cp's OEIS Frontend

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

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A034275 a(n) = f(n,n-2) where f is given in A034261.

Original entry on oeis.org

1, 3, 14, 65, 294, 1302, 5676, 24453, 104390, 442442, 1864356, 7818538, 32657884, 135950700, 564306840, 2336457645, 9652643910, 39800950530, 163830074100, 673327275390, 2763494696820, 11327881630260, 46381659765480, 189711966348450, 775232392541724, 3165127107345252
Offset: 1

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Comments

Divisible by the Catalan numbers, by the explicit formula. - F. Chapoton, Jun 24 2021

Crossrefs

Programs

  • Mathematica
    a[n_] := Binomial[2*n-2,n-1] * (n^2-n+1) / n; Array[a, 25] (* Amiram Eldar, Sep 04 2025 *)
  • PARI
    a(n) = binomial(2*n-2,n-1)/n * (n^2-n+1); \\ Michel Marcus, Jun 24 2021
  • Sage
    [binomial(2*n-2,n-1)//n * (n**2-n+1) for n in range(1,8)]
    

Formula

a(n) = binomial(2*n-2,n-1)/n * (n^2-n+1).
a(n) = binomial(2*n-2,n-1) + (n-1)*binomial(2*n-2,n).
D-finite with recurrence n*a(n) + 2*(-6*n+7)*a(n-1) + 4*(11*n-24)*a(n-2) + 24*(-2*n+7)*a(n-3) = 0. - R. J. Mathar, Feb 10 2025
a(n) ~ 2^(2*n-2) * sqrt(n/Pi). - Amiram Eldar, Sep 04 2025

Extensions

Corrected and extended by N. J. A. Sloane, Apr 21 2000

A034267 a(n) = f(n,n) where f is given in A034261.

Original entry on oeis.org

0, 1, 7, 39, 196, 930, 4257, 19019, 83512, 361998, 1553630, 6615686, 27992472, 117823940, 493768485, 2061580275, 8580127920, 35611376790, 147447066690, 609200868210, 2512267906200, 10342848445020, 42516521451690, 174535009040574, 715599131576976, 2930673247541900
Offset: 0

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Crossrefs

Cf. A034261.

Programs

  • Maple
    A034261 := proc(n, k) binomial(n+k, k+1)*(n*k+n+1)/(k+2); end;
    seq( A034261(n,n),n=0..40) ; # R. J. Mathar, Feb 10 2025
  • Mathematica
    a[n_] := Binomial[2*n, n+1]*(n^2 + n + 1)/(n + 2); Array[a, 25, 0] (* Amiram Eldar, Sep 04 2025 *)

Formula

Conjecture D-finite with recurrence -(n+2)*(11*n-7)*a(n) + 2*(23*n^2+44*n+30)*a(n-1) - 4*(n+5)*(2*n-3)*a(n-2) = 0. - R. J. Mathar, Feb 10 2025
From Amiram Eldar, Sep 04 2025: (Start)
a(n) = binomial(2*n, n+1)*(n^2+n+1)/(n+2);
a(n) ~ 2^(2*n) * sqrt(n/Pi). (End)

Extensions

Corrected and extended by N. J. A. Sloane, Apr 21 2000

A034269 a(n) = f(n,n+2) where f is given in A034261.

Original entry on oeis.org

0, 1, 11, 76, 435, 2255, 11011, 51688, 236028, 1056210, 4654430, 20266312, 87402185, 374008635, 1590144675, 6724160400, 28303504680, 118665923310, 495820606410, 2065492595400, 8581809094950, 35572994871414, 147149351758926, 607555878211376, 2504288265509800
Offset: 0

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Crossrefs

Cf. A034261.

Programs

  • Mathematica
    Table[n*(n^2+3*n+1)*Binomial[2*n+2,n]/((n+3)*(n+4)),{n,0,20}] (* Vaclav Kotesovec, Oct 27 2012 *)
  • PARI
    a(n)=n*(n^2+3*n+1)*binomial(2*n+2,n)/((n+3)*(n+4)); \\ Joerg Arndt, May 04 2013

Formula

Appears to be a(n) = 4^(n+1)*Pi^(-1/2)*Gamma(n+3/2)/Gamma(n+5)*(n+1)*n*(n^2+3*n+1). - Mark van Hoeij, Oct 30 2011
a(n) ~ 2^(2*n+2) * sqrt(n/Pi). - Amiram Eldar, Sep 04 2025

Extensions

Corrected and extended by N. J. A. Sloane, Apr 21 2000

A034270 f(n,n+3) where f is given in A034261.

Original entry on oeis.org

0, 1, 13, 99, 605, 3289, 16653, 80444, 375972, 1715130, 7681586, 33913385, 148018455, 640049865, 2746422525, 11708971560, 49647131160, 209523969630, 880668859590, 3688555690950, 15401089588794, 64128925255626, 266376231244178, 1104047701891624, 4566956385797400
Offset: 0

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Crossrefs

Programs

  • Magma
    [n*(n^2+4*n+1)*Binomial(2*n+3, n)/((n+4)*(n+5)): n in [0..25]]; // Vincenzo Librandi, May 04 2013
    
  • Mathematica
    Table[n*(n^2+4*n+1)*Binomial[2*n+3,n]/((n+4)*(n+5)),{n,0,20}] (* Vaclav Kotesovec, Oct 27 2012 *)
  • PARI
    a(n)=n*(n^2+4*n+1)*binomial(2*n+3,n)/((n+4)*(n+5)); \\ Joerg Arndt, May 04 2013

Formula

a(n) ~ c*sqrt(n)*4^n, with c = 8/sqrt(Pi) = 2*A249521. - Stefano Spezia, Oct 19 2023

Extensions

Corrected and extended by N. J. A. Sloane, Apr 21 2000

A034273 a(n) = binomial(2*n+6,n+7)*(n^2+7*n+1)/(n+8) = f(n,n+6) where f is given in A034261.

Original entry on oeis.org

0, 1, 19, 186, 1365, 8540, 48348, 255816, 1289739, 6273135, 29683225, 137447310, 625482585, 2806282440, 12443418600, 54633668400, 237871030860, 1028260382994, 4417404967206, 18874729444340, 80265980340370, 339907420551336, 1434074601137640, 6030288337651760
Offset: 0

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Crossrefs

Cf. A034261 (main entry), A034263 - A034275 (other columns and diagonals n -> f(n,n+k)).

Programs

  • Mathematica
    f[a_, b_] := Binomial[a + b, b + 1] (a b + a + 1)/(b + 2); Array[f[#, # + 6] &, 22, 0] (* Michael De Vlieger, Nov 08 2017 *)
  • PARI
    A034273(n)=binomial(2*n+6, n+7)*(n^2+7*n+1)/(n+8) \\ M. F. Hasler, Nov 08 2017

Formula

a(n) ~ 2^(2*n+6) * sqrt(n/Pi). - Amiram Eldar, Sep 04 2025

Extensions

Corrected and extended by N. J. A. Sloane, Apr 21 2000
Edited by M. F. Hasler, Nov 08 2017

A034274 a(n)=f(n,n-1) where f is given in A034261.

Original entry on oeis.org

1, 5, 25, 119, 546, 2442, 10725, 46475, 199342, 848198, 3585946, 15080870, 63146500, 263432340, 1095517485, 4543460595, 18798494550, 77616288750, 319874637390, 1316106144210, 5407045011420, 22184521682700, 90910797617250, 372137346502974, 1521789223654476, 6217349014923452
Offset: 1

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Programs

Formula

From Peter Bala, Aug 19 2025: (Start)
a(n) = (n^2 + 1)/2 * A000108(n).
a(n) = (1/2) * A180266(n+1).
a(n) = Sum_{k = 1..n} k^2/(n+k-1) * binomial(n+k-1, k). Cf. Sum_{k = 1..n} k/(n+k-1) * binomial(n+k-1, k) = 1/2 * binomial(2*n, n) = 1/2 * A000984(n).
a(n) = 2*(n^2 + 1)*(2*n - 1)/((n + 1)*(n^2 - 2*n + 2)) * a(n-1) with a(1) = 1. (End)
a(n) ~ 2^(2*n-1) * sqrt(n/Pi). - Amiram Eldar, Sep 04 2025

Extensions

Corrected and extended by N. J. A. Sloane, Apr 21 2000

A034271 a(n) = f(n,n+4) where f is given in A034261.

Original entry on oeis.org

0, 1, 15, 125, 814, 4641, 24388, 121380, 581400, 2707386, 12339569, 55304535, 244567050, 1069784625, 4637437560, 19951650840, 85290143760, 362608278750, 1534330716750, 6465583372794, 27147098904156, 113618646478106, 474177584413720, 1973918305023000, 8198353999348240
Offset: 0

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Keywords

Crossrefs

Cf. A034261.

Programs

  • Mathematica
    a[n_] := Binomial[2*n + 4, n + 5]*(n^2 + 5*n + 1)/(n + 6); Array[a, 25, 0] (* Amiram Eldar, Sep 04 2025 *)

Formula

From Amiram Eldar, Sep 04 2025: (Start)
a(n) = binomial(2*n+4, n+5)*(n^2+5*n+1)/(n+6);
a(n) ~ 2^(2*n+4) * sqrt(n/Pi). (End)

Extensions

Corrected and extended by N. J. A. Sloane, Apr 21 2000

A034272 a(n) = f(n,n+5) where f is given in A034261.

Original entry on oeis.org

0, 1, 17, 154, 1066, 6370, 34748, 178296, 875976, 4167669, 19348175, 88112310, 395137470, 1749894510, 7669512840, 33323053200, 143720109360, 615950596170, 2625443201394, 11137650267156, 47051359045460, 198038873490436, 830820684414952, 3475320352223440, 14499176256634160
Offset: 0

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Author

Keywords

Crossrefs

Cf. A034261.

Programs

  • Mathematica
    a[n_] := Binomial[2*n + 5, n + 6]*(n^2 + 6*n + 1)/(n + 7); Array[a, 25, 0] (* Amiram Eldar, Sep 04 2025 *)

Formula

From Amiram Eldar, Sep 04 2025: (Start)
a(n) = binomial(2*n+5, n+6)*(n^2+6*n+1)/(n+7);
a(n) ~ 2^(2*n+5) * sqrt(n/Pi). (End)

Extensions

Corrected and extended by N. J. A. Sloane, Apr 21 2000

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

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Keywords

Comments

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

Examples

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

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 159.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 83, Problem 25; p. 168, #30.
  • W. Feller, An Introduction to Probability Theory and Its Applications, Vol. I.
  • C. Jordan, Calculus of Finite Differences. Röttig and Romwalter, Budapest, 1939; Chelsea, NY, 1965, p. 449.
  • M. Klamkin, ed., Problems in Applied Mathematics: Selections from SIAM Review, SIAM, 1990; see pp. 127-129.
  • C. Lanczos, Applied Analysis. Prentice-Hall, Englewood Cliffs, NJ, 1956, p. 514.
  • A. P. Prudnikov, Yu. A. Brychkov and O.I. Marichev, "Integrals and Series", Volume 1: "Elementary Functions", Chapter 4: "Finite Sums", New York, Gordon and Breach Science Publishers, 1986-1992.
  • J. Ser, Les Calculs Formels des Séries de Factorielles. Gauthier-Villars, Paris, 1933, p. 92.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • J. Wallis, Operum Mathematicorum, pars altera, Oxford, 1656, pp 31,34 [Marc van Leeuwen, Apr 14 2010]

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

A001296 4-dimensional pyramidal numbers: a(n) = (3*n+1)*binomial(n+2, 3)/4. Also Stirling2(n+2, n).

Original entry on oeis.org

0, 1, 7, 25, 65, 140, 266, 462, 750, 1155, 1705, 2431, 3367, 4550, 6020, 7820, 9996, 12597, 15675, 19285, 23485, 28336, 33902, 40250, 47450, 55575, 64701, 74907, 86275, 98890, 112840, 128216, 145112, 163625, 183855, 205905, 229881, 255892, 284050, 314470
Offset: 0

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Author

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Comments

Permutations avoiding 12-3 that contain the pattern 31-2 exactly once.
Kekulé numbers for certain benzenoids. - Emeric Deutsch, Nov 18 2005
Partial sums of A002411. - Jonathan Vos Post, Mar 16 2006
If Y is a 3-subset of an n-set X then, for n>=6, a(n-5) is the number of 6-subsets of X having at least two elements in common with Y. - Milan Janjic, Nov 23 2007
Starting with 1 = binomial transform of [1, 6, 12, 10, 3, 0, 0, 0, ...]. Equals row sums of triangle A143037. - Gary W. Adamson, Jul 18 2008
Rephrasing the Perry formula of 2003: a(n) is the sum of all products of all two numbers less than or equal to n, including the squares. Example: for n=3 the sum of these products is 1*1 + 1*2 + 1*3 + 2*2 + 2*3 + 3*3 = 25. - J. M. Bergot, Jul 16 2011
Half of the partial sums of A011379. [Jolley, Summation of Series, Dover (1961), page 12 eq (66).] - R. J. Mathar, Oct 03 2011
Also the number of (w,x,y,z) with all terms in {1,...,n+1} and w < x >= y > z (see A211795). - Clark Kimberling, May 19 2012
Convolution of A000027 with A000326. - Bruno Berselli, Dec 06 2012
This sequence is related to A000292 by a(n) = n*A000292(n) - Sum_{i=0..n-1} A000292(i) for n>0. - Bruno Berselli, Nov 23 2017
a(n-2) is the maximum number of intersections made from the perpendicular bisectors of all pair combinations of n points. - Ian Tam, Dec 22 2020

Examples

			G.f. = x + 7*x^2 + 25*x^3 + 65*x^4 + 140*x^5 + 266*x^6 + 462*x^7 + 750*x^8 + 1155*x^9 + ...
		

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. 835.
  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 195.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 227, #16.
  • 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/3).
  • F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 223.
  • 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

a(n)=f(n, 2) where f is given in A034261.
a(n)= A093560(n+3, 4), (3, 1)-Pascal column.
Cf. A220212 for a list of sequences produced by the convolution of the natural numbers with the k-gonal numbers.
Cf. similar sequences listed in A241765 and A254142.
Cf. A000914.

Programs

  • Magma
    /* A000027 convolved with A000326: */ A000326:=func; [&+[(n-i+1)*A000326(i): i in [0..n]]: n in [0..40]]; // Bruno Berselli, Dec 06 2012
    
  • Magma
    [(3*n+1)*Binomial(n+2,3)/4: n in [0..40]]; // Vincenzo Librandi, Jul 30 2014
  • Maple
    A001296:=-(1+2*z)/(z-1)**5; # Simon Plouffe in his 1992 dissertation for sequence without the leading zero
  • Mathematica
    Table[n*(1+n)*(2+n)*(1+3*n)/24, {n, 0, 100}]
    CoefficientList[Series[x (1 + 2 x)/(1 - x)^5, {x, 0, 40}], x] (* Vincenzo Librandi, Jul 30 2014 *)
    Table[StirlingS2[n+2, n], {n, 0, 40}] (* Jean-François Alcover, Jun 24 2015 *)
    Table[ListCorrelate[Accumulate[Range[n]],Range[n]],{n,0,40}]//Flatten (* or *) LinearRecurrence[{5,-10,10,-5,1},{0,1,7,25,65},40] (* Harvey P. Dale, Aug 14 2017 *)
  • PARI
    t(n)=n*(n+1)/2
    for(i=1,30,print1(","sum(j=1,i,j*t(j))))
    
  • PARI
    {a(n) = n * (n+1) * (n+2) * (3*n+1) / 24}; /* Michael Somos, Sep 04 2017 */
    
  • Sage
    [stirling_number2(n+2,n) for n in range(0,38)] # Zerinvary Lajos, Mar 14 2009
    

Formula

a(n) = n*(1+n)*(2+n)*(1+3*n)/24. - T. D. Noe, Jan 21 2008
G.f.: x*(1+2*x)/(1-x)^5. - Paul Barry, Jul 23 2003
a(n) = Sum_{j=0..n} j*A000217(j). - Jon Perry, Jul 28 2003
E.g.f. with offset -1: exp(x)*(1*(x^2)/2! + 4*(x^3)/3! + 3*(x^4)/4!). For the coefficients [1, 4, 3] see triangle A112493.
E.g.f. x*exp(x)*(24 + 60*x + 28*x^2 + 3*x^3)/24 (above e.g.f. differentiated).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) + 3. - Kieren MacMillan, Sep 29 2008
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Jaume Oliver Lafont, Nov 23 2008
O.g.f. is D^2(x/(1-x)) = D^3(x), where D is the operator x/(1-x)*d/dx. - Peter Bala, Jul 02 2012
a(n) = A153978(n)/2. - J. M. Bergot, Aug 09 2013
a(n) = A002817(n) + A000292(n-1). - J. M. Bergot, Aug 29 2013; [corrected by Cyril Damamme, Feb 26 2018]
a(n) = A000914(n+1) - 2 * A000330(n+1). - Antal Pinter, Dec 31 2015
a(n) = A080852(3,n-1). - R. J. Mathar, Jul 28 2016
a(n) = 1*(1+2+...+n) + 2*(2+3+...+n) + ... + n*n. For example, a(6) = 266 = 1(1+2+3+4+5+6) + 2*(2+3+4+5+6) + 3*(3+4+5+6) + 4*(4+5+6) + 5*(5+6) + 6*(6).- J. M. Bergot, Apr 20 2017
a(n) = A000914(-2-n) for all n in Z. - Michael Somos, Sep 04 2017
a(n) = A000292(n) + A050534(n+1). - Cyril Damamme, Feb 26 2018
From Amiram Eldar, Jul 02 2020: (Start)
Sum_{n>=1} 1/a(n) = (6/5) * (47 - 3*sqrt(3)*Pi - 27*log(3)).
Sum_{n>=1} (-1)^(n+1)/a(n) = (6/5) * (16*log(2) + 6*sqrt(3)*Pi - 43). (End)
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