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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A000574 Coefficient of x^5 in expansion of (1 + x + x^2)^n.

Original entry on oeis.org

3, 16, 51, 126, 266, 504, 882, 1452, 2277, 3432, 5005, 7098, 9828, 13328, 17748, 23256, 30039, 38304, 48279, 60214, 74382, 91080, 110630, 133380, 159705, 190008, 224721, 264306, 309256, 360096, 417384, 481712, 553707, 634032, 723387, 822510
Offset: 3

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Author

Keywords

Comments

If Y is a 3-subset of an n-set X then, for n>=7, a(n-4) is the number of 5-subsets of X having at most one element in common with Y. - Milan Janjic, Nov 23 2007

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 78.
  • 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

Column m=5 of (1, 3) Pascal triangle A095660.

Programs

  • Magma
    [3*Binomial(n+2,5)-2*Binomial(n+1,5): n in [3..50]]; // Vincenzo Librandi, Jun 10 2012
    
  • Maple
    A000574:=-(-3+2*z)/(z-1)**6; # conjectured by Simon Plouffe in his 1992 dissertation
    seq(3*binomial(n+2,5)-2*binomial(n+1,5),n=3..100); # Robert Israel, Aug 04 2015
    A000574 := n -> GegenbauerC(`if`(5A000574(n)), n=3..20); # Peter Luschny, May 10 2016
  • Mathematica
    CoefficientList[Series[(3-2*x)/(1-x)^6,{x,0,40}],x] (* Vincenzo Librandi, Jun 10 2012 *)
  • PARI
    x='x+O('x^50); Vec(x^3*(3-2*x)/(1-x)^6) \\ G. C. Greubel, Nov 22 2017

Formula

G.f.: x^3*(3-2*x)/(1-x)^6.
a(n) = 3*binomial(n+2,5) - 2*binomial(n+1,5).
a(n) = A111808(n,5) for n>4. - Reinhard Zumkeller, Aug 17 2005
a(n) = binomial(n+1, 4)*(n+12)/5 = 3*b(n-3)-2*b(n-4), with b(n)=binomial(n+5, 5); cf. A000389.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6). - Vincenzo Librandi, Jun 10 2012
a(n) = 3*binomial(n, 3) + 4*binomial(n, 4) + binomial(n, 5). - Vladimir Shevelev and Peter J. C. Moses, Jun 22 2012
a(n) = GegenbauerC(N, -n, -1/2) where N = 5 if 5Peter Luschny, May 10 2016
a(n) = Sum_{i=1..n-1} A000217(i)*A055998(n-1-i). - Bruno Berselli, Mar 05 2018
E.g.f.: exp(x)*x^3*(60 + 20*x + x^2)/120. - Stefano Spezia, Jul 09 2023

Extensions

More terms from Vladeta Jovovic, Oct 02 2000

A005716 Coefficient of x^8 in expansion of (1+x+x^2)^n.

Original entry on oeis.org

1, 15, 90, 357, 1107, 2907, 6765, 14355, 28314, 52624, 93093, 157950, 258570, 410346, 633726, 955434, 1409895, 2040885, 2903428, 4065963, 5612805, 7646925, 10293075, 13701285, 18050760, 23554206, 30462615, 39070540, 49721892, 62816292, 78816012, 98253540
Offset: 4

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Author

Keywords

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 78.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Magma
    I:=[1, 15, 90, 357, 1107, 2907, 6765, 14355, 28314]; [n le 9 select I[n] else 9*Self(n-1)-36*Self(n-2)+84*Self(n-3)-126*Self(n-4)+126*Self(n-5)-84*Self(n-6)+36*Self(n-7)-9*Self(n-8)+Self(n-9): n in [1..40]]; // Vincenzo Librandi, Jun 16 2012
    
  • Magma
    /* By definition: */ P:=PolynomialRing(Integers()); [ Coefficients((1+x+x^2)^n)[9]: n in [4..32] ]; // Bruno Berselli, Jun 17 2012
  • Maple
    A005716:=-(6*z-9*z**2+3*z**3+1)/(z-1)**9; # Conjectured by Simon Plouffe in his 1992 dissertation.
    A005716 := n -> GegenbauerC(`if`(8A005716(n)), n=4..20); # Peter Luschny, May 10 2016
  • Mathematica
    CoefficientList[Series[(1+6*x-9*x^2+3*x^3)/(1-x)^9,{x,0,40}],x] (* Vincenzo Librandi, Jun 16 2012 *)

Formula

a(n) = binomial(n+1, 5)*(n^2+23*n-84)*(n+10)/336, n >= 4.
G.f.: (x^4)*(1+6*x-9*x^2+3*x^3)/(1-x)^9. (Numerator polynomial is N3(8, x) from A063420).
a(n) = A027907(n, 8), n >= 4 (ninth column of trinomial coefficients).
a(n) = A111808(n,8) for n>7. - Reinhard Zumkeller, Aug 17 2005
a(n) = 9*a(n-1) - 36*a(n-2) + 84*a(n-3) - 126*a(n-4) + 126*a(n-5) - 84*a(n-6) + 36*a(n-7) - 9*a(n-8) + a(n-9). Vincenzo Librandi, Jun 16 2012
a(n) = binomial(n,4) + 10*binomial(n,5) + 15*binomial(n,6) + 7*binomial(n,7) + binomial(n,8) (see our comment in A026729). - Vladimir Shevelev and Peter J. C. Moses, Jun 22 2012
a(n) = GegenbauerC(N, -n, -1/2) where N = 8 if 8Peter Luschny, May 10 2016

Extensions

More terms from Vladeta Jovovic, Oct 02 2000

A014532 Form array in which n-th row is obtained by expanding (1+x+x^2)^n and taking the 3rd column from the center.

Original entry on oeis.org

1, 4, 15, 50, 161, 504, 1554, 4740, 14355, 43252, 129844, 388752, 1161615, 3465840, 10329336, 30759120, 91538523, 272290140, 809676735, 2407049106, 7154586747, 21263575256, 63191778950, 187790510700, 558069593445, 1658498131836
Offset: 1

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Author

Keywords

Comments

Number of Dyck paths of semilength n+2 having exactly one occurrence of UUU, where U=(1,1). E.g. a(2)=4 because we have UDUUUDDD, UUUDDDUD, UUUDDUDD and UUUDUDDD, where U=(1,1) and D=(1,-1). - Emeric Deutsch, Dec 05 2003
a(n) is the number of Motzkin (2n+2)-paths whose longest basin has length n-1. A basin is a sequence of contiguous flatsteps preceded by a down step and followed by an up step. Example: a(2) counts FUDFUD, UDFUDF, UDFUFD, UFDFUD. - David Callan, Jul 15 2004
a(n) is the total number of valleys (DUs) in all Motzkin (n+3)-paths. Example: a(2)=4 counts the valleys (indicated by *) in FUD*UD, UD*UDF, UD*UFD, UFD*UD; the remaining 17 Motzkin 5-paths contain no valleys. - David Callan, Jul 03 2006
a(n) is the number of lattice paths from (0,0) to (n+1,n-1) taking north and east steps avoiding north^{>=3}. - Shanzhen Gao, Apr 20 2010
a(n) is the number of paths in the half-plane x>=0, from (0,0) to (n+2,3), and consisting of steps U=(1,1), D=(1,-1) and H=(1,0). For example, for n=2, we have the 4 paths: HUUU, UHUU, UUHU, UUUH. - José Luis Ramírez Ramírez, Apr 19 2015

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 78.

Crossrefs

First differences are in A025181.

Programs

  • Maple
    a := n -> simplify(GegenbauerC(n-1, -n-2, -1/2)):
    seq(a(n), n=1..26); # Peter Luschny, May 09 2016
  • Mathematica
    Table[GegenbauerC[n - 1, -n - 2, -1/2], {n,1,50}] (* G. C. Greubel, Feb 28 2017 *)
  • PARI
    z='z+O('z^50); Vec(2*z/(1-4*z+z^2+6*z^3+(1-3*z+2*z^3)*sqrt(1-2*z-3*z^2))) \\ G. C. Greubel, Feb 28 2017

Formula

G.f.: 2*z/(1-4*z+z^2+6*z^3+(1-3*z+2*z^3)*sqrt(1-2*z-3*z^2)). - Emeric Deutsch, Dec 05 2003
E.g.f.: exp(x)*BesselI(3, 2x) [0, 0, 0, 1, 4, 15..]. - Paul Barry, Sep 21 2004
a(n-2) = A111808(n,n-3) for n>2. - Reinhard Zumkeller, Aug 17 2005
a(n) = Sum_{i=0..floor((n-1)/2)} binomial(n+2,n-1-i) * binomial(n-1-i,i). - Shanzhen Gao, Apr 20 2010
a(n) = -(1/(162*(n+5)*(n+3)))*(9*n+18)*(-1)^n*(-3)^(1/2) * ((n+7)*hypergeom([1/2, n+5],[1],4/3) + hypergeom([1/2, n+4],[1],4/3) * (5*n+19)). - Mark van Hoeij, Oct 30 2011
D-finite with recurrence -(n+5)*(n-1)*a(n) +(n+2)*(2*n+3)*a(n-1) +3*(n+2)*(n+1)*a(n-2)=0. - R. J. Mathar, Dec 02 2012
a(n) ~ 3^(n+5/2)/(2*sqrt(Pi*n)). - Vaclav Kotesovec, Aug 10 2013
G.f.: z*M(z)^3/(1-z-2*z^2*M(z)), where M(z) is the g.f. of Motzkin paths (A001006). - José Luis Ramírez Ramírez, Apr 19 2015
From Peter Luschny, May 09 2016: (Start)
a(n) = C(4+2*n, n-1)*hypergeom([-n+1, -n-5], [-3/2-n], 1/4).
a(n) = GegenbauerC(n-1, -n-2, -1/2). (End)

Extensions

More terms from James Sellers, Feb 05 2000

A005714 Coefficient of x^6 in expansion of (1+x+x^2)^n.

Original entry on oeis.org

1, 10, 45, 141, 357, 784, 1554, 2850, 4917, 8074, 12727, 19383, 28665, 41328, 58276, 80580, 109497, 146490, 193249, 251713, 324093, 412896, 520950, 651430, 807885, 994266, 1214955, 1474795, 1779121, 2133792, 2545224, 3020424, 3567025, 4193322, 4908309, 5721717
Offset: 3

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Author

Keywords

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 78.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Magma
    I:=[1, 10, 45, 141, 357, 784, 1554]; [n le 7 select I[n] else 7*Self(n-1)-21*Self(n-2)+35*Self(n-3)-35*Self(n-4)+21*Self(n-5)-7*Self(n-6)+Self(n-7): n in [1..40]]; // Vincenzo Librandi, Jun 16 2012
    
  • Magma
    /* By definition: */ P:=PolynomialRing(Integers()); [ Coefficients((1+x+x^2)^n)[7]: n in [3..35] ]; // Bruno Berselli, Jun 17 2012
  • Maple
    A005714:=-(1+3*z-4*z**2+z**3)/(z-1)**7; # Conjectured by Simon Plouffe in his 1992 dissertation.
    A005714 := n -> GegenbauerC(`if`(6A005714(n)), n=3..20); # Peter Luschny, May 10 2016
  • Mathematica
    a[n_] := Coefficient[(1 + x + x^2)^n, x, 6]; Table[a[n], {n, 3, 35}]
    CoefficientList[Series[(1+3*x-4*x^2+x^3)/(1-x)^7,{x,0,40}],x] (* Vincenzo Librandi, Jun 16 2012 *)

Formula

a(n) = binomial(n, 3)*(n^3+18*n^2+17*n-120) /120.
G.f.: (x^3)*(1+3*x-4*x^2+x^3)/(1-x)^7. (Numerator polynomial is N3(6, x) from A063420).
a(n) = A027907(n, 6), n >= 3 (seventh column of trinomial coefficients).
a(n) = A111808(n,6) for n>5. - Reinhard Zumkeller, Aug 17 2005
a(n) = 7*a(n-1) -21*a(n-2) +35*a(n-3) -35*a(n-4) +21*a(n-5) -7*a(n-6) +a(n-7). Vincenzo Librandi, Jun 16 2012
a(n) = binomial(n,3) + 6*binomial(n,4) + 5*binomial(n,5) + binomial(n,6) (see our comment in A026729). - Vladimir Shevelev and Peter J. C. Moses, Jun 22 2012
a(n) = GegenbauerC(N, -n, -1/2) where N = 6 if 6Peter Luschny, May 10 2016
E.g.f.: exp(x)*x^3*(120 + 180*x + 30*x^2 + x^3)/720. - Stefano Spezia, Mar 28 2023

Extensions

More terms from Vladeta Jovovic, Oct 02 2000

A027908 a(n) = T(2*n, n), T given by A027907.

Original entry on oeis.org

1, 2, 10, 50, 266, 1452, 8074, 45474, 258570, 1481108, 8533660, 49402850, 287134346, 1674425300, 9792273690, 57407789550, 337281021450, 1985342102964, 11706001102180, 69124774458092, 408737856117916, 2419833655003752, 14341910428953018, 85087759173024870
Offset: 0

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Author

Keywords

Comments

Central terms of the triangle in A111808. - Reinhard Zumkeller, Aug 17 2005
Number of paths of semilength n starting at (0,0) and ending on the X-axis using steps (1,1), (1,-1) and (1,3). - David Scambler, Jun 21 2013

Crossrefs

Programs

  • Maple
    ogf := series( RootOf( (144*x^2+140*x-27)*g^4+(18-12*x)*g^2+8*g+1, g), x=0, 20); # Mark van Hoeij, Nov 16 2011
    a := n -> simplify(GegenbauerC(n, -2*n, -1/2)):
    seq(a(n), n=0..23); # Peter Luschny, May 09 2016
  • Mathematica
    Table[Binomial[4 n, n] Hypergeometric2F1[-3 n, -n, 1/2 - 2 n, 1/4], {n, 0, 20}] (* or *) Table[GegenbauerC[3 n, -2 n, -1/2] + KroneckerDelta[n], {n, 0, 20}] (* Vladimir Reshetnikov, May 07 2016 *)
  • Maxima
    makelist(ultraspherical(n,-2*n,-1/2),n,0,12); /* Emanuele Munarini, Oct 18 2016 */

Formula

G.f.: -(g^2+g+1)/(3*g^2+g-1) where g = x times the g.f. of A143927. - Mark van Hoeij, Nov 16 2011
a(n) = GegenbauerC(n, -2*n, -1/2). - Peter Luschny, May 09 2016
From Peter Bala, Jan 26 2020: (Start)
a(n) = [x^(2*n)](1 + x^2 + x^4)^(2*n).
a(n) = Sum_{k = 0..floor(n/2)} C(2*n, n-k)*C(n-k, k).
a(n) = C(2*n,n) * hypergeom([-n/2, (1 - n)/2], [n + 1], 4)
Conjectural: a(n*p^k) == a(n*p^(k-1)) ( mod p^(2*k) ) for all primes p >= 5 and positive integers n and k. (End)
From Peter Bala, Aug 03 2023: (Start)
P-recursive: 3*n*(13*n - 17)*(3*n - 1)*(3*n - 2)*a(n) = 2*(2*n - 1)*(455*n^3 - 1050*n^2 + 691*n - 120)*a(n-1) + 36*(n - 1)*(13*n - 4)*(2*n - 1)*(2*n - 3)*a(n-2) with a(0) = 1 and a(1) = 2.
exp(Sum_{n >= 0} a(n)*x^n/n) = 1 + 2*x + 7*x^2 + 28*x^3 + 123*x^4 + ... is the g.f. of A143927.
a(n) = 2*A344396(n-1) for n >= 1. (End)

A094531 Array read by rows: right-hand side of triangle A027907 of trinomial coefficients.

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 7, 6, 3, 1, 19, 16, 10, 4, 1, 51, 45, 30, 15, 5, 1, 141, 126, 90, 50, 21, 6, 1, 393, 357, 266, 161, 77, 28, 7, 1, 1107, 1016, 784, 504, 266, 112, 36, 8, 1, 3139, 2907, 2304, 1554, 882, 414, 156, 45, 9, 1, 8953, 8350, 6765, 4740, 2850, 1452, 615, 210, 55
Offset: 0

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Author

Paul Barry, May 07 2004

Keywords

Comments

Sometimes called a Motzkin triangle, although that name is usually reserved for A026300.
Expand (1+x+x^2)^n and take last (nonzero) coefficient of first row, last two coefficients of second row, etc.
Equals A094531*(1,xc(-x^2)) where c(x) is the g.f. of A000108. - Paul Barry, May 12 2009
Coefficients of Faber polynomials for (1/x+1+x): Fa(n,x) = Sum_{k=0..n} T(n,k)*x^k, g.f.: -log((sqrt(-3*t^2-2*t+1)-t+1)/2-t*x) = Sum_{n>0} Fa(n,x)*t^n/n. - Vladimir Kruchinin, Jul 01 2013

Examples

			Triangle begins:
    1;
    1,   1;
    3,   2,   1;
    7,   6,   3,   1;
   19,  16,  10,   4,   1;
   51,  45,  30,  15,   5,   1;
  141, 126,  90,  50,  21,   6,   1;
  393, 357, 266, 161,  77,  28,   7,   1;
  ...
		

Crossrefs

Binomial transform is triangle A094527. Row sums are A027914.
Cf. A111808 (row reversed).

Programs

  • Maple
    T := (n, k) -> simplify(GegenbauerC(n-k, -n, -1/2)):
    for n from 0 to 9 do seq(T(n,k), k=0..n) od; # Peter Luschny, May 12 2016
  • Mathematica
    max = 10; se = Series[ -Log[ (Sqrt[-3*t^2 - 2*t + 1] - t + 1)/2 - t*x], {t, 0, max + 1}, {x, 0, max}]; a[n_, k_] := SeriesCoefficient[se, {t, 0, n}, {x, 0, k}]*n; a[0, 0] = 1; Table[a[n, k], {n, 0, max }, {k, 0, n}] // Flatten  (* Jean-François Alcover, Jul 02 2013, after Vladimir Kruchinin *)
    Table[Binomial[n, k] Hypergeometric2F1[(k - n)/2, (k - n + 1)/2, k + 1, 4], {n, 0, 9}, {k, 0, n}] // Flatten (* or *)
    Table[If[n == 0, 1, GegenbauerC[n - k, -n, -1/2]], {n, 0, 9}, {k, 0, n}] // Flatten (* Michael De Vlieger, May 12 2016 *)

Formula

Riordan array ( 1/sqrt(1-2*x-3*x^2), (1-x-sqrt(1-2*x-3*x^2))/(2*x) ). - N. J. A. Sloane, Jun 02 2005
Product of Riordan arrays (1/(1-x), x/(1-x)) (Pascal's triangle, A007318) and (1/sqrt(1-4x^2), (1-sqrt(1-4*x^2))/(2*x)) (A108044). Inverse is A102587. - Paul Barry, Jul 14 2005
Column k has e.g.f. exp(x)*Bessel_I(k, 2x). - Paul Barry, Jul 14 2005
T(n, k) = Sum_{i=0..n} C(n-k-i, i)*C(n, k+i). - Paul Barry, Nov 04 2005
T(n, k) = Sum_{j=0..n} C(n,j)*C(j,n-k-j). - Paul Barry, Oct 25 2006
From Paul Barry, May 12 2009: (Start)
Production matrix is
1, 1;
2, 1, 1;
0, 1, 1, 1;
0, 0, 1, 1, 1;
0, 0, 0, 1, 1, 1; (End)
From Peter Bala, Jun 29 2015: (Start)
Riordan array has the form ( x*h'(x)/h(x), h(x) ) with h(x) = (1 - x -sqrt(1 - 2*x - 3*x^2))/(2*x) and so belongs to the hitting time subgroup H of the Riordan group (see Peart and Woan, Example 1.1).
T(n,k) = [x^(n-k)] f(x)^n with f(x) = 1 + x + x^2. In general the (n,k)th entry of the hitting time array ( x*h'(x)/h(x), h(x) ) has the form [x^(n-k)] f(x)^n, where f(x) = x/( series reversion of h(x) ). (End)
From Peter Luschny, May 12 2016: (Start)
T(n,k) = binomial(n, k)*hypergeom([(k-n)/2, (k-n+1)/2], [k+1], 4):
T(n,k) = GegenbauerC(n-k, -n, -1/2). (End)

A213742 Triangle of numbers C^(3)(n,k) of combinations with repetitions from n different elements over k for each of them not more than three appearances allowed.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 6, 10, 1, 4, 10, 20, 31, 1, 5, 15, 35, 65, 101, 1, 6, 21, 56, 120, 216, 336, 1, 7, 28, 84, 203, 413, 728, 1128, 1, 8, 36, 120, 322, 728, 1428, 2472, 3823, 1, 9, 45, 165, 486, 1206, 2598, 4950, 8451, 13051, 1, 10
Offset: 0

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Author

Keywords

Comments

The left side of triangle consists of 1's, while the right side is formed by A005725. Further, T(n,0)=1, T(n,1)=n, T(n,2)=A000217(n) for n>1, T(n,3)=A000292(n) for n>=3, T(n,4)=A005718(n) for n>=2, T(n,5)=A005719(n) for n>=5, T(n,6)=A005720(n) for n>=6, T(n,7)=A001919(n) for n>=7, T(n,8)=A064055(n) for n>=5.

Examples

			Triangle begins
n/k.|..0.....1.....2.....3.....4.....5.....6.....7
==================================================
.0..|..1
.1..|..1.....1
.2..|..1.....2.....3
.3..|..1.....3.....6....10
.4..|..1.....4....10....20....31
.5..|..1.....5....15....35....65....101
.6..|..1.....6....21....56...120....216...336
.7..|..1.....7....28....84...203....413...728....1128
		

Crossrefs

Programs

  • Mathematica
    Flatten[Table[Sum[(-1)^r Binomial[n,r] Binomial[n-# r+k-1,n-1],{r,0,Floor[k/#]}],{n,0,15},{k,0,n}]/.{0}->{1}]&[4] (* Peter J. C. Moses, Apr 16 2013 *)

Formula

C^(3)(n,k)=sum{r=0,...,floor(k/4)}(-1)^r*C(n,r)*C(n-4*r+k-1, n-1)

A005715 Coefficient of x^7 in expansion of (1+x+x^2)^n.

Original entry on oeis.org

4, 30, 126, 393, 1016, 2304, 4740, 9042, 16236, 27742, 45474, 71955, 110448, 165104, 241128, 344964, 484500, 669294, 910822, 1222749, 1621224, 2125200, 2756780, 3541590, 4509180, 5693454, 7133130, 8872231, 10960608, 13454496
Offset: 4

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Author

Keywords

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 78.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Magma
    I:=[4, 30, 126, 393, 1016, 2304, 4740, 9042]; [n le 8 select I[n] else 8*Self(n-1)-28*Self(n-2)+56*Self(n-3)-70*Self(n-4)+56*Self(n-5)-28*Self(n-6)+8*Self(n-7)-Self(n-8): n in [1..40]]; // Vincenzo Librandi, Jun 16 2012
    
  • Magma
    /* By definition: */ P:=PolynomialRing(Integers()); [ Coefficients((1+x+x^2)^n)[8]: n in [4..33] ]; // Bruno Berselli, Jun 17 2012
  • Maple
    A005715:=(z-2)*(z**2-2)/(z-1)**8; # Conjectured by Simon Plouffe in his 1992 dissertation.
    A005715 := n -> GegenbauerC(`if`(7A005715(n)), n=4..20); # Peter Luschny, May 10 2016
  • Mathematica
    CoefficientList[Series[(x-2)*(x^2-2)/(1-x)^8,{x,0,40}],x] (* Vincenzo Librandi, Jun 16 2012 *)

Formula

a(n) = binomial(n, 4)*(n^3+27*n^2+116*n-120)/210, n >= 4.
G.f.: (x^4)*(x-2)*(x^2-2)/(1-x)^8. (Numerator polynomial is N3(7, x) from A063420).
a(n) = A027907(n, 7), n >= 4 (eighth column of trinomial coefficients).
a(n) = A111808(n,7) for n>6. - Reinhard Zumkeller, Aug 17 2005
a(n) = 8*a(n-1) -28*a(n-2) +56*a(n-3) -70*a(n-4) +56*a(n-5) -28*a(n-6) +8*a(n-7) -a(n-8). Vincenzo Librandi, Jun 16 2012
a(n) = 4*binomial(n,4) + 10*binomial(n,5) + 6*binomial(n,6) + binomial(n,7) (see our comment in A026729). - Vladimir Shevelev and Peter J. C. Moses, Jun 22 2012
a(n) = GegenbauerC(N, -n, -1/2) where N = 7 if 7Peter Luschny, May 10 2016

Extensions

More terms from Vladeta Jovovic, Oct 02 2000

A014533 Form array in which n-th row is obtained by expanding (1 + x + x^2)^n and taking the 4th column from the center.

Original entry on oeis.org

1, 5, 21, 77, 266, 882, 2850, 9042, 28314, 87802, 270270, 827190, 2520336, 7651632, 23162976, 69954048, 210859245, 634569201, 1907165337, 5725520801, 17172595110, 51465297950, 154135675070, 461366154990, 1380317174145
Offset: 1

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Keywords

Comments

First differences seem to be in A025182.
a(n-3) = A111808(n, n-4) for n > 3. - Reinhard Zumkeller, Aug 17 2005
a(n-4) = number of paths in the half-plane x >= 0, from (0,0) to (n,4), and consisting of steps U=(1,1), D=(1,-1) and H=(1,0). For example, for n=5, we have the 5 paths HUUUU, UHUUU, UUHUU, UUUHU, UUUUH. - José Luis Ramírez Ramírez, Apr 19 2015

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 78.

Programs

  • Maple
    a := n -> simplify(GegenbauerC(n-1, -n-3, -1/2)):
    seq(a(n), n=1..25); # Peter Luschny, May 09 2016
  • Mathematica
    Rest[CoefficientList[Series[x*((1-x-Sqrt[1-2*x-3*x^2])/(2*x^2))^4/(1-x-2*x^2*(1-x-Sqrt[1-2*x-3*x^2])/(2*x^2)), {x, 0, 20}], x]] (* Vaclav Kotesovec, Apr 20 2015 *)
    Table[GegenbauerC[n-1, -n - 3, -1/2], {n,0,50}] (* G. C. Greubel, Feb 28 2017 *)
  • PARI
    x='x + O('x^50); Vec(x*((1-x-sqrt(1-2*x-3*x^2))/(2*x^2))^4/(1-x-2*x^2*(1-x-sqrt(1-2*x-3*x^2))/(2*x^2))) \\ G. C. Greubel, Feb 28 2017

Formula

Conjecture: -(n+7)*(n-1)*a(n) + (n+3)*(2*n+5)*a(n-1) + 3*(n+3)*(n+2)*a(n-2) = 0. - R. J. Mathar, Feb 25 2015
G.f.: z*M(z)^4/(1-z-2*z^2*M(z)), where M(z) is the g.f. of Motzkin paths. - José Luis Ramírez Ramírez, Apr 19 2015
a(n) ~ 3^(n+7/2) / (2*sqrt(Pi*n)). - Vaclav Kotesovec, Apr 20 2015
From Peter Luschny, May 09 2016: (Start)
a(n) = C(6+2*n, n-1)*hypergeom([-n+1, -n-7], [-5/2-n], 1/4).
a(n) = GegenbauerC(n-1, -n-3, -1/2). (End)

Extensions

More terms from James Sellers, Feb 05 2000

A213743 Triangle T(n,k), read by rows, of numbers T(n,k)=C^(4)(n,k) of combinations with repetitions from n different elements over k for each of them not more than four appearances allowed.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 6, 10, 1, 4, 10, 20, 35, 1, 5, 15, 35, 70, 121, 1, 6, 21, 56, 126, 246, 426, 1, 7, 28, 84, 210, 455, 875, 1520, 1, 8, 36, 120, 330, 784, 1652, 3144, 5475, 1, 9, 45, 165, 495, 1278, 2922, 6030, 11385, 19855, 1, 10, 55, 220, 715, 1992, 4905, 10890, 22110, 41470, 72403
Offset: 0

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Comments

The left side of triangle consists of 1's, while the right side is formed by A187925. Further, T(n,0)=1, T(n,1)=n, T(n,2)=A000217(n) for n>1, T(n,3)=A000292(n) for n>=3, T(n,4)=A000332(n) for n>=7, T(n,5)=A027659(n) for n>=3, T(n,6)=A064056(n) for n>=4, T(n,7)=A064057(n) for n>=5, T(n,8)=A064058(n) for n>=6, T(n,9)=A000575(n) for n>=6.

Examples

			Triangle begins
  n/k.|..0.....1.....2.....3.....4.....5.....6.....7
  ==================================================
  .0..|..1
  .1..|..1.....1
  .2..|..1.....2.....3
  .3..|..1.....3.....6....10
  .4..|..1.....4....10....20....35
  .5..|..1.....5....15....35....70....121
  .6..|..1.....6....21....56...126....246...426
  .7..|..1.....7....28....84...210....455...875....1520
T(4,2)=C^(4)(4,2): From 4 elements {1,2,3,4}, we have the following 10 allowed combinations of 2 elements: {1,1}, {1,2}, {1,3}, {1,4}, {2,2}, {2,3}, {2,4}, {3,3}, {3,4}, {4,4}.
		

Crossrefs

Programs

  • Mathematica
    Flatten[Table[Sum[(-1)^r Binomial[n,r] Binomial[n-# r+k-1,n-1],{r,0,Floor[k/#]}],{n,0,15},{k,0,n}]/.{0}->{1}]&[5] (* Peter J. C. Moses, Apr 16 2013 *)

Formula

C^(4)(n,k) = Sum_{r=0...floor(k/5)} (-1)^r*C(n,r)*C(n-5*r+k-1, n-1).
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