A167713 -id:A167713 - cp's OEIS frontend

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

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

Offset: 0

Alexander Adamchuk, Nov 13 2009

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.

Vincenzo Librandi, Table of n, a(n) for n = 0..300

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

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

Table[4^n*Sum[Binomial[2*k,k]^2/4^k,{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 20 2012 *)

a(n) = 4^n*sum(k=0,n, binomial(2*k,k)^2/4^k) \\ Charles R Greathouse IV, Sep 21 2016

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

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

Alexander Adamchuk, Nov 14 2009

nonn

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).

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A079727, A167867, A167868, A167869, A167870, A167872.

Cf. A000984, A066796, A006134, A082590, A132310, A002457, A144635, A167713, A167859.

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

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

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

Alexander Adamchuk, Nov 14 2009

nonn

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).

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A079727, A167867, A167868, A167869, A167870, A167872.

Cf. A000984, A066796, A006134, A082590, A132310, A002457, A144635, A167713, A167859.

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

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

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

Alexander Adamchuk, Nov 14 2009

nonn

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).

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A079727, A167867, A167868, A167869, A167870, A167872.

Cf. A000984, A066796, A006134, A082590, A132310, A002457, A144635, A167713, A167859.

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

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

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

Alexander Adamchuk, Nov 14 2009

nonn

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).

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A079727, A167867, A167868, A167869, A167870, A167872.

Cf. A000984, A066796, A006134, A082590, A132310, A002457, A144635, A167713, A167859.

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

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

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

Alexander Adamchuk, Nov 14 2009

nonn

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.

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

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A079727, A167867, A167868, A167869, A167870, A167872.

Cf. A000984, A066796, A006134, A082590, A132310, A002457, A144635, A167713, A167859, A002145.

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

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

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

Alexander Adamchuk, Nov 13 2009

nonn

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.

R. J. Mathar, Table of n, a(n) for n = 1..57

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

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

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

More terms from Jinyuan Wang, Jul 24 2022

cp's OEIS Frontend

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

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

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

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

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

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

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

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