n!^3*polcoeff(sum(m=0,n,x^m/m!^3+x*O(x^n))^2,n)} \\ Paul D. Hanna, Jan 19 2011
  • PARI
    A000172(n)={sum(k=0,(n-1)\2,binomial(n,k)^3)*2+if(!bittest(n,0),binomial(n,n\2)^3)} \\ M. F. Hasler, Sep 21 2015
  • Sage
    def A000172():
    x, y, n = 1, 2, 1
    while True:
        yield x
        n += 1
        x, y = y, (8*(n-1)^2*x + (7*n^2-7*n + 2)*y) // n^2
a = A000172()
[next(a) for i in range(21)]   # Peter Luschny, Oct 12 2013

Formula

A002893(n) = Sum_{m = 0..n} binomial(n, m)*a(m) [Barrucand].
Sum_{k = 0..n} C(n, k)^3 = (-1)^n*Integral_{x = 0..infinity} L_k(x)^3 exp(-x) dx. - from Askey's book, p. 43.
D-finite with recurrence (n + 1)^2*a(n+1) = (7*n^2 + 7*n + 2)*a(n) + 8*n^2*a(n-1) [Franel]. - Felix Goldberg (felixg(AT)tx.technion.ac.il), Jan 31 2001
a(n) ~ 2*3^(-1/2)*Pi^-1*n^-1*2^(3*n). - Joe Keane (jgk(AT)jgk.org), Jun 21 2002
O.g.f.: A(x) = Sum_{n >= 0} (3*n)!/n!^3 * x^(2*n)/(1 - 2*x)^(3*n+1). - Paul D. Hanna, Oct 30 2010
G.f.: hypergeom([1/3, 2/3], [1], 27 x^2 / (1 - 2x)^3) / (1 - 2x). - Michael Somos, Dec 17 2010
G.f.: Sum_{n >= 0} a(n)*x^n/n!^3 = [ Sum_{n >= 0} x^n/n!^3 ]^2. - Paul D. Hanna, Jan 19 2011
G.f.: A(x) = 1/(1-2*x)*(1+6*(x^2)/(G(0)-6*x^2)),
with G(k) = 3*(x^2)*(3*k+1)*(3*k+2) + ((1-2*x)^3)*((k+1)^2) - 3*(x^2)*((1-2*x)^3)*((k+1)^2)*(3*k+4)*(3*k+5)/G(k+1) ; (continued fraction). - Sergei N. Gladkovskii, Dec 03 2011
In 2011 Zhi-Wei Sun found the formula Sum_{k = 0..n} C(2*k,n)*C(2*k,k)*C(2*(n-k),n-k) = (2^n)*a(n) and proved it via the Zeilberger algorithm. - Zhi-Wei Sun, Mar 20 2013
0 = a(n)*(a(n+1)*(-2048*a(n+2) - 3392*a(n+3) + 768*a(n+4)) + a(n+2)*(-1280*a(n+2) - 2912*a(n+3) + 744*a(n+4)) + a(n+3)*(+288*a(n+3) - 96*a(n+4))) + a(n+1)*(a(n+1)*(-704*a(n+2) - 1232*a(n+3) + 288*a(n+4)) + a(n+2)*(-560*a(n+2) - 1372*a(n+3) + 364*a(n+4)) + a(n+3)*(+154*a(n+3) - 53*a(n+4))) + a(n+2)*(a(n+2)*(+24*a(n+2) + 70*a(n+3) - 20*a(n+4)) + a(n+3)*(-11*a(n+3) + 4*a(n+4))) for all n in Z. - Michael Somos, Jul 16 2014
For r a nonnegative integer, Sum_{k = r..n} C(k,r)^3*C(n,k)^3 = C(n,r)^3*a(n-r), where we take a(n) = 0 for n < 0. - Peter Bala, Jul 27 2016
a(n) = (n!)^3 * [x^n] hypergeom([], [1, 1], x)^2. - Peter Luschny, May 31 2017
From Gheorghe Coserea, Jul 04 2018: (Start)
a(n) = Sum_{k=0..floor(n/2)} (n+k)!/(k!^3*(n-2*k)!) * 2^(n-2*k).
G.f. y=A(x) satisfies: 0 = x*(x + 1)*(8*x - 1)*y'' + (24*x^2 + 14*x - 1)*y' + 2*(4*x + 1)*y. (End)
a(n) = [x^n] (1 - x^2)^n*P(n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. See Gould, p. 56. - Peter Bala, Mar 24 2022
a(n) = (2^n/(4*Pi^2)) * Integral_{x,y=0..2*Pi} (1+cos(x)+cos(y)+cos(x+y))^n dx dy = (8^n/(Pi^2)) * Integral_{x,y=0..Pi} (cos(x)*cos(y)*cos(x+y))^n dx dy (Pla, 1995). - Amiram Eldar, Jul 16 2022
a(n) = Sum_{k = 0..n} m^(n-k)*binomial(n,k)*binomial(n+2*k,n)*binomial(2*k,k) at m = -4. Cf. A081798 (m = 1), A006480 (m = 0), A124435 (m = -1), A318109 (m = -2) and A318108 (m = -3). - Peter Bala, Mar 16 2023
From Bradley Klee, Jun 05 2023: (Start)
The g.f. T(x) obeys a period-annihilating ODE:
0=2*(1 + 4*x)*T(x) + (-1 + 14*x + 24*x^2)*T'(x) + x*(1 + x)*(-1 + 8*x)*T''(x).
The periods ODE can be derived from the following Weierstrass data:
g2 = (4/243)*(1 - 8*x + 240*x^2 - 464*x^3 + 16*x^4);
g3 = -(8/19683)*(1 - 12*x - 480*x^2 + 3080*x^3 - 12072*x^4 + 4128*x^5 +
64*x^6);
which determine an elliptic surface with four singular fibers. (End)
From Peter Bala, Oct 31 2024: (Start)
For n >= 1, a(n) = 2 * Sum_{k = 0..n-1} binomial(n, k)^2 * binomial(n-1, k). Cf. A361716.
For n >= 1, a(n) = 2 * hypergeom([-n, -n, -n + 1], [1, 1], -1). (End)

A001813 Quadruple factorial numbers: a(n) = (2n)!/n!.

Original entry on oeis.org

1, 2, 12, 120, 1680, 30240, 665280, 17297280, 518918400, 17643225600, 670442572800, 28158588057600, 1295295050649600, 64764752532480000, 3497296636753920000, 202843204931727360000, 12576278705767096320000, 830034394580628357120000, 58102407620643984998400000
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Counts binary rooted trees (with out-degree <= 2), embedded in plane, with n labeled end nodes of degree 1. Unlabeled version gives Catalan numbers A000108.
Define a "downgrade" to be the permutation which places the items of a permutation in descending order. We are concerned with permutations that are identical to their downgrades. Only permutations of order 4n and 4n+1 can have this property; the number of permutations of length 4n having this property are equinumerous with those of length 4n+1. If a permutation p has this property then the reversal of this permutation also has it. a(n) = number of permutations of length 4n and 4n+1 that are identical to their downgrades. - Eugene McDonnell (eemcd(AT)mac.com), Oct 26 2003
Number of broadcast schemes in the complete graph on n+1 vertices, K_{n+1}. - Calin D. Morosan (cd_moros(AT)alumni.concordia.ca), Nov 28 2008
Hankel transform is A137565. - Paul Barry, Nov 25 2009
The e.g.f. of 1/a(n) = n!/(2*n)! is (exp(sqrt(x)) + exp(-sqrt(x)) )/2. - Wolfdieter Lang, Jan 09 2012
From Tom Copeland, Nov 15 2014: (Start)
Aerated with intervening zeros (1,0,2,0,12,0,120,...) = a(n) (cf. A123023 and A001147), the e.g.f. is e^(t^2), so this is the base for the Appell sequence with e.g.f. e^(t^2) e^(x*t) = exp(P(.,x),t) (reverse A059344, cf. A099174, A066325 also). P(n,x) = (a. + x)^n with (a.)^n = a_n and comprise the umbral compositional inverses for e^(-t^2)e^(x*t) = exp(UP(.,x),t), i.e., UP(n,P(.,t)) = x^n = P(n,UP(.,t)), e.g., (P(.,t))^n = P(n,t).
Equals A000407*2 with leading 1 added. (End)
a(n) is also the number of square roots of any permutation in S_{4*n} whose disjoint cycle decomposition consists of 2*n transpositions. - Luis Manuel Rivera Martínez, Mar 04 2015
Self-convolution gives A076729. - Vladimir Reshetnikov, Oct 11 2016
For n > 1, it follows from the formula dated Aug 07 2013 that a(n) is a Zumkeller number (A083207). - Ivan N. Ianakiev, Feb 28 2017
For n divisible by 4, a(n/4) is the number of ways to place n points on an n X n grid with pairwise distinct abscissae, pairwise distinct ordinates, and 90-degree rotational symmetry. For n == 1 (mod 4), the number of ways is a((n-1)/4) because the center point can be considered "fixed". For 180-degree rotational symmetry see A006882, for mirror symmetry see A000085, A135401, and A297708. - Manfred Scheucher, Dec 29 2017

Examples

			The following permutations of order 8 and their reversals have this property:
  1 7 3 5 2 4 0 6
  1 7 4 2 5 3 0 6
  2 3 7 6 1 0 4 5
  2 4 7 1 6 0 3 5
  3 2 6 7 0 1 5 4
  3 5 1 7 0 6 2 4

References

  • D. E. Knuth, The Art of Computer Programming, Vol. 4, Section 7.2.1.6, Eq. 32.
  • L. C. Larson, The number of essentially different nonattacking rook arrangements, J. Recreat. Math., 7 (No. 3, 1974), circa pages 180-181.
  • Eugene McDonnell, "Magic Squares and Permutations" APL Quote-Quad 7.3 (Fall, 1976)
  • R. W. Robinson, Counting arrangements of bishops, pp. 198-214 of Combinatorial Mathematics IV (Adelaide 1975), Lect. Notes Math., 560 (1976).
  • 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).

Links

Crossrefs

Cf. A037224, A048854, A001147, A007696, A008545, A122670 (essentially the same sequence), A000165, A047055, A047657, A084947, A084948, A084949, A010050, A000142, A008275, A000108, A000984, A008276, A000680, A094216.
Catalan(n-1)*M^(n-1)*n! for M=1,2,3,4,5,6: A001813, A052714 (or A144828), A221954, A052734, A221953, A221955.
Cf. A123023, A059344, A099174, A066325, A001700, A000407, A006882.
Cf. A000085, A135401, A297708. - Manfred Scheucher, Jan 07 2018

Programs

  • GAP
    List([0..20],n->Factorial(2*n)/Factorial(n)); # Muniru A Asiru, Nov 01 2018
  • Magma
    [Factorial(2*n)/Factorial(n): n in [0..20]]; // Vincenzo Librandi, Oct 09 2018
  • Maple
    A001813 := n->(2*n)!/n!;
A001813 := n -> mul(k, k = select(k-> k mod 4 = 2,[$1 .. 4*n])):
seq(A001813(n), n=0..16);  # Peter Luschny, Jun 23 2011
  • Mathematica
    Table[(2n)!/n!, {n,0,20}] (* Harvey P. Dale, May 02 2011 *)
  • Maxima
    makelist(binomial(n+n, n)*n!,n,0,30); /* Martin Ettl, Nov 05 2012 */
  • PARI
    a(n)=binomial(n+n,n)*n! \\ Charles R Greathouse IV, Jun 15 2011
  • PARI
    first(n) = x='x+O('x^n); Vec(serlaplace((1 - 4*x)^(-1/2))) \\ Iain Fox, Jan 01 2018 (corrected by Iain Fox, Jan 11 2018)
  • Python
    from math import factorial
def A001813(n): return factorial(n<<1)//factorial(n) # Chai Wah Wu, Feb 14 2023
  • Sage
    [binomial(2*n,n)*factorial(n) for n in range(0, 17)] # Zerinvary Lajos, Dec 03 2009

Formula

E.g.f.: (1-4*x)^(-1/2).
a(n) = (2*n)!/n! = Product_{k=0..n-1} (4*k + 2) = A081125(2*n).
Integral representation as n-th moment of a positive function on a positive half-axis: a(n) = Integral_{x=0..oo} x^n*exp(-x/4)/(sqrt(x)*2*sqrt(Pi)) dx, n >= 0. This representation is unique. - Karol A. Penson, Sep 18 2001
Define a'(1)=1, a'(n) = Sum_{k=1..n-1} a'(n-k)*a'(k)*C(n, k); then a(n)=a'(n+1). - Benoit Cloitre, Apr 27 2003
With interpolated zeros (1, 0, 2, 0, 12, ...) this has e.g.f. exp(x^2). - Paul Barry, May 09 2003
a(n) = A000680(n)/A000142(n)*A000079(n) = Product_{i=0..n-1} (4*i + 2) = 4^n*Pochhammer(1/2, n) = 4^n*GAMMA(n+1/2)/sqrt(Pi). - Daniel Dockery (peritus(AT)gmail.com), Jun 13 2003
For asymptotics, see the Robinson paper.
a(k) = (2*k)!/k! = Sum_{i=1..k+1} |A008275(i,k+1)| * k^(i-1). - André F. Labossière, Jun 21 2007
a(n) = 12*A051618(a) n >= 2. - Zerinvary Lajos, Feb 15 2008
a(n) = A000984(n)*A000142(n). - Zerinvary Lajos, Mar 25 2008
a(n) = A016825(n-1)*a(n-1). - Roger L. Bagula, Sep 17 2008
a(n) = (-1)^n*A097388(n). - D. Morosan (cd_moros(AT)alumni.concordia.ca), Nov 28 2008
From Paul Barry, Jan 15 2009: (Start)
G.f.: 1/(1-2x/(1-4x/(1-6x/(1-8x/(1-10x/(1-... (continued fraction);
a(n) = (n+1)!*A000108(n). (End)
a(n) = Sum_{k=0..n} A132393(n,k)*2^(2n-k). - Philippe Deléham, Feb 10 2009
G.f.: 1/(1-2x-8x^2/(1-10x-48x^2/(1-18x-120x^2/(1-26x-224x^2/(1-34x-360x^2/(1-42x-528x^2/(1-... (continued fraction). - Paul Barry, Nov 25 2009
a(n) = A173333(2*n,n) for n>0; cf. A006963, A001761. - Reinhard Zumkeller, Feb 19 2010
From Gary W. Adamson, Jul 19 2011: (Start)
a(n) = upper left term of M^n, M = an infinite square production matrix as follows:
2, 2, 0, 0, 0, 0, ...
4, 4, 4, 0, 0, 0, ...
6, 6, 6, 6, 0, 0, ...
8, 8, 8, 8, 8, 0, ...
...
(End)
a(n) = (-2)^n*Sum_{k=0..n} 2^k*s(n+1,n+1-k), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012
G.f.: 1/Q(0), where Q(k) = 1 + x*(4*k+2) - x*(4*k+4)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 18 2013
G.f.: 2/G(0), where G(k) = 1 + 1/(1 - x*(8*k+4)/(x*(8*k+4) - 1 + 8*x*(k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 30 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - 2*x/(2*x + 1/(2*k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013
D-finite with recurrence: a(n) = (4*n-6)*a(n-2) + (4*n-3)*a(n-1), n>=2. - Ivan N. Ianakiev, Aug 07 2013
Sum_{n>=0} 1/a(n) = (exp(1/4)*sqrt(Pi)*erf(1/2) + 2)/2 = 1 + A214869, where erf(x) is the error function. - Ilya Gutkovskiy, Nov 10 2016
Sum_{n>=0} (-1)^n/a(n) = 1 - sqrt(Pi)*erfi(1/2)/(2*exp(1/4)), where erfi(x) is the imaginary error function. - Amiram Eldar, Feb 20 2021
a(n) = 1/([x^n] hypergeom([1], [1/2], x/4)). - Peter Luschny, Sep 13 2024
a(n) = 2^n*n!*JacobiP(n, -1/2, -n, 3). - Peter Luschny, Jan 22 2025
G.f.: 2F0(1,1/2;;4x). - R. J. Mathar, Jun 07 2025

Extensions

More terms from James Sellers, May 01 2000
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