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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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A001109 a(n)^2 is a triangular number: a(n) = 6*a(n-1) - a(n-2) with a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 6, 35, 204, 1189, 6930, 40391, 235416, 1372105, 7997214, 46611179, 271669860, 1583407981, 9228778026, 53789260175, 313506783024, 1827251437969, 10650001844790, 62072759630771, 361786555939836, 2108646576008245, 12290092900109634, 71631910824649559, 417501372047787720
Offset: 0

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Comments

8*a(n)^2 + 1 = 8*A001110(n) + 1 = A055792(n+1) is a perfect square. - Gregory V. Richardson, Oct 05 2002
For n >= 2, A001108(n) gives exactly the positive integers m such that 1,2,...,m has a perfect median. The sequence of associated perfect medians is the present sequence. Let a_1,...,a_m be an (ordered) sequence of real numbers, then a term a_k is a perfect median if Sum_{j=1..k-1} a_j = Sum_{j=k+1..m} a_j. See Puzzle 1 in MSRI Emissary, Fall 2005. - Asher Auel, Jan 12 2006
(a(n), b(n)) where b(n) = A082291(n) are the integer solutions of the equation 2*binomial(b,a) = binomial(b+2,a). - Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de); comment revised by Michael Somos, Apr 07 2003
This sequence gives the values of y in solutions of the Diophantine equation x^2 - 8y^2 = 1. It also gives the values of the product xy where (x,y) satisfies x^2 - 2y^2 = +-1, i.e., a(n) = A001333(n)*A000129(n). a(n) also gives the inradius r of primitive Pythagorean triangles having legs whose lengths are consecutive integers, with corresponding semiperimeter s = a(n+1) = {A001652(n) + A046090(n) + A001653(n)}/2 and area rs = A029549(n) = 6*A029546(n). - Lekraj Beedassy, Apr 23 2003 [edited by Jon E. Schoenfield, May 04 2014]
n such that 8*n^2 = floor(sqrt(8)*n*ceiling(sqrt(8)*n)). - Benoit Cloitre, May 10 2003
For n > 0, ratios a(n+1)/a(n) may be obtained as convergents to continued fraction expansion of 3+sqrt(8): either successive convergents of [6;-6] or odd convergents of [5;1, 4]. - Lekraj Beedassy, Sep 09 2003
a(n+1) + A053141(n) = A001108(n+1). Generating floretion: - 2'i + 2'j - 'k + i' + j' - k' + 2'ii' - 'jj' - 2'kk' + 'ij' + 'ik' + 'ji' + 'jk' - 2'kj' + 2e ("jes" series). - Creighton Dement, Dec 16 2004
Kekulé numbers for certain benzenoids (see the Cyvin-Gutman reference). - Emeric Deutsch, Jun 19 2005
Number of D steps on the line y=x in all Delannoy paths of length n (a Delannoy path of length n is a path from (0,0) to (n,n), consisting of steps E=(1,0), N=(0,1) and D=(1,1)). Example: a(2)=6 because in the 13 (=A001850(2)) Delannoy paths of length 2, namely (DD), (D)NE, (D)EN, NE(D), NENE, NEEN, NDE, NNEE, EN(D), ENNE, ENEN, EDN and EENN, we have altogether six D steps on the line y=x (shown between parentheses). - Emeric Deutsch, Jul 07 2005
Define a T-circle to be a first-quadrant circle with integral radius that is tangent to the x- and y-axes. Such a circle has coordinates equal to its radius. Let C(0) be the T-circle with radius 1. Then for n > 0, define C(n) to be the smallest T-circle that does not intersect C(n-1). C(n) has radius a(n+1). Cf. A001653. - Charlie Marion, Sep 14 2005
Numbers such that there is an m with t(n+m)=2t(m), where t(n) are the triangular numbers A000217. For instance, t(20)=2*t(14)=210, so 6 is in the sequence. - Floor van Lamoen, Oct 13 2005
One half the bisection of the Pell numbers (A000129). - Franklin T. Adams-Watters, Jan 08 2006
Pell trapezoids: for n > 0, a(n) = (A000129(n-1)+A000129(n+1))*A000129(n)/2; see also A084158. - Charlie Marion, Apr 01 2006
Tested for 2 < p < 27: If and only if 2^p - 1 (the Mersenne number M(p)) is prime then M(p) divides a(2^(p-1)). - Kenneth J Ramsey, May 16 2006
If 2^p - 1 is prime then M(p) divides a(2^(p-1)-1). - Kenneth J Ramsey, Jun 08 2006; comment corrected by Robert Israel, Mar 18 2007
If 8*n+5 and 8*n+7 are twin primes then their product divides a(4*n+3). - Kenneth J Ramsey, Jun 08 2006
If p is an odd prime, then if p == 1 or 7 (mod 8), then a((p-1)/2) == 0 (mod p) and a((p+1)/2) == 1 (mod p); if p == 3 or 5 (mod 8), then a((p-1)/2) == 1 (mod p) and a((p+1)/2) == 0 (mod p). Kenneth J Ramsey's comment about twin primes follows from this. - Robert Israel, Mar 18 2007
a(n)*(a(n+b) - a(b-2)) = (a(n+1)+1)*(a(n+b-1) - a(b-1)). This identity also applies to any series a(0) = 0 a(1) = 1 a(n) = b*a(n-1) - a(n-2). - Kenneth J Ramsey, Oct 17 2007
For n < 0, let a(n) = -a(-n). Then (a(n+j) + a(k+j)) * (a(n+b+k+j) - a(b-j-2)) = (a(n+j+1) + a(k+j+1)) * (a(n+b+k+j-1) - a(b-j-1)). - Charlie Marion, Mar 04 2011
Sequence gives y values of the Diophantine equation: 0+1+2+...+x = y^2. If (a,b) and (c,d) are two consecutive solutions of the Diophantine equation: 0+1+2+...+x = y^2 with aMohamed Bouhamida, Aug 29 2009
If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: 0+1+2+...+x = y^2 with p < r then r = 3*p+4*q+1 and s = 2*p+3*q+1. - Mohamed Bouhamida, Sep 02 2009
a(n)/A002315(n) converges to cos^2(Pi/8) (see A201488). - Gary Detlefs, Nov 25 2009
Binomial transform of A086347. - Johannes W. Meijer, Aug 01 2010
If x=a(n), y=A055997(n+1) and z = x^2+y, then x^4 + y^3 = z^2. - Bruno Berselli, Aug 24 2010
In general, if b(0)=1, b(1)=k and for n > 1, b(n) = 6*b(n-1) - b(n-2), then
for n > 0, b(n) = a(n)*k-a(n-1); e.g.,
for k=2, when b(n) = A038725(n), 2 = 1*2 - 0, 11 = 6*2 - 1, 64 = 35*2 - 6, 373 = 204*2 - 35;
for k=3, when b(n) = A001541(n), 3 = 1*3 - 0, 17 = 6*3 - 1; 99 = 35*3 - 6; 577 = 204*3 - 35;
for k=4, when b(n) = A038723(n), 4 = 1*4 - 0, 23 = 6*4 - 1; 134 = 35*4 - 6; 781 = 204*4 - 35;
for k=5, when b(n) = A001653(n), 5 = 1*5 - 0, 29 = 6*5 - 1; 169 = 35*5 - 6; 985 = 204*5 - 35.
- Charlie Marion, Dec 08 2010
See a Wolfdieter Lang comment on A001653 on a sequence of (u,v) values for Pythagorean triples (x,y,z) with x=|u^2-v^2|, y=2*u*v and z=u^2+v^2, with u odd and v even, generated from (u(0)=1,v(0)=2), the triple (3,4,5), by a substitution rule given there. The present a(n) appears there as b(n). The corresponding generated triangles have catheti differing by one length unit. - Wolfdieter Lang, Mar 06 2012
a(n)*a(n+2k) + a(k)^2 and a(n)*a(n+2k+1) + a(k)*a(k+1) are triangular numbers. Generalizes description of sequence. - Charlie Marion, Dec 03 2012
a(n)*a(n+2k) + a(k)^2 is the triangular square A001110(n+k). a(n)*a(n+2k+1) + a(k)*a(k+1) is the triangular oblong A029549(n+k). - Charlie Marion, Dec 05 2012
From Richard R. Forberg, Aug 30 2013: (Start)
The squares of a(n) are the result of applying triangular arithmetic to the squares, using A001333 as the "guide" on what integers to square, as follows:
a(2n)^2 = A001333(2n)^2 * (A001333(2n)^2 - 1)/2;
a(2n+1)^2 = A001333(2n+1)^2 * (A001333(2n+1)^2 + 1)/2. (End)
For n >= 1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,5}. - Milan Janjic, Jan 25 2015
Panda and Rout call these "balancing numbers" and note that the period of the sequence modulo a prime p is the same as that modulo p^2 when p = 13, 31, 1546463. But these are precisely the p in A238736 such that p^2 divides A000129(p - (2/p)), where (2/p) is a Jacobi symbol. In light of the above observation by Franklin T. Adams-Watters that the present sequence is one half the bisection of the Pell numbers, i.e., a(n) = A000129(2*n)/2, it follows immediately that modulo a fixed prime p, or any power thereof, the period of a(n) is half that of A000129(n). - John Blythe Dobson, Mar 06 2015
The triangular number = square number identity Tri((T(n, 3) - 1)/2) = S(n-1, 6)^2 with Tri, T, and S given in A000217, A053120 and A049310, is the special case k = 1 of the k-family of identities Tri((T(n, 2*k+1) - 1)/2) = Tri(k)*S(n-1, 2*(2*k+1))^2, k >= 0, n >= 0, with S(-1, x) = 0. For k=2 see A108741(n) for S(n-1, 10)^2. This identity boils down to the identities S(n-1, 2*x)^2 = (T(2*n, x) - 1)/(2*(x^2-1)) and 2*T(n, x)^2 - 1 = T(2*n, x) with x = 2*k+1. - Wolfdieter Lang, Feb 01 2016
a(2)=6 is perfect. For n=2*k, k > 0, k not equal to 1, a(n) is a multiple of a(2) and since every multiple (beyond 1) of a perfect number is abundant, then a(n) is abundant. sigma(a(4)) = 504 > 408 = 2*a(4). For n=2*k+1, k > 0, a(n) mod 10 = A000012(n), so a(n) is odd. If a(n) is a prime number, it is deficient; otherwise a(n) has one or two distinct prime factors and is therefore deficient again. So for n=2k+1, k > 0, a(n) is deficient. sigma(a(5)) = 1260 < 2378 = 2*a(5). - Muniru A Asiru, Apr 14 2016
Behera & Panda call these the balancing numbers, and A001541 are the balancers. - Michel Marcus, Nov 07 2017
In general, a second-order linear recurrence with constant coefficients having a signature of (c,d) will be duplicated by a third-order recurrence having a signature of (x,c^2-c*x+d,-d*x+c*d). The formulas of Olivares and Bouhamida in the formula section which have signatures of (7,-7,1) and (5,5,-1), respectively, are specific instances of this general rule for x = 7 and x = 5. - Gary Detlefs, Jan 29 2021
Note that 6 is the largest triangular number in the sequence, because it is proved that 8 and 9 are the largest perfect powers which are consecutive (Catalan's conjecture). 0 and 1 are also in the sequence because they are also perfect powers and 0*1/2 = 0^2 and 8*9/2 = (2*3)^2. - Metin Sariyar, Jul 15 2021

Examples

			G.f. = x + 6*x^2 + 35*x^3 + 204*x^4 + 1189*x^5 + 6930*x^6 + 40391*x^7 + ...
6 is in the sequence since 6^2 = 36 is a triangular number: 36 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8. - _Michael B. Porter_, Jul 02 2016
		

References

  • Julio R. Bastida, Quadratic properties of a linearly recurrent sequence. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 163--166, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561042 (81e:10009) - From N. J. A. Sloane, May 30 2012
  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, pp. 193, 197.
  • D. M. Burton, The History of Mathematics, McGraw Hill, (1991), p. 213.
  • L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 10.
  • P. Franklin, E. F. Beckenbach, H. S. M Coxeter, N. H. McCoy, K. Menger, and J. L. Synge, Rings And Ideals, No 8, The Carus Mathematical Monographs, The Mathematical Association of America, (1967), pp. 144-146.
  • A. Patra, G. K. Panda, and T. Khemaratchatakumthorn. "Exact divisibility by powers of the balancing and Lucas-balancing numbers." Fibonacci Quart., 59:1 (2021), 57-64; see B(n).
  • 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 257-258.
  • P.-F. Teilhet, Query 2376, L'Intermédiaire des Mathématiciens, 11 (1904), 138-139. - N. J. A. Sloane, Mar 08 2022

Crossrefs

Chebyshev sequence U(n, m): A000027 (m=1), A001353 (m=2), this sequence (m=3), A001090 (m=4), A004189 (m=5), A004191 (m=6), A007655 (m=7), A077412 (m=8), A049660 (m=9), A075843 (m=10), A077421 (m=11), A077423 (m=12), A097309 (m=13), A097311 (m=14), A097313 (m=15), A029548 (m=16), A029547 (m=17), A144128 (m=18), A078987 (m=19), A097316 (m=33).
Cf. A323182.

Programs

  • GAP
    a:=[0,1];; for n in [3..25] do a[n]:=6*a[n-1]-a[n-2]; od; a; # Muniru A Asiru, Dec 18 2018
  • Haskell
    a001109 n = a001109_list !! n :: Integer
    a001109_list = 0 : 1 : zipWith (-)
       (map (* 6) $ tail a001109_list) a001109_list
    -- Reinhard Zumkeller, Dec 17 2011
    
  • Magma
    [n le 2 select n-1 else 6*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jul 25 2015
    
  • Maple
    a[0]:=1: a[1]:=6: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n],n=0..26); # Emeric Deutsch
    with (combinat):seq(fibonacci(2*n,2)/2, n=0..20); # Zerinvary Lajos, Apr 20 2008
  • Mathematica
    Transpose[NestList[Flatten[{Rest[#],ListCorrelate[{-1,6},#]}]&, {0,1}, 30]][[1]]  (* Harvey P. Dale, Mar 23 2011 *)
    CoefficientList[Series[x/(1-6x+x^2),{x,0,30}],x]  (* Harvey P. Dale, Mar 23 2011 *)
    LinearRecurrence[{6, -1}, {0, 1}, 50] (* Vladimir Joseph Stephan Orlovsky, Feb 12 2012 *)
    a[ n_]:= ChebyshevU[n-1, 3]; (* Michael Somos, Sep 02 2012 *)
    Table[Fibonacci[2n, 2]/2, {n, 0, 20}] (* Vladimir Reshetnikov, Sep 16 2016 *)
    TrigExpand@Table[Sinh[2 n ArcCsch[1]]/(2 Sqrt[2]), {n, 0, 10}] (* Federico Provvedi, Feb 01 2021 *)
  • PARI
    {a(n) = imag((3 + quadgen(32))^n)}; /* Michael Somos, Apr 07 2003 */
    
  • PARI
    {a(n) = subst( poltchebi( abs(n+1)) - 3 * poltchebi( abs(n)), x, 3) / 8}; /* Michael Somos, Apr 07 2003 */
    
  • PARI
    {a(n) = polchebyshev( n-1, 2, 3)}; /* Michael Somos, Sep 02 2012 */
    
  • PARI
    is(n)=ispolygonal(n^2,3) \\ Charles R Greathouse IV, Nov 03 2016
    
  • Sage
    [lucas_number1(n,6,1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008
    
  • Sage
    [chebyshev_U(n-1,3) for n in (0..20)] # G. C. Greubel, Dec 23 2019
    

Formula

G.f.: x / (1 - 6*x + x^2). - Simon Plouffe in his 1992 dissertation.
a(n) = S(n-1, 6) = U(n-1, 3) with U(n, x) Chebyshev's polynomials of the second kind. S(-1, x) := 0. Cf. triangle A049310 for S(n, x).
a(n) = sqrt(A001110(n)).
a(n) = A001542(n)/2.
a(n) = sqrt((A001541(n)^2-1)/8) (cf. Richardson comment).
a(n) = 3*a(n-1) + sqrt(8*a(n-1)^2+1). - R. J. Mathar, Oct 09 2000
a(n) = A000129(n)*A001333(n) = A000129(n)*(A000129(n)+A000129(n-1)) = ceiling(A001108(n)/sqrt(2)). - Henry Bottomley, Apr 19 2000
a(n) ~ (1/8)*sqrt(2)*(sqrt(2) + 1)^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
Limit_{n->oo} a(n)/a(n-1) = 3 + 2*sqrt(2). - Gregory V. Richardson, Oct 05 2002
a(n) = ((3 + 2*sqrt(2))^n - (3 - 2*sqrt(2))^n) / (4*sqrt(2)). - Gregory V. Richardson, Oct 13 2002. Corrected for offset 0, and rewritten. - Wolfdieter Lang, Feb 10 2015
a(2*n) = a(n)*A003499(n). 4*a(n) = A005319(n). - Mario Catalani (mario.catalani(AT)unito.it), Mar 21 2003
a(n) = floor((3+2*sqrt(2))^n/(4*sqrt(2))). - Lekraj Beedassy, Apr 23 2003
a(-n) = -a(n). - Michael Somos, Apr 07 2003
For n >= 1, a(n) = Sum_{k=0..n-1} A001653(k). - Charlie Marion, Jul 01 2003
For n > 0, 4*a(2*n) = A001653(n)^2 - A001653(n-1)^2. - Charlie Marion, Jul 16 2003
For n > 0, a(n) = Sum_{k = 0..n-1}((2*k+1)*A001652(n-1-k)) + A000217(n). - Charlie Marion, Jul 18 2003
a(2*n+1) = a(n+1)^2 - a(n)^2. - Charlie Marion, Jan 12 2004
a(k)*a(2*n+k) = a(n+k)^2 - a(n)^2; e.g., 204*7997214 = 40391^2 - 35^2. - Charlie Marion, Jan 15 2004
For j < n+1, a(k+j)*a(2*n+k-j) - Sum_{i = 0..j-1} a(2*n-(2*i+1)) = a(n+k)^2 - a(n)^2. - Charlie Marion, Jan 18 2004
From Paul Barry, Feb 06 2004: (Start)
a(n) = A000129(2*n)/2;
a(n) = ((1+sqrt(2))^(2*n) - (1-sqrt(2))^(2*n))*sqrt(2)/8;
a(n) = Sum_{i=0..n} Sum_{j=0..n} A000129(i+j)*n!/(i!*j!*(n-i-j)!)/2. (End)
E.g.f.: exp(3*x)*sinh(2*sqrt(2)*x)/(2*sqrt(2)). - Paul Barry, Apr 21 2004
A053141(n+1) + A055997(n+1) = A001541(n+1) + a(n+1). - Creighton Dement, Sep 16 2004
a(n) = Sum_{k=0..n} binomial(2*n, 2*k+1)*2^(k-1). - Paul Barry, Oct 01 2004
a(n) = A001653(n+1) - A038723(n); (a(n)) = chuseq[J]( 'ii' + 'jj' + .5'kk' + 'ij' - 'ji' + 2.5e ), apart from initial term. - Creighton Dement, Nov 19 2004, modified by Davide Colazingari, Jun 24 2016
a(n+1) = Sum_{k=0..n} A001850(k)*A001850(n-k), self convolution of central Delannoy numbers. - Benoit Cloitre, Sep 28 2005
a(n) = 7*(a(n-1) - a(n-2)) + a(n-3), a(1) = 0, a(2) = 1, a(3) = 6, n > 3. Also a(n) = ( (1 + sqrt(2) )^(2*n) - (1 - sqrt(2) )^(2*n) ) / (4*sqrt(2)). - Antonio Alberto Olivares, Oct 23 2003
a(n) = 5*(a(n-1) + a(n-2)) - a(n-3). - Mohamed Bouhamida, Sep 20 2006
Define f(x,s) = s*x + sqrt((s^2-1)*x^2+1); f(0,s)=0. a(n) = f(a(n-1),3), see second formula. - Marcos Carreira, Dec 27 2006
The perfect median m(n) can be expressed in terms of the Pell numbers P() = A000129() by m(n) = P(n + 2) * (P(n + 2) + P(n + 1)) for n >= 0. - Winston A. Richards (ugu(AT)psu.edu), Jun 11 2007
For k = 0..n, a(2*n-k) - a(k) = 2*a(n-k)*A001541(n). Also, a(2*n+1-k) - a(k) = A002315(n-k)*A001653(n). - Charlie Marion, Jul 18 2007
[A001653(n), a(n)] = [1,4; 1,5]^n * [1,0]. - Gary W. Adamson, Mar 21 2008
a(n) = Sum_{k=0..n-1} 4^k*binomial(n+k,2*k+1). - Paul Barry, Apr 20 2009
a(n+1)^2 - 6*a(n+1)*a(n) + a(n)^2 = 1. - Charlie Marion, Dec 14 2010
a(n) = A002315(m)*A011900(n-m-1) + A001653(m)*A001652(n-m-1) - a(m) = A002315(m)*A053141(n-m-1) + A001653(m)*A046090(n-m-1) + a(m) with m < n; otherwise a(n) = A002315(m)*A053141(m-n) - A001653(m)*A011900(m-n) + a(m) = A002315(m)*A053141(m-n) - A001653(m)*A046090(m-n) - a(m) = (A002315(n) - A001653(n))/2. - Kenneth J Ramsey, Oct 12 2011
16*a(n)^2 + 1 = A056771(n). - James R. Buddenhagen, Dec 09 2011
A010054(A000290(a(n))) = 1. - Reinhard Zumkeller, Dec 17 2011
In general, a(n+k)^2 - A003499(k)*a(n+k)*a(n) + a(n)^2 = a(k)^2. - Charlie Marion, Jan 11 2012
a(n+1) = Sum_{k=0..n} A101950(n,k)*5^k. - Philippe Deléham, Feb 10 2012
PSUM transform of a(n+1) is A053142. PSUMSIGN transform of a(n+1) is A084158. BINOMIAL transform of a(n+1) is A164591. BINOMIAL transform of A086347 is a(n+1). BINOMIAL transform of A057087(n-1). - Michael Somos, May 11 2012
a(n+k) = A001541(k)*a(n) + sqrt(A132592(k)*a(n)^2 + a(k)^2). Generalizes formula dated Oct 09 2000. - Charlie Marion, Nov 27 2012
a(n) + a(n+2*k) = A003499(k)*a(n+k); a(n) + a(n+2*k+1) = A001653(k+1)*A002315(n+k). - Charlie Marion, Nov 29 2012
From Peter Bala, Dec 23 2012: (Start)
Product_{n >= 1} (1 + 1/a(n)) = 1 + sqrt(2).
Product_{n >= 2} (1 - 1/a(n)) = (1/3)*(1 + sqrt(2)). (End)
G.f.: G(0)*x/(2-6*x), where G(k) = 1 + 1/(1 - x*(8*k-9)/( x*(8*k-1) - 3/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 12 2013
G.f.: H(0)*x/2, where H(k) = 1 + 1/( 1 - x*(6-x)/(x*(6-x) + 1/H(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Feb 18 2014
a(n) = (a(n-1)^2 - a(n-3)^2)/a(n-2) + a(n-4) for n > 3. - Patrick J. McNab, Jul 24 2015
a(n-k)*a(n+k) + a(k)^2 = a(n)^2, a(n+k) + a(n-k) = A003499(k)*a(n), for n >= k >= 0. - Alexander Samokrutov, Sep 30 2015
Dirichlet g.f.: (PolyLog(s,3+2*sqrt(2)) - PolyLog(s,3-2*sqrt(2)))/(4*sqrt(2)). - Ilya Gutkovskiy, Jun 27 2016
4*a(n)^2 - 1 = A278310(n) for n > 0. - Bruno Berselli, Nov 24 2016
From Klaus Purath, Jan 18 2020: (Start)
a(n) = (a(n-3) + a(n+3))/198.
a(n) = Sum_{i=1..n} A001653(i), n>=1.
a(n) = sinh( 2 * n * arccsch(1) ) / ( 2 * sqrt(2) ). - Federico Provvedi, Feb 01 2021
(End)
a(n) = A002965(2*n)*A002965(2*n+1). - Jon E. Schoenfield, Jan 08 2022
a(n) = A002965(4*n)/2. - Gerry Martens, Jul 14 2023
a(n) = Sum_{k = 0..n-1} (-1)^(n+k+1)*binomial(n+k, 2*k+1)*8^k. - Peter Bala, Jul 17 2023

Extensions

Additional comments from Wolfdieter Lang, Feb 10 2000
Duplication of a formula removed by Wolfdieter Lang, Feb 10 2015

A005153 Practical numbers: positive integers m such that every k <= sigma(m) is a sum of distinct divisors of m. Also called panarithmic numbers.

Original entry on oeis.org

1, 2, 4, 6, 8, 12, 16, 18, 20, 24, 28, 30, 32, 36, 40, 42, 48, 54, 56, 60, 64, 66, 72, 78, 80, 84, 88, 90, 96, 100, 104, 108, 112, 120, 126, 128, 132, 140, 144, 150, 156, 160, 162, 168, 176, 180, 192, 196, 198, 200, 204, 208, 210, 216, 220, 224, 228, 234, 240, 252
Offset: 1

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Comments

Equivalently, positive integers m such that every number k <= m is a sum of distinct divisors of m.
2^r is a member for all r as every number < = sigma(2^r) = 2^(r+1)-1 is a sum of a distinct subset of divisors {1, 2, 2^2, ..., 2^m}. - Amarnath Murthy, Apr 23 2004
Also, numbers m such that A030057(m) > m. This is a consequence of the following theorem (due to Stewart), found at the McLeman link: An integer m >= 2 with factorization Product_{i=1..k} p_i^e_i with the p_i in ascending order is practical if and only if p_1 = 2 and, for 1 < i <= k, p_i <= sigma(Product_{j < i} p_j^e_j) + 1. - Franklin T. Adams-Watters, Nov 09 2006
Practical numbers first appear in Srinivasan's short paper, which contains terms up to 200. Let m be a practical number. He states that (1) if m>2, m is a multiple of 4 or 6; (2) sigma(m) >= 2*m-1 (A103288); and (3) 2^t*m is practical. He also states that highly composite numbers (A002182), perfect numbers (A000396), and primorial numbers (A002110) are practical. - T. D. Noe, Apr 02 2010
Conjecture: The sequence a(n)^(1/n) (n=3,4,...) is strictly decreasing to the limit 1. - Zhi-Wei Sun, Jan 12 2013
Conjecture: For any positive rational number r, there are finitely many pairwise distinct practical numbers q(1)..q(k) such that r = Sum_{j=1..k} 1/q(j). For example, 2 = 1/1 + 1/2 + 1/4 + 1/6 + 1/12 with 1, 2, 4, 6 and 12 all practical, and 10/11 = 1/2 + 1/4 + 1/8 + 1/48 + 1/132 + 1/176 with 2, 4, 8, 48, 132 and 176 all practical. - Zhi-Wei Sun, Sep 12 2015
Analogous with the {1 union primes} (A008578), practical numbers form a complete sequence. This is because it contains all powers of 2 as a subsequence. - Frank M Jackson, Jun 21 2016
Sun's 2015 conjecture on the existence of Egyptian fractions with practical denominators for any positive rational number is true. See the link "Egyptian fractions with practical denominators". - David Eppstein, Nov 20 2016
Conjecture: if all divisors of m are 1 = d_1 < d_2 < ... < d_k = m, then m is practical if and only if d_(i+1)/d_i <= 2 for 1 <= i <= k-1. - Jianing Song, Jul 18 2018
The above conjecture is incorrect. The smallest counterexample is 78 (for which one of these quotients is 13/6; see A174973). m is practical if and only if the divisors of m form a complete subsequence. See Wikipedia links. - Frank M Jackson, Jul 25 2018
Reply to the comment above: Yes, and now I can show the opposite: The largest value of d_(i+1)/d_i is not bounded for practical numbers. Note that sigma(n)/n is not bounded for primorials, and primorials are practical numbers. For any constant c >= 2, let k be a practical number such that sigma(k)/k > 2c. By Bertrand's postulate there exists some prime p such that c*k < p < 2c*k < sigma(k), so k*p is a practical number with consecutive divisors k and p where p/k > c. For example, for k = 78 we have 13/6 > 2, and for 97380 we have 541/180 > 3. - Jianing Song, Jan 05 2019
Erdős (1950) and Erdős and Loxton (1979) proved that the asymptotic density of practical numbers is 0. - Amiram Eldar, Feb 13 2021
Let P(x) denote the number of practical numbers up to x. P(x) has order of magnitude x/log(x) (see Saias 1997). Moreover, we have P(x) = c*x/log(x) + O(x/(log(x))^2), where c = 1.33607... (see Weingartner 2015, 2020 and Remark 1 of Pomerance & Weingartner 2021). As a result, a(n) = k*n*log(n*log(n)) + O(n), where k = 1/c = 0.74846... - Andreas Weingartner, Jun 26 2021
From Hal M. Switkay, Dec 22 2022: (Start)
Every number of least prime signature (A025487) is practical, thereby including two classes of number mentioned in Noe's comment. This follows from Stewart's characterization of practical numbers, mentioned in Adams-Watters's comment, combined with Bertrand's postulate (there is a prime between every natural number and its double, inclusive).
Also, the first condition in Stewart's characterization (p_1 = 2) is equivalent to the second condition with index i = 1, given that an empty product is equal to 1. (End)
Conjecture: every odd number, beginning with 3, is the sum of a prime number and a practical number. Note that this conjecture occupies the space between the unproven Goldbach conjecture and the theorem that every even number, beginning with 2, is the sum of two practical numbers (Melfi's 1996 proof of Margenstern's conjecture). - Hal M. Switkay, Jan 28 2023

References

  • H. Heller, Mathematical Buds, Vol. 1, Chap. 2, pp. 10-22, Mu Alpha Theta OK, 1978.
  • Malcolm R. Heyworth, More on Panarithmic Numbers, New Zealand Math. Mag., Vol. 17 (1980), pp. 28-34 [ ISSN 0549-0510 ].
  • Ross Honsberger, Mathematical Gems, M.A.A., 1973, p. 113.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • A. K. Srinivasan, Practical numbers, Current Science, 17 (1948), 179-180.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 146-147.

Crossrefs

Subsequence of A103288.
Cf. A002093, A007620 (second definition), A030057, A033630, A119348, A174533, A174973.
Cf. A027750.

Programs

  • Haskell
    a005153 n = a005153_list !! (n-1)
    a005153_list = filter (\x -> all (p $ a027750_row x) [1..x]) [1..]
       where p _  0 = True
             p [] _ = False
             p ds'@(d:ds) m = d <= m && (p ds (m - d) || p ds m)
    -- Reinhard Zumkeller, Feb 23 2014, Oct 27 2011
    
  • Maple
    isA005153 := proc(n)
        local ifs,pprod,p,i ;
        if n = 1 then
            return true;
        elif type(n,'odd') then
            return false ;
        end if;
        # not using ifactors here directly because no guarantee primes are sorted...
        ifs := ifactors(n)[2] ;
        pprod := 1;
        for p in sort(numtheory[factorset](n) ) do
            for i in ifs do
                if op(1,i) = p then
                    if p > 2 and p > 1+numtheory[sigma](pprod) then
                        return false ;
                    end if;
                    pprod := pprod*p^op(2,i) ;
                end if;
            end do:
        end do:
        return true ;
    end proc:
    for n from 1 to 300 do
        if isA005153(n)  then
            printf("%d,",n) ;
        end if;
    end do: # R. J. Mathar, Jul 07 2023
  • Mathematica
    PracticalQ[n_] := Module[{f,p,e,prod=1,ok=True}, If[n<1 || (n>1 && OddQ[n]), False, If[n==1, True, f=FactorInteger[n]; {p,e} = Transpose[f]; Do[If[p[[i]] > 1+DivisorSigma[1,prod], ok=False; Break[]]; prod=prod*p[[i]]^e[[i]], {i,Length[p]}]; ok]]]; Select[Range[200], PracticalQ] (* T. D. Noe, Apr 02 2010 *)
  • PARI
    is_A005153(n)=bittest(n,0) && return(n==1); my(P=1); n && !for(i=2,#n=factor(n)~,n[1,i]>1+(P*=sigma(n[1,i-1]^n[2,i-1])) && return) \\ M. F. Hasler, Jan 13 2013
    
  • Python
    from sympy import factorint
    def is_A005153(n):
        if n & 1: return n == 1
        f = factorint(n) ; P = (2 << f.pop(2)) - 1
        for p in f: # factorint must have prime factors in increasing order
            if p > 1 + P: return
            P *= p**(f[p]+1)//(p-1)
        return True # M. F. Hasler, Jan 02 2023
    
  • Python
    from sympy import divisors;from more_itertools import powerset
    [i for i in range(1,253) if (lambda x:len(set(map(sum,powerset(x))))>sum(x))(divisors(i))] # Nicholas Stefan Georgescu, May 20 2023

Formula

Weingartner proves that a(n) ~ k*n log n, strengthening an earlier result of Saias. In particular, a(n) = k*n log n + O(n log log n). - Charles R Greathouse IV, May 10 2013
More precisely, a(n) = k*n*log(n*log(n)) + O(n), where k = 0.74846... (see comments). - Andreas Weingartner, Jun 26 2021

Extensions

More terms from Pab Ter (pabrlos(AT)yahoo.com), May 09 2004
Erroneous comment removed by T. D. Noe, Nov 14 2010
Definition changed to exclude n = 0 explicitly by M. F. Hasler, Jan 19 2013

A001158 sigma_3(n): sum of cubes of divisors of n.

Original entry on oeis.org

1, 9, 28, 73, 126, 252, 344, 585, 757, 1134, 1332, 2044, 2198, 3096, 3528, 4681, 4914, 6813, 6860, 9198, 9632, 11988, 12168, 16380, 15751, 19782, 20440, 25112, 24390, 31752, 29792, 37449, 37296, 44226, 43344, 55261, 50654, 61740, 61544, 73710, 68922, 86688
Offset: 1

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Comments

If the canonical factorization of n into prime powers is the product of p^e(p) then sigma_k(n) = Product_p ((p^((e(p)+1)*k))-1)/(p^k-1).
Sum_{d|n} 1/d^k is equal to sigma_k(n)/n^k. So sequences A017665-A017712 also give the numerators and denominators of sigma_k(n)/n^k for k = 1..24. The power sums sigma_k(n) are in sequences A000203 (k=1), A001157-A001160 (k=2,3,4,5), A013954-A013972 for k = 6..24. - Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 05 2001
Also the eigenvalues of the Hecke operator T_n for the entire modular normalized Eisenstein form E_4(z) (see A004009): T_n E_4 = a(n) E_4, n >= 1. For the Hecke operator T_n and eigenforms see, e.g., the Koecher-Krieg reference, p. 207, eq. (5) and p. 211, section 4, or the Apostol reference p. 120, eq. (13) and pp. 129 - 133. - Wolfdieter Lang, Jan 28 2016

Examples

			G.f. = x + 9*x^2 + 28*x^3 + 73*x^4 + 126*x^5 + 252*x^6 + 344*x^7 + ...
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math.Series 55, Tenth Printing, 1972, p. 827.
  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 38.
  • T. M. Apostol, Modular Functions and Dirichlet Series in Number Theory, Second edition, Springer, 1990, pp. 120, 129 - 133.
  • G. H. Hardy, Ramanujan: twelve lectures on subjects suggested by his life and work, AMS Chelsea Publishing, Providence, Rhode Island, 2002, p. 166.
  • Max Koecher and Aloys Krieg, Elliptische Funktionen und Modulformen, 2. Auflage, Springer, 2007, pp. 207, 211.
  • 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).
  • Zagier, Don. "Elliptic modular forms and their applications." The 1-2-3 of modular forms. Springer Berlin Heidelberg, 2008. 1-103. See p. 17, G_4(z).

Crossrefs

Cf. A004009, A064603 (partial sums).

Programs

  • Haskell
    a001158 n = product $ zipWith (\p e -> (p^(3*e + 3) - 1) `div` (p^3 - 1))
                          (a027748_row n) (a124010_row n)
    -- Reinhard Zumkeller, Jun 30 2013
    
  • Magma
    [DivisorSigma(3,n): n in [1..40]]; // Bruno Berselli, Apr 10 2013
    
  • Maple
    seq(numtheory:-sigma[3](n),n=1..100); # Robert Israel, Feb 05 2016
  • Mathematica
    Table[DivisorSigma[3,n],{n,100}] (* corrected by T. D. Noe, Mar 22 2009 *)
  • Maxima
    makelist(divsum(n,3),n,1,100); /* Emanuele Munarini, Mar 26 2011 */
    
  • PARI
    N=99; q='q+O('q^N);
    Vec(sum(n=1,N,n^3*q^n/(1-q^n))) /* Joerg Arndt, Feb 04 2011 */
    
  • PARI
    {a(n) = if( n<1, 0, sumdiv(n, d, d^3))}; /* Michael Somos, Jan 07 2017 */
    
  • Python
    from sympy import divisor_sigma
    def a(n): return divisor_sigma(n, 3)
    print([a(n) for n in range(1, 43)]) # Michael S. Branicky, Jan 09 2021
  • Sage
    [sigma(n, 3) for n in range(1, 40)]  # Zerinvary Lajos, Jun 04 2009
    

Formula

Multiplicative with a(p^e) = (p^(3e+3)-1)/(p^3-1). - David W. Wilson, Aug 01 2001
Dirichlet g.f. zeta(s)*zeta(s-3). - R. J. Mathar, Mar 04 2011
G.f.: sum(k>=1, k^3*x^k/(1-x^k)). - Benoit Cloitre, Apr 21 2003
Equals A051731 * [1, 8, 27, 64, 125, ...] = A127093 * [1, 4, 9, 16, 25, ...]. - Gary W. Adamson, Nov 02 2007
L.g.f.: -log(Product_{j>=1} (1-x^j)^(j^2)) = (1/1)*z^1 + (9/2)*z^2 + (28/3)*z^3 + (73/4)*z^4 + ... + (a(n)/n)*z^n + ... - Joerg Arndt, Feb 04 2011
a(n) = Sum{d|n} tau_{-2}^d*J_3(n/d), where tau_{-2} is A007427 and J_3 is A059376. - Enrique Pérez Herrero, Jan 19 2013
a(n) = A004009(n)/240. - Artur Jasinski, Sep 06 2016. See, e.g., Hardy, p. 166, (10.5.6), with Q = E_4, and with present offset 0. - Wolfdieter Lang, Jan 31 2017
8*a(n) = sum of cubes of even divisors of 2*n. - Wolfdieter Lang, Jan 07 2017
G.f.: Sum_{n >= 1} x^n*(1 + 4*x^n + x^(2*n))/(1 - x^n)^4. - Peter Bala, Jan 11 2021
Faster converging g.f.: Sum_{n >= 1} q^(n^2)*( n^3 + ((n + 1)^3 - 3*n^3)*q^n + (4 - 6*n^2)*q^(2*n) + (3*n^3 - (n - 1)^3)*q^(3*n) - n^3*q^(4*n) )/(1 - q^n)^4 - apply the operator x*d/dx three times to equation 5 in Arndt and then set x = 1. - Peter Bala, Jan 21 2021
a(n) = Sum_{1 <= i, j, k <= n} tau(gcd(i, j, k, n)) = Sum_{d divides n} tau(d)* J_3(n/d), where the divisor function tau(n) = A000005(n) and the Jordan totient function J_3(n) = A059376(n). - Peter Bala, Jan 22 2024

A005251 a(0) = 0, a(1) = a(2) = a(3) = 1; thereafter, a(n) = a(n-1) + a(n-2) + a(n-4).

Original entry on oeis.org

0, 1, 1, 1, 2, 4, 7, 12, 21, 37, 65, 114, 200, 351, 616, 1081, 1897, 3329, 5842, 10252, 17991, 31572, 55405, 97229, 170625, 299426, 525456, 922111, 1618192, 2839729, 4983377, 8745217, 15346786, 26931732, 47261895, 82938844, 145547525, 255418101, 448227521
Offset: 0

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a(n+3) is the number of n-bit sequences that avoid 010. Example: For n=4 the 12 sequences are all 4-bit sequences except 0100, 0101, 0010, 1010. - David Callan, Mar 25 2004
a(n+2) is the number of compositions (ordered partitions) of n where no two adjacent parts are != 1, see example. - Joerg Arndt, Jan 26 2013
a(n+1) is the number of compositions of n avoiding the part 2. - Joerg Arndt, Jul 13 2014
Number of different positive braids with n crossings of 3 strands.
This is a_2(n) in the Doroslovacki reference. Note that there is a typo in the paper in the formula for a_2(n): the upper bound in the inner sum should be "n-i" not "i-1". - Max Alekseyev, Jun 26 2007
a(n) is the number of peakless Motzkin paths of length n-1 with no UHH...HD's starting at level > 0 (here n > 0 and U=(1,1), H=(1,0), D=(1,-1)). Example: a(5)=7 because from all 8 peakless Motzkin paths of length 5 (see A004148) only UUHDD does not qualify. - Emeric Deutsch, Sep 13 2004
Equals the INVERT transform of (1, 0, 1, 1, 1, ...); equivalent to a(n) = a(n-1) + a(n-3) + a(n-4) + ... - Gary W. Adamson, Apr 27 2009
a(n) is the number of length n-1 words on {0,1} such that each string of 1's is followed by a string of at least two 0's. For example, a(5) = 4 because we have: 0000, 0100, 1000, and 1100. - Geoffrey Critzer, Aug 09 2013
a(n+1) is the top left entry of the n-th power of any of the 3 X 3 matrices [1, 1, 0; 0, 1, 1; 1, 0, 0] or [1, 0, 1; 1, 1, 0; 0, 1, 0] or [1, 1, 0; 0, 0, 1; 1, 0, 1] or [1, 0, 1; 1, 0, 0; 0, 1, 1]. - R. J. Mathar, Feb 03 2014
For n >= 2, a(n) is the number of (n-2)-length binary words with no isolated zeros. - Milan Janjic, Mar 07 2015
Apart from the first three terms, the total number of bargraphs of semiperimeter n of height at most two for n >= 2 starts 1, 2, 4, 7, 12, ... - Arnold Knopfmacher, Nov 02 2016
Number of DD-equivalence classes of Łukasiewicz paths. Łukasiewicz paths are DD-equivalent iff the positions of pattern DD are identical in these paths. - Sergey Kirgizov, Apr 08 2018
From Gus Wiseman, Nov 25 2019: (Start)
For n > 0, also the number of subsets of {1, ..., n - 3} such that if x and x + 2 are both in the subset, then so is x + 1. For example, the a(3) = 1 through a(7) = 12 subsets are:
{} {} {} {} {}
{1} {1} {1} {1}
{2} {2} {2}
{1,2} {3} {3}
{1,2} {4}
{2,3} {1,2}
{1,2,3} {1,4}
{2,3}
{3,4}
{1,2,3}
{2,3,4}
{1,2,3,4}
(End)
The two-dimensional version, which counts sets of pairs where, if two pairs are separated by graph-distance 2, then the intermediate pair or pairs are also in the set, is A329871. - Gus Wiseman, Nov 30 2019
a(n+1) is the number of ways to tile a strip of length n with squares, dominoes, and tetrominoes, where the first tile cannot be a domino. - Greg Dresden and Myanna Nash, Aug 18 2020
For n>=3, a(n) is the number of binary strings of length n-2 without any maximal runs of ones of length 1. - Félix Balado, Aug 25 2025

Examples

			From _Joerg Arndt_, Jan 26 2013: (Start)
The a(5+2) = 12 compositions of 5 where no two adjacent parts are != 1 are
  [ 1]  [ 1 1 1 1 1 ]
  [ 2]  [ 1 1 1 2 ]
  [ 3]  [ 1 1 2 1 ]
  [ 4]  [ 1 1 3 ]
  [ 5]  [ 1 2 1 1 ]
  [ 6]  [ 1 3 1 ]
  [ 7]  [ 1 4 ]
  [ 8]  [ 2 1 1 1 ]
  [ 9]  [ 2 1 2 ]
  [10]  [ 3 1 1 ]
  [11]  [ 4 1 ]
  [12]  [ 5 ]
(End)
G.f. = x + x^2 + x^3 + 2*x^4 + 4*x^5 + 7*x^6 + 12*x^7 + 21*x^8 + 37*x^9 + ...
		

References

  • S. Burckel, Efficient methods for three strand braids (submitted). [Apparently unpublished]
  • P. Chinn and S. Heubach, "Compositions of n with no occurrence of k", Congressus Numeratium, 2002, v. 162, pp. 33-51.
  • John H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, p. 205.
  • R. K. Guy, "Anyone for Twopins?" in D. A. Klarner, editor, The Mathematical Gardner. Prindle, Weber and Schmidt, Boston, 1981, pp. 2-15.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Bisection of Padovan sequence A000931.
Partial sums of A005314 shifted 3 times to the right, if we assume A005314(0) = 1.
Compositions without adjacent equal parts are A003242.
Compositions without isolated parts are A114901.
Row sums of A097230(n-2) for n>1.

Programs

  • Haskell
    a005251 n = a005251_list !! n
    a005251_list = 0 : 1 : 1 : 1 : zipWith (+) a005251_list
       (drop 2 $ zipWith (+) a005251_list (tail a005251_list))
    -- Reinhard Zumkeller, Dec 28 2011
    
  • Magma
    I:=[0,1,1,1]; [n le 4 select I[n] else Self(n-1)+Self(n-2)+Self(n-4): n in [1..45]]; // Vincenzo Librandi, Nov 30 2018
    
  • Magma
    R:=PowerSeriesRing(Integers(), 40); [0] cat Coefficients(R!( x*(1-x)/(1-2*x + x^2 - x^3) )); // Marius A. Burtea, Oct 24 2019
    
  • Maple
    A005251 := proc(n) option remember; if n <= 2 then n elif n = 3 then 4 else 2*A005251(n - 1) - A005251(n - 2) + A005251(n - 3); fi; end;
    A005251:=(-1+z)/(-1+2*z-z**2+z**3); # Simon Plouffe in his 1992 dissertation
    a := n -> `if`(n<=1, n, hypergeom([(2-n)/3, 1-n/3, (1-n)/3], [1/2, -n+1], 27/4)):
    seq(simplify(a(n)), n=0..36); # Peter Luschny, Apr 08 2018
  • Mathematica
    LinearRecurrence[{2,-1,1},{0,1,1},40]  (* Harvey P. Dale, May 05 2011 *)
    a[ n_]:= If[n<0, SeriesCoefficient[ -x(1-x)/(1 -x + 2x^2 -x^3), {x, 0, -n}], SeriesCoefficient[ x(1-x)/(1 -2x +x^2 -x^3), {x, 0, n}]] (* Michael Somos, Dec 13 2013 *)
    a[0] = 1; a[1] = a[2] = 0; a[n_] := a[n] = a[n-2] + a[n-3]; Table[a[2 n-1], {n, 1, 20}] (* Rigoberto Florez, Oct 15 2019 *)
    Table[If[n==0,0,Length[DeleteCases[Subsets[Range[n-3]],{_,x_,y_,_}/;x+2==y]]],{n,0,10}] (* Gus Wiseman, Nov 25 2019 *)
  • PARI
    Vec((1-x)/(1-2*x+x^2-x^3)+O(x^99)) /* Charles R Greathouse IV, Nov 20 2012 */
    
  • PARI
    {a(n) = if( n<0, polcoeff( -x*(1-x)/(1 -x +2*x^2 -x^3) + x*O(x^-n), -n), polcoeff( x*(1-x)/(1 -2*x +x^2 -x^3) + x*O(x^n), n))} /* Michael Somos, Dec 13 2013 */
    
  • SageMath
    [sum( binomial(n-j-1, 2*j) for j in (0..floor((n-1)/3)) ) for n in (0..50)] # G. C. Greubel, Apr 13 2022

Formula

a(n) = 2*a(n-1) - a(n-2) + a(n-3).
G.f.: z*(1-z)/(1 - 2*z + z^2 - z^3). - Emeric Deutsch, Sep 13 2004
23*a_n = 3*P_{2n+1} + 7*P_{2n} - 2*P_{2n-1}, where P_n are the Perrin numbers, A001608. - Don Knuth, Dec 09 2008
a(n+1) = Sum_{k=0..n} binomial(n-k, 2k). - Richard L. Ollerton, May 12 2004
From Henry Bottomley, Feb 21 2001: (Start)
a(n) = (Sum_{j
a(n) = A005314(n) - A005314(n-1).
a(n) = A049853(n-1) - a(n-1).
a(n) = A005314(n) - a(n-2). (End)
Conjecture: a(n+1) + |A078065(n)| = 2*A005314(n+1). - Creighton Dement, Dec 21 2004
a(n+2) has g.f. (F_3(-x) + F_2(-x))/(F_4(-x) + F_3(-x)) = 1/(-x+1/(-x+1/(-x+1))) where F_n(x) is the n-th Fibonacci polynomial; see A011973. - Qiaochu Yuan (qchu(AT)mit.edu), Feb 19 2009
a(n) = A173022(2^(n-2) - 1) for n > 1. - Reinhard Zumkeller, Feb 07 2010
BINOMIAL transform of A176971 is a(n+1). - Michael Somos, Dec 13 2013
a(n) = hypergeom([(2-n)/3, 1-n/3, (1-n)/3], [1/2, -n+1], 27/4) for n > 1. - Peter Luschny, Apr 08 2018
G.f.: z/(1-z-z^3-z^4-z^5-...) for the compositions of n-1 avoiding 2. The g.f. for the number of compositions of n avoiding the part k is 1/(1-z-...-z^(k-1) - z^(k+1)-...). - Gregory L. Simay, Sep 09 2018
If p,q,r are the three solutions to x^3 = 2x^2 - x + 1, then a(n) = (p-1)*p^n/((p-q)*(p-r)) + (q-1)*q^n/((q-p)*(q-r)) + (r-1)*r^n/((r-p)*(r-q)). - Greg Dresden and AnXing Yang, Aug 12 2025

A047999 Sierpiński's [Sierpinski's] triangle (or gasket): triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 2.

Original entry on oeis.org

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

Comments

Restored the alternative spelling of Sierpinski to facilitate searching for this triangle using regular-expression matching commands in ASCII. - N. J. A. Sloane, Jan 18 2016
Also triangle giving successive states of cellular automaton generated by "Rule 60" and "Rule 102". - Hans Havermann, May 26 2002
Also triangle formed by reading triangle of Eulerian numbers (A008292) mod 2. - Philippe Deléham, Oct 02 2003
Self-inverse when regarded as an infinite lower triangular matrix over GF(2).
Start with [1], repeatedly apply the map 0 -> [00/00], 1 -> [10/11] [Allouche and Berthe]
Also triangle formed by reading triangles A011117, A028338, A039757, A059438, A085881, A086646, A086872, A087903, A104219 mod 2. - Philippe Deléham, Jun 18 2005
J. H. Conway writes (in Math Forum): at least the first 31 rows give odd-sided constructible polygons (sides 1, 3, 5, 15, 17, ... see A001317). The 1's form a Sierpiński sieve. - M. Dauchez (mdzzdm(AT)yahoo.fr), Sep 19 2005
When regarded as an infinite lower triangular matrix, its inverse is a (-1,0,1)-matrix with zeros undisturbed and the nonzero entries in every column form the Prouhet-Thue-Morse sequence (1,-1,-1,1,-1,1,1,-1,...) A010060 (up to relabeling). - David Callan, Oct 27 2006
Triangle read by rows: antidiagonals of an array formed by successive iterates of running sums mod 2, beginning with (1, 1, 1, ...). - Gary W. Adamson, Jul 10 2008
T(n,k) = A057427(A143333(n,k)). - Reinhard Zumkeller, Oct 24 2010
The triangle sums, see A180662 for their definitions, link Sierpiński’s triangle A047999 with seven sequences, see the crossrefs. The Kn1y(n) and Kn2y(n), y >= 1, triangle sums lead to the Sierpiński-Stern triangle A191372. - Johannes W. Meijer, Jun 05 2011
Used to compute the total Steifel-Whitney cohomology class of the Real Projective space. This was an essential component of the proof that there are no product operations without zero divisors on R^n for n not equal to 1, 2, 4 or 8 (real numbers, complex numbers, quaternions, Cayley numbers), proved by Bott and Milnor. - Marcus Jaiclin, Feb 07 2012
T(n,k) = A134636(n,k) mod 2. - Reinhard Zumkeller, Nov 23 2012
T(n,k) = 1 - A219463(n,k), 0 <= k <= n. - Reinhard Zumkeller, Nov 30 2012
From Vladimir Shevelev, Dec 31 2013: (Start)
Also table of coefficients of polynomials s_n(x) of degree n which are defined by formula s_n(x) = Sum_{i=0..n} (binomial(n,i) mod 2)*x^k. These polynomials we naturally call Sierpiński's polynomials. They also are defined by the recursion: s_0(x)=1, s_(2*n+1)(x) = (x+1)*s_n(x^2), n>=0, and s_(2*n)(x) = s_n(x^2), n>=1.
Note that: s_n(1) = A001316(n),
s_n(2) = A001317(n),
s_n(3) = A100307(n),
s_n(4) = A001317(2*n),
s_n(5) = A100308(n),
s_n(6) = A100309(n),
s_n(7) = A100310(n),
s_n(8) = A100311(n),
s_n(9) = A100307(2*n),
s_n(10) = A006943(n),
s_n(16) = A001317(4*n),
s_n(25) = A100308(2*n), etc.
The equality s_n(10) = A006943(n) means that sequence A047999 is obtained from A006943 by the separation by commas of the digits of its terms. (End)
Comment from N. J. A. Sloane, Jan 18 2016: (Start)
Take a diamond-shaped region with edge length n from the top of the triangle, and rotate it by 45 degrees to get a square S_n. Here is S_6:
[1, 1, 1, 1, 1, 1]
[1, 0, 1, 0, 1, 0]
[1, 1, 0, 0, 1, 1]
[1, 0, 0, 0, 1, 0]
[1, 1, 1, 1, 0, 0]
[1, 0, 1, 0, 0, 0].
Then (i) S_n contains no square (parallel to the axes) with all four corners equal to 1 (cf. A227133); (ii) S_n can be constructed by using the greedy algorithm with the constraint that there is no square with that property; and (iii) S_n contains A064194(n) 1's. Thus A064194(n) is a lower bound on A227133(n). (End)
See A123098 for a multiplicative encoding of the rows, i.e., product of the primes selected by nonzero terms; e.g., 1 0 1 => 2^1 * 3^0 * 5^1. - M. F. Hasler, Sep 18 2016
From Valentin Bakoev, Jul 11 2020: (Start)
The Sierpinski's triangle with 2^n rows is a part of a lower triangular matrix M_n of dimension 2^n X 2^n. M_n is a block matrix defined recursively: M_1= [1, 0], [1, 1], and for n>1, M_n = [M_(n-1), O_(n-1)], [M_(n-1), M_(n-1)], where M_(n-1) is a block matrix of the same type, but of dimension 2^(n-1) X 2^(n-1), and O_(n-1) is the zero matrix of dimension 2^(n-1) X 2^(n-1). Here is how M_1, M_2 and M_3 look like:
1 0 1 0 0 0 1 0 0 0 0 0 0 0
1 1 1 1 0 0 1 1 0 0 0 0 0 0 - It is seen the self-similarity of the
1 0 1 0 1 0 1 0 0 0 0 0 matrices M_1, M_2, ..., M_n, ...,
1 1 1 1 1 1 1 1 0 0 0 0 analogously to the Sierpinski's fractal.
1 0 0 0 1 0 0 0
1 1 0 0 1 1 0 0
1 0 1 0 1 0 1 0
1 1 1 1 1 1 1 1
M_n can also be defined as M_n = M_1 X M_(n-1) where X denotes the Kronecker product. M_n is an important matrix in coding theory, cryptography, Boolean algebra, monotone Boolean functions, etc. It is a transformation matrix used in computing the algebraic normal form of Boolean functions. Some properties and links concerning M_n can be seen in LINKS. (End)
Sierpinski's gasket has fractal (Hausdorff) dimension log(A000217(2))/log(2) = log(3)/log(2) = 1.58496... (and cf. A020857). This gasket is the first of a family of gaskets formed by taking the Pascal triangle (A007318) mod j, j >= 2 (see CROSSREFS). For prime j, the dimension of the gasket is log(A000217(j))/log(j) = log(j(j + 1)/2)/log(j) (see Reiter and Bondarenko references). - Richard L. Ollerton, Dec 14 2021

Examples

			Triangle begins:
              1,
             1,1,
            1,0,1,
           1,1,1,1,
          1,0,0,0,1,
         1,1,0,0,1,1,
        1,0,1,0,1,0,1,
       1,1,1,1,1,1,1,1,
      1,0,0,0,0,0,0,0,1,
     1,1,0,0,0,0,0,0,1,1,
    1,0,1,0,0,0,0,0,1,0,1,
   1,1,1,1,0,0,0,0,1,1,1,1,
  1,0,0,0,1,0,0,0,1,0,0,0,1,
  ...
		

References

  • Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
  • Brand, Neal; Das, Sajal; Jacob, Tom. The number of nonzero entries in recursively defined tables modulo primes. Proceedings of the Twenty-first Southeastern Conference on Combinatorics, Graph Theory, and Computing (Boca Raton, FL, 1990). Congr. Numer. 78 (1990), 47--59. MR1140469 (92h:05004).
  • John W. Milnor and James D. Stasheff, Characteristic Classes, Princeton University Press, 1974, pp. 43-49 (sequence appears on p. 46).
  • H.-O. Peitgen, H. Juergens and D. Saupe: Chaos and Fractals (Springer-Verlag 1992), p. 408.
  • Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.
  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; Chapter 3.

Crossrefs

Sequences based on the triangles formed by reading Pascal's triangle mod m: (this sequence) (m = 2), A083093 (m = 3), A034931 (m = 4), A095140 (m = 5), A095141 (m = 6), A095142 (m = 7), A034930(m = 8), A095143 (m = 9), A008975 (m = 10), A095144 (m = 11), A095145 (m = 12), A275198 (m = 14), A034932 (m = 16).
Other versions: A090971, A038183.
From Johannes W. Meijer, Jun 05 2011: (Start)
A106344 is a skew version of this triangle.
Triangle sums (see the comments): A001316 (Row1; Related to Row2), A002487 (Related to Kn11, Kn12, Kn13, Kn21, Kn22, Kn23), A007306 (Kn3, Kn4), A060632 (Fi1, Fi2), A120562 (Ca1, Ca2), A112970 (Gi1, Gi2), A127830 (Ze3, Ze4). (End)

Programs

  • Haskell
    import Data.Bits (xor)
    a047999 :: Int -> Int -> Int
    a047999 n k = a047999_tabl !! n !! k
    a047999_row n = a047999_tabl !! n
    a047999_tabl = iterate (\row -> zipWith xor ([0] ++ row) (row ++ [0])) [1]
    -- Reinhard Zumkeller, Dec 11 2011, Oct 24 2010
    
  • Magma
    A047999:= func< n,k | BitwiseAnd(n-k, k) eq 0 select 1 else 0 >;
    [A047999(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Dec 03 2024
  • Maple
    # Maple code for first M rows (here M=10) - N. J. A. Sloane, Feb 03 2016
    ST:=[1,1,1]; a:=1; b:=2; M:=10;
    for n from 2 to M do ST:=[op(ST),1];
    for i from a to b-1 do ST:=[op(ST), (ST[i+1]+ST[i+2]) mod 2 ]; od:
    ST:=[op(ST),1];
    a:=a+n; b:=a+n; od:
    ST; # N. J. A. Sloane
    # alternative
    A047999 := proc(n,k)
        modp(binomial(n,k),2) ;
    end proc:
    seq(seq(A047999(n,k),k=0..n),n=0..12) ; # R. J. Mathar, May 06 2016
  • Mathematica
    Mod[ Flatten[ NestList[ Prepend[ #, 0] + Append[ #, 0] &, {1}, 13]], 2] (* Robert G. Wilson v, May 26 2004 *)
    rows = 14; ca = CellularAutomaton[60, {{1}, 0}, rows-1]; Flatten[ Table[ca[[k, 1 ;; k]], {k, 1, rows}]] (* Jean-François Alcover, May 24 2012 *)
    Mod[#,2]&/@Flatten[Table[Binomial[n,k],{n,0,20},{k,0,n}]] (* Harvey P. Dale, Jun 26 2019 *)
    A047999[n_,k_]:= Boole[BitAnd[n-k,k]==0];
    Table[A047999[n,k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Sep 03 2025 *)
  • PARI
    \\ Recurrence for Pascal's triangle mod p, here p = 2.
    p = 2; s=13; T=matrix(s,s); T[1,1]=1;
    for(n=2,s, T[n,1]=1; for(k=2,n, T[n,k] = (T[n-1,k-1] + T[n-1,k])%p ));
    for(n=1,s,for(k=1,n,print1(T[n,k],", "))) \\ Gerald McGarvey, Oct 10 2009
    
  • PARI
    A011371(n)=my(s);while(n>>=1,s+=n);s
    T(n,k)=A011371(n)==A011371(k)+A011371(n-k) \\ Charles R Greathouse IV, Aug 09 2013
    
  • PARI
    T(n,k)=bitand(n-k,k)==0 \\ Charles R Greathouse IV, Aug 11 2016
    
  • Python
    def A047999_T(n,k):
        return int(not ~n & k) # Chai Wah Wu, Feb 09 2016
    

Formula

Lucas's Theorem is that T(n,k) = 1 if and only if the 1's in the binary expansion of k are a subset of the 1's in the binary expansion of n; or equivalently, k AND NOT n is zero, where AND and NOT are bitwise operators. - Chai Wah Wu, Feb 09 2016 and N. J. A. Sloane, Feb 10 2016
Sum_{k>=0} T(n, k) = A001316(n) = 2^A000120(n).
T(n,k) = T(n-1,k-1) XOR T(n-1,k), 0 < k < n; T(n,0) = T(n,n) = 1. - Reinhard Zumkeller, Dec 13 2009
T(n,k) = (T(n-1,k-1) + T(n-1,k)) mod 2 = |T(n-1,k-1) - T(n-1,k)|, 0 < k < n; T(n,0) = T(n,n) = 1. - Rick L. Shepherd, Feb 23 2018
From Vladimir Shevelev, Dec 31 2013: (Start)
For polynomial {s_n(x)} we have
s_0(x)=1; for n>=1, s_n(x) = Product_{i=1..A000120(n)} (x^(2^k_i) + 1),
if the binary expansion of n is n = Sum_{i=1..A000120(n)} 2^k_i;
G.f. Sum_{n>=0} s_n(x)*z^n = Product_{k>=0} (1 + (x^(2^k)+1)*z^(2^k)) (0
Let x>1, t>0 be real numbers. Then
Sum_{n>=0} 1/s_n(x)^t = Product_{k>=0} (1 + 1/(x^(2^k)+1)^t);
Sum_{n>=0} (-1)^A000120(n)/s_n(x)^t = Product_{k>=0} (1 - 1/(x^(2^k)+1)^t).
In particular, for t=1, x>1, we have
Sum_{n>=0} (-1)^A000120(n)/s_n(x) = 1 - 1/x. (End)
From Valentin Bakoev, Jul 11 2020: (Start)
(See my comment about the matrix M_n.) Denote by T(i,j) the number in the i-th row and j-th column of M_n (0 <= i, j < 2^n). When i>=j, T(i,j) is the j-th number in the i-th row of the Sierpinski's triangle. For given i and j, we denote by k the largest integer of the type k=2^m and k
T(i,0) = T(i,i) = 1, or
T(i,j) = 0 if i < j, or
T(i,j) = T(i-k,j), if j < k, or
T(i,j) = T(i-k,j-k), if j >= k.
Thus, for given i and j, T(i,j) can be computed in O(log_2(i)) steps. (End)

Extensions

Additional links from Lekraj Beedassy, Jan 22 2004

A019538 Triangle of numbers T(n,k) = k!*Stirling2(n,k) read by rows (n >= 1, 1 <= k <= n).

Original entry on oeis.org

1, 1, 2, 1, 6, 6, 1, 14, 36, 24, 1, 30, 150, 240, 120, 1, 62, 540, 1560, 1800, 720, 1, 126, 1806, 8400, 16800, 15120, 5040, 1, 254, 5796, 40824, 126000, 191520, 141120, 40320, 1, 510, 18150, 186480, 834120, 1905120, 2328480, 1451520, 362880, 1, 1022, 55980, 818520, 5103000, 16435440, 29635200, 30240000, 16329600, 3628800
Offset: 1

Author

N. J. A. Sloane and Manfred Goebel (goebel(AT)informatik.uni-tuebingen.de), Dec 11 1996

Comments

Number of ways n labeled objects can be distributed into k nonempty parcels. Also number of special terms in n variables with maximal degree k.
In older terminology these are called differences of 0. - Michael Somos, Oct 08 2003
Number of surjections (onto functions) from an n-element set to a k-element set.
Also coefficients (in ascending order) of so-called ordered Bell polynomials.
(k-1)!*Stirling2(n,k-1) is the number of chain topologies on an n-set having k open sets [Stephen].
Number of set compositions (ordered set partitions) of n items into k parts. Number of k dimensional 'faces' of the n dimensional permutohedron (see Simion, p. 162). - Mitch Harris, Jan 16 2007
Correction of comment before: Number of (n-k)-dimensional 'faces' of the permutohedron of order n (an (n-1)-dimensional polytope). - Tilman Piesk, Oct 29 2014
This array is related to the reciprocal of an e.g.f. as sketched in A133314. For example, the coefficient of the fourth-order term in the Taylor series expansion of 1/(a(0) + a(1) x + a(2) x^2/2! + a(3) x^3/3! + ...) is a(0)^(-5) * {24 a(1)^4 - 36 a(1)^2 a(2) a(0) + [8 a(1) a(3) + 6 a(2)^2] a(0)^2 - a(4) a(0)^3}. The unsigned coefficients characterize the P3 permutohedron depicted on page 10 in the Loday link with 24 vertices (0-D faces), 36 edges (1-D faces), 6 squares (2-D faces), 8 hexagons (2-D faces) and 1 3-D permutohedron. Summing coefficients over like dimensions gives A019538 and A090582. Compare to A133437 for the associahedron. - Tom Copeland, Sep 29 2008, Oct 07 2008
Further to the comments of Tom Copeland above, the permutohedron of type A_3 can be taken as the truncated octahedron. Its dual is the tetrakis hexahedron, a simplicial polyhedron, with f-vector (1,14,36,24) giving the fourth row of this triangle. See the Wikipedia entry and [Fomin and Reading p. 21]. The corresponding h-vectors of permutohedra of type A give the rows of the triangle of Eulerian numbers A008292. See A145901 and A145902 for the array of f-vectors for type B and type D permutohedra respectively. - Peter Bala, Oct 26 2008
Subtriangle of triangle in A131689. - Philippe Deléham, Nov 03 2008
Since T(n,k) counts surjective functions and surjective functions are "consistent", T(n,k) satisfies a binomial identity, namely, T(n,x+y) = Sum_{j=0..n} C(n,j)*T(j,x)*T(n-j,y). For definition of consistent functions and a generalized binomial identity, see "Toy stories and combinatorial identities" in the link section below. - Dennis P. Walsh, Feb 24 2012
T(n,k) is the number of labeled forests on n+k vertices satisfying the following two conditions: (i) each forest consists of exactly k rooted trees with roots labeled 1, 2, ..., k; (ii) every root has at least one child vertex. - Dennis P. Walsh, Feb 24 2012
The triangle is the inverse binomial transform of triangle A028246, deleting the left column and shifting up one row. - Gary W. Adamson, Mar 05 2012
See A074909 for associations among this array and the Bernoulli polynomials and their umbral compositional inverses. - Tom Copeland, Nov 14 2014
E.g.f. for the shifted signed polynomials is G(x,t) = (e^t-1)/[1+(1+x)(e^t-1)] = 1-(1+x)(e^t-1) + (1+x)^2(e^t-1)^2 - ... (see also A008292 and A074909), which has the infinitesimal generator g(x,u)d/du = [(1-x*u)(1-(1+x)u)]d/du, i.e., exp[t*g(x,u)d/du]u eval. at u=0 gives G(x,t), and dG(x,t)/dt = g(x,G(x,t)). The compositional inverse is log((1-xt)/(1-(1+x)t)). G(x,t) is a generating series associated to the generalized Hirzebruch genera. See the G. Rzadowski link for the relation of the derivatives of g(x,u) to solutions of the Riccatt differential equation, soliton solns. to the KdV equation, and the Eulerian and Bernoulli numbers. In addition A145271 connects products of derivatives of g(x,u) and the refined Eulerian numbers to the inverse of G(x,t), which gives the normalized, reverse face polynomials of the simplices (A135278, divided by n+1). See A028246 for the generator g(x,u)d/dx. - Tom Copeland, Nov 21 2014
For connections to toric varieties and Eulerian polynomials, see the Dolgachev and Lunts and the Stembridge links. - Tom Copeland, Dec 31 2015
See A008279 for a relation between the e.g.f.s enumerating the faces of permutahedra (this entry) and stellahedra. - Tom Copeland, Nov 14 2016
T(n, k) appears in a Worpitzky identity relating monomials to binomials: x^n = Sum_{k=1..n} T(n, k)*binomial(x,k), n >= 1. See eq. (11.) of the Worpitzky link on p. 209. The relation to the Eulerian numbers is given there in eqs. (14.) and (15.). See the formula below relating to A008292. See also Graham et al. eq. (6.10) (relating monomials to falling factorials) on p. 248 (2nd ed. p. 262). The Worpitzky identity given in the Graham et al. reference as eq. (6.37) (2nd ed. p. 269) is eq. (5.), p. 207, of Worpitzky. - Wolfdieter Lang, Mar 10 2017
T(n, m) is also the number of minimum clique coverings and minimum matchings in the complete bipartite graph K_{m,n}. - Eric W. Weisstein, Apr 26 2017
From the Hasan and Franco and Hasan papers: The m-permutohedra for m=1,2,3,4 are the line segment, hexagon, truncated octahedron and omnitruncated 5-cell. The first three are well-known from the study of elliptic models, brane tilings and brane brick models. The m+1 torus can be tiled by a single (m+2)-permutohedron. Relations to toric Calabi-Yau Kahler manifolds are also discussed. - Tom Copeland, May 14 2020
From Manfred Boergens, Jul 25 2021: (Start)
Number of n X k binary matrices with row sums = 1 and no zero columns. These matrices are a subset of the matrices defining A183109.
The distribution into parcels in the leading comment can be regarded as a covering of [n] by tuples (A_1,...,A_k) in P([n])^k with nonempty and disjoint A_j, with P(.) denoting the power set (corrected for clarity by Manfred Boergens, May 26 2024). For the non-disjoint case see A183109 and A218695.
For tuples with "nonempty" dropped see A089072. For tuples with "nonempty and disjoint" dropped see A092477 and A329943 (amendment by Manfred Boergens, Jun 24 2024). (End)

Examples

			The triangle T(n, k) begins:
  n\k 1    2     3      4       5        6        7        8        9      10
  1:  1
  2:  1    2
  3:  1    6     6
  4:  1   14    36     24
  5:  1   30   150    240     120
  6:  1   62   540   1560    1800      720
  7:  1  126  1806   8400   16800    15120     5040
  8:  1  254  5796  40824  126000   191520   141120    40320
  9:  1  510 18150 186480  834120  1905120  2328480  1451520   362880
  10: 1 1022 55980 818520 5103000 16435440 29635200 30240000 16329600 3628800
  ... Reformatted and extended - _Wolfdieter Lang_, Oct 04 2014
---------------------------------------------------------------------------
T(4,1) = 1: {1234}. T(4,2) = 14: {1}{234} (4 ways), {12}{34} (6 ways), {123}{4} (4 ways). T(4,3) = 36: {12}{3}{4} (12 ways), {1}{23}{4} (12 ways), {1}{2}{34} (12 ways). T(4,4) = 1: {1}{2}{3}{4} (1 way).
		

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 89, ex. 1; also p. 210.
  • Miklos Bona, Combinatorics of Permutations, Chapman and Hall,2004, p.12.
  • G. Boole, A Treatise On The Calculus of Finite Differences, Dover Publications, 1960, p. 20.
  • H. T. Davis, Tables of the Mathematical Functions. Vols. 1 and 2, 2nd ed., 1963, Vol. 3 (with V. J. Fisher), 1962; Principia Press of Trinity Univ., San Antonio, TX, Vol. 2, p. 212.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, 1989, p. 155. Also eqs.(6.10) and (6.37).
  • Kiran S. Kedlaya and Andrew V. Sutherland, Computing L -Series of Hyperelliptic Curves in Algorithmic Number Theory Lecture Notes in Computer Science Volume 5011/2008.
  • T. K. Petersen, Eulerian Numbers, Birkhauser, 2015, Section 5.6.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 33.
  • J. F. Steffensen, Interpolation, 2nd ed., Chelsea, NY, 1950, see p. 54.
  • A. H. Voigt, Theorie der Zahlenreihen und der Reihengleichungen, Goschen, Leipzig, 1911, p. 31.
  • E. Whittaker and G. Robinson, The Calculus of Observations, Blackie, London, 4th ed., 1949; p. 7.

Crossrefs

Row sums give A000670. Maximal terms in rows give A002869. Central terms T(2k-1,k) give A233734.
Diagonal is n! (A000142). 2nd diagonal is A001286. 3rd diagonal is A037960.
Reflected version of A090582. A371568 is another version.
See also the two closely related triangles: A008277(n, k) = T(n, k)/k! (Stirling numbers of second kind) and A028246(n, k) = T(n, k)/k.
Cf. A033282 'faces' of the associahedron.
Cf. A008292, A047969, A145901, A145902. - Peter Bala, Oct 26 2008
Visible in the 3-D array in A249042.
See also A000182.

Programs

  • Haskell
    a019538 n k = a019538_tabl !! (n-1) !! (k-1)
    a019538_row n = a019538_tabl !! (n-1)
    a019538_tabl = iterate f [1] where
       f xs = zipWith (*) [1..] $ zipWith (+) ([0] ++ xs) (xs ++ [0])
    -- Reinhard Zumkeller, Dec 15 2013
    
  • Maple
    with(combinat): A019538 := (n,k)->k!*stirling2(n,k);
  • Mathematica
    Table[k! StirlingS2[n, k], {n, 9}, {k, n}] // Flatten
  • PARI
    {T(n, k) = if( k<0 || k>n, 0, sum(i=0, k, (-1)^i * binomial(k, i) * (k-i)^n))}; /* Michael Somos, Oct 08 2003 */
    
  • Sage
    def T(n, k): return factorial(k)*stirling_number2(n,k) # Danny Rorabaugh, Oct 10 2015

Formula

T(n, k) = k*(T(n-1, k-1)+T(n-1, k)) with T(0, 0) = 1 [or T(1, 1) = 1]. - Henry Bottomley, Mar 02 2001
E.g.f.: (y*(exp(x)-1) - exp(x))/(y*(exp(x)-1) - 1). - Vladeta Jovovic, Jan 30 2003
Equals [0, 1, 0, 2, 0, 3, 0, 4, 0, 5, ...] DELTA [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, ...] where DELTA is Deléham's operator defined in A084938.
T(n, k) = Sum_{j=0..k} (-1)^(k-j)*j^n*binomial(k, j). - Mario Catalani (mario.catalani(AT)unito.it), Nov 28 2003. See Graham et al., eq. (6.19), p. 251. For a proof see Bert Seghers, Jun 29 2013.
Sum_{k=0..n} T(n, k)(-1)^(n-k) = 1, Sum_{k=0..n} T(n, k)(-1)^k = (-1)^n. - Mario Catalani (mario.catalani(AT)unito.it), Dec 11 2003
O.g.f. for n-th row: polylog(-n, x/(1+x))/(x+x^2). - Vladeta Jovovic, Jan 30 2005
E.g.f.: 1 / (1 + t*(1-exp(x))). - Tom Copeland, Oct 13 2008
From Peter Bala, Oct 26 2008: (Start)
O.g.f. as a continued fraction: 1/(1 - x*t/(1 - (x + 1)*t/(1 - 2*x*t/(1 - 2*(x + 1)*t/(1 - ...))))) = 1 + x*t + (x + 2*x^2)*t^2 + (x + 6*x^2 + 6*x^3)*t^3 + ... .
The row polynomials R(n,x), which begin R(1,x) = x, R(2,x) = x + 2*x^2, R(3,x) = x + 6*x^2 + 6*x^3, satisfy the recurrence x*d/dx ((x + 1)*R(n,x)) = R(n+1,x). It follows that the zeros of R(n,x) are real and negative (apply Corollary 1.2 of [Liu and Wang]).
Since this is the triangle of f-vectors of the (simplicial complexes dual to the) type A permutohedra, whose h-vectors form the Eulerian number triangle A008292, the coefficients of the polynomial (x-1)^n*R(n,1/(x-1)) give the n-th row of A008292. For example, from row 3 we have x^2 + 6*x + 6 = 1 + 4*y + y^2, where y = x + 1, producing [1,4,1] as the third row of A008292. The matrix product A008292 * A007318 gives the mirror image of this triangle (see A090582).
For n,k >= 0, T(n+1,k+1) = Sum_{j=0..k} (-1)^(k-j)*binomial(k,j)*[(j+1)^(n+1) - j^(n+1)]. The matrix product of Pascal's triangle A007318 with the current array gives (essentially) A047969. This triangle is also related to triangle A047969 by means of the S-transform of [Hetyei], a linear transformation of polynomials whose value on the basis monomials x^k is given by S(x^k) = binomial(x,k). The S-transform of the shifted n-th row polynomial Q(n,x) := R(n,x)/x is S(Q(n,x)) = (x+1)^n - x^n. For example, from row 3 we obtain S(1 + 6*x + 6*x^2) = 1 + 6*x + 6*x*(x-1)/2 = 1 + 3*x + 3*x^2 = (x+1)^3 - x^3. For fixed k, the values S(Q(n,k)) give the nonzero entries in column (k-1) of the triangle A047969 (the Hilbert transform of the Eulerian numbers). (End)
E.g.f.: (exp(x)-1)^k = sum T(n,k)x^n/n!. - Vladimir Kruchinin, Aug 10 2010
T(n,k) = Sum_{i=1..k} A(n,i)*Binomial(n-i,k-i) where A(n,i) is the number of n-permutations that have i ascending runs, A008292.
From Tom Copeland, Oct 11 2011: (Start)
With e.g.f. A(x,t) = -1 + 1/(1+t*(1-exp(x))), the comp. inverse in x is B(x,t) = log(((1+t)/t) - 1/(t(1+x))).
With h(x,t) = 1/(dB/dx)= (1+x)((1+t)(1+x)-1), the row polynomial P(n,t) is given by (h(x,t)*d/dx)^n x, eval. at x=0, A=exp(x*h(y,t)*d/dy) y, eval. at y=0, and dA/dx = h(A(x,t),t), with P(0,t)=0.
(A factor of -1/n! was removed by Copeland on Aug 25 2016.) (End)
The term linear in x of [x*h(d/dx,t)]^n 1 gives the n-th row polynomial. (See A134685.) - Tom Copeland, Nov 07 2011
Row polynomials are given by D^n(1/(1-x*t)) evaluated at x = 0, where D is the operator (1+x)*d/dx. - Peter Bala, Nov 25 2011
T(n,x+y) = Sum_{j=0..n} binomial(n,j)*T(j,x)*T(n-j,y). - Dennis P. Walsh, Feb 24 2012
Let P be a Rota-Baxter operator of weight 1 satisfying the identity P(x)*P(y) = P(P(x)*y) + P(x*P(y)) + P(x*y). Then P(1)^2 = P(1) + 2*P^2(1). More generally, Guo shows that P(1)^n = Sum_{k=1..n} T(n,k)*P^k(1). - Peter Bala, Jun 08 2012
Sum_{i=1..n} (-1)^i*T(n,i)/i = 0, for n > 1. - Leonid Bedratyuk, Aug 09 2012
T(n, k) = Sum_{j=0..k} (-1)^j*binomial(k, j)*(k-j)^n. [M. Catalani's re-indexed formula from Nov 28 2003] Proof: count the surjections of [n] onto [k] with the inclusion-exclusion principle, as an alternating sum of the number of functions from [n] to [k-j]. - Bert Seghers, Jun 29 2013
n-th row polynomial = 1/(1 + x)*( Sum_{k>=0} k^n*(x/(1 + x))^k ), valid for x in the open interval (-1/2, inf). See Tanny link. Cf. A145901. - Peter Bala, Jul 22 2014
T(n,k) = k * A141618(n,k-1) / binomial(n,k-1). - Tom Copeland, Oct 25 2014
Sum_{n>=0} n^k*a^n = Sum_{i=1..k} (a / (1 - a))^i * T(k, i)/(1-a) for |a| < 1. - David A. Corneth, Mar 09 2015
From Peter Bala, May 26 2015: (Start)
The row polynomials R(n,x) satisfy (1 + x)*R(n,x) = (-1)^n*x*R(n,-(1 + x)).
For a fixed integer k, the expansion of the function A(k,z) := exp( Sum_{n >= 1} R(n,k)*z^n/n ) has integer coefficients and satisfies the functional equation A(k,z)^(k + 1) = BINOMIAL(A(k,z))^k, where BINOMIAL(F(z))= 1/(1 - z)*F(z/(1 - z)) denotes the binomial transform of the o.g.f. F(z). Cf. A145901. For cases see A084784 (k = 1), A090352 (k = 2), A090355 (k = 3), A090357 (k = 4), A090362 (k = 5) and A084785 (k = -2 with z -> -z).
A(k,z)^(k + 1) = A(-(k + 1),-z)^k and hence BINOMIAL(A(k,z)) = A(-(k + 1),-z). (End)
From Tom Copeland, Oct 19 2016: (Start)
Let a(1) = 1 + x + B(1) = x + 1/2 and a(n) = B(n) = (B.)^n, where B(n) are the Bernoulli numbers defined by e^(B.t) = t / (e^t-1), then t / e^(a.t) = t / [(x + 1) * t + exp(B.t)] = (e^t - 1) /[ 1 + (x + 1) (e^t - 1)] = exp(p.(x)t), where (p.(x))^n = p_n(x) are the shifted, signed row polynomials of this array: p_0(x) = 0, p_1(x) = 1, p_2(x) = -(1 + 2 x), p_3(x) = 1 + 6 x + 6 x^2, ... and p_n(x) = n * b(n-1), where b(n) are the partition polynomials of A133314 evaluated with these a(n).
Sum_{n > 0} R(n,-1/2) x^n/n! = 2 * tanh(x/2), where R(n,x) = Sum_{k = 1..n} T(n,k) x^(k-1) are the shifted row polynomials of this entry, so R(n,-1/2) = 4 * (2^(n+1)-1) B(n+1)/(n+1). (Cf. A000182.)
(End)
Also the Bernoulli numbers are given by B(n) = Sum_{k =1..n} (-1)^k T(n,k) / (k+1). - Tom Copeland, Nov 06 2016
G.f. for column k: k! x^k / Product_{i=1..k} (1-i*x). - Robert A. Russell, Sep 25 2018
a(j) <= A183109(j). - Manfred Boergens, Jul 25 2021

A010892 Inverse of 6th cyclotomic polynomial. A period 6 sequence.

Original entry on oeis.org

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

Author

Keywords

Comments

Any sequence b(n) satisfying the recurrence b(n) = b(n-1) - b(n-2) can be written as b(n) = b(0)*a(n) + (b(1)-b(0))*a(n-1).
a(n) is the determinant of the n X n matrix M with m(i,j)=1 if |i-j| <= 1 and 0 otherwise. - Mario Catalani (mario.catalani(AT)unito.it), Jan 25 2003
Also row sums of triangle in A108299; a(n)=L(n-1,1), where L is also defined as in A108299; see A061347 for L(n,-1). - Reinhard Zumkeller, Jun 01 2005
Pisano period lengths: 1, 3, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, ... - R. J. Mathar, Aug 10 2012
Periodic sequences of this type can also be calculated as a(n) = c + floor(q/(p^m-1)*p^n) mod p, where c is a constant, q is the number representing the periodic digit pattern and m is the period. c, p and q can be calculated as follows: Let D be the array representing the number pattern to be repeated, m = size of D, max = maximum value of elements in D, min = minimum value of elements in D. Then c := min, p := max - min + 1 and q := p^m*Sum_{i=1..m} (D(i)-min)/p^i. Example: D = (1, 1, 0, -1, -1, 0), c = -1, m = 6, p = 3 and q = 676 for this sequence. - Hieronymus Fischer, Jan 04 2013
B(n) = a(n+5) = S(n-1, 1) appears, together with a(n) = A057079(n+1), in the formula 2*exp(Pi*n*i/3) = A(n) + B(n)*sqrt(3)*i with i = sqrt(-1). For S(n, x) see A049310. See also a Feb 27 2014 comment on A099837. - Wolfdieter Lang, Feb 27 2014
a(n) (for n>=1) is the difference between numbers of even and odd permutations p of 1,2,...,n such that |p(i)-i|<=1 for i=1,2,...,n. - Dmitry Efimov, Jan 08 2016
From Tom Copeland, Jan 31 2016: (Start)
Specialization of the o.g.f. 1 / ((x - w1)(x-w2)) = (1/(w1-w2)) ((w1-w2) + (w1^2 - w2^2) x + (w1^3-w2^3) x^2 + ...) with w1*w2 = (1/w1) + (1/w2) = 1. Then w1 = q = e^(i*Pi/3) and w2 = 1/q = e^(-i*Pi/3), giving the o.g.f. 1 /(1-x+x^2) for this entry with a(n) = (2/sqrt(3)) sin((n+1)Pi/3). See the Copeland link for more relations.
a(n) = (q^(n+1) - q^(-(n+1))) / (q - q^(-1)), so this entry gives the o.g.f. for an instance of the quantum integers denoted by [m]_q in Morrison et al. and Tingley. (End)

Examples

			G.f. = 1 + x - x^3 - x^4 + x^6 + x^7 - x^9 - x^10 + x^12 + x^13 - x^15 + ...
		

Crossrefs

a(n) = row sums of signed triangle A049310.
Differs only by a shift from A128834.
a(n+1) = row sums of triangle A130777: repeat(1,0,-1,-1,0,1).

Programs

  • Magma
    &cat[[1,1,0,-1,-1,0]: n in [0..20]]; // Vincenzo Librandi, Apr 03 2014
  • Maple
    a:=n->coeftayl(1/(x^2-x+1), x=0, n);
    a:=n->2*sin(Pi*(n+1)/3)/sqrt(3);
    A010892:=n->[1,1,0,-1,-1,0][irem(n,6)+1];
    A010892:=n->Array(0..5,[1,1,0,-1,-1,0])[irem(n,6)];
    A010892:=n->table([0=1,1=1,2=0,3=-1,4=-1,5=0])[irem(n,6)];
    with(numtheory,cyclotomic); c := series(1/cyclotomic(6,x),x,102): seq(coeff(c,x,n),n=0..101); # Rainer Rosenthal, Jan 01 2007
  • Mathematica
    a[n_] := {1, 1, 0, -1, -1, 0}[[Mod[n, 6] + 1]]; Table[a[n], {n, 0, 101}] (* Jean-François Alcover, Jul 19 2013 *)
    CoefficientList[Series[1/Cyclotomic[6, x], {x, 0, 100}], x] (* Vincenzo Librandi, Apr 03 2014 *)
    PadRight[{},120,{1,1,0,-1,-1,0}] (* Harvey P. Dale, Jul 07 2020 *)
  • PARI
    {a(n) = (-1)^(n\3) * sign((n + 1)%3)}; /* Michael Somos, Sep 23 2005 */
    
  • PARI
    {a(n) = subst( poltchebi(n) + poltchebi(n-1), 'x, 1/2) * 2/3}; /* Michael Somos, Sep 23 2005 */
    
  • PARI
    {a(n) = [1, 1, 0, -1, -1, 0][n%6 + 1]}; /* Michael Somos, Feb 14 2006 */
    
  • PARI
    {a(n) = my(A, p, e); if( n<0, 0, n++; A = factor(n); prod( k=1, matsize(A)[1], [p, e] = A[k, ]; if( p==2, -(-1)^e, p==3, 0, p%6 == 1, 1, (-1)^e)))}; /* Michael Somos, Oct 29 2006 */
    
  • Python
    def A010892(n): return [1,1,0,-1,-1,0][n%6] # Alec Mihailovs, Jan 01 2007
    
  • Sage
    [lucas_number1(n,1,+1) for n in range(-5, 97)] # Zerinvary Lajos, Apr 22 2009
    
  • Sage
    def A010892():
        x, y = -1, -1
        while True:
            yield -x
            x, y = y, -x + y
    a = A010892()
    [next(a) for i in range(40)]  # Peter Luschny, Jul 11 2013
    

Formula

G.f.: 1 / (1 - x + x^2).
a(n) = a(n-1) - a(n-2), a(0)=1, a(1)=1.
a(n) = ((-1)^floor(n/3) + (-1)^floor((n+1)/3))/2.
a(n) = 0 if n mod 6 = 2 or 5, a(n) = +1 if n mod 6 = 0 or 1, a(n) = -1 otherwise. a(n) = S(n, 1) = U(n, 1/2) (Chebyshev U(n, x) polynomials).
a(n) = sqrt(4/3)*Im((1/2 + i*sqrt(3/4))^(n+1)). - Henry Bottomley, Apr 12 2000
Binomial transform of A057078. a(n) = Sum_{k=0..n} C(k, n-k)*(-1)^(n-k). - Paul Barry, Sep 13 2003
a(n) = 2*sin(Pi*n/3 + Pi/3)/sqrt(3). - Paul Barry, Jan 28 2004
a(n) = Sum_{k=0..floor(n/2)} C(n-k, k)*(-1)^k. - Paul Barry, Jul 28 2004
Euler transform of length 6 sequence [1, -1, -1, 0, 0, 1]. - Michael Somos, Sep 23 2005
a(n) = a(1 - n) = -a(-2 - n) for all n in Z. - Michael Somos, Feb 14 2006
a(n) = Sum_{k=0..n} (-2)^(n-k) * A085838(n,k). - Philippe Deléham, Oct 26 2006
a(n) = b(n+1) where b(n) is multiplicative with b(2^e) = -(-1)^e if e>0, b(3^e) = 0^e, b(p^e) = 1 if p == 1 (mod 6), b(p^e) = (-1)^e if p == 5 (mod 6). - Michael Somos, Oct 29 2006
Given g.f. A(x), then, B(x) = x * A(x) satisfies 0 = f(B(x), B(x^2)) where f(u, v) = u^2 - v - 2*u*v * (1 - u). - Michael Somos, Oct 29 2006
a(2*n) = A057078(n), a(2*n+1) = A049347(n).
a(n) = Sum_{k=0..n} A109466(n,k). - Philippe Deléham, Nov 14 2006
a(n) = Sum_{k=0..n} A133607(n,k). - Philippe Deléham, Dec 30 2007
a(n) = A128834(n+1). - Jaume Oliver Lafont, Dec 05 2008
a(n) = Sum_{k=0..n} C(n+k+1,2k+1) * (-1)^k. - Paul Barry, Jun 03 2009
a(n) = A101950(n,0) = (-1)^n * A049347(n). - Philippe Deléham, Feb 10 2012
a(n) = Product_{k=1..floor(n/2)} 1 - 4*(cos(k*Pi/(n+1)))^2. - Mircea Merca, Apr 01 2012
G.f.: 1 / (1 - x / (1 + x / (1 - x))). - Michael Somos, Apr 02 2012
a(n) = -1 + floor(181/819*10^(n+1)) mod 10. - Hieronymus Fischer, Jan 03 2013
a(n) = -1 + floor(13/14*3^(n+1)) mod 3. - Hieronymus Fischer, Jan 04 2013
a(n) = 1/(1+r2)*(1/r1)^n + 1/(1+r1)*(1/r2)^n, with r1=(1-i*sqrt(3))/2 and r2=(1+i*sqrt(3))/2. - Ralf Stephan, Jul 19 2013
a(n) = ((n+1)^2 mod 3) * (-1)^floor((n+1)/3). - Wesley Ivan Hurt, Mar 15 2015
a(n-1) = n - Sum_{i=1..n-1} i*a(n-i). - Derek Orr, Apr 28 2015
a(n) = S(2*n+1, sqrt(3))/sqrt(3) = S(n, 1) with S(n, x) coefficients given in A049310. The S(n, 1) formula appeared already above. S(2*n, sqrt(3)) = A057079(n). See also a Feb 27 2014 comment above. - Wolfdieter Lang, Jan 16 2018
E.g.f.: sqrt(exp(x)*4/3) * cos(x*sqrt(3/4) - Pi/6). - Michael Somos, Jul 05 2018
a(n) = Determinant(Tri(n)), for n >= 1, with Tri(n) the n X n tridiagonal matrix with entries 1 (a special Toeplitz matrix). - Wolfdieter Lang, Sep 20 2019
a(n) = Product_{k=1..n}(1 + 2*cos(k*Pi/(n+1))). - Peter Luschny, Nov 28 2019

Extensions

Entry revised by N. J. A. Sloane, Jul 16 2004

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

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).

Crossrefs

Diagonal 3 of triangle A100257.
First differences are in A076540.
A345197 counts compositions by length and alternating sum.

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

A002411 Pentagonal pyramidal numbers: a(n) = n^2*(n+1)/2.

Original entry on oeis.org

0, 1, 6, 18, 40, 75, 126, 196, 288, 405, 550, 726, 936, 1183, 1470, 1800, 2176, 2601, 3078, 3610, 4200, 4851, 5566, 6348, 7200, 8125, 9126, 10206, 11368, 12615, 13950, 15376, 16896, 18513, 20230, 22050, 23976, 26011, 28158, 30420, 32800, 35301, 37926, 40678
Offset: 0

Comments

a(n) = n^2(n+1)/2 is half the number of colorings of three points on a line with n+1 colors. - R. H. Hardin, Feb 23 2002
Sum of n smallest multiples of n. - Amarnath Murthy, Sep 20 2002
a(n) = number of (n+6)-bit binary sequences with exactly 7 1's none of which is isolated. A 1 is isolated if its immediate neighbor(s) are 0. - David Callan, Jul 15 2004
Also as a(n) = (1/6)*(3*n^3+3*n^2), n > 0: structured trigonal prism numbers (cf. A100177 - structured prisms; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Kekulé numbers for certain benzenoids. - Emeric Deutsch, Nov 18 2005
If Y is a 3-subset of an n-set X then, for n >= 5, a(n-4) is the number of 5-subsets of X having at least two elements in common with Y. - Milan Janjic, Nov 23 2007
a(n-1), n >= 2, is the number of ways to have n identical objects in m=2 of altogether n distinguishable boxes (n-2 boxes stay empty). - Wolfdieter Lang, Nov 13 2007
a(n+1) is the convolution of (n+1) and (3n+1). - Paul Barry, Sep 18 2008
The number of 3-character strings from an alphabet of n symbols, if a string and its reversal are considered to be the same.
Partial sums give A001296. - Jonathan Vos Post, Mar 26 2011
a(n-1):=N_1(n), n >= 1, is the number of edges of n planes in generic position in three-dimensional space. See a comment under A000125 for general arrangement. Comment to Arnold's problem 1990-11, see the Arnold reference, p.506. - Wolfdieter Lang, May 27 2011
Partial sums of pentagonal numbers A000326. - Reinhard Zumkeller, Jul 07 2012
From Ant King, Oct 23 2012: (Start)
For n > 0, the digital roots of this sequence A010888(A002411(n)) form the purely periodic 9-cycle {1,6,9,4,3,9,7,9,9}.
For n > 0, the units' digits of this sequence A010879(A002411(n)) form the purely periodic 20-cycle {1,6,8,0,5,6,6,8,5,0,6,6,3,0,0,6,1,8,0,0}.
(End)
a(n) is the number of inequivalent ways to color a path graph having 3 nodes using at most n colors. Note, here there is no restriction on the color of adjacent nodes as in the above comment by R. H. Hardin (Feb 23 2002). Also, here the structures are counted up to graph isomorphism, where as in the above comment the "three points on a line" are considered to be embedded in the plane. - Geoffrey Critzer, Mar 20 2013
After 0, row sums of the triangle in A101468. - Bruno Berselli, Feb 10 2014
Latin Square Towers: Take a Latin square of order n, with symbols from 1 to n, and replace each symbol x with a tower of height x. Then the total number of unit cubes used is a(n). - Arun Giridhar, Mar 29 2015
This is the case k = n+4 of b(n,k) = n*((k-2)*n-(k-4))/2, which is the n-th k-gonal number. Therefore, this is the 3rd upper diagonal of the array in A139600. - Luciano Ancora, Apr 11 2015
For n > 0, a(n) is the number of compositions of n+7 into n parts avoiding the part 2. - Milan Janjic, Jan 07 2016
Also the Wiener index of the n-antiprism graph. - Eric W. Weisstein, Sep 07 2017
For n > 0, a(2n+1) is the number of non-isomorphic 5C_m-snakes, where m = 2n+1 or m = 2n (for n >= 2). A kC_n-snake is a connected graph in which the k >= 2 blocks are isomorphic to the cycle C_n and the block-cutpoint graph is a path. - Christian Barrientos, May 15 2019
For n >= 1, a(n-1) is the number of 0°- and 45°-tilted squares that can be drawn by joining points in an n X n lattice. - Paolo Xausa, Apr 13 2021
a(n) is the number of all possible products of n rolls of a six-sided die. This can be easily seen by the recursive formula a(n) = a(n - 1) + 2 * binomial(n, 2) + binomial(n + 1, 2). - Rafal Walczak, Jun 15 2024
a(n) is the number of all triples consisting of nonnegative integers smaller than n such that the sum of the first two integers is less than n. - Ruediger Jehn, Aug 17 2025

Examples

			a(3)=18 because 4 identical balls can be put into m=2 of n=4 distinguishable boxes in binomial(4,2)*(2!/(1!*1!) + 2!/2!) = 6*(2+1) = 18 ways. The m=2 part partitions of 4, namely (1,3) and (2,2), specify the filling of each of the 6 possible two-box choices. - _Wolfdieter Lang_, Nov 13 2007
		

References

  • V. I. Arnold (ed.), Arnold's Problems, Springer, 2004, comments on Problem 1990-11 (p. 75), pp. 503-510. Numbers N_1.
  • Christian Barrientos, Graceful labelings of cyclic snakes, Ars Combin., Vol. 60 (2001), pp. 85-96.
  • Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 194.
  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (see p. 166, Table 10.4/I/5).
  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 93.
  • L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see Vol. 2, p. 2.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

A006002(n) = -a(-1-n).
a(n) = A093560(n+2, 3), (3, 1)-Pascal column.
A row or column of A132191.
Second column of triangle A103371.
Cf. similar sequences listed in A237616.

Programs

  • GAP
    List([0..45], n->n^2*(n+1)/2); # Muniru A Asiru, Feb 19 2018
  • Haskell
    a002411 n = n * a000217 n  -- Reinhard Zumkeller, Jul 07 2012
    
  • Magma
    [n^2*(n+1)/2: n in [0..40]]; // Wesley Ivan Hurt, May 25 2014
    
  • Maple
    seq(n^2*(n+1)/2, n=0..40);
  • Mathematica
    Table[n^2 (n + 1)/2, {n, 0, 40}]
    LinearRecurrence[{4, -6, 4, -1}, {0, 1, 6, 18}, 50] (* Harvey P. Dale, Oct 20 2011 *)
    Nest[Accumulate, Range[1, 140, 3], 2] (* Vladimir Joseph Stephan Orlovsky, Jan 21 2012 *)
    CoefficientList[Series[x (1 + 2 x) / (1 - x)^4, {x, 0, 45}], x] (* Vincenzo Librandi, Jan 08 2016 *)
  • PARI
    a(n)=n^2*(n+1)/2
    
  • PARI
    concat(0, Vec(x*(1+2*x)/(1-x)^4 + O(x^100))) \\ Altug Alkan, Jan 07 2016
    

Formula

Average of n^2 and n^3.
G.f.: x*(1+2*x)/(1-x)^4. - Simon Plouffe in his 1992 dissertation
a(n) = n*Sum_{k=0..n} (n-k) = n*Sum_{k=0..n} k. - Paul Barry, Jul 21 2003
a(n) = n*A000217(n). - Xavier Acloque, Oct 27 2003
a(n) = (1/2)*Sum_{j=1..n} Sum_{i=1..n} (i+j) = (1/2)*(n^2+n^3) = (1/2)*A011379(n). - Alexander Adamchuk, Apr 13 2006
Row sums of triangle A127739, triangle A132118; and binomial transform of [1, 5, 7, 3, 0, 0, 0, ...] = (1, 6, 18, 40, 75, ...). - Gary W. Adamson, Aug 10 2007
G.f.: x*F(2,3;1;x). - Paul Barry, Sep 18 2008
Sum_{j>=1} 1/a(j) = hypergeom([1, 1, 1], [2, 3], 1) = -2 + 2*zeta(2) = A195055 - 2. - Stephen Crowley, Jun 28 2009
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4); a(0)=0, a(1)=1, a(2)=6, a(3)=18. - Harvey P. Dale, Oct 20 2011
From Ant King, Oct 23 2012: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 3.
a(n) = (n+1)*(2*A000326(n)+n)/6 = A000292(n) + 2*A000292(n-1).
a(n) = A000330(n)+A000292(n-1) = A000217(n) + 3*A000292(n-1).
a(n) = binomial(n+2,3) + 2*binomial(n+1,3).
(End)
a(n) = (A000330(n) + A002412(n))/2 = (A000292(n) + A002413(n))/2. - Omar E. Pol, Jan 11 2013
a(n) = (24/(n+3)!)*Sum_{j=0..n} (-1)^(n-j)*binomial(n,j)*j^(n+3). - Vladimir Kruchinin, Jun 04 2013
Sum_{n>=1} a(n)/n! = (7/2)*exp(1). - Richard R. Forberg, Jul 15 2013
E.g.f.: x*(2 + 4*x + x^2)*exp(x)/2. - Ilya Gutkovskiy, May 31 2016
From R. J. Mathar, Jul 28 2016: (Start)
a(n) = A057145(n+4,n).
a(n) = A080851(3,n-1). (End)
For n >= 1, a(n) = (Sum_{i=1..n} i^2) + Sum_{i=0..n-1} i^2*((i+n) mod 2). - Paolo Xausa, Apr 13 2021
a(n) = Sum_{k=1..n} GCD(k,n) * LCM(k,n). - Vaclav Kotesovec, May 22 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = 2 + Pi^2/6 - 4*log(2). - Amiram Eldar, Jan 03 2022

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

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]

Crossrefs

Cf. A000531 (Banach's original match problem).
Cf. A033876, A000984, A001803, A132818, A046521 (second column).
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
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