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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A090181 Triangle of Narayana (A001263) with 0 <= k <= n, read by rows.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 6, 6, 1, 0, 1, 10, 20, 10, 1, 0, 1, 15, 50, 50, 15, 1, 0, 1, 21, 105, 175, 105, 21, 1, 0, 1, 28, 196, 490, 490, 196, 28, 1, 0, 1, 36, 336, 1176, 1764, 1176, 336, 36, 1, 0, 1, 45, 540, 2520, 5292, 5292, 2520, 540, 45, 1, 0, 1, 55, 825, 4950, 13860
Offset: 0

Views

Author

Philippe Deléham, Jan 19 2004

Keywords

Comments

Number of Dyck n-paths with exactly k peaks. - Peter Luschny, May 10 2014

Examples

			Triangle starts:
[0] 1;
[1] 0, 1;
[2] 0, 1,  1;
[3] 0, 1,  3,   1;
[4] 0, 1,  6,   6,    1;
[5] 0, 1, 10,  20,   10,    1;
[6] 0, 1, 15,  50,   50,   15,    1;
[7] 0, 1, 21, 105,  175,  105,   21,   1;
[8] 0, 1, 28, 196,  490,  490,  196,  28,  1;
[9] 0, 1, 36, 336, 1176, 1764, 1176, 336, 36, 1;
		

Crossrefs

Mirror image of triangle A131198. A000108 (row sums, Catalan).
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A000108(n), A006318(n), A047891(n+1), A082298(n), A082301(n), A082302(n), A082305(n), A082366(n), A082367(n) for x=0,1,2,3,4,5,6,7,8,9. - Philippe Deléham, Aug 10 2006
Sum_{k=0..n} x^(n-k)*T(n,k) = A090192(n+1), A000012(n), A000108(n), A001003(n), A007564(n), A059231(n), A078009(n), A078018(n), A081178(n), A082147(n), A082181(n), A082148(n), A082173(n) for x = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. - Philippe Deléham, Oct 21 2006
Sum_{k=0..n} T(n,k)*x^k*(x-1)^(n-k) = A000012(n), A006318(n), A103210(n), A103211(n), A133305(n), A133306(n), A133307(n), A133308(n), A133309(n) for x = 1, 2, 3, 4, 5, 6, 7, 8, 9, respectively. - Philippe Deléham, Oct 20 2007

Programs

  • Magma
    [[(&+[(-1)^(j-k)*Binomial(2*n-j,j)*Binomial(j,k)*Binomial(2*n-2*j,n-j)/(n-j+1): j in [0..n]]): k in [0..n]]: n in [0..10]];
  • Maple
    A090181 := (n,k) -> binomial(n,n-k)*binomial(n-1,n-k)/(n-k+1):
    seq(print( seq(A090181(n,k),k=0..n)),n=0..5); # Peter Luschny, May 10 2014
    egf := 1+int((sqrt(t)*exp((1+t)*x)*BesselI(1,2*sqrt(t)*x))/x,x);
    s := n -> n!*coeff(series(egf,x,n+2),x,n);
    seq(print(seq(coeff(s(n),t,j),j=0..n)),n=0..9); # Peter Luschny, Oct 30 2014
    T := proc(n, k) option remember; if k = n or k = 1 then 1 elif k < 1 then 0 else (2*n/k - 1) * T(n-1, k-1) + T(n-1, k) fi end:
    for n from 0 to 8 do seq(T(n, k), k = 0..n) od;  # Peter Luschny, Dec 31 2024
  • Mathematica
    Flatten[Table[Sum[(-1)^(j-k) * Binomial[2n-j,j] * Binomial[j,k] * CatalanNumber[n-j], {j, 0, n}], {n,0,11},{k,0,n}]] (* Indranil Ghosh, Mar 05 2017 *)
    p[0, ] := 1; p[1, x] := x; p[n_, x_] := ((2 n - 1) (1 + x) p[n - 1, x] - (n - 2) (x - 1)^2 p[n - 2, x]) / (n + 1);
    Table[CoefficientList[p[n, x], x], {n, 0, 9}] // TableForm (* Peter Luschny, Apr 26 2022 *)
  • PARI
    c(n) = binomial(2*n,n)/ (n+1);
    tabl(nn) = {for(n=0, nn, for(k=0, n, print1(sum(j=0, n, (-1)^(j-k) * binomial(2*n-j,j) * binomial(j,k) * c(n-j)),", ");); print(););};
    tabl(11); \\ Indranil Ghosh, Mar 05 2017
    
  • Python
    from functools import cache
    @cache
    def Trow(n):
        if n == 0: return [1]
        if n == 1: return [0, 1]
        if n == 2: return [0, 1, 1]
        A = Trow(n - 2) + [0, 0]
        B = Trow(n - 1) + [1]
        for k in range(n - 1, 1, -1):
            B[k] = (((B[k] + B[k - 1]) * (2 * n - 1)
                   - (A[k] - 2 * A[k - 1] + A[k - 2]) * (n - 2)) // (n + 1))
        return B
    for n in range(10): print(Trow(n)) # Peter Luschny, May 02 2022
    
  • Sage
    def A090181_row(n):
        U = [0]*(n+1)
        for d in DyckWords(n):
            U[d.number_of_peaks()] +=1
        return U
    for n in range(8): A090181_row(n) # Peter Luschny, May 10 2014
    

Formula

Triangle T(n, k), read by rows, given by [0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] where DELTA is the operator defined in A084938. T(0, 0) = 1, T(n, 0) = 0 for n>0, T(n, k) = C(n-1, k-1)*C(n, k-1)/k for k>0.
Sum_{j>=0} T(n,j)*binomial(j,k) = A060693(n,k). - Philippe Deléham, May 04 2007
Sum_{k=0..n} T(n,k)*10^k = A143749(n+1). - Philippe Deléham, Oct 14 2008
From Paul Barry, Nov 10 2008: (Start)
Coefficient array of the polynomials P(n,x) = x^n*2F1(-n,-n+1;2;1/x).
T(n,k) = Sum_{j=0..n} (-1)^(j-k)*C(2n-j,j)*C(j,k)*A000108(n-j). (End)
Sum_{k=0..n} T(n,k)*5^k*3^(n-k) = A152601(n). - Philippe Deléham, Dec 10 2008
Sum_{k=0..n} T(n,k)*(-2)^k = A152681(n); Sum_{k=0..n} T(n,k)*(-1)^k = A105523(n). - Philippe Deléham, Feb 03 2009
Sum_{k=0..n} T(n,k)*2^(n+k) = A156017(n). - Philippe Deléham, Nov 27 2011
T(n, k) = C(n,n-k)*C(n-1,n-k)/(n-k+1). - Peter Luschny, May 10 2014
E.g.f.: 1+Integral((sqrt(t)*exp((1+t)*x)*BesselI(1,2*sqrt(t)*x))/x dx). - Peter Luschny, Oct 30 2014
G.f.: (1+x-x*y-sqrt((1-x*(1+y))^2-4*y*x^2))/(2*x). - Alois P. Heinz, Nov 28 2021, edited by Ron L.J. van den Burg, Dec 19 2021
T(n, k) = [x^k] (((2*n - 1)*(1 + x)*p(n-1, x) - (n - 2)*(x - 1)^2*p(n-2, x))/(n + 1)) with p(0, x) = 1 and p(1, x) = x. - Peter Luschny, Apr 26 2022
Recursion based on rows (see the Python program):
T(n, k) = (((B(k) + B(k-1))*(2*n - 1) - (A(k) - 2*A(k-1) + A(k-2))*(n-2))/(n+1)), where A(k) = T(n-2, k) and B(k) = T(n-1, k), for n >= 3. # Peter Luschny, May 02 2022

A132813 Triangle read by rows: A001263 * A127648 as infinite lower triangular matrices.

Original entry on oeis.org

1, 1, 2, 1, 6, 3, 1, 12, 18, 4, 1, 20, 60, 40, 5, 1, 30, 150, 200, 75, 6, 1, 42, 315, 700, 525, 126, 7, 1, 56, 588, 1960, 2450, 1176, 196, 8, 1, 72, 1008, 4704, 8820, 7056, 2352, 288, 9, 1, 90, 1620, 10080, 26460, 31752, 17640, 4320, 405, 10
Offset: 0

Views

Author

Gary W. Adamson, Sep 01 2007

Keywords

Comments

Also T(n,k) = binomial(n-1, k-1)*binomial(n, k-1), related to Narayana polynomials (see Sulanke reference). - Roger L. Bagula, Apr 09 2008
h-vector for cluster complex associated to the root system B_n. See p. 8, Athanasiadis and C. Savvidou. - Tom Copeland, Oct 19 2014

Examples

			First few rows of the triangle are:
  1;
  1,  2;
  1,  6,   3;
  1, 12,  18,   4;
  1, 20,  60,  40,   5;
  1, 30, 150, 200,  75,   6;
  1, 42, 315, 700, 525, 126, 7;
  ...
		

Crossrefs

Family of polynomials (see A062145): A008459 (c=1), this sequence (c=2), A062196 (c=3), A062145 (c=4), A062264 (c=5), A062190 (c=6).
Columns: A000012 (k=0), A002378 (k=1), A006011 (k=2), 4*A006542 (k=3), 5*A006857 (k=4), 6*A108679 (k=5), 7*A134288 (k=6), 8*A134289 (k=7), 9*A134290 (k=8), 10*A134291 (k=9).
Diagonals: A000027 (k=n), A002411 (k=n-1), A004302 (k=n-2), A108647 (k=n-3), A134287 (k=n-4).
Main diagonal: A000894.
Sums: (-1)^floor((n+1)/2)*A001405 (signed row), A001700 (row), A203611 (diagonal).
Cf. A103371 (mirrored).

Programs

  • GAP
    Flat(List([0..10],n->List([0..n], k->(k+1)*Binomial(n+1,k+1)*Binomial(n+1,k)/(n+1)))); # Muniru A Asiru, Feb 26 2019
    
  • Haskell
    a132813 n k = a132813_tabl !! n !! k
    a132813_row n = a132813_tabl !! n
    a132813_tabl = zipWith (zipWith (*)) a007318_tabl $ tail a007318_tabl
    -- Reinhard Zumkeller, Apr 04 2014
    
  • Magma
    /* triangle */ [[(k+1)*Binomial(n+1,k+1)*Binomial(n+1,k)/(n+1): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Oct 19 2014
    
  • Maple
    P := (n, x) -> hypergeom([1-n, -n], [1], x): for n from 1 to 9 do PolynomialTools:-CoefficientList(simplify(P(n,x)),x) od; # Peter Luschny, Nov 26 2014
  • Mathematica
    T[n_,k_]=Binomial[n-1,k-1]*Binomial[n,k-1]; Table[Table[T[n,k],{k,1,n}],{n,1,11}]; Flatten[%] (* Roger L. Bagula, Apr 09 2008 *)
    P[n_, x_] := HypergeometricPFQ[{1-n, -n}, {1}, x]; Table[CoefficientList[P[n, x], x], {n, 1, 10}] // Flatten (* Jean-François Alcover, Nov 27 2014, after Peter Luschny *)
  • PARI
    tabl(nn) = {for (n = 1, nn, for (k = 1, n, print1(binomial(n-1, k-1)*binomial(n, k-1) , ", ");););} \\ Michel Marcus, Feb 12 2014
    
  • SageMath
    def A132813(n,k): return binomial(n,k)*binomial(n+1,k)
    print(flatten([[A132813(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Mar 12 2025

Formula

T(n,k) = (k+1)*binomial(n+1,k+1)*binomial(n+1,k)/(n+1), n >= k >= 0.
From Roger L. Bagula, May 14 2010: (Start)
T(n, m) = coefficients(p(x,n)), where
p(x,n) = (1-x)^(2*n)*Sum_{k >= 0} binomial(k+n-1, k)*binomial(n+k, k)*x^k,
or p(x,n) = (1-x)^(2*n)*Hypergeometric2F1([n, n+1], [1], x). (End)
T(n,k) = binomial(n,k) * binomial(n+1,k). - Reinhard Zumkeller, Apr 04 2014
These are the coefficients of the polynomials Hypergeometric2F1([1-n,-n], [1], x). - Peter Luschny, Nov 26 2014
G.f.: A(x,y) = A281260(x,y)/(1-A281260(x,y))/x. - Vladimir Kruchinin, Oct 10 2020

A078009 a(0)=1, for n>=1 a(n) = Sum_{k=0..n} 5^k*N(n,k) where N(n,k) = C(n,k)*C(n,k+1)/n are the Narayana numbers (A001263).

Original entry on oeis.org

1, 1, 6, 41, 306, 2426, 20076, 171481, 1500666, 13386206, 121267476, 1112674026, 10318939956, 96572168916, 910896992856, 8650566601401, 82644968321226, 793753763514806, 7659535707782916, 74225795172589006, 722042370787826076
Offset: 0

Views

Author

Benoit Cloitre, May 10 2003

Keywords

Comments

More generally coefficients of (1 + m*x - sqrt(m^2*x^2 - (2*m+2)*x + 1) )/( 2*m*x ) are given by a(n) = Sum_{k=0..n} (m+1)^k * N(n,k).
a(n) is the series reversion of x*(1-5*x)/(1-4*x). a(n+1) is the series reversion of x/(1 + 6*x + 5*x^2). a(n+1) counts (6,5)-Motzkin paths of length n, where there are 6 colors available for the H(1,0) steps and 5 for the U(1,1) steps. - Paul Barry, May 19 2005
The Hankel transform of this sequence is 5^C(n+1,2). - Philippe Deléham, Oct 29 2007
a(n) is the number of Schröder paths of semilength n in which there are no (2,0)-steps at level 0 and at a higher level they come in 4 colors. Example: a(2)=6 because we have UDUD, UUDD, UBD, UGD, URD, and UYD, where U=(1,1), D=(1,-1), while B, G, R, and Y are, respectively, blue, green, red, and yellow (2,0)-steps. - Emeric Deutsch, May 02 2011
Shifts left when INVERT transform applied five times. - Benedict W. J. Irwin, Feb 03 2016

Crossrefs

Programs

  • Magma
    R:=PowerSeriesRing(Rationals(), 30); Coefficients(R!( (1+4*x - Sqrt(16*x^2-12*x+1))/(10*x) )); // G. C. Greubel, Jun 28 2019
    
  • Magma
    [1] cat [&+[5^k*Binomial(n,k)*Binomial(n,k+1)/n:k in [0..n]]:n in [1..20]]; // Marius A. Burtea, Jan 21 2020
    
  • Maple
    A078009_list := proc(n) local j, a, w; a := array(0..n); a[0] := 1;
    for w from 1 to n do a[w] := a[w-1]+5*add(a[j]*a[w-j-1],j=1..w-1) od;
    convert(a, list) end: A078009_list(20); # Peter Luschny, May 19 2011
  • Mathematica
    Table[SeriesCoefficient[(1+4*x-Sqrt[16*x^2-12*x+1])/(10*x),{x,0,n}],{n,0,30}] (* Vaclav Kotesovec, Oct 13 2012 *)
    a[n_] := Hypergeometric2F1[1 - n, -n, 2, 5];
    Table[a[n], {n, 0, 30}] (* Peter Luschny, Mar 19 2018 *)
  • PARI
    a(n)=sum(k=0,n,5^k/n*binomial(n,k)*binomial(n,k+1))
    
  • Sage
    a=((1+4*x -sqrt(16*x^2-12*x+1))/(10*x)).series(x, 30).coefficients(x, sparse=False); [1]+a[1:] # G. C. Greubel, Jun 28 2019

Formula

G.f.: (1 + 4*x - sqrt(16*x^2 - 12*x + 1))/(10*x).
a(n) = Sum_{k=0..n} A088617(n, k)*5^k*(-4)^(n-k). - Philippe Deléham, Jan 21 2004
With offset 1 : a(1)=1, a(n) = -4*a(n-1) + 5*Sum_{i=1..n-1} a(i)*a(n-i). - Benoit Cloitre, Mar 16 2004
a(n+1) = Sum_{k=0..floor(n/2)} C(n, 2*k)*C(k)*6^(n-2k)*5^k; - Paul Barry, May 19 2005
a(n) = ( 6*(2*n-1)*a(n-1) - 16*(n-2)*a(n-2) ) / (n+1) for n >= 2, a(0) = a(1) = 1. - Philippe Deléham, Aug 19 2005
From Gary W. Adamson, Jul 08 2011: (Start)
a(n) = upper left term in M^n, M = the production matrix:
1, 1
5, 5, 5
1, 1, 1, 1
5, 5, 5, 5, 5
1, 1, 1, 1, 1, 1
... (End)
a(n) ~ sqrt(10+6*sqrt(5))*(6+2*sqrt(5))^n/(10*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 13 2012. Equivalently, a(n) ~ 2^(2*n) * phi^(2*n + 1) / (5^(3/4) * sqrt(Pi) * n^(3/2)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 08 2021
a(n) = A127848(n) for n > 0. - Philippe Deléham, Apr 03 2013
G.f.: 1/(1 - x/(1 - 5*x/(1 - x/(1 - 5*x/(1 - x/(1 - ...)))))), a continued fraction. - Ilya Gutkovskiy, Apr 21 2017
a(n) = hypergeom([1 - n, -n], [2], 5). - Peter Luschny, Mar 19 2018

A103365 First column of triangle A103364, which equals the matrix inverse of the Narayana triangle (A001263).

Original entry on oeis.org

1, -1, 2, -7, 39, -321, 3681, -56197, 1102571, -27036487, 810263398, -29139230033, 1238451463261, -61408179368043, 3513348386222286, -229724924077987509, 17023649385410772579, -1419220037471837658603, 132236541042728184852942, -13690229149108218523467549
Offset: 1

Views

Author

Paul D. Hanna, Feb 02 2005

Keywords

Examples

			From _Paul D. Hanna_, Jan 31 2009: (Start)
G.f.: A(x) = 1 - x + 2*x^2/3 - 7*x^3/18 + 39*x^4/180 - 321*x^5/2700 +...
G.f.: A(x) = 1/B(x) where:
B(x) = 1 + x + x^2/3 + x^3/18 + x^4/180 + x^5/2700 +...+ x^n/[n!*(n+1)!/2^n] +... (End)
		

Crossrefs

Programs

  • Mathematica
    Table[(-1)^((n-1)/2) * (CoefficientList[Series[x/BesselJ[1,2*x],{x,0,40}],x])[[n]] * ((n+1)/2)! * ((n-1)/2)!,{n,1,41,2}] (* Vaclav Kotesovec, Mar 01 2014 *)
  • PARI
    a(n)=if(n<1,0,(matrix(n,n,m,j,binomial(m-1,j-1)*binomial(m,j-1)/j)^-1)[n,1])
    
  • PARI
    {a(n)=local(B=sum(k=0,n,x^k/(k!*(k+1)!/2^k))+x*O(x^n));polcoeff(1/B,n)*n!*(n+1)!/2^n} \\ Paul D. Hanna, Jan 31 2009

Formula

From Paul D. Hanna, Jan 31 2009: (Start)
G.f.: A(x) = 1/B(x) where A(x) = Sum_{n>=0} (-1)^n*a(n)*x^n/[n!*(n+1)!/2^n] and B(x) = Sum_{n>=0} x^n/[n!*(n+1)!/2^n].
G.f. satisfies: A(x) = 1/F(x*A(x)) and F(x) = 1/A(x*F(x)) where F(x) = Sum_{n>=0} A155926(n)*x^n/[n!*(n+1)!/2^n].
G.f. satisfies: A(x) = 1/G(x/A(x)) and G(x) = 1/A(x/G(x)) where G(x) = Sum_{n>=0} A155927(n)*x^n/[n!*(n+1)!/2^n]. (End)
a(n) ~ (-1)^(n+1) * c * n! * (n-1)! * d^n, where d = 4/BesselJZero[1, 1]^2 = 0.2724429913055159309179376055957891881897555639652..., and c = 9.11336321311226744479181866135367355200240221549667284076... = BesselJZero[1, 1]^2 / (4*BesselJ[2, BesselJZero[1, 1]]). - Vaclav Kotesovec, Mar 01 2014, updated Apr 01 2018

A078018 a(n) = Sum_{k=0..n} 6^k*N(n,k), with a(0)=1, where N(n,k) = C(n,k) * C(n,k+1)/n are the Narayana numbers (A001263).

Original entry on oeis.org

1, 1, 7, 55, 469, 4237, 39907, 387739, 3858505, 39130777, 402972031, 4202705311, 44299426717, 471189693925, 5051001609115, 54513542257795, 591858123926545, 6459813793353265, 70837427884259575, 780073647992404615
Offset: 0

Views

Author

Benoit Cloitre, May 10 2003

Keywords

Comments

More generally, coefficients of (1 + m*x - sqrt(m^2*x^2 - (2*m+4)*x + 1) )/( (2*m+2)*x ) are given by a(n) = Sum_{k=0..n} (m+1)^k*N(n,k).
The Hankel transform of this sequence is 6^C(n+1,2). - Philippe Deléham, Oct 29 2007
Shifts left when INVERT transform applied six times. - Benedict W. J. Irwin, Feb 07 2016

Crossrefs

Programs

  • Magma
    R:=PowerSeriesRing(Rationals(), 30); Coefficients(R!( (1 + 5*x - Sqrt(25*x^2-14*x+1))/(12*x) )); // G. C. Greubel, Jun 29 2019
    
  • Maple
    A078018_list := proc(n) local j, a, w; a := array(0..n); a[0] := 1;
    for w from 1 to n do a[w] := a[w-1]+6*add(a[j]*a[w-j-1],j=1..w-1) od;
    convert(a, list) end: A078018_list(19);
    # Peter Luschny, May 19 2011
  • Mathematica
    Table[SeriesCoefficient[(1+5*x-Sqrt[25*x^2-14*x+1])/(12*x),{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 13 2012 *)
    a[n_]:= Hypergeometric2F1[1 - n, -n, 2, 6]; Table[a[n], {n, 0, 20}] (* Peter Luschny, Mar 19 2018 *)
  • PARI
    a(n)=if(n<1,1,sum(k=0,n,6^k/n*binomial(n,k)*binomial(n,k+1)))
    
  • Sage
    a=((1 + 5*x - sqrt(25*x^2-14*x+1))/(12*x)).series(x, 30).coefficients(x, sparse=False); [1]+a[1:] # G. C. Greubel, Jun 29 2019

Formula

G.f.: (1 + 5*x - sqrt(25*x^2-14*x+1))/(12*x).
a(n) = Sum_{k=0..n} A088617(n, k)*6^k*(-5)^(n-k). - Philippe Deléham, Jan 21 2004
a(n) = ( 7*(2*n-1)*a(n-1) - 25*(n-2)*a(n-2) ) / (n+1) for n>=2, a(0) = a(1) = 1. - Philippe Deléham, Aug 19 2005
From Gary W. Adamson, Jul 08 2011: (Start)
a(n) = upper left term in M^n, M = the production matrix:
1, 1;
6, 6, 6;
1, 1, 1, 1;
6, 6, 6, 6, 6;
1, 1, 1, 1, 1, 1;
... (End)
a(n) ~ sqrt(12+7*sqrt(6))*(7+2*sqrt(6))^n/(12*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 13 2012
G.f.: 1/(1 - x/(1 - 6*x/(1 - x/(1 - 6*x/(1 - x/(1 - ...)))))), a continued fraction. - Ilya Gutkovskiy, Apr 21 2017
a(n) = hypergeom([1 - n, -n], [2], 6). - Peter Luschny, Mar 19 2018

A081178 a(0) = 1; for n>=1, a(n) = Sum_{k=0..n} 7^k*N(n,k), where N(n,k)=(1/n)*C(n,k)*C(n,k+1) are the Narayana numbers (A001263).

Original entry on oeis.org

1, 1, 8, 71, 680, 6882, 72528, 788019, 8766248, 99362894, 1143498224, 13326176998, 156950554384, 1865210341828, 22338852956064, 269355965364459, 3267146912972328, 39837475762660374, 488032452193307568
Offset: 0

Views

Author

Benoit Cloitre, May 10 2003

Keywords

Comments

More generally, coefficients of (1+m*x - sqrt(m^2*x^2-(2*m+4)*x+1) )/((2*m+2)*x) are given by: a(n) = Sum_{k=0..n} (m+1)^k*N(n,k).
The Hankel transform of this sequence is 7^C(n+1,2). - Philippe Deléham, Oct 29 2007
From Gary W. Adamson, Jul 08 2011: (Start)
a(n) = upper left term in M^n, M = the production matrix:
1, 1
7, 7, 7
1, 1, 1, 1
7, 7, 7, 7, 7
1, 1, 1, 1, 1, 1
...
(End)
Shifts left when INVERT transform applied seven times. - Benedict W. J. Irwin, Feb 07 2016
G.f.: 1/(1 - x/(1 - 7*x/(1 - x/(1 - 7*x/(1 - x/(1 - ...)))))), a continued fraction. - Ilya Gutkovskiy, Apr 21 2017

Crossrefs

Programs

  • Magma
    B:=Binomial;
    A081178:= func< n | n eq 0 select 1 else (&+[7^k*B(n,k)*B(n,k+1): k in [0..n]])/n >;
    [A081178(n): n in [0..40]]; // G. C. Greubel, Jan 15 2024
    
  • Maple
    A081178_list := proc(n) local j, a, w; a := array(0..n); a[0] := 1;
    for w from 1 to n do a[w] := a[w-1]+7*add(a[j]*a[w-j-1],j=1..w-1) od;
    convert(a, list) end: A081178_list(18); # Peter Luschny, May 19 2011
  • Mathematica
    Table[SeriesCoefficient[(1+6*x-Sqrt[36*x^2-16*x+1])/(14*x),{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 13 2012 *)
    a[n_] := Hypergeometric2F1[1 - n, -n, 2, 7];
    Table[a[n], {n, 0, 18}] (* Peter Luschny, Mar 19 2018 *)
  • PARI
    a(n)=if(n<1,1,sum(k=0,n,7^k/n*binomial(n,k)*binomial(n,k+1)))
    
  • SageMath
    def A081178(n):
        b=binomial;
        if n==0: return 1
        else: return (1/n)*sum(7^k*b(n,k)*b(n,k+1) for k in range(n+1))
    [A081178(n) for n in range(41)] # G. C. Greubel, Jan 15 2024

Formula

G.f.: (1+6*x-sqrt(36*x^2-16*x+1))/(14*x).
a(n) = (8*(2*n-1)*a(n-1) - 36*(n-2)*a(n-2))/(n+1) for n>=2, a(0) = a(1) = 1. - Philippe Deléham, Aug 19 2005
a(n) ~ sqrt(14+8*sqrt(7))*(8+2*sqrt(7))^n/(14*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 13 2012
a(n) = hypergeom([1 - n, -n], [2], 7). - Peter Luschny, Mar 19 2018

A134291 Tenth column (and diagonal) of Narayana triangle A001263.

Original entry on oeis.org

1, 55, 1210, 15730, 143143, 1002001, 5725720, 27810640, 118195220, 449141836, 1551580888, 4936848280, 14620666060, 40648664980, 106847919376, 267119798440, 638337753625, 1464421905375, 3237143159250, 6917263803450
Offset: 0

Views

Author

Wolfdieter Lang, Nov 13 2007

Keywords

Comments

See a comment under A134288 on the coincidence of column and diagonal sequences.
Kekulé numbers K(O(1,9,n)) for certain benzenoids (see the Cyvin-Gutman reference, p. 105, eq. (i)).

References

  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988.

Crossrefs

Cf. A134290 (ninth column of Narayana triangle).

Programs

  • GAP
    List([0..30], n-> Binomial(n+10,10)*Binomial(n+9,8)/9); # G. C. Greubel, Aug 28 2019
  • Magma
    [Binomial(n+10,10)*Binomial(n+9,8)/9: n in [0..30]]; // G. C. Greubel, Aug 28 2019
    
  • Maple
    a := n -> ((n+1)*((n+2)*(n+3)*(n+4)*(n+5)*(n+6)*(n+7)*(n+8)*(n+9))^2*(n+10))/ 1316818944000:
    seq(a(n), n=0..19); # Peter Luschny, Sep 01 2016
  • Mathematica
    Table[Binomial[n + 10, 10]*Binomial[n + 10, 9]/(n + 10), {n, 0, 30}] (* Wesley Ivan Hurt, Apr 25 2017 *)
  • PARI
    vector(30, n, binomial(n+9,10)*binomial(n+8,8)/9) \\ G. C. Greubel, Aug 28 2019
    
  • Sage
    [binomial(n+10,10)*binomial(n+9,8)/9 for n in (0..30)] # G. C. Greubel, Aug 28 2019
    

Formula

a(n) = A001263(n+10,10) = binomial(n+10,10)*binomial(n+10,9)/(n+10).
O.g.f.: P(9,x)/(1-x)^19 with the numerator polynomial P(9,x) = Sum_{k=1..9} A001263(9,k)*x^(k-1), the ninth row polynomial of the Narayana triangle: P(9,x) = 1 + 36*x + 336*x^2 + 1176*x^3 + 1764*x^4 + 1176*x^5 + 336*x^6 + 36*x^7 + x^8.
a(n) = Product_{i=1..9} A002378(n+i)/A002378(i). - Bruno Berselli, Sep 01 2016
From Amiram Eldar, Oct 19 2020: (Start)
Sum_{n>=0} 1/a(n) = 2987553139/196 - 1544400*Pi^2.
Sum_{n>=0} (-1)^n/a(n) = 1179648*log(2)/7 - 114472793/980. (End)

A082147 a(0)=1; for n >= 1, a(n) = Sum_{k=0..n} 8^k*N(n,k) where N(n,k) = (1/n)*C(n,k)*C(n,k+1) are the Narayana numbers (A001263).

Original entry on oeis.org

1, 1, 9, 89, 945, 10577, 123129, 1476841, 18130401, 226739489, 2878666857, 37006326777, 480750990993, 6301611631473, 83240669582937, 1106980509493641, 14808497812637121, 199138509770855489, 2690461489090104009
Offset: 0

Views

Author

Benoit Cloitre, May 10 2003

Keywords

Comments

More generally coefficients of (1 + m*x - sqrt(m^2*x^2 - (2*m+4)*x+1))/( (2*m+2)*x) are given by a(n) = Sum_{k=0..n} (m+1)^k*N(n,k)).
The Hankel transform of this sequence is 8^C(n+1,2). - Philippe Deléham, Oct 29 2007
Shifts left when INVERT transform applied eight times. - Benedict W. J. Irwin, Feb 07 2016

Crossrefs

Programs

  • GAP
    a:=n->Sum([0..n],k->8^k*(1/n)*Binomial(n,k)*Binomial(n,k+1));;
    Concatenation([1],List([1..18],n->a(n))); # Muniru A Asiru, Feb 10 2018
  • Magma
    Q:=Rationals(); R:=PowerSeriesRing(Q, 40); Coefficients(R!((1+7*x-Sqrt(49*x^2-18*x+1))/(16*x))) // G. C. Greubel, Feb 05 2018
    
  • Maple
    A082147_list := proc(n) local j, a, w; a := array(0..n); a[0] := 1;
    for w from 1 to n do a[w] := a[w-1]+8*add(a[j]*a[w-j-1],j=1..w-1) od;
    convert(a, list) end: A082147_list(18); # Peter Luschny, May 19 2011
  • Mathematica
    Table[SeriesCoefficient[(1+7*x-Sqrt[49*x^2-18*x+1])/(16*x),{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 14 2012 *)
    f[n_] := Sum[ 8^k*Binomial[n, k]*Binomial[n, k + 1]/n, {k, 0, n}]; f[0] = 1; Array[f, 21, 0] (* Robert G. Wilson v, Feb 24 2018 *)
    a[n_] := Hypergeometric2F1[1 - n, -n, 2, 8];
    Table[a[n], {n, 0, 18}] (* Peter Luschny, Mar 19 2018 *)
  • PARI
    a(n)=if(n<1,1,sum(k=0,n,8^k/n*binomial(n,k)*binomial(n,k+1)))
    

Formula

G.f.: (1 + 7*x - sqrt(49*x^2-18*x+1))/(16*x).
a(n) = Sum_{k=0..n} A088617(n, k)*8^k*(-7)^(n-k). - Philippe Deléham, Jan 21 2004
a(n) = (9(2n-1)a(n-1) - 49(n-2)a(n-2)) / (n+1) for n >= 2, a(0) = a(1) = 1. - Philippe Deléham, Aug 19 2005
a(n) = upper left term in M^n, M = the production matrix:
1, 1
8, 8, 8
1, 1, 1, 1
8, 8, 8, 8, 8
1, 1, 1, 1, 1, 1
...
- Gary W. Adamson, Jul 08 2011
a(n) ~ sqrt(16+18*sqrt(2))*(9+4*sqrt(2))^n/(16*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 14 2012
G.f.: 1/(1 - x/(1 - 8*x/(1 - x/(1 - 8*x/(1 - x/(1 - ...)))))), a continued fraction. - Ilya Gutkovskiy, Apr 21 2017
a(n) = hypergeom([1 - n, -n], [2], 8). - Peter Luschny, Mar 19 2018

A103364 Matrix inverse of the Narayana triangle A001263.

Original entry on oeis.org

1, -1, 1, 2, -3, 1, -7, 12, -6, 1, 39, -70, 40, -10, 1, -321, 585, -350, 100, -15, 1, 3681, -6741, 4095, -1225, 210, -21, 1, -56197, 103068, -62916, 19110, -3430, 392, -28, 1, 1102571, -2023092, 1236816, -377496, 68796, -8232, 672, -36, 1, -27036487, 49615695, -30346380, 9276120, -1698732, 206388
Offset: 1

Views

Author

Paul D. Hanna, Feb 02 2005

Keywords

Comments

The first column is A103365. The second column is A103366. Row sums are all zeros (for n > 1). Absolute row sums form A103367.
Let E(y) = Sum_{n >= 0} y^n/(n!*(n+1)!) = (1/sqrt(y))*BesselI(1,2*sqrt(y)). Then this triangle is the generalized Riordan array (1/E(y), y) with respect to the sequence n!*(n+1)! as defined in Wang and Wang. - Peter Bala, Aug 07 2013

Examples

			Rows begin:
        1;
       -1,        1;
        2,       -3,       1;
       -7,       12,      -6,       1;
       39,      -70,      40,     -10,     1;
     -321,      585,    -350,     100,   -15,     1;
     3681,    -6741,    4095,   -1225,   210,   -21,   1;
   -56197,   103068,  -62916,   19110, -3430,   392, -28,   1;
  1102571, -2023092, 1236816, -377496, 68796, -8232, 672, -36, 1;
  ...
From _Peter Bala_, Aug 09 2013: (Start)
The real zeros of the row polynomials R(n,x) appear to converge to zeros of E(alpha*x) as n increases, where alpha = -3.67049 26605 ... ( = -(A115369/2)^2).
Polynomial | Real zeros to 5 decimal places
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
R(5,x)     | 1, 3.57754, 3.81904
R(10,x)    | 1, 3.35230, 7.07532,  9.14395
R(15,x)    | 1, 3.35231, 7.04943, 12.09668, 15.96334
R(20,x)    | 1, 3.35231, 7.04943, 12.09107, 18.47845, 24.35255
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Function   |
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
E(alpha*x) | 1, 3.35231, 7.04943, 12.09107, 18.47720, 26.20778, ...
Note: The n-th zero of E(alpha*x) may be calculated in Maple 17 using the instruction evalf( BesselJZeros(1,n)/ BesselJZeros(1,1))^2 ). (End)
		

Crossrefs

Programs

  • Mathematica
    T[n_, 1]:= Last[Table[(-1)^(n - 1)*(CoefficientList[Series[x/BesselJ[1, 2*x], {x, 0, 1000}], x])[[k]]*(n)!*(n - 1)!, {k, 1, 2*n - 1, 2}]]
    T[n_, n_] := 1; T[2, 1] := -1; T[3, 1] := 2; T[n_, k_] := T[n, k] = T[n - 1, k - 1]*n*(n - 1)/(k*(k - 1)); Table[T[n, k], {n, 1, 50}, {k, 1, n}] // Flatten (* G. C. Greubel, Jan 04 2016 *)
    T[n_, n_] := 1; T[n_, 1]/;n>1 := T[n, 1] = -Sum[T[n, j], {j, 2, n}]; T[n_, k_]/;1Oliver Seipel, Jan 01 2025 *)
  • PARI
    T(n,k)=if(n
    				

Formula

From Peter Bala, Aug 07 2013: (Start)
Let E(y) = Sum_{n >= 0} y^n/(n!*(n+1)!) = (1/sqrt(y))* BesselI(1,2*sqrt(y)). Generating function: E(x*y)/E(y) = 1 + (-1 + x)*y/(1!*2!) + (2 - 3*x + x^2)*y^2/(2!*3!) + (-7 + 12*x - 6*x^2 + x^3)*y^3/(3!*4!) + .... The n-th power of this array has a generating function E(x*y)/E(y)^n. In particular, the matrix inverse A001263 has a generating function E(y)*E(x*y).
Recurrence equation for the row polynomials: R(n,x) = x^n - Sum_{k = 0..n-1} 1/(n-k+1)*binomial(n,k)*binomial(n+1,k+1) *R(k,x) with initial value R(0,x) = 1.
Let alpha denote the root of E(x) = 0 that is smallest in absolute magnitude. Numerically, alpha = -3.67049 26605 ... = -(A115369/2)^2. It appears that for arbitrary complex x we have lim_{n->oo} R(n,x)/R(n,0) = E(alpha*x). Cf. A055133, A086646 and A104033.
A stronger result than pointwise convergence may hold: the convergence may be uniform on compact subsets of the complex plane. This would explain the observation that the real zeros of the polynomials R(n,x) seem to converge to the zeros of E(alpha*x) as n increases. Some numerical examples are given below. (End)
From Werner Schulte, Jan 04 2017, corrected May 05 2025: (Start)
T(n,k) = T(n-1,k-1)*n*(n-1)/(k*(k-1)) for 1 < k <= n;
T(n,k) = T(n+1-k,1)*A001263(n,k) for 1 <= k <= n;
Sum_{k=1..n} T(n,k)*A000108(k) = 1 for n > 0. (End)

A134289 Eighth column (and diagonal) of Narayana triangle A001263.

Original entry on oeis.org

1, 36, 540, 4950, 32670, 169884, 736164, 2760615, 9202050, 27810640, 77364144, 200443464, 488259720, 1126753200, 2478857040, 5226256926, 10606227291, 20796524100, 39525557500, 73018266750, 131432880150, 231003243900, 397179490500, 669161098125, 1106346348900
Offset: 0

Views

Author

Wolfdieter Lang, Nov 13 2007

Keywords

Comments

See a comment under A134288 on the coincidence of column and diagonal sequences.
Kekulé numbers K(O(1,7,n)) for certain benzenoids (see the Cyvin-Gutman reference, p. 105, eq. (i)).

References

  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988.

Crossrefs

Cf. A002378.
Cf. A134288 (seventh column of Narayana triangle).
Cf. A134290 (ninth column of Narayana triangle).

Programs

  • GAP
    List([0..30], n-> Binomial(n+8,8)*Binomial(n+7,6)/7); # G. C. Greubel, Aug 28 2019
  • Magma
    [Binomial(n+8,8)*Binomial(n+7,6)/7: n in [0..30]]; // G. C. Greubel, Aug 28 2019
    
  • Maple
    a := n -> ((n+1)*((n+2)*(n+3)*(n+4)*(n+5)*(n+6)*(n+7))^2*(n+8))/203212800;
    seq(a(n), n=0..24); # Peter Luschny, Sep 01 2016
  • Mathematica
    Table[(Binomial[n + 8, 8] Binomial[n + 8, 7])/(n + 8), {n, 0, 30}] (* or *) LinearRecurrence[{15, -105, 455, -1365, 3003, -5005, 6435, -6435, 5005, -3003, 1365, -455, 105, -15, 1},{1, 36, 540, 4950, 32670, 169884, 736164, 2760615, 9202050, 27810640, 77364144, 200443464, 488259720, 1126753200, 2478857040}, 30] (* Harvey P. Dale, Jul 23 2012 *)
  • PARI
    vector(30, n, binomial(n+7,8)*binomial(n+6,6)/7) \\ G. C. Greubel, Aug 28 2019
    
  • Sage
    [binomial(n+8,8)*binomial(n+7,6)/7 for n in (0..30)] # G. C. Greubel, Aug 28 2019
    

Formula

a(n) = A001263(n+8,8) = binomial(n+8,8)*binomial(n+8,7)/(n+8).
O.g.f.: P(7,x)/(1-x)^15 with the numerator polynomial P(7,x) = Sum_{k=1..7} A001263(7,k)*x^(k-1), the seventh row polynomial of the Narayana triangle: P(7,x) = 1 + 21*x + 105*x^2 + 175*x^3 + 105*x^4 + 21*x^5 + x^6.
For n>14: a(n) = 15*a(n-1) - 105*a(n-2) + 455*a(n-3) - 1365*a(n-4) + 3003*a(n-5) - 5005*a(n-6) + 6435*a(n-7) - 6435*a(n-8) + 5005*a(n-9) - 3003*a(n-10) + 1365*a(n-11) - 455*a(n-12) + 105*a(n-13) - 15*a(n-14) + a(n-15). - Harvey P. Dale, Jul 23 2012
a(n) = Product_{i=1..7} A002378(n+i)/A002378(i). - Bruno Berselli, Sep 01 2016
From Amiram Eldar, Oct 19 2020: (Start)
Sum_{n>=0} 1/a(n) = 12767346/25 - 51744*Pi^2.
Sum_{n>=0} (-1)^n/a(n) = 1192508/75 - 114688*log(2)/5. (End)
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