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.

Showing 1-10 of 15 results. Next

A094346 Another version of triangular array in A036970: triangle T(n,k), 0<=k<=n, read by rows; given by [0, 1, 2, 4, 6, 9, 12, 16, 20, 25, 30, ...] DELTA [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, ...] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 0, 1, 0, 1, 2, 0, 3, 8, 6, 0, 17, 54, 60, 24, 0, 155, 556, 762, 480, 120, 0, 2073, 8146, 12840, 10248, 4200, 720, 0, 38227, 161424, 282078, 263040, 139440, 40320, 5040, 0, 929569, 4163438, 7886580, 8240952, 5170800, 1965600, 423360, 40320
Offset: 0

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Author

Philippe Deléham, Jun 08 2004, Jun 13 2007

Keywords

Comments

Diagonals: A000007, A001469, A005440; A000182, A005990. Row sums: A001469.

Examples

			Triangle begins:
1;
0, 1;
0, 1, 2;
0, 3, 8, 6;
0, 17, 54, 60, 24;
0, 155, 556, 762, 480, 120;
0, 2073, 8146, 12840, 10248, 4200, 720;
0, 38227, 161424, 282078, 263040, 139440, 40320, 5040;
0, 929569, 4163438, 7886580, 8240952, 5170800, 1965600, 423360, 40320; ...
		

Crossrefs

Programs

  • Mathematica
    G[_, 1] = 1;
    G[x_, n_] := G[x, n] = (x+1)^2 G[x+1, n-1] - x^2 G[x, n-1] // Expand;
    row[0] = {1};
    row[n_] := CoefficientList[x G[x, n], x];
    Table[row[n], {n, 0, 8}] // Flatten (* Jean-François Alcover, Aug 17 2018 *)
  • PARI
    {T(n, k) = local( A = x); if( k<0 || k>n, 0, for( j = 1, n, A = x^2 * ( subst(A, x, x+1) - A)); polcoeff( A, k+1))} /* Michael Somos, Apr 10 2011 */

Formula

For n>=1, Sum_{k =1..n} T(n, k)*x^(k-1) = G(x, n), n-th Gandhi polynomial; the Gandhi polynomials are defined by G(x, n) = (x+1)^2*G(x+1, n-1) - x^2*G(x, n-1), G(x, 1) = 1. Sum_{k =0..n} T(n, k)*2^(2n-k) = A000182(n+1), tangent numbers. Sum_{k =0..n} T(n, k) = A001469(n+1), Genocchi numbers of first kind.
Sum_{k = 0..n} T(n, k)*2^(n-k) = A002105(n+1). - Philippe Deléham, Jun 10 2004

A001008 a(n) = numerator of harmonic number H(n) = Sum_{i=1..n} 1/i.

Original entry on oeis.org

1, 3, 11, 25, 137, 49, 363, 761, 7129, 7381, 83711, 86021, 1145993, 1171733, 1195757, 2436559, 42142223, 14274301, 275295799, 55835135, 18858053, 19093197, 444316699, 1347822955, 34052522467, 34395742267, 312536252003, 315404588903, 9227046511387
Offset: 1

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Author

Keywords

Comments

H(n)/2 is the maximal distance that a stack of n cards can project beyond the edge of a table without toppling.
By Wolstenholme's theorem, p^2 divides a(p-1) for all primes p > 3.
From Alexander Adamchuk, Dec 11 2006: (Start)
p divides a(p^2-1) for all primes p > 3.
p divides a((p-1)/2) for primes p in A001220.
p divides a((p+1)/2) or a((p-3)/2) for primes p in A125854.
a(n) is prime for n in A056903. Corresponding primes are given by A067657. (End)
a(n+1) is the numerator of the polynomial A[1, n](1) where the polynomial A[genus 1, level n](m) is defined to be Sum_{d = 1..n - 1} m^(n - d)/d. (See the Mathematica procedure generating A[1, n](m) below.) - Artur Jasinski, Oct 16 2008
Better solutions to the card stacking problem have been found by M. Paterson and U. Zwick (see link). - Hugo Pfoertner, Jan 01 2012
a(n) = A213999(n, n-1). - Reinhard Zumkeller, Jul 03 2012
a(n) coincides with A175441(n) if and only if n is not from the sequence A256102. The quotient a(n) / A175441(n) for n in A256102 is given as corresponding entry of A256103. - Wolfdieter Lang, Apr 23 2015
For a very short proof that the Harmonic series diverges, see the Goldmakher link. - N. J. A. Sloane, Nov 09 2015
All terms are odd while corresponding denominators (A002805) are all even for n > 1 (proof in Pólya and Szegő). - Bernard Schott, Dec 24 2021

Examples

			H(n) = [ 1, 3/2, 11/6, 25/12, 137/60, 49/20, 363/140, 761/280, 7129/2520, ... ].
Coincidences with A175441: the first 19 entries coincide because 20 is the first entry of A256102. Indeed, a(20)/A175441(20) = 55835135 / 11167027 = 5 = A256103(1). - _Wolfdieter Lang_, Apr 23 2015
		

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 258-261.
  • H. W. Gould, Combinatorial Identities, Morgantown Printing and Binding Co., 1972, # 1.45, page 6, #3.122, page 36.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 259.
  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, page 347.
  • D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 1, p. 615.
  • G. Pólya and G. Szegő, Problems and Theorems in Analysis, volume II, Springer, reprint of the 1976 edition, 1998, problem 251, p. 154.
  • 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

Cf. A145609-A145640. - Artur Jasinski, Oct 16 2008
Cf. A003506. - Paul Curtz, Nov 30 2013
The following fractions are all related to each other: Sum 1/n: A001008/A002805, Sum 1/prime(n): A024451/A002110 and A106830/A034386, Sum 1/nonprime(n): A282511/A282512, Sum 1/composite(n): A250133/A296358.
Cf. A195505.

Programs

  • GAP
    List([1..30],n->NumeratorRat(Sum([1..n],i->1/i))); # Muniru A Asiru, Dec 20 2018
  • Haskell
    import Data.Ratio ((%), numerator)
    a001008 = numerator . sum . map (1 %) . enumFromTo 1
    a001008_list = map numerator $ scanl1 (+) $ map (1 %) [1..]
    -- Reinhard Zumkeller, Jul 03 2012
    
  • Magma
    [Numerator(HarmonicNumber(n)): n in [1..30]]; // Bruno Berselli, Feb 17 2016
    
  • Maple
    A001008 := proc(n)
        add(1/k,k=1..n) ;
        numer(%) ;
    end proc:
    seq( A001008(n),n=1..40) ; # Zerinvary Lajos, Mar 28 2007; R. J. Mathar, Dec 02 2016
  • Mathematica
    Table[Numerator[HarmonicNumber[n]], {n, 30}]
    (* Procedure generating A[1,n](m) (see Comments section) *) m =1; aa = {}; Do[k = 0; Do[k = k + m^(r - d)/d, {d, 1, r - 1}]; AppendTo[aa, k], {r, 1, 20}]; aa (* Artur Jasinski, Oct 16 2008 *)
    Numerator[Accumulate[1/Range[25]]] (* Alonso del Arte, Nov 21 2018 *)
    Numerator[Table[((n - 1)/2)*HypergeometricPFQ[{1, 1, 2 - n}, {2, 3}, 1] + 1, {n, 1, 29}]] (* Artur Jasinski, Jan 08 2021 *)
  • PARI
    A001008(n) = numerator(sum(i=1,n,1/i)) \\ Michael B. Porter, Dec 08 2009
    
  • PARI
    H1008=List(1); A001008(n)={for(k=#H1008,n-1,listput(H1008,H1008[k]+1/(k+1))); numerator(H1008[n])} \\ about 100x faster for n=1..1500. - M. F. Hasler, Jul 03 2019
    
  • Python
    from sympy import Integer
    [sum(1/Integer(i) for i in range(1, n + 1)).numerator() for n in range(1, 31)]  # Indranil Ghosh, Mar 23 2017
    
  • Sage
    def harmonic(a, b):  # See the F. Johansson link.
        if b - a == 1:
            return 1, a
        m = (a+b)//2
        p, q = harmonic(a,m)
        r, s = harmonic(m,b)
        return p*s+q*r, q*s
    def A001008(n): H = harmonic(1,n+1); return numerator(H[0]/H[1])
    [A001008(n) for n in (1..29)] # Peter Luschny, Sep 01 2012
    

Formula

H(n) ~ log n + gamma + O(1/n). [See Hardy and Wright, Th. 422.]
log n + gamma - 1/n < H(n) < log n + gamma + 1/n [follows easily from Hardy and Wright, Th. 422]. - David Applegate and N. J. A. Sloane, Oct 14 2008
G.f. for H(n): log(1-x)/(x-1). - Benoit Cloitre, Jun 15 2003
H(n) = sqrt(Sum_{i = 1..n} Sum_{j = 1..n} 1/(i*j)). - Alexander Adamchuk, Oct 24 2004
a(n) is the numerator of Gamma/n + Psi(1 + n)/n = Gamma + Psi(n), where Psi is the digamma function. - Artur Jasinski, Nov 02 2008
H(n) = 3/2 + 2*Sum_{k = 0..n-3} binomial(k+2, 2)/((n-2-k)*(n-1)*n), n > 1. - Gary Detlefs, Aug 02 2011
H(n) = (-1)^(n-1)*(n+1)*n*Sum_{k = 0..n-1} k!*Stirling2(n-1, k) * Stirling1(n+k+1,n+1)/(n+k+1)!. - Vladimir Kruchinin, Feb 05 2013
H(n) = n*Sum_{k = 0..n-1} (-1)^k*binomial(n-1,k)/(k+1)^2. (Wenchang Chu) - Gary Detlefs, Apr 13 2013
H(n) = (1/2)*Sum_{k = 1..n} (-1)^(k-1)*binomial(n,k)*binomial(n+k, k)/k. (H. W. Gould) - Gary Detlefs, Apr 13 2013
E.g.f. for H(n) = a(n)/A002805(n): (gamma + log(x) - Ei(-x)) * exp(x), where gamma is the Euler-Mascheroni constant, and Ei(x) is the exponential integral. - Vladimir Reshetnikov, Apr 24 2013
H(n) = residue((psi(-s)+gamma)^2/2, {s, n}) where psi is the digamma function and gamma is the Euler-Mascheroni constant. - Jean-François Alcover, Feb 19 2014
H(n) = Sum_{m >= 1} n/(m^2 + n*m) = gamma + digamma(1+n), numerators and denominators. (see Mathworld link on Digamma). - Richard R. Forberg, Jan 18 2015
H(n) = (1/2) Sum_{j >= 1} Sum_{k = 1..n} ((1 - 2*k + 2*n)/((-1 + k + j*n)*(k + j*n))) + log(n) + 1/(2*n). - Dimitri Papadopoulos, Jan 13 2016
H(n) = (n!)^2*Sum_{k = 1..n} 1/(k*(n-k)!*(n+k)!). - Vladimir Kruchinin, Mar 31 2016
a(n) = Stirling1(n+1, 2) / gcd(Stirling1(n+1, 2), n!) = A000254(n) / gcd(A000254(n), n!). - Max Alekseyev, Mar 01 2018
From Peter Bala, Jan 31 2019: (Start)
H(n) = 1 + (1 + 1/2)*(n-1)/(n+1) + (1/2 + 1/3)*(n-1)*(n-2)/((n+1)*(n+2)) + (1/3 + 1/4)*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ... .
H(n)/n = 1 + (1/2^2 - 1)*(n-1)/(n+1) + (1/3^2 - 1/2^2)*(n-1)*(n-2)/((n+1)*(n+2)) + (1/4^2 - 1/3^2)*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ... .
For odd n >= 3, (1/2)*H((n-1)/2) = (n-1)/(n+1) + (1/2)*(n-1)*(n-3)/((n+1)*(n+3)) + 1/3*(n-1)*(n-3)*(n-5)/((n+1)*(n+3)*(n+5)) + ... . Cf. A195505. See the Bala link in A036970. (End)
H(n) = ((n-1)/2) * hypergeom([1,1,2-n], [2,3], 1) + 1. - Artur Jasinski, Jan 08 2021
Conjecture: for nonzero m, H(n) = (1/m)*Sum_{k = 1..n} ((-1)^(k+1)/k) * binomial(m*k,k)*binomial(n+(m-1)*k,n-k). The case m = 1 is well-known; the case m = 2 is given above by Detlefs (dated Apr 13 2013). - Peter Bala, Mar 04 2022
a(n) = the (reduced) numerator of the continued fraction 1/(1 - 1^2/(3 - 2^2/(5 - 3^2/(7 - ... - (n-1)^2/(2*n-1))))). - Peter Bala, Feb 18 2024
H(n) = Sum_{k=1..n} (-1)^(k-1)*binomial(n,k)/k (H. W. Gould). - Gary Detlefs, May 28 2024

Extensions

Edited by Max Alekseyev, Oct 21 2011
Changed title, deleting the incorrect name "Wolstenholme numbers" which conflicted with the definition of the latter in both Weisstein's World of Mathematics and in Wikipedia, as well as with OEIS A007406. - Stanislav Sykora, Mar 25 2016

A015237 a(n) = (2*n - 1)*n^2.

Original entry on oeis.org

0, 1, 12, 45, 112, 225, 396, 637, 960, 1377, 1900, 2541, 3312, 4225, 5292, 6525, 7936, 9537, 11340, 13357, 15600, 18081, 20812, 23805, 27072, 30625, 34476, 38637, 43120, 47937, 53100, 58621, 64512
Offset: 0

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Author

Keywords

Comments

Structured hexagonal prism numbers. - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Number of divisors of 60^(n-1) for n>0. - J. Lowell, Aug 30 2008
The sum of the 2*n+1 numbers between n*(n+1) and (n+1)*(n+2) gives a(n+1). - J. M. Bergot, Apr 17 2013
Partial sums of A080859. - J. M. Bergot, Jul 03 2013
a(n) = number of 2 X 2 matrices having all elements in {0..n} with determinant = permanent. - Indranil Ghosh, Dec 26 2016
Number of additions and multiplications needed to multiply two n X n matrices naively. - Charles R Greathouse IV, Jan 19 2018

Crossrefs

Cf. A100177 (structured prisms); A100145 (more on structured numbers).
Cf. A000578, A045991, A000384, A080859 (first diffs), A103220 (partial sums).
Cf. similar sequences, with the formula (k*n-k+2)*n^2/2, listed in A262000.

Programs

Formula

a(n) = A000578(n) + A045991(n). - Zerinvary Lajos, Jun 11 2008
a(n) = A199771(2*n-1) for n > 0. - Reinhard Zumkeller, Nov 23 2011
G.f.: x*(1+8*x+3*x^2)/(1-x)^4. - Colin Barker, Jun 08 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 12, a(0)=1, a(1)=1, a(2)=12. - G. C. Greubel, Jul 31 2015
E.g.f.: x*(2*x^2 + 5*x + 1)*exp(x). - G. C. Greubel, Jul 31 2015
a(n) = Sum_{i=0..n-1} n*(4*i+1) for n>0. - Bruno Berselli, Sep 08 2015
Sum_{n>=1} 1/a(n) = 4*log(2) - Pi^2/6. - Vaclav Kotesovec, Oct 04 2016
a(n) = Sum_{i=n^2-n+1..n^2+n-1} i. - Wesley Ivan Hurt, Dec 27 2016
From Peter Bala, Jan 30 2019: (Start)
Let a(n,x) = Product_{k = 0..n} (x - k)/(x + k). Then for positive integer x we have (2*x - 1)*x^2 = Sum_{n >= 0} ((n+1)^5 + n^5)*a(n,x) and (2*x - 1)*x = Sum_{n >= 0} ((n+1)^4 - n^4)*a(n,x). Both identities are also valid for complex x in the half-plane Re(x) > 2. See the Bala link in A036970. Cf. A272378. (End)
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi - Pi^2/12 - 2*log(2). - Amiram Eldar, Jul 12 2020

A083061 Triangle of coefficients of a companion polynomial to the Gandhi polynomial.

Original entry on oeis.org

1, 1, 3, 4, 15, 15, 34, 147, 210, 105, 496, 2370, 4095, 3150, 945, 11056, 56958, 111705, 107415, 51975, 10395, 349504, 1911000, 4114110, 4579575, 2837835, 945945, 135135, 14873104, 85389132, 197722980, 244909665, 178378200, 77567490
Offset: 0

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Author

Hans J. H. Tuenter, Apr 19 2003

Keywords

Comments

This polynomial arises in the setting of a symmetric Bernoulli random walk and occurs in an expression for the even moments of the absolute distance from the origin after an even number of timesteps. The Gandhi polynomial, sequence A036970, occurs in an expression for the odd moments.
When formatted as a square array, first row is A002105, first column is A001147, second column is A001880.
Another version of the triangle T(n,k), 0<=k<=n, read by rows; given by [0, 1, 3, 6, 10, 15, 21, 28, ...] DELTA [1, 2, 3, 4, 5, 6, 7, 8, 9, ...] = 1; 0, 1; 0, 1, 3; 0, 4, 15, 15; 0, 34, 147, 210, 105; ... where DELTA is the operator defined in A084938. - Philippe Deléham, Jun 07 2004
In A160464 we defined the coefficients of the ES1 matrix. Our discovery that the n-th term of the row coefficients ES1[1-2*m,n] for m>=1, can be generated with rather simple polynomials led to triangle A094665 and subsequently to this one. - Johannes W. Meijer, May 24 2009
Related to polynomials defined in A160485 by a shift of +-1/2 and scaling by a power of 2. - Richard P. Brent, Jul 15 2014

Examples

			Triangle starts (with an additional first column 1,0,0,...):
[1]
[0,      1]
[0,      1,       3]
[0,      4,      15,      15]
[0,     34,     147,     210,     105]
[0,    496,    2370,    4095,    3150,     945]
[0,  11056,   56958,  111705,  107415,   51975,  10395]
[0, 349504, 1911000, 4114110, 4579575, 2837835, 945945, 135135]
		

Crossrefs

From Johannes W. Meijer, May 24 2009 and Jun 27 2009: (Start)
A002105 equals the row sums (n>=2) and the first left hand column (n>=1).
A001147, A001880, A160470, A160471 and A160472 are the first five right hand columns.
Appears in A162005, A162006 and A162007.
(End)

Programs

  • Maple
    imax := 6;
    T1(0, x) := 1:
    T1(0, x+1) := 1:
    for i from 1 to imax do
        T1(i, x) := expand((2*x+1) * (x+1) * T1(i-1, x+1) - 2*x^2*T1(i-1, x)):
        dx := degree(T1(i, x)):
        for k from 0 to dx do
            c(k) := coeff(T1(i, x), x, k)
        od:
        T1(i, x+1) := sum(c(j1)*(x+1)^(j1), j1 = 0..dx):
    od:
    for i from 0 to imax do
        for j from 0 to i do
            a(i, j) := coeff(T1(i, x), x, j)
        od:
    od:
    seq(seq(a(i, j), j = 0..i), i = 0..imax);
    # Johannes W. Meijer, Jun 27 2009, revised Sep 23 2012
  • Mathematica
    b[0, 0] = 1;
    b[n_, k_] := b[n, k] = Sum[2^j*(Binomial[k + j, 1 + j] + Binomial[k + j + 1, 1 + j])*b[n - 1, k - 1 + j], {j, Max[0, 1 - k], n - k}];
    a[0, 0] = 1;
    a[n_, k_] := b[n, k]/2^(n - k);
    Table[a[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 19 2018, after Philippe Deléham *)
  • Sage
    # uses[fr2_row from A088874]
    A083061_row = lambda n: [(-1)^(n-k)*m*2^(-n+k) for k,m in enumerate(fr2_row(n))]
    for n in (0..7): print(A083061_row(n)) # Peter Luschny, Sep 19 2017

Formula

Let T(i, x)=(2x+1)(x+1)T(i-1, x+1)-2x^2T(i-1, x), T(0, x)=1; so that T(1, x)=1+3x; T(2, x)=4+15x+15x^2; T(3, x)=34+147x+210x^2+105x^3, etc. Then the (i, j)-th entry in the table is the coefficient of x^j in T(i, x).
a(n, k)*2^(n-k) = A085734(n, k). - Philippe Deléham, Feb 27 2005

A160485 Triangle of the RBS1 polynomial coefficients.

Original entry on oeis.org

1, 1, -2, 1, -8, 12, 1, -2, 60, -120, 1, -128, -168, 0, 1680, 1, 2638, 7320, -5040, -25200, -30240, 1, -98408, -300828, 52800, 1053360, 1330560, 665280, 1, 5307118, 17914260, 2522520, -56456400, -90810720, -60540480, -17297280
Offset: 1

Views

Author

Johannes W. Meijer, May 24 2009, Jul 06 2009, Sep 19 2012

Keywords

Comments

In A160480 we defined the BS1 matrix by BS1[2*m-1,n=1] = 2*beta(2*m) and the recurrence relation BS1 [2*m-1,n] = (2*n-3)/(2*n-2)*(BS1[2*m-1,n-1]- BS1[2*m-3,n-1]/(2*n-3)^2), for positive and negative values of m and n= 1, 2, .. . As usual beta(m) = sum((-1)^k/(1+2*k)^m, k=0..infinity). It is well-known that BS1[1-2*m,n=1] = euler(2*m-2) for m = 1, 2, .., with euler(2*m-2) the Euler numbers A000364. These values together with the recurrence relation lead to BS1[ -1,n] = 1 for n = 1, 2, .. .
We discovered that the n-th term of the row coefficients BS1[1-2*m,n] for m = 1, 2, .., can be generated with the rather simple polynomials RBS1(1-2*m,n). Our discovery was enabled by the recurrence relation for the RBS1(1-2*m,n) polynomials which we derived from the recurrence relation for the BS[2*m-1,n] coefficients and the fact that RBS1(-1,n) = 1.
The RBS1 polynomials and the polynomials defined by sequence A083061 are related by a shift of +-1/2 and scaling by a power of 2 (see arXiv link). - Richard P. Brent, Jul 15 2014

Examples

			The first few rows of the triangle are:
[1]
[1, -2]
[1, -8, 12]
[1, -2, 60, -120]
[1, -128, -168, 0, 1680]
The first few RBS1(1-2*m,n) polynomials are:
RBS1(-1,n) = 1
RBS1(-3,n) = 1 - 2*n
RBS1(-5,n) = 1 - 8*n + 12*n^2
RBS1(-7,n) = 1 - 2*n + 60*n^2 - 120*n^3
From _Peter Bala_, Jan 22 2019: (Start)
Qbar(r,n) = binomial(2*n+2,n+1)/(2^(2*n+1)) * Sum_{k = 0..n} binomial(n,k)/binomial(n+k+1,k)*(2*k + 1)^(2*r):
Case r = 2: Qbar(2,n) = binomial(2*n+2,n+1)/2^(2*n+1) * ( 1 + 3^4*n/(n+2) + 5^4*n*(n-1)/((n+2)*(n+3)) + 7^4*n*(n-1)*(n-2)/((n+2)*(n+3)*(n+4)) + ... ) = 12*n^2 + 8*n + 1, valid for n a nonnegative integer (when the series terminates). The identity is also valid for complex n with real part greater than 1 (provided the factor binomial(2*n,n) is replaced with the appropriate expression involving the gamma function).
Case r = 3: Qbar(3,n) = binomial(2*n+2,n+1)/(2^(2*n+1)) * ( 1 + 3^6*n/(n+2) + 5^6*n*(n-1)/((n+2)*(n+3)) + 7^6*n*(n-1)*(n-2)/((n+2)*(n+3)*(n+4)) + ... ) = 120*n^3 + 60*n^2 + 2*n + 1, valid for n a nonnegative integer. The identity is also valid for complex n with real part greater than 2.
Note, the case r = 0 is equivalent to the identity 1 = binomial(2*n,n)/2^(2*n-1) * ( 1 + (n-1)/(n+1) + (n-1)*(n-2)/((n+1)*(n+2)) + (n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ... ), which is valid for complex n with real part greater than 0. This identity was found by Ramanujan. See Example 6, Chapter 10 in Berndt. (End)
		

References

  • B. C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag, Chapter 10, p. 21.

Crossrefs

A160480 is the Beta triangle.
A009389(2*n) equals the second left hand column divided by 2.
A001813 equals the first right hand column.
The absolute values of the row sums equal the Euler numbers A000364.

Programs

  • Maple
    nmax := 8; mmax := nmax: A(1, 1) := 1: RBS1(n, 2) := (2*n-1)^2*1-(2*n)*(2*n-1)*1: for m from 3 to mmax do for k from 0 to m-1 do A(m-1, k+1) := coeff(RBS1(n, m-1), n, k) od; RBS1(n+1, m-1) := 0: for k from 0 to m-1 do RBS1(n+1, m-1) := RBS1(n+1, m-1) + A(m-1, k+1)*(n+1)^k od: RBS1(n, m) := (2*n-1)^2*RBS1(n, m-1)-(2*n)*(2*n-1) * RBS1(n+1, m-1) od: for k from 0 to nmax-1 do A(nmax, k+1) := coeff(RBS1(n, nmax), n, k) od: seq(seq(A(n, m), m=1..n), n=1..nmax);

Formula

RBS1(1-2*m,n) = (2*n-1)^2*RBS1(3-2*m,n)-(2*n)*(2*n-1)*RBS1(3-2*m,n+1) for m = 2, 3, .., with RBS1(-1,n) =1 for n = 1, 2, .. .
From Peter Bala, Jan 22 2019: (Start)
The row polynomials RBS1 of the triangle are related to the polynomials Qbar(r,n), r = 0,1,2,..., introduced by Brent by Qbar(r,n) = RBS1(-2*r-1,-n).
Recurrence: Qbar(r+1,n) = (2*n + 1)^2*Qbar(r,n) - 2*n(2*n + 1)*Qbar(r,n-1) with Qbar(0,n) = 1 (Brent, equation 19, p.7).
Qbar(r,n) = binomial(2*n + 2,n + 1)/(2^(2*n + 1)) * Sum_{k = 0..n} binomial(n,k)/binomial(n + k + 1,k)*(2*k + 1)^(2*r); this follows easily from the above recurrence. Two examples are given below.
Qbar(r,n) = 1/4^n * Sum_{k = 0..n} binomial(2*n + 1,n - k)*(2*k + 1)^(2*r) For related polynomial sequences see A036970, A083061 and A245244. (End)

A245244 Triangle of coefficients of the Pbar polynomials, read by rows.

Original entry on oeis.org

1, -3, 4, 25, -56, 32, -427, 1228, -1184, 384, 12465, -41840, 52416, -29184, 6144, -555731, 2079892, -3076288, 2258688, -829440, 122880, 35135945, -142843304, 237829600, -208562688, 102279168, -26787840, 2949120, -2990414715, 12987478876, -23672564832, 23581133952, -13947525120, 4929576960, -970260480
Offset: 0

Views

Author

Richard P. Brent, Jul 14 2014

Keywords

Comments

Pbar(r,n) is a polynomial of degree r defined by the recurrence
Pbar(r+1,n) = (2*n-1)^2 * Pbar(r,n) - 4*(n-1)^2 * Pbar(r,n-1)
with initial condition Pbar(0,n) = 1.

Examples

			Pbar(1,n) = 4*n-3, Pbar(2,n) = 32*n^2 - 56*n + 25.
Triangle begins:
1,
-3, 4,
25, -56, 32,
-427, 1228, -1184, 384,
12465, -41840, 52416, -29184, 6144,
...
From _Peter Bala_, Jan 22 2018: (Start)
The polynomials Pbar(r,n) as hypergeometric series:
r = 0: n*Pbar(0,n) = n = 1 + 3*(n-1)/(n+1) + 5*(n-1)*(n-2)/((n+1)*(n+2)) + 7*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ..., for n a positive integer (when the series terminates). The identity is also valid for complex n with real part greater than 1/2.
r = 1: n*Pbar(1,n) = n*(4*n - 3) = 1 + 3^3*(n-1)/(n+1) + 5^3*(n-1)*(n-2)/((n+1)*(n+2)) + 7^3*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ..., for n a positive integer (when the series terminates). The identity is also valid for complex n with real part greater than 3/2.
r = 2: n*Pbar(2,n) = n*(32*n^2 - 56*n + 25) = 1 + 3^5*(n-1)/(n+1) + 5^5*(n-1)*(n-2)/((n+1)*(n+2)) + 7^5*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ..., for n a positive integer (when the series terminates). The identity is also valid for complex n with real part greater than 5/2.
The above identities when r = 0 and r = 1 were found by Ramanujan. See Example 5 and Example 13 in Chapter 10 of Berndt. (End)
		

References

  • B. C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag, Chapter 10, p. 20 and p. 23.

Crossrefs

(-1)^r Pbar(r,0) is sequence A009843. The leading coefficient of Pbar(r,n) is sequence A047053. Cf. also A036970, A083061, A160485 for analogous moments of Bernoulli random walks.

Programs

  • PARI
    N=10; P=vector(N+2); P[1]=1;
    Pbar(r)=P[r+1];
    for (r=0, N, P[r+2] = (2*n-1)^2 * Pbar(r) - 4*(n-1)^2 * subst(Pbar(r),n,n-1) );
    seq=[];  for(r=1,N, seq=concat(seq, Vecrev(P[r])); );  seq
    \\ Joerg Arndt, Jan 27 2015

Formula

In terms of Dumont-Foata polynomials F(r,x,y,z),
Pbar(r,n) = (-4)^r F(r+1,1/2-n,1/2,1/2).
In terms of odd absolute moments of a symmetric Bernoulli random walk with an odd number of steps,
n*C(2*n,n)*Pbar(r,n) = Sum_{k} C(2*n-1,k) * |2*n-1-2*k|^(2*r+1).
In terms of the Pochhammer symbol or ascending factorial (x)_k,
Pbar(r,n) = Sum_{1 <= j <= k <= r+1} (-1)^(j+1)*(1-n)_{k-1}*(2j-1)^(2r+1)/((k-j)!(k)_j).
n*Pbar(r,n) = 1 + 3^(2*r+1)*(n-1)/(n+1) + 5^(2*r+1)*(n-1)*(n-2)/((n+1)*(n+2)) + 7^(2*r+1)*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ... = Sum_{k = 0..n-1} binomial(n-1,k)/binomial(n+k,k)*(2*k + 1)^(2*r+1); this follows easily from the above recurrence. Examples are given below. - Peter Bala, Jan 22 2018

Extensions

More terms from Joerg Arndt, Jan 27 2015

A272378 a(n) = n*(6*n^2 - 8*n + 3).

Original entry on oeis.org

0, 1, 22, 99, 268, 565, 1026, 1687, 2584, 3753, 5230, 7051, 9252, 11869, 14938, 18495, 22576, 27217, 32454, 38323, 44860, 52101, 60082, 68839, 78408, 88825, 100126, 112347, 125524, 139693, 154890, 171151, 188512, 207009, 226678, 247555, 269676, 293077
Offset: 0

Views

Author

Vincenzo Librandi, Apr 29 2016

Keywords

Crossrefs

Programs

  • Magma
    [n*(6*n^2 - 8*n + 3): n in [0..50]];
    
  • Mathematica
    Table[n (6 n^2 - 8 n + 3), {n, 0, 50}]
    LinearRecurrence[{4,-6,4,-1},{0,1,22,99},40] (* Harvey P. Dale, Dec 29 2017 *)
  • PARI
    vector(100, n, n--; n*(6*n^2 - 8*n + 3)) \\ Altug Alkan, Apr 29 2016
    
  • Python
    for n in range(0,10**3):print(n*(6*n**2-8*n+3),end=", ") # Soumil Mandal, Apr 30 2016

Formula

G.f.: x*(1 + 18*x + 17*x^2)/(1 - x)^4.
E.g.f.: x*(1 + 10*x + 6*x^2)*exp(x).
a(n) = n*A080859(n+1).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), for n>3.
See page 7 in Brent's paper:
a(n) = n^2*A000384(n) - n*(n-1)*A000384(n-1).
A272379(n) = n^2*a(n) - n*(n-1)*a(n-1).
From Peter Bala, Jan 30 2019: (Start)
Let a(n,x) = Product_{k = 0..n} (x - k)/(x + k). Then for positive integer x we have x^2*(6*x^2 - 8*x + 3) = Sum_{n >= 0} ((n+1)^7 + n^7)*a(n,x) and x*(6*x^2 - 8*x + 3) = Sum_{n >= 0} ((n+1)^6 - n^6)*a(n,x). Both identities are also valid for complex x in the half-plane Re(x) > 7/2. See the Bala link in A036970. Cf. A272379. (End)

A272379 a(n) = n*(24*n^3 - 60*n^2 + 54*n - 17).

Original entry on oeis.org

0, 1, 86, 759, 3100, 8765, 19986, 39571, 70904, 117945, 185230, 277871, 401556, 562549, 767690, 1024395, 1340656, 1725041, 2186694, 2735335, 3381260, 4135341, 5009026, 6014339, 7163880, 8470825, 9948926, 11612511, 13476484, 15556325, 17868090, 20428411
Offset: 0

Views

Author

Vincenzo Librandi, Apr 29 2016

Keywords

Crossrefs

Programs

  • Magma
    [n*(24*n^3 - 60*n^2 + 54*n - 17): n in [0..50]];
    
  • Mathematica
    Table[n (24 n^3 - 60 n^2 + 54 n - 17), {n, 0, 50}]
    LinearRecurrence[{5,-10,10,-5,1},{0,1,86,759,3100},40] (* Harvey P. Dale, Mar 24 2021 *)
  • PARI
    vector(100, n, n--; n*(24*n^3 - 60*n^2 + 54*n - 17)) \\ Altug Alkan, Apr 29 2016

Formula

G.f.: x*(1 + 81*x + 339*x^2 + 155*x^3)/(1 - x)^5.
E.g.f.: x*(1 + 42*x + 84*x^2 + 24*x^3)*exp(x).
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
See page 7 in Brent's paper:
a(n) = n^2*A272378(n) - n*(n-1)*A272378(n-1),
A272380(n) = n^2*a(n) - n*(n-1)*a(n-1).
From Peter Bala, Jan 30 2019: (Start)
Let a(n,x) = Product_{k = 0..n} (x - k)/(x + k). Then for positive integer x we have x^2*(24*x^3 - 60*x^2 + 54*x - 17) = Sum_{n >= 0} ((n+1)^9 + n^9)*a(n,x) and x*(24*x^3 - 60*x^2 + 54*x - 17) = Sum_{n >= 0} ((n+1)^8 - n^8)*a(n,x). Both identities are also valid for complex x in the half-plane Re(x) > 9/2. See the Bala link in A036970. Cf. A272378 and A272380. (End)

A272380 a(n) = n*(120*n^4 - 480*n^3 + 762*n^2 - 556*n + 155).

Original entry on oeis.org

0, 1, 342, 6315, 40492, 157125, 456546, 1099567, 2321880, 4448457, 7907950, 13247091, 21145092, 32428045, 48083322, 69273975, 97353136, 133878417, 180626310, 239606587, 313076700, 403556181, 513841042, 647018175, 806479752, 995937625, 1219437726, 1481374467
Offset: 0

Views

Author

Vincenzo Librandi, Apr 29 2016

Keywords

Crossrefs

Programs

  • Magma
    [n*(120*n^4 - 480*n^3 + 762*n^2 - 556*n + 155): n in [0..50]];
    
  • Mathematica
    Table[n (120 n^4 - 480 n^3 + 762 n^2 - 556 n + 155), {n, 0, 50}]
    LinearRecurrence[{6,-15,20,-15,6,-1},{0,1,342,6315,40492,157125},40] (* Harvey P. Dale, Mar 15 2018 *)
  • PARI
    vector(100, n, n--; n*(120*n^4 - 480*n^3 + 762*n^2 - 556*n + 155)) \\ Altug Alkan, Apr 29 2016

Formula

O.g.f.: x*(1 + 336*x + 4278*x^2 + 7712*x^3 + 2073*x^4)/(1-x)^6.
E.g.f.: x*(1 + 170*x + 882*x^2 + 720*x^3 + 120*x^4)*exp(x).
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), for n>5.
a(n) = n^2*A272379(n) - n*(n-1)*A272379(n-1), see page 7 in Brent's paper.
From Peter Bala, Jan 30 2019: (Start)
Let a(n,x) = Product_{k = 0..n} (x - k)/(x + k). Then for positive integer x we have x^2*(120*x^4 - 480*x^3 + 762*x^2 - 556*x + 155) = Sum_{n >= 0} ((n+1)^11 + n^11)*a(n,x) and x*(120*x^4 - 480*x^3 + 762*x^2 - 556*x + 155) = Sum_{n >= 0} ((n+1)^10 - n^10)*a(n,x). Both identities are also valid for complex x in the half-plane Re(x) > 11/2. See the Bala link in A036970. Cf. A272378 and A272379. (End)

A005440 Coefficients of Gandhi polynomials.

Original entry on oeis.org

2, 8, 54, 556, 8146, 161424, 4163438, 135634292, 5448798090, 264689281240, 15296907175462, 1037373202178748, 81588771795362114, 7366855482991121696, 756909709680583939806, 87807399365909591247364
Offset: 2

Views

Author

Keywords

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Second column in table A036970.

Programs

  • Maple
    B[1]:= X -> X^2:
    for n from 2 to 50 do B[n]:= unapply(expand(X^2*(B[n-1](X+1)-B[n-1](X))), X) od:
    seq(D(B[n])(1),n=1..50); # Robert Israel, Apr 21 2016

Extensions

More terms from David W. Wilson, Jan 12 2001
Showing 1-10 of 15 results. Next