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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A260793 Primes p such that p does not divide any term of the Apéry-like sequence A000172 (also known as Type I primes).

Original entry on oeis.org

3, 11, 17, 19, 43, 83, 89, 97, 113, 137, 139, 163, 193, 211, 233, 241, 283, 307, 313, 331, 347, 353, 379, 401, 409, 419, 433, 443, 491, 499, 523, 547, 569, 587, 601, 617, 619, 641, 643, 673, 811, 827, 859, 881, 929, 947, 953, 977, 1009, 1019, 1033, 1049, 1051
Offset: 1

Views

Author

N. J. A. Sloane, Aug 05 2015

Keywords

Comments

See Schulte et al. (2014) for the precise definition of Type I primes.

Crossrefs

For primes that do not divide the terms of the sequences A000172, A005258, A002893, A081085, A006077, A093388, A125143, A229111, A002895, A290575, A290576, A005259 see this sequence, A291275-A291284 and A133370 respectively.

Programs

  • Mathematica
    maxPrime = 1051;
    maxPi = PrimePi @ maxPrime;
    okQ[p_] := AllTrue[Range[3 maxPi (* coeff 3 is empirical *)], GCD[HypergeometricPFQ[{-#, -#, -#}, {1, 1}, -1], p] == 1&];
    Select[Prime[Range[maxPi]], okQ] (* Jean-François Alcover, Jan 13 2020 *)

Extensions

Edited by N. J. A. Sloane, Aug 22 2017

A133370 Primes p such that p does not divide any term of the Apery sequence A005259 .

Original entry on oeis.org

2, 3, 7, 13, 23, 29, 43, 47, 53, 67, 71, 79, 83, 89, 101, 103, 107, 109, 113, 127, 131, 137, 149, 157, 167, 173, 199, 223, 229, 239, 263, 269, 277, 281, 311, 313, 317, 337, 349, 353, 359, 373, 383, 389, 397, 401, 409, 421, 449, 457, 461, 467, 479, 487, 491
Offset: 1

Views

Author

Philippe Deléham, Oct 27 2007

Keywords

Comments

Malik and Straub give arguments suggesting that this sequence is infinite. - N. J. A. Sloane, Aug 06 2017

Crossrefs

For primes that do not divide the terms of the sequences A000172, A005258, A002893, A081085, A006077, A093388, A125143, A229111, A002895, A290575, A290576, A005259 see A260793, A291275-A291284 and A133370 respectively.

Programs

  • Mathematica
    NeverDividesLucasSeqQ[a_, p_] := And @@ Table[Mod[a[n], p]>0, {n, 0, p-1}];
    A3[a_, b_, c_, n_ /; n < 0] = 0;
    A3[a_, b_, c_, 0] = 1;
    A3[a_, b_, c_, n_] := A3[a, b, c, n] = (((2n - 1)(a (n-1)^2 + a (n-1) + b)) A3[a, b, c, n-1] - c (n-1)^3 A3[a, b, c, n-2])/n^3;
    A3[a_, b_, c_, d_, n_ /; n < 0] = 0;
    Agamma[n_] := A3[17, 5, 1, n];
    Select[Range[1000], PrimeQ[#] && NeverDividesLucasSeqQ[Agamma, #]&] (* Jean-François Alcover, Aug 05 2018, copied from Amita Malik's notebook *)

Extensions

Terms a(16) onwards computed by Amita Malik - N. J. A. Sloane, Aug 21 2017

A291284 Primes p such that p does not divide any term of the Apery-like sequence A290576.

Original entry on oeis.org

2, 5, 7, 13, 17, 19, 29, 37, 43, 47, 59, 61, 67, 71, 83, 89, 101, 109, 127, 139, 149, 167, 173, 191, 211, 233, 239, 241, 251, 257, 271, 277, 281, 307, 311, 313, 331, 337, 347, 349, 353, 359, 373, 379, 383, 409, 419, 421, 431, 433, 443
Offset: 1

Views

Author

N. J. A. Sloane, Aug 21 2017

Keywords

Crossrefs

For primes that do not divide the terms of the sequences A000172, A005258, A002893, A081085, A006077, A093388, A125143, A229111, A002895, A290575, A290576, A005259 see A260793, A291275-A291284 and A133370 respectively.

A002736 Apéry numbers: a(n) = n^2*C(2n,n).

Original entry on oeis.org

0, 2, 24, 180, 1120, 6300, 33264, 168168, 823680, 3938220, 18475600, 85357272, 389398464, 1757701400, 7862853600, 34901442000, 153876579840, 674412197580, 2940343837200, 12759640231800, 55138611528000, 237371722628040, 1018383898440480
Offset: 0

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Author

Keywords

Comments

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-1 of B equals -a(n-1). - T. D. Noe, May 01 2011

References

  • J. Ser, Les Calculs Formels des Séries de Factorielles, Gauthier-Villars, Paris, 1933, p. 93.
  • 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

Programs

  • Magma
    [n^2*Binomial(2*n, n): n in [0..30]]; // Vincenzo Librandi, Aug 08 2014
    
  • Maple
    seq(n^2*binomial(2*n,n), n=0..50); # Robert Israel, Aug 07 2014
  • Mathematica
    CoefficientList[ Series[x (4 x + 2)/(1 - 4 x)^(5/2), {x, 0, 20}], x] (* Robert G. Wilson v, Aug 08 2011 *)
    Table[n^2 Binomial[2n,n],{n,0,30}] (* Harvey P. Dale, Jun 21 2017 *)
  • MuPAD
    combinat::catalan(n)*(n+1)*n^2 $ n = 0..36 // Zerinvary Lajos, Apr 17 2007
    
  • PARI
    my(x='x+O('x^100)); concat(0, Vec(x*(4*x+2)/((1-4*x)^(5/2)))) \\ Altug Alkan, Mar 21 2016
    
  • PARI
    a(n) = n^2*binomial(2*n, n); \\ Michel Marcus, Mar 21 2016
    
  • Sage
    [n^2*(n+1)*catalan_number(n) for n in (0..30)] # G. C. Greubel, Mar 23 2022

Formula

G.f.: x*(4*x+2)/((1-4*x)^(5/2)). - Marco A. Cisneros Guevara, Jul 25 2011
Sum_{n>=1} 1/a(n) = Pi^2/18 (Euler). - Benoit Cloitre, Apr 07 2002
From Ilya Gutkovskiy, Jan 17 2017: (Start)
a(n) ~ 4^n*n^(3/2)/sqrt(Pi).
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*log(phi)^2 = A086467, where phi is the golden ratio. (End)
D-finite with recurrence: (-n+1)*a(n) +2*(n+4)*a(n-1) +4*(2*n-3)*a(n-2)=0. - R. J. Mathar, Jan 21 2020
a(n) = (2n)!/(Gamma(n))^2. - Diego Rattaggi, Mar 30 2020
a(n) = Sum_{k=0..2*n} binomial(2*n,k)*abs(n-k)^3 (Bruckman, 1999; Strazdins, 2000). - Amiram Eldar, Jan 12 2022
Sum_{n>=1} x^n/a(n) = 2*arcsin(sqrt(x)/2)^2, for abs(x) < 4 (Adegoke et al., 2022, section 5, p. 10). - Amiram Eldar, Dec 07 2024
From Peter Bala, Aug 02 2025: (Start)
For n >= 1,
a(n) = 2*n*(2*n-1)/(n-1)^2 * a(n-1) with a(1) = 2 and
1/a(n) = Sum_{k = 0..n} (-1)^(n+k+1) * binomial(n, k)*binomial(n+k, k)/(n+k)^2. (End)
a(n) = 2 * A002544(n-1) for n>=1. - Alois P. Heinz, Aug 03 2025

A099601 Quotient of de Bruijn sums S(4,n)/S(2,n).

Original entry on oeis.org

1, 7, 131, 3067, 79459, 2181257, 62165039, 1818812387, 54257991011, 1642977121597, 50344383988381, 1557608560147757, 48577698917598031, 1525245771206644117, 48165918788138198759, 1528611371067309862067
Offset: 0

Views

Author

Michael Somos, Oct 24 2004

Keywords

Comments

The de Bruijn sum S(s,n) = Sum_{k=0..2n} (-1)^(k+n) binomial(2n,k)^s.
Also the n-th term of the crystal ball sequence for the A_{2n} lattice.

Examples

			A003215(1)=7=a(1), A008384(2)=131=a(2), A008388(3)=3067=a(3), ...
		

References

  • G. E. Andrews, Application of SCRATCHPAD to Problems in Special Functions and Combinatorics, in Trends in Computer Algebra, Springer-Verlag, 1988, pp. 158-166 MR0935413 (89c:05010).

Crossrefs

Programs

  • Maple
    a := n -> hypergeom([-2*n, -n, 2*n+1], [1, 1], 1):
    seq(simplify(a(n)), n=0..15); # Peter Luschny, Feb 13 2018
  • Mathematica
    Table[Sum[(-1)^k*Binomial[2*n,k]^4,{k,0,2*n}] / Sum[(-1)^k*Binomial[2*n, k]^2,{k,0,2*n}],{n,0,20}] (* Vaclav Kotesovec, Mar 07 2014 *)
  • PARI
    a(n)=if(n<0,0, sum(k=0,2*n, (-1)^k*binomial(2*n,k)^4)/ sum(k=0,2*n, (-1)^k*binomial(2*n,k)^2))

Formula

Recurrence: n^2*(2*n-1)^2*(48*n^2 - 126*n + 83)*a(n) = (6528*n^6 - 30192*n^5 + 55040*n^4 - 50406*n^3 + 24465*n^2 - 5985*n + 585)*a(n-1) - (n-1)^2*(2*n-3)^2*(48*n^2 - 30*n + 5)*a(n-2). - Vaclav Kotesovec, Mar 07 2014
a(n) ~ (1+sqrt(2))^(4*n+3/2) / (2^(9/4)*Pi*n). - Vaclav Kotesovec, Mar 07 2014
a(n) = Sum_{k=0..n} binomial(n,k)*binomial(2*n+k,2*n-k)*binomial(2*k,k). - Ilya Gutkovskiy, Nov 24 2017
a(n) = hypergeom([-2*n, -n, 2*n+1], [1, 1], 1). - Peter Luschny, Feb 13 2018
From Peter Bala, Dec 21 2020: (Start)
a(n) = Sum_{k = 0..n} C(2*n,n-k)^2 * C(2*n+k,k). Cf. A005258.
a(n) = Sum_{k = 0..n} (-1)^(n+k)*C(2*n,n-k)*C(2*n+k,k)^2.
a(n) = C(2*n,n)^2 * hypergeom([-n, -n, 2*n+1], [n+1, n+1], 1).
a(n) = (-1)^n* C(2*n,n) * hypergeom([-n, 2*n+1, 2*n+1], [1, n+1], 1). (End)
a(n) = [x^n] 1/(1 - x)*( Legendre_P(m*n,(1 + x)/(1 - x)) ) at m = 2. At m = 1 we get the Apéry numbers A005258. - Peter Bala, Dec 23 2020
a(n) = A108625(2*n, n). - Peter Bala, Jun 20 2023

A103882 a(n) = Sum_{i=0..n} C(n+1,i)*C(n-1,i-1)*C(2n-i,n).

Original entry on oeis.org

1, 2, 12, 92, 780, 7002, 65226, 623576, 6077196, 60110030, 601585512, 6078578508, 61908797418, 634756203018, 6545498596110, 67830161708592, 705951252118284, 7375213677918294, 77310179609631564, 812839595630249540, 8569327862277434280, 90562666977432643862
Offset: 0

Views

Author

Ralf Stephan, Feb 20 2005

Keywords

Comments

Number of permutations of n copies of 1..3 with all adjacent differences <= 1 in absolute value. - R. H. Hardin, May 06 2010 [Cf. A177316. - Peter Bala, Jan 14 2020]

Crossrefs

Equals A103881(n, n).
Row n=3 of A331562.

Programs

  • Magma
    [1] cat [&+[Binomial(n+1, i)*Binomial(n-1, i-1) * Binomial(2*n-i, n): i in [0..n]]:n in  [1..21]]; // Marius A. Burtea, Jan 19 2020
    
  • Magma
    [&+[Binomial(n, k)^2*Binomial(n+k-1, k): k in [0..n]]:n in  [0..21]]; // Marius A. Burtea, Jan 19 2020
    
  • Maple
    a:= proc(n) option remember; `if`(n<2, n+1,
          ((n-1)*(55*n^3-143*n^2+102*n-24)*a(n-1)+
          n*(5*n-3)*(n-2)^2*a(n-2))/((n-1)*(5*n-8)*n^2))
        end:
    seq(a(n), n=0..30);  # Alois P. Heinz, Jun 29 2015
    # Alternative:
    a := n -> hypergeom([-n, -n, n], [1, 1], 1):
    seq(simplify(a(n)), n=0..21); # Peter Luschny, Jan 19 2020
  • Mathematica
    Drop[Table[Sum[Sum[Multinomial[r, g, n + 1 - r - g] Binomial[n - 1,n - r] Binomial[n - 1, n - g], {g, 1, n}], {r, 1, n}], {n, 0, 18}], 1] (* Geoffrey Critzer, Jun 29 2015 *)
    Table[Sum[Binomial[n+1,k]Binomial[n-1,k-1]Binomial[2n-k,n],{k,0,n}],{n,0,30}] (* Harvey P. Dale, Jun 19 2021 *)
  • PARI
    a(n) = polcoef(pollegendre(n, (1 + x)/(1 - x)) + O(x^(n+1)), n); \\ Michel Marcus, Dec 20 2020
    
  • Python
    def A103882(n):
        if n == 0: return 1
        m, g = 1, 0
        for k in range(n+1):
            g += m*n//(n+k)
            m *= (n+k+1)*(n-k)**2
            m //= (k+1)**3
        return g # Chai Wah Wu, Oct 04 2022
    
  • SageMath
    def A103882(n): return hypergeometric([-n,-n,n], [1,1], 1).simplify()
    [A103882(n) for n in range(31)] # G. C. Greubel, May 24 2023

Formula

a(n) = (A005258(n-1) + 3*A005258(n))/5 (Apéry numbers). - Mark van Hoeij, Jul 13 2010
n^2*(n-1)*(5*n-8)*a(n) = (n-1)*(55*n^3-143*n^2+102*n-24)*a(n-1) + n*(n-2)^2*(5*n-3)*a(n-2). - Alois P. Heinz, Jun 29 2015
a(n) ~ phi^(5*n + 3/2) / (2*Pi*5^(1/4)*n), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Jul 21 2019
From Peter Bala, Jan 14 2020: (Start)
a(n) = Sum_{k = 0..n} C(n,k)^2*C(n+k-1,k). Cf. A005258.
For any prime p >= 5, a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for all positive integers n and k (follows from known supercongruences satisfied by the Apéry numbers A005258 - see Straub, Example 3.4). (End)
a(n) = hypergeometric([-n, -n, n], [1, 1], 1). - Peter Luschny, Jan 19 2020
From Peter Bala, Dec 19 2020: (Start)
a(n) = Sum_{k = 1..n} C(n,k)*C(n+k,k)*C(n-1,k-1) for n >= 1.
a(n) = [x^n] P(n, (1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Cf. A156554. (End)
a(n) = Sum_{k = 0..n} binomial(2*n-k-1,n-k)*binomial(n,k)^2. Cf. A108628. - Peter Bala, Mar 24 2022
From Peter Bala, Apr 15 2022: (Start)
a(-n) = (-1)^n*A352654(n).
a(n) = [x^n*y^n*z^(n-1)] 1/(1 - x - y - z + x*z + y*z - x*y*z) for n >= 1.
a(n) = B(n,n,n-1) in the notation of Straub, see equation 24.
a(n) = [x^n*y^n*z^(n-1)] (x + y + z)^n*(x + y)^n*(y + z)^(n-1) for n >= 1. (End)
D-finite with recurrence 9*n^2*a(n) -3*(31*n^2-27*n+6)*a(n-1) -2*(37*n^2-138*n+108)*a(n-2) -(n-3)*(17*n-56)*a(n-3) -(n-4)^2*a(n-4) = 0. - R. J. Mathar, Aug 01 2022
a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n-1, n-k)*binomial(n+k, k)*binomial(n+k-1, k). - Peter Bala, Aug 13 2023
a(n) = Sum_{k = 0..n} (-1)^k * binomial(n+1, k)*binomial(2*n-k, n-k)^2. - Peter Bala, Oct 05 2024

Extensions

a(0)=1 prepended by Alois P. Heinz, Jun 29 2015

A088917 Central Delannoy numbers (mod 3); Characteristic function for Cantor set.

Original entry on oeis.org

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

Views

Author

Benoit Cloitre, Nov 30 2003

Keywords

Comments

Also Apery numbers (mod 3).
More generally also (Sum_{k=0..n} binomial(n,k)^x*binomial(n+k,k)^y) (mod 3) for any x >= 1 in N and any odd y >= 1.
a(n) = 0 if the ternary expansion of n contains one or more 1-digits, otherwise 1. - Antti Karttunen, Aug 23 2019
Main diagonal of the Sierpinski carpet (A153490). - Paolo Xausa, May 19 2023

Crossrefs

Characteristic function of A005823, and with offset 1, characteristic function of A191106.

Programs

  • Mathematica
    Nest[ Flatten[# /. {0 -> {0, 0, 0}, 1 -> {1, 0, 1}}] &, {1}, 5] (* Or *)
    f[n_] := Mod[LegendreP[n, 3], 3]; Array[f, 111, 0] (* Or *)
    f[n_] := If[ FreeQ[ IntegerDigits[n, 3], 1], 1, 0]; Array[f, 111, 0] (* also from Mathematica v8.0 Mathematical Functions Help section for "IntegerDigits" "Cantor set construction:" *) (* Robert G. Wilson v, Jun 16 2011 *)
    Nest[Join[#, 0 #, #] &, {1}, 5] (* IWABUCHI Yu(u)ki, Sep 08 2012 *)
  • PARI
    a(n)=sum(k=0,n,binomial(n,k)*binomial(n+k,k))%3
    
  • PARI
    A088917(n) = { while(n, if(n%3==1, return(0), n\=3)); (1); }; \\ Antti Karttunen, Aug 23 2019 (copied from A005823)
    
  • PARI
    A088917(n) = abs(factorback(apply(d -> d-1,digits(n,3)))); \\ Antti Karttunen, Aug 23 2019

Formula

a(A005823(n)) = 1; a(A081606(n)) = 0.
a(n) = A001850(n) - 3*floor(A001850(n)/3).
a(n) = 2 - A105220(n) = 1 - A316829(n). - Antti Karttunen and Jon Maiga, Aug 24 2019
G.f.: Product_{k>=0} (1 + x^(2*3^k)). - Ilya Gutkovskiy, Jun 05 2021

Extensions

Secondary name added by Antti Karttunen, Aug 23 2019

A108628 n-th term of the crystal ball sequence for A_{n+1} lattice for n >= 0.

Original entry on oeis.org

1, 7, 55, 471, 4251, 39733, 380731, 3716695, 36808723, 368750757, 3728940249, 38003358693, 389866749975, 4022124746409, 41697566691555, 434124925278807, 4536783726146499, 47569453938399445, 500266519237489357, 5275183203229043221, 55760274296452936741
Offset: 0

Views

Author

Paul D. Hanna, Jun 14 2005

Keywords

Comments

Equals the secondary diagonal of square array A108625, in which row n equals the crystal ball sequence for A_n lattice. Main diagonal of square array A108625 equals the Apery numbers (A005258).

Crossrefs

Programs

  • Magma
    A108628:= func< n | (&+[Binomial(n+1,k)^2*Binomial(n+k,k-1): k in [0..n+1]]) >;
    [A108628(n): n in [0..30]]; // G. C. Greubel, Oct 06 2023
    
  • Maple
    seq(add(binomial(n,k)*binomial(n+1,k)*binomial(n+k+1,k), k = 0..n), n = 0..20); # Peter Bala, Apr 14 2022
  • Mathematica
    Table[Sum[Binomial[n+1,k]^2 Binomial[n+k,k-1],{k,0,n+1}],{n,0,20}] (* Harvey P. Dale, Apr 01 2013 *)
  • PARI
    a(n)=sum(k=0,n+1,binomial(n+1,k)^2*binomial(n+k,k-1))
    
  • Python
    def A108628(n):
        m, g = 1, 0
        for k in range(n+1):
            g += m
            m *= (n+k+2)*(n-k)*(n-k+1)
            m //= (k+1)**3
        return g # Chai Wah Wu, Oct 03 2022
    
  • SageMath
    def A108628(n): return sum(binomial(n+1,k)^2*binomial(n+k,k-1) for k in range(n+2))
    [A108628(n) for n in range(31)] # G. C. Greubel, Oct 06 2023

Formula

a(n) = Sum_{k = 0..n+1} C(n+1, k)^2 * C(n+k, k-1).
a(n) = A108625(n+1, n).
exp( Sum_{n >= 1} a(n-1)*x^n/n ) = 1 + x + 4*x^2 + 22*x^3 + 144*x^4 + 1048*x^5 + 8189*x^6 + 67325*x^7 + 574999*x^8 + ... appears to have integer coefficients. Cf. A208675. - Peter Bala, Jan 12 2016
Recurrence: (n+1)^2*(5*n^2 - 6*n + 2)*a(n) = (55*n^4 - 11*n^3 - 26*n^2 + 5*n + 5)*a(n-1) + (n-1)^2*(5*n^2 + 4*n + 1)*a(n-2). - Vaclav Kotesovec, Jan 13 2016
a(n) ~ sqrt(89/8 + 199/(8*sqrt(5))) * ((1+sqrt(5))/2)^(5*n) / (Pi*n). - Vaclav Kotesovec, Jan 13 2016
Equivalently, a(n) ~ phi^(5*n + 11/2) / (2*5^(1/4)*Pi*n), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 06 2021
From Peter Bala, Mar 24 2022: (Start)
a(n) = Sum_{k = 0..n} binomial(n,k)*binomial(n+1,k)*binomial(n+k+1,k).
Using binomial(-n,k) = (-1)^k*binomial(n+k-1,k) for nonnegative k, we have a(-n) = Sum_{k} binomial(-n,k)*binomial(-n+1,k)*binomial(-n+k+1,k) = Sum_{k} (-1)^k* binomial(n+k-1,k)*binomial(n+k-2,k)*binomial(n-2,k) = (-1)^n*A208675(n-1) for n >= 2.
a(n-1) = (1/2)*Sum_{k = 0..floor(n/2)} binomial(n,k)^2 * binomial(3*n-2*k-1,n-2*k) for n >= 1. Cf. A103882.
a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n+1,k+1)*binomial(n+k+1,k)^2.
Equivalently, a(n) = [(x*z)^(n+1)*y^n] ( (x + y + z)^(n+1) * (x + y)^n * (y + z)^(n+1) ).
a(n) = (1/5)*( 2*A005258(n+1) - A005258(n) ).
a(n) = hypergeometric3F2([n + 2, -n, -n - 1], [1, 1], 1).
a(n) = (-1)^n*(n+1)*hypergeom([n + 2, n + 2, -n ], [1, 2], 1).
a(n) = [x^n] 1/(1 - x)*P(n+1,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Compare with A208675(n) = [x^n] 1/(1 - x)*P(n-1,(1 + x)/(1 - x)) for n >= 1 and A005258(n) = [x^n] 1/(1 - x)*P(n,(1 + x)/(1 - x)).
a(n) = B(n+1,n,n+1) in the notation of Straub, equation 24. Hence
a(n) = [(x*z)^(n+1)*y^n] 1/(1 - x - y - z + x*z + y*z - x*y*z).
The following takes the sequence offset to be 1: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and all positive integers n and k.
Conjectures: a((p-1)/2) == 0 (mod p) and a((p^3-1)/2) == 0 (mod p^3) for primes p of the form 4*m + 1; a((p^3-1)/2) == 0 (mod p^2) for primes p of the form 4*m + 3; a((p^2-1)/2) == 0 (mod p^2) for primes p >= 5. (End)
From Peter Bala, Sep 11 2024: (Start)
a(n) = Sum_{k = 0..n+1} (-1)^(n+k+1)*binomial(n+1, k)*binomial(n+k, k)* binomial(n+k+1, k).
a(n) = (-1)^(n+1) * hypergeom([n+1, n+2, -n-1], [1, 1], 1). (End)
a(n)^2 = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A143007(n-1, k). Cf. A005258(n)^2 = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A143007(n, k). - Peter Bala, Sep 25 2024

A208675 Number of words, either empty or beginning with the first letter of the ternary alphabet, where each letter of the alphabet occurs n times and letters of neighboring word positions are equal or neighbors in the alphabet.

Original entry on oeis.org

1, 1, 5, 37, 309, 2751, 25493, 242845, 2360501, 23301307, 232834755, 2349638259, 23905438725, 244889453043, 2523373849701, 26132595017037, 271826326839477, 2838429951771795, 29740725671232119, 312573076392760183, 3294144659048391059, 34802392680979707121
Offset: 0

Views

Author

Alois P. Heinz, Feb 29 2012

Keywords

Comments

Also the number of (3*n-1)-step walks on 3-dimensional cubic lattice from (1,0,0) to (n,n,n) with positive unit steps in all dimensions such that the absolute difference of the dimension indices used in consecutive steps is <= 1.

Examples

			a(2) = 5 = |{aabbcc, aabcbc, aabccb, ababcc, abccba}|.
a(3) = 37 = |{aaabbbccc, aaabbcbcc, aaabbccbc, aaabbcccb, aaabcbbcc, aaabcbcbc, aaabcbccb, aaabccbbc, aaabccbcb, aaabcccbb, aababbccc, aababcbcc, aababccbc, aababcccb, aabbabccc, aabbcccba, aabcbabcc, aabcbccba, aabccbabc, aabccbcba, aabcccbab, aabcccbba, abaabbccc, abaabcbcc, abaabccbc, abaabcccb, abababccc, ababcccba, abbaabccc, abbcccbaa, abcbaabcc, abcbccbaa, abccbaabc, abccbcbaa, abcccbaab, abcccbaba, abcccbbaa}|.
		

Crossrefs

Programs

  • Magma
    A208675:= func< n | (&+[Binomial(n,j)*Binomial(n-1,j)*Binomial(n+j-1,j): j in [0..2*n]]) >;
    [A208675(n): n in [0..30]]; // G. C. Greubel, Oct 05 2023
    
  • Maple
    a:= n-> add(binomial(n-1, k)^2 *binomial(2*n-1-k, n-k), k=0..n):
    seq(a(n), n=0..30);  # Alois P. Heinz, Jun 26 2012
  • Mathematica
    a[n_]:= HypergeometricPFQ[{1-n,-n,n}, {1,1}, 1] (* Michael Somos, Jun 03 2012 *)
  • SageMath
    def A208675(n): return sum(binomial(n,j)*binomial(n-1,j)*binomial(n+j-1,j) for j in range(n+1))
    [A208675(n) for n in range(31)] # G. C. Greubel, Oct 05 2023

Formula

From Michael Somos, Jun 03 2012: (Start)
a(n) = A108625(n-1, n).
a(n) = Hypergeometric3F2([1-n, -n, n], [1, 1], 1).
(n+1)^2 * (1 -4*n +5*n^2) * a(n+1) = (5 -5*n -26*n^2 +11*n^3 +55*n^4) * a(n) + (n-1)^2 * (2 +6*n +5*n^2) * a(n-1). (End)
a(n) ~ sqrt((5-sqrt(5))/10)/(2*Pi*n) * ((1+sqrt(5))/2)^(5*n). - Vaclav Kotesovec, Dec 06 2012. Equivalently, a(n) ~ phi^(5*n - 1/2) / (2 * 5^(1/4) * Pi * n), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 07 2021
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 3*x^2 + 15*x^3 + 94*x^4 + 668*x^5 + 5144*x^6 + 41884*x^7 + 355307*x^8 + ... appears to have integer coefficients. Cf. A108628. - Peter Bala, Jan 12 2016
From Peter Bala, Apr 05 2022: (Start)
a(n) = Sum_{k = 0..n} binomial(n,k)*binomial(n-1,k)*binomial(n+k-1,k).
Using binomial(-n,k) = (-1)^k*binomial(n+k-1,k) for nonnegative k, we have:
a(-n) = Sum_{k = 0..n} binomial(-n,k)*binomial(-n-1,k)*binomial(-n+k-1,k).
a(-n) = Sum_{k = 0..n} (-1)^k* binomial(n+k-1,k)*binomial(n+k,k)*binomial(n,k)
a(-n) = (-1)^n*A108628(n-1), for n >= 1.
a(n) = Sum_{k = 1..n} binomial(n,k)*binomial(n-1,k-1)*binomial(n+k-1,k-1) for n >= 1.
Equivalently, a(n) = [(x^n)*(y*z)^(n-1)] (x + y + z)^n*(x + y)^(n-1)*(y + z)^(n-1) for n >= 1.
a(n) = Sum_{k = 0..n-1} (-1)^k*binomial(n-1,k)*binomial(2*n-k-1,n-k)^2.
a(n) = (1/5)*(A005258(n) + 2*A005258(n-1)) for n >= 1.
a(n) = [x^n] 1/(1 - x)*P(n-1,(1 + x)/(1 - x)) for n >= 1, where P(n,x) denotes the n-th Legendre polynomial. Compare with A005258(n) = [x^n] 1/(1 - x)*P(n,(1 + x)/(1 - x)).
a(n) = B(n,n-1,n-1) in the notation of Straub, equation 24. Hence
a(n) = [(x^n)*(y*z)^(n-1)] 1/(1 - x - y - z + x*z + y*z - x*y*z) for n >= 1.
The supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and all positive integers n and k.
Conjectures:
1) a(n) = [(x*y)^n*z^(n-1)] 1/(1 - x - y - z + x*y + x*y*z) for n >= 1.
2) a(n) = - [(x*z)^(n-1)*(y^n)] 1/(1 + y + z + x*y + y*z + x*z + x*y*z) for n >= 1.
3) a(n) = [x^(n-1)*(y*z)^n] 1/(1 - x - x*y - y*z - x*z - x*y*z) for n >= 1. (End)
From Peter Bala, Mar 17 2023: (Start)
For n >= 1:
a(n) = Sum_{k = 0..n} ((n-k)/(n+k))*binomial(n,k)^2*binomial(n+k,k).
a(n) = Sum_{k = 0..n} (-1)^(n+k-1) * ((n-k)/(n+k)) * binomial(n,k) * binomial(n+k,k)^2. (End)

A363871 a(n) = A108625(2*n, 3*n).

Original entry on oeis.org

1, 37, 5321, 980407, 201186025, 43859522037, 9939874413899, 2314357836947571, 549694303511409641, 132569070434503802605, 32360243622138480889321, 7977001183875449759759807, 1982402220908671654519130731, 496031095735572731850517509727
Offset: 0

Views

Author

Peter Bala, Jun 27 2023

Keywords

Comments

a(n) = B(2*n, 3*n, 2*n) in the notation of Straub, equation 24. It follows from Straub, Theorem 3.2, that the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and all positive integers n and k.
More generally, for positive integers r and s the sequence {A108625(r*n, s*n) : n >= 0} satisfies the same supercongruences.
For other cases, see A099601 (r = 2, s = 1), A363867 (r = 1, s = 2), A363868 (r = 3, s = 1), A363869 (r = 3, s = 2) and A363870 (r = 1, s = 3).

Crossrefs

Programs

  • Magma
    A363871:= func< n | (&+[Binomial(2*n,j)^2*Binomial(5*n-j,2*n): j in [0..2*n]]) >;
    [A363871(n): n in [0..30]]; // G. C. Greubel, Oct 05 2023
    
  • Maple
    A108625 := (n, k) -> hypergeom([-n, -k, n+1], [1, 1], 1):
    seq(simplify(A108625(2*n, 3*n)), n = 0..20);
  • Mathematica
    Table[HypergeometricPFQ[{-2*n,-3*n,2*n+1}, {1,1}, 1], {n,0,30}] (* G. C. Greubel, Oct 05 2023 *)
  • SageMath
    def A363871(n): return sum(binomial(2*n,j)^2*binomial(5*n-j,2*n) for j in range(2*n+1))
    [A363871(n) for n in range(31)] # G. C. Greubel, Oct 05 2023

Formula

a(n) = Sum_{k = 0..2*n} binomial(2*n, k)^2 * binomial(5*n-k, 2*n).
a(n) = Sum_{k = 0..2*n} (-1)^k * binomial(2*n, k)*binomial(5*n-k, 2*n)^2.
a(n) = hypergeometric3F2([-2*n, -3*n, 2*n+1], [1, 1], 1).
a(n) = [x^(3*n)] 1/(1 - x)*Legendre_P(2*n, (1 + x)/(1 - x)).
a(n) ~ sqrt(1700 + 530*sqrt(10)) * (98729 + 31220*sqrt(10))^n / (120 * Pi * n * 3^(6*n)). - Vaclav Kotesovec, Feb 17 2024
a(n) = Sum_{k = 0..2*n} binomial(2*n, k) * binomial(3*n, k) * binomial(2*n+k, k). - Peter Bala, Feb 26 2024
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