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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A006134 a(n) = Sum_{k=0..n} binomial(2*k,k).

Original entry on oeis.org

1, 3, 9, 29, 99, 351, 1275, 4707, 17577, 66197, 250953, 956385, 3660541, 14061141, 54177741, 209295261, 810375651, 3143981871, 12219117171, 47564380971, 185410909791, 723668784231, 2827767747951, 11061198475551, 43308802158651, 169719408596403, 665637941544507
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

The expression a(n) = B^n*Sum_{ k=0..n } binomial(2*k,k)/B^k gives A006134 for B=1, A082590 (B=2), A132310 (B=3), A002457 (B=4), A144635 (B=5). - N. J. A. Sloane, Jan 21 2009
T(n+1,1) from table A045912 of characteristic polynomial of negative Pascal matrix. - Michael Somos, Jul 24 2002
p divides a((p-3)/2) for p=11, 13, 23, 37, 47, 59, 61, 71, 73, 83, 97, 107, 109, 131, 157, 167, ...: A097933. Also primes congruent to {1, 2, 3, 11} mod 12 or primes p such that 3 is a square mod p (excluding 2 and 3) A038874. - Alexander Adamchuk, Jul 05 2006
Partial sums of the even central binomial coefficients. For p prime >=5, a(p-1) = 1 or -1 (mod p) according as p = 1 or -1 (mod 3) (see Pan and Sun link). - David Callan, Nov 29 2007
First column of triangle A187887. - Michel Marcus, Jun 23 2013
From Gus Wiseman, Apr 20 2023: (Start)
Also the number of nonempty subsets of {1,...,2n+1} with median n+1, where the median of a multiset is either the middle part (for odd length), or the average of the two middle parts (for even length). The odd/even-length cases are A000984 and A006134(n-1). For example, the a(0) = 1 through a(2) = 9 subsets are:
{1} {2} {3}
{1,3} {1,5}
{1,2,3} {2,4}
{1,3,4}
{1,3,5}
{2,3,4}
{2,3,5}
{1,2,4,5}
{1,2,3,4,5}
Alternatively, a(n-1) is the number of nonempty subsets of {1,...,2n-1} with median n.
(End)

Examples

			1 + 3*x + 9*x^2 + 29*x^3 + 99*x^4 + 351*x^5 + 1275*x^6 + 4707*x^7 + 17577*x^8 + ...

References

  • Marko Petkovsek, Herbert Wilf and Doron Zeilberger, A=B, A K Peters, 1996, p. 22.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Cf. A000984 (first differences), A097933, A038874, A132310.
Cf. A006135, A006136, A045912.
Equals A066796 + 1.
Odd bisection of A100066.
Row sums of A361654 (also column k = 2).
A007318 counts subsets by length, A231147 by median, A013580 by integer median.
A359893 and A359901 count partitions by median.
Cf. A000975, A024718, A057552, A079309, A327481.

Programs

  • MATLAB
    n=10; x=pascal(n); trace(x)
  • Magma
    &cat[ [&+[ Binomial(2*k, k): k in [0..n]]]: n in [0..30]]; // Vincenzo Librandi, Aug 13 2015
  • Maple
    A006134 := proc(n) sum(binomial(2*k,k),k=0..n); end;
a := n -> -binomial(2*(n+1),n+1)*hypergeom([1,n+3/2],[n+2], 4) - I/sqrt(3):
seq(simplify(a(n)), n=0..24); # Peter Luschny, Oct 29 2015
# third program:
A006134 := series(exp(2*x)*BesselI(0, 2*x) + exp(x)*int(BesselI(0, 2*x)*exp(x), x), x = 0, 25):
seq(n!*coeff(A006134, x, n), n=0..24); # Mélika Tebni, Feb 27 2024
  • Mathematica
    Table[Sum[((2k)!/(k!)^2),{k,0,n}], {n,0,50}] (* Alexander Adamchuk, Jul 05 2006 *)
a[ n_] := (4/3) Binomial[ 2 n, n] Hypergeometric2F1[ 1/2, 1, -n + 1/2, -1/3] (* Michael Somos, Jun 20 2012 *)
Accumulate[Table[Binomial[2n,n],{n,0,30}]] (* Harvey P. Dale, Jan 11 2015 *)
CoefficientList[Series[1/((1 - x) Sqrt[1 - 4 x]), {x, 0, 33}], x] (* Vincenzo Librandi, Aug 13 2015 *)
  • Maxima
    makelist(sum(binomial(2*k,k),k,0,n),n,0,12); /* Emanuele Munarini, Mar 15 2011 */
  • PARI
    {a(n) = if( n<0, 0, polcoeff( charpoly( matrix( n+1, n+1, i, j, -binomial( i+j-2, i-1))), 1))} \\ Michael Somos, Jul 10 2002
  • PARI
    {a(n)=binomial(2*n,n)*sum(k=0,2*n,(-1)^k*polcoeff((1+x+x^2)^n,k)/binomial(2*n,k))} \\ Paul D. Hanna, Aug 21 2007
  • PARI
    my(x='x+O('x^100)); Vec(1/((1-x)*sqrt(1-4*x))) \\ Altug Alkan, Oct 29 2015

Formula

From Alexander Adamchuk, Jul 05 2006: (Start)
a(n) = Sum_{k=0..n} (2k)!/(k!)^2.
a(n) = A066796(n) + 1, n>0. (End)
G.f.: 1/((1-x)*sqrt(1-4*x)).
D-finite with recurrence: (n+2)*a(n+2) - (5*n+8)*a(n+1) + 2*(2*n+3)*a(n) = 0. - Emanuele Munarini, Mar 15 2011
a(n) = C(2n,n) * Sum_{k=0..2n} (-1)^k*trinomial(n,k)/C(2n,k) where trinomial(n,k) = [x^k] (1 + x + x^2)^n. E.g. a(2) = C(4,2)*(1/1 - 2/4 + 3/6 - 2/4 + 1/1) = 6*(3/2) = 9 ; a(3) = C(6,3)*(1/1 - 3/6 + 6/15 - 7/20 + 6/15 - 3/6 + 1/1) = 20*(29/20) = 29. - Paul D. Hanna, Aug 21 2007
From Alzhekeyev Ascar M, Jan 19 2012: (Start)
a(n) = Sum_{ k=0..n } b(k)*binomial(n+k,k), where b(k)=0 for n-k == 2 (mod 3), b(k)=1 for n-k == 0 or 1 (mod 6), and b(k)=-1 for n-k== 3 or 4 (mod 6).
a(n) = Sum_{ k=0..n-1 } c(k)*binomial(2n,k) + binomial(2n,n), where c(k)=0 for n-k == 0 (mod 3), c(k)=1 for n-k== 1 (mod 3), and c(k)=-1 for n-k==2 (mod 3). (End)
a(n) ~ 2^(2*n+2)/(3*sqrt(Pi*n)). - Vaclav Kotesovec, Nov 06 2012
G.f.: G(0)/2/(1-x), where G(k)= 1 + 1/(1 - 2*x*(2*k+1)/(2*x*(2*k+1) + (k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 24 2013
G.f.: G(0)/(1-x), where G(k)= 1 + 4*x*(4*k+1)/( (4*k+2) - x*(4*k+2)*(4*k+3)/(x*(4*k+3) + (k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 26 2013
a(n) = Sum_{k = 0..n} binomial(n+1,k+1)*A002426(k). - Peter Bala, Oct 29 2015
a(n) = -binomial(2*(n+1),n+1)*hypergeom([1,n+3/2],[n+2], 4) - i/sqrt(3). - Peter Luschny, Oct 29 2015
a(n) = binomial(2*n, n)*hypergeom([1,-n], [1/2-n], 1/4). - Peter Luschny, Mar 16 2016
From Gus Wiseman, Apr 20 2023: (Start)
a(n+1) - a(n) = A000984(n).
a(n) = A013580(2n+1,n+1) (conjectured).
a(n) = 2*A024718(n) - 1.
a(n) = A100066(2n+1).
a(n) = A231147(2n+1,n+1) (conjectured). (End)
a(n) = Sum_{k=0..floor(n/3)} 3^(n-3*k) * binomial(n-k,2*k) * binomial(2*k,k) (Sawhney, 2017). - Amiram Eldar, Feb 24 2024
From Mélika Tebni, Feb 27 2024: (Start)
Limit_{n -> oo} a(n) / A281593(n) = 2.
E.g.f.: exp(2*x)*BesselI(0,2*x) + exp(x)*integral( BesselI(0,2*x)*exp(x) ) dx. (End)
a(n) = [(x*y)^n] 1/((1 - (x + y))*(1 - x*y)). - Stefano Spezia, Feb 16 2025
a(n) = Sum_{k=0..floor(n/2)} (-1)^k*binomial(2*n+1-k, n-2*k). - Michael Weselcouch, Jun 17 2025
a(n) = binomial(1+2*n, n)*hypergeom([1, (1-n)/2, -n/2], [-1-2*n, 2+n], 4). - Stefano Spezia, Jun 18 2025

Extensions

Simpler definition from Alexander Adamchuk, Jul 05 2006

A167713 a(n) = 16^n * Sum_{k=0..n} binomial(2*k, k) / 16^k.

Original entry on oeis.org

1, 18, 294, 4724, 75654, 1210716, 19372380, 309961512, 4959397062, 79350401612, 1269606610548, 20313706474200, 325019306291356, 5200308911062296, 83204942617113336, 1331279082028930896, 21300465313063974726, 340807445011357201836, 5452919120190790364676
Offset: 0

Views

Author

Alexander Adamchuk, Nov 10 2009

Keywords

Comments

p^2 divides a((p-3)/2) for prime p of the form p = 6k + 1 (A002476).
The expression a(n) = B^n*Sum_{k=0..n} binomial(2*k,k)/B^k gives A006134 for B=1, A082590 (B=2), A132310 (B=3), A002457 (B=4), A144635 (B=5), a(n) = A167713 (B=16).

Links

Crossrefs

Cf. A000984, A002476, A066796, A006134, A082590, A132310, A002457, A144635.

Programs

  • Maple
    A167713 := proc(n) coeftayl( 1/(1-16*x)/sqrt(1-4*x),x=0,n) ; end proc: seq(A167713(n),n=0..40) ; # R. J. Mathar, Nov 13 2009
  • Mathematica
    Table[ 16^n * Sum[ (2k)!/(k!)^2 / 16^k, {k,0,n} ], {n,0,20} ]
CoefficientList[Series[1 / ((1 - 16 x) Sqrt[1 - 4 x]), {x, 0, 20}], x] (* Vincenzo Librandi, May 27 2013 *)

Formula

a(n) = 16^n * Sum_{k=0..n} ((2k)!/(k!)^2) / 16^k.
a(n) = 16^n * Sum_{k=0..n} binomial(2k,k) / 16^k.
G.f.: 1/((1-16*x)*sqrt(1-4*x)). - R. J. Mathar, Nov 13 2009
From Vaclav Kotesovec, Oct 20 2012: (Start)
Recurrence: n*a(n) = 2*(10*n-1)*a(n-1) - 32*(2*n-1)*a(n-2).
a(n) ~ 2^(4*n+1)/sqrt(3). (End)

Extensions

Extended by R. J. Mathar, Nov 13 2009

A100095 An inverse Chebyshev transform of the Fibonacci numbers.

Original entry on oeis.org

0, 1, 1, 5, 7, 25, 41, 125, 225, 625, 1195, 3125, 6227, 15625, 32059, 78125, 163727, 390625, 831505, 1953125, 4206145, 9765625, 21215481, 48828125, 106782837, 244140625, 536618341, 1220703125, 2693492305, 6103515625, 13507578125
Offset: 0

Views

Author

Paul Barry, Nov 03 2004

Keywords

Comments

Image of x/(1-x-x^2) under the transform g(x)->(1/sqrt(1-4*x^2))*g(x*c(x^2)), where c(x) is the g.f. of the Catalan numbers A000108. This is the inverse of the Chebyshev transform which takes A(x) to ((1-x^2)/(1+x^2))*A(x/(1+x^2)).
Hankel transform is (-1)^n*(2^n-0^n)/2. Hankel transform of a(n+1) is A141125. - Paul Barry, Jun 05 2008
Basically A000351 interleaved with A144635. - Peter Luschny, May 31 2014

Links

Crossrefs

Cf. A000351, A144635, A000045, A100096, A100097.

Programs

  • Mathematica
    CoefficientList[Series[(x^2*Sqrt[1-4*x^2]+x*(1-4*x^2))/((1-4*x^2)*(1-5*x^2)), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 12 2014 *)
  • Maxima
    a(n):=sum(4^(j)*binomial((n-1)/2,j),j,0,(n-1)/2); /* Vladimir Kruchinin, May 31 2014 */

Formula

G.f.: (x^2*sqrt(1-4*x^2)+x*(1-4*x^2))/((1-4*x^2)*(1-5*x^2)).
a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)*Fibonacci(n-2*k).
Conjecture: (-n+2)*a(n) +(-n+3)*a(n-1) +(9*n-22)*a(n-2) +(9*n-31)*a(n-3) +20*(-n+3)*a(n-4) +20*(-n+4)*a(n-5)=0. - R. J. Mathar, Nov 24 2012
Recurrence: (n-2)*a(n) = (9*n-22)*a(n-2) - 20*(n-3)*a(n-4). - Vaclav Kotesovec, Feb 12 2014
a(n) ~ 5^((n-1)/2). - Vaclav Kotesovec, Feb 12 2014
a(n) = sum(j=0..(n-1)/2, 4^(j)*binomial((n-1)/2,j)). - Vladimir Kruchinin, May 31 2014
a(2*n) = 5^n/sqrt(5) - 2^n * (2*n-1)!! * hypergeom([1, n+1/2], [n+1], 4/5)/(5*n!), a(2*n+1) = 5^n. - Vladimir Reshetnikov, Oct 13 2016

A167859 a(n) = 4^n * Sum_{k=0..n} binomial(2*k, k)^2 / 4^k.

Original entry on oeis.org

1, 8, 68, 672, 7588, 93856, 1229200, 16695424, 232418596, 3293578784, 47309094672, 686870685312, 10059942413584, 148412250014336, 2202990595617344, 32873407393419776, 492791264816231204
Offset: 0

Views

Author

Alexander Adamchuk, Nov 13 2009

Keywords

Comments

Every a(n) from a((p-1)/2) to a(p-1) is divisible by prime p for p = {7, 47, 191, 383, 439, 1151, 1399, 2351, 2879, 3119, 3511, 3559, ...} = A167860, apparently a subset of primes of the form 8n+7 (A007522).
7^3 divides a(13) and 7^2 divides a(10)-a(13).
Every a(n) from a(kp-1 - (p-1)/2) to a(kp-1) is divisible by prime p from A167860.
Every a(n) from a((p^2-1)/2) to a(p^2-1) is divisible by prime p from A167860. For p=7 every a(n) from a((p^3-1)/2) to a(p^3-1) and from a((p^4-1)/2) to a(p^4-1)is divisible by p^2.

Links

Crossrefs

Cf. A000984, A066796, A006134, A082590, A132310, A002457, A144635, A167713, A167860, A007522.

Programs

  • Maple
    A167859 := proc(n)
    add( (binomial(2*k,k)/2^k)^2,k=0..n) ;
    4^n*% ;
end proc:
seq(A167859(n),n=0..20) ; # R. J. Mathar, Sep 21 2016
  • Mathematica
    Table[4^n*Sum[Binomial[2*k,k]^2/4^k,{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 20 2012 *)
  • PARI
    a(n) = 4^n*sum(k=0,n, binomial(2*k,k)^2/4^k) \\ Charles R Greathouse IV, Sep 21 2016

Formula

Recurrence: n^2*a(n) = 4*(5*n^2 - 4*n + 1)*a(n-1) - 16*(2*n - 1)^2*a(n-2). - Vaclav Kotesovec, Oct 20 2012
a(n) ~ 2^(4*n+2)/(3*Pi*n). - Vaclav Kotesovec, Oct 20 2012
G.f.: 2*EllipticK(4*sqrt(x))/(Pi*(1-4*x)), where EllipticK is the complete elliptic integral of the first kind, using the Gradshteyn and Ryzhik convention, also used by Maple. In the convention of Abramowitz and Stegun, used by Mathematica, this would be written as 2*K(16*x)/(Pi*(1-4*x)). - Robert Israel, Sep 21 2016

Extensions

More terms from Sean A. Irvine, Apr 14 2010
Further terms from Jon E. Schoenfield, May 09 2010

A167867 a(n) = 2^n * Sum_{k=0..n} binomial(2*k,k)^3 / 2^k.

Original entry on oeis.org

1, 10, 236, 8472, 359944, 16722896, 822334816, 42068907200, 2215884717400, 119364801362800, 6545334930678816, 364137834051739200, 20502307365808906816, 1166063313963833813632, 66893439680369963627264
Offset: 0

Views

Author

Alexander Adamchuk, Nov 14 2009

Keywords

Comments

The expression a(n) = B^n*Sum_{ k=0..n } binomial(2*k,k)/B^k gives A006134 for B=1, A082590 (B=2), A132310 (B=3), A002457 (B=4), A144635 (B=5), A167713 (B=16).
The expression a(n) = B^n*Sum_{ k=0..n } binomial(2*k,k)^3/B^k gives A079727 for B=1, A167867 (B=2), A167868 (B=3), A167869 (B=4), A167870 (B=16), A167871 (B=64).

Links

Crossrefs

Cf. A079727, A167867, A167868, A167869, A167870, A167872.
Cf. A000984, A066796, A006134, A082590, A132310, A002457, A144635, A167713, A167859.

Programs

  • Mathematica
    Table[2^n Sum[Binomial[2k,k]^3/2^k,{k,0,n}],{n,0,30}] (* Vincenzo Librandi, Mar 26 2012 *)

Formula

a(n) = 2^n * Sum_{k=0..n} binomial(2*k,k)^3 / 2^k.
Recurrence: n^3*a(n) = 2*(33*n^3 - 48*n^2 + 24*n - 4)*a(n-1) - 16*(2*n-1)^3*a(n-2). - Vaclav Kotesovec, Aug 13 2013
a(n) ~ 2^(6*n+5)/(31*(Pi*n)^(3/2)). - Vaclav Kotesovec, Aug 13 2013

Extensions

More terms from Sean A. Irvine, Apr 27 2010

A167868 a(n) = 3^n * Sum_{k=0..n} binomial(2*k,k)^3 / 3^k.

Original entry on oeis.org

1, 11, 249, 8747, 369241, 17110731, 840221217, 42944901219, 2260581606657, 121714776747971, 6671749658197129, 371062413164972955, 20887218937200347281, 1187720356043817041843, 68124474120573747125529
Offset: 0

Views

Author

Alexander Adamchuk, Nov 14 2009

Keywords

Comments

The expression a(n) = B^n*Sum_{k=0..n} binomial(2*k,k)/B^k gives A006134 for B=1, A082590 (B=2), A132310 (B=3), A002457 (B=4), A144635 (B=5), A167713 (B=16).
The expression a(n) = B^n*Sum_{k=0..n} binomial(2*k,k)^3/B^k gives A079727 for B=1, A167867 (B=2), A167868 (B=3), A167869 (B=4), A167870 (B=16), A167871 (B=64).

Links

Crossrefs

Cf. A079727, A167867, A167868, A167869, A167870, A167872.
Cf. A000984, A066796, A006134, A082590, A132310, A002457, A144635, A167713, A167859.

Programs

  • Mathematica
    Table[3^n Sum[Binomial[2k,k]^3/3^k,{k,0,n}],{n,0,20}] (* Vincenzo Librandi, Mar 26 2012 *)

Formula

a(n) = 3^n * Sum_{k=0..n} binomial(2*k,k)^3 / 3^k.
Recurrence: n^3*a(n) = (67*n^3 - 96*n^2 + 48*n - 8)*a(n-1) - 24*(2*n-1)^3*a(n-2). - Vaclav Kotesovec, Aug 13 2013
a(n) ~ 2^(6*n+6)/(61*(Pi*n)^(3/2)). - Vaclav Kotesovec, Aug 13 2013

Extensions

More terms from Sean A. Irvine, Apr 27 2010

A167869 a(n) = 4^n * Sum_{k=0..n} binomial(2*k,k)^3 / 4^k.

Original entry on oeis.org

1, 12, 264, 9056, 379224, 17519904, 858968640, 43860112128, 2307187351512, 124161781334048, 6803252453289408, 378260174003539200, 21287072393719585216, 1210206988807094340864, 69402141007670673363456
Offset: 0

Views

Author

Alexander Adamchuk, Nov 14 2009

Keywords

Comments

The expression a(n) = B^n*Sum_{k=0..n} binomial(2*k,k)/B^k gives A006134 for B=1, A082590 (B=2), A132310 (B=3), A002457 (B=4), A144635 (B=5), A167713 (B=16).
The expression a(n) = B^n*Sum_{k=0..n} binomial(2*k,k)^3/B^k gives A079727 for B=1, A167867 (B=2), A167868 (B=3), A167869 (B=4), A167870 (B=16), A167871 (B=64).

Links

Crossrefs

Cf. A079727, A167867, A167868, A167869, A167870, A167872.
Cf. A000984, A066796, A006134, A082590, A132310, A002457, A144635, A167713, A167859.

Programs

  • Mathematica
    Table[4^n Sum[Binomial[2k,k]^3/4^k,{k,0,n}],{n,0,30}] (* Vincenzo Librandi, Mar 26 2012 *)

Formula

a(n) = 4^n * Sum_{k=0..n} binomial(2*k,k)^3 / 4^k.
Recurrence: n^3*a(n) = 4*(17*n^3 - 24*n^2 + 12*n - 2)*a(n-1) - 32*(2*n-1)^3*a(n-2). - Vaclav Kotesovec, Aug 13 2013
a(n) ~ 2^(6*n+4)/(15*(Pi*n)^(3/2)). - Vaclav Kotesovec, Aug 13 2013

Extensions

More terms from Sean A. Irvine, Apr 27 2010

A167870 a(n) = 16^n * Sum_{k=0..n} binomial(2*k,k)^3 / 16^k.

Original entry on oeis.org

1, 24, 600, 17600, 624600, 25996608, 1204834752, 59701593600, 3086972400600, 164324590337600, 8935798773354816, 494019944564058624, 27678350810730366400, 1567912312203901862400, 89647910047704725798400
Offset: 0

Views

Author

Alexander Adamchuk, Nov 14 2009

Keywords

Comments

The expression a(n) = B^n*Sum_{k=0..n} binomial(2*k,k)/B^k gives A006134 for B=1, A082590 (B=2), A132310 (B=3), A002457 (B=4), A144635 (B=5), A167713 (B=16).
The expression a(n) = B^n*Sum_{k=0..n} binomial(2*k,k)^3/B^k gives A079727 for B=1, A167867 (B=2), A167868 (B=3), A167869 (B=4), A167870 (B=16), A167871 (B=64).

Links

Crossrefs

Cf. A079727, A167867, A167868, A167869, A167870, A167872.
Cf. A000984, A066796, A006134, A082590, A132310, A002457, A144635, A167713, A167859.

Programs

  • Mathematica
    Table[16^n Sum[Binomial[2k,k]^3/16^k,{k,0,n}],{n,0,20}] (* Harvey P. Dale, Jan 21 2012 *)

Formula

a(n) = 16^n * Sum_{k=0..n} binomial(2*k,k)^3 / 16^k.
Recurrence: n^3*a(n) = 8*(10*n^3 - 12*n^2 + 6*n - 1)*a(n-1) - 128*(2*n-1)^3*a(n-2). - Vaclav Kotesovec, Aug 13 2013
a(n) ~ 2^(6*n+2)/(3*(Pi*n)^(3/2)). - Vaclav Kotesovec, Aug 13 2013

Extensions

More terms from Sean A. Irvine, Apr 27 2010

A167871 a(n) = 64^n * Sum_{k=0..n} binomial(2*k,k)^3 / 64^k.

Original entry on oeis.org

1, 72, 4824, 316736, 20614104, 1335305664, 86248451520, 5560325134848, 357992555533272, 23026456586057408, 1479999826835627328, 95071036081670530560, 6104320340924619384256, 391801560518407856592384
Offset: 0

Views

Author

Alexander Adamchuk, Nov 14 2009

Keywords

Comments

The expression a(n) = B^n*Sum_{k=0..n} binomial(2*k,k)/B^k gives A006134 for B=1, A082590 (B=2), A132310 (B=3), A002457 (B=4), A144635 (B=5), A167713 (B=16).
The expression a(n) = B^n*Sum_{k=0..n} binomial(2*k,k)^3/B^k gives A079727 for B=1, A167867 (B=2), A167868 (B=3), A167869 (B=4), A167870 (B=16), A167871 (B=64).
p^2 divides all a(n) from n = (p-1)/2 to n = p-1 for prime p of the form p = 4k+3, p = {3,7,11,19,23,31,43,47,59,...} = A002145.
p^2 divides all a(n) from n = (2p-1 - (p-1)/2) to n = 2p-1 for prime p of the form p = 4k+3.
p^2 divides all a(n) from n = (3p-1 - (p-1)/2) to n = 3p-1 for prime p of the form p = 4k+3.
p^2 divides all a(n) from n = (p^2-1)/2 to n = p^2-1 for prime p of the form p = 4k+3.

References

  • W. Feller, An Introduction to Probability Theory and Its Applications, Vol. 1, 3rd ed., Wiley, 1968, p. 361.

Links

Crossrefs

Cf. A079727, A167867, A167868, A167869, A167870, A167872.
Cf. A000984, A066796, A006134, A082590, A132310, A002457, A144635, A167713, A167859, A002145.

Programs

  • Mathematica
    Table[64^n Sum[Binomial[2k,k]^3/64^k,{k,0,n}],{n,0,20}] (* Vincenzo Librandi, Mar 26 2012 *)

Formula

a(n) = 64^n * Sum_{k=0..n} binomial(2*k,k)^3 / 64^k.
Recurrence: n^3*a(n) = 8*(4*n-1)*(4*n^2 - 2*n + 1)*a(n-1) - 512*(2*n-1)^3 *a(n-2). - Vaclav Kotesovec, Aug 14 2013
a(n) ~ 64^n*(Pi/GAMMA(3/4)^4 - 2/(Pi^(3/2)*sqrt(n))). - Vaclav Kotesovec, Aug 14 2013

Extensions

More terms from Sean A. Irvine, Apr 25 2010

A167860 Primes p dividing every A167859(m) from m=(p-1)/2 to m=(p-1).

Original entry on oeis.org

7, 47, 191, 383, 439, 1151, 1399, 2351, 2879, 3119, 3511, 3559, 4127, 5087, 5431, 6911, 8887, 9127, 9791, 9887, 12391, 13151, 14407, 15551, 16607, 19543, 20399, 21031, 21319, 21839, 23039, 25391, 26399, 28087, 28463, 28711, 29287, 33223, 39551, 43103, 44879, 46271
Offset: 1

Views

Author

Alexander Adamchuk, Nov 13 2009

Keywords

Comments

Apparently A167860 is a subset of primes of the form 8*k + 7 (A007522).
Every A167859(m) from m=(p-1)/2 to m=(p-1) is divisible by prime p belonging to A167860.
7^3 divides A167859(13) and 7^2 divides A167859(10)-A167859(13).
Every A167859(m) from m=(kp-1 - (p-1)/2) to m=(kp-1) is divisible by prime p from A167860.
Every A167859(m) from m=((p^2-1)/2) to m=(p^2-1) is divisible by prime p from A167860. For p=7 every A167859(m) from m=((p^3-1)/2) to m=(p^3-1) and from m=((p^4-1)/2) to m(p^4-1)is divisible by p^2.

Links

Crossrefs

Cf. A000984, A006134, A007522, A066796, A082590, A132310, A144635, A167713, A167859, A276649.

Programs

  • Maple
    A167859 := proc(n)
    option remember;
    if n <= 1 then
        add( (binomial(2*k, k)/2^k)^2, k=0..n) ;
        4^n*% ;
    else
        4*(5*n^2 - 4*n + 1)*procname(n-1) - 16*(2*n - 1)^2*procname(n-2) ;
        %/n^2 ;
    end if;
end proc:
isA167860 := proc(p)
    local m ;
    for m from (p-1)/2 to p-1 do
        if modp(A167859(m),p) > 0 then
            return false;
        end if;
    end do:
    true ;
end proc:
A167860 := proc(n)
    option remember ;
    if n = 0 then
        2;
    else
        p := nextprime(procname(n-1)) ;
        while not isA167860(p) do
            p := nextprime(p) ;
        end do ;
        return p;
    end if;
end proc:
seq(A167860(n),n=1..10) ; # R. J. Mathar, Jan 22 2025
  • PARI
    is(p) = if(isprime(p)&&p%2, my(m=Mod(1, p), s=m); for(k=1, p\2, s+=(m*=(2*k-1)/k)^2); !s, 0); \\ Jinyuan Wang, Jul 24 2022

Extensions

More terms from Jinyuan Wang, Jul 24 2022
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