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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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).

Crossrefs

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).

Crossrefs

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.

Crossrefs

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.

Crossrefs

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

A146977 a(n) = Sum_{k=1..prime(n)} binomial(2k,k).

Original entry on oeis.org

8, 28, 350, 4706, 956384, 14061140, 3143981870, 47564380970, 11061198475550, 40328534467494098, 624021469287616610, 2338958497948612884092, 568740383597968804344752, 8885203954833615367662872, 2175457720411301277807088352, 8390179243143767727343229863076
Offset: 1

Views

Author

N. J. A. Sloane, Apr 25 2009

Keywords

Crossrefs

A subsequence of A066796.

Programs

  • Magma
    [ &+[Binomial(2*k,k): k in [1..NthPrime(n)]]: n in [1..15]]; // G. C. Greubel, Jan 09 2020
    
  • Maple
    seq( add(binomial(2*k,k), k=1..ithprime(n)), n=1..15); # G. C. Greubel, Jan 09 2020
  • Mathematica
    Table[Sum[Binomial[2k,k],{k,Prime[n]}],{n,15}] (* Harvey P. Dale, Dec 31 2012 *)
  • PARI
    vector(15, n, sum(k=1,prime(n), binomial(2*k,k)) ) \\ G. C. Greubel, Jan 09 2020
    
  • Sage
    [sum(binomial(2*k,k) for k in (1..nth_prime(n))) for n in (1..15)] # G. C. Greubel, Jan 09 2020

A066798 a(n) = Sum_{i=1..n} binomial(6*i,3*i).

Original entry on oeis.org

20, 944, 49564, 2753720, 157871240, 9233006540, 547490880980, 32795094564080, 1979734520212192, 120244316085073616, 7339672750101339356, 449852213026938118560, 27666867082225970134160
Offset: 1

Views

Author

Benoit Cloitre, Jan 18 2002

Keywords

Crossrefs

Programs

  • Maple
    s := RootOf((s+8)*s^3*x-s+1, s):
    series( (1+8/s)^(3/2)*(s-4)*s^5/(3*(s^4+8*s^3-s+1)*(s^2+4*s-8)) - 1/(1-x), x=0, 30); # Mark van Hoeij, May 02 2013
  • Mathematica
    Accumulate[Table[Binomial[6n,3n],{n,20}]] (* Harvey P. Dale, Apr 04 2020 *)
  • PARI
    { a=0; for (n=1, 100, write("b066798.txt", n, " ", a+=binomial(6*n, 3*n)) ) } \\ Harry J. Smith, Mar 28 2010

Formula

G.f.: (1+8/s)^(3/2)*(s-4)*s^5/(3*(s^4+8*s^3-s+1)*(s^2+4*s-8)) - 1/(1-x) where (s+8)*s^3*x-s+1 = 0. - Mark van Hoeij, May 02 2013
a(n) ~ sqrt(3) * 64^(n+1) / (189*sqrt(Pi*n)). - Vaclav Kotesovec, Jun 07 2019
D-finite with recurrence n*(3*n-1)*(3*n-2)*a(n) +(-585*n^3+873*n^2-370*n+40)*a(n-1) +8*(6*n-5)*(6*n-1)*(2*n-1)*a(n-2)=0. - R. J. Mathar, Jan 11 2025

A120279 a(n) = Sum[Sum[(i+j)!/i!/j!,{i,1,j}],{j,1,n}].

Original entry on oeis.org

2, 11, 45, 170, 631, 2346, 8780, 33089, 125466, 478181, 1830258, 7030557, 27088856, 104647615, 405187809, 1571990918, 6109558567, 23782190466, 92705454875, 361834392094, 1413883873953, 5530599237752, 21654401079301, 84859704298176
Offset: 1

Views

Author

Alexander Adamchuk, Jul 05 2006

Keywords

Comments

p divides a(p-1) and a(p-2) for prime p=5,11,17,23,29,41,47,53,59,71..=A007528[n] Primes of form 6n-1.
p divides a([(2p-1)/2]) for prime p=5,11,17,23,29,41,47,53,59,71..=A007528[n] Primes of form 6n-1.
p divides a((p-5)/2) for prime p=17,29,41,53,89,101.. =A040115[n] Primes of form 12n+5. Primes congruent to 5 (mod 12) excluding 5.
p divides a((p-5)/3) for prime p=11,17,23,29,41,47,53,59,71..=A007528[n] Primes of form 6n-1 excluding 5.
p divides a([(p-3)/3]) for prime p=11,17,23,29,41,47,53,59,71..=A007528[n] Primes of form 6n-1 excluding 5.

Crossrefs

Programs

  • Mathematica
    Table[Sum[Sum[(i+j)!/i!/j!,{i,1,j}],{j,1,n}],{n,1,50}]

Formula

a(n) = Sum[Sum[(i+j)!/i!/j!,{i,1,j}],{j,1,n}]. a(n) = A079309(n+1) - (n+1). a(n) = A066796(n+1)/2 - (n+1).
Recurrence: (n+1)*(3*n-2)*a(n) = 6*(3*n^2-1)*a(n-1) - 3*(9*n^2-n-2)*a(n-2) + 2*(2*n-1)*(3*n+1)*a(n-3). - Vaclav Kotesovec, Oct 19 2012
a(n) ~ 2^(2*n+3)/(3*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 19 2012
a(n) = Sum_{k=1..n} Sum_{i=1..k} C(k+i,i). - Wesley Ivan Hurt, Sep 19 2017

A122485 Values of A083097(k) such that A083097(k) = A083097(k+1) - 1.

Original entry on oeis.org

5, 14, 41, 59, 122, 140, 167, 176, 365, 383, 410, 419, 491, 500, 527, 545, 1094, 1112, 1139, 1148, 1220, 1229, 1256, 1274, 1463, 1472, 1499, 1517, 1580, 1598, 1625, 1634, 3281, 3299, 3326, 3335, 3407, 3416, 3443, 3461, 3650, 3659, 3686, 3704, 3767, 3785, 3812
Offset: 1

Views

Author

Alexander Adamchuk, Sep 15 2006

Keywords

Comments

A083097(n) = A083095(n) = A083096(n)/6 = A083094(n)/4, where A083096 are the Numbers k such that 3 divides Sum_{j=1..k} C(2*j,j) = A066796(k).
All terms are of the form 9*m + 5 and belong to A017221 with m = {0, 1, 4, 6, 13, 15, 18, 19, 40, 42, ...}.
Corresponding numbers m such that a(m) = A083097(m) are A129771 (evil odd numbers).

Examples

			A083097 begins {0, 2, 5, 6, 14, 15, 18, 20, 41, 42, 45, 47, 54, 56, 59, 60, ...}.
So a(1) = 5 because 5 = A083097(3) = A083097(3+1) - 1.
a(2) = 14 because 14 = A083097(5) = A083097(5+1) - 1.
		

Crossrefs

Formula

a(n) = A083097(A129771(n)).

Extensions

More terms from R. J. Mathar, Jan 17 2008
More terms from Jinyuan Wang, Jan 22 2022
Edited by Michel Marcus, Jan 22 2022

A147304 a(n) = Sum_{k=1..prime(n)^2-1} binomial(2k,k).

Original entry on oeis.org

28, 17576, 43308802158650, 8610524734277600186228691452, 121374542758943982922417964798154019940274699584207321286055873543631298, 8126392396649531937838689708830356413772063825711016912849229977138431439363305375418692100492504264
Offset: 1

Views

Author

N. J. A. Sloane, Apr 25 2009

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[2k,k],{k,Prime[n]^2-1}],{n,7}] (* Harvey P. Dale, Dec 26 2014 *)
  • PARI
    a(n) = sum(k=1, prime(n)^2-1, binomial(2*k,k)); \\ Michel Marcus, Jul 07 2018

A054114 T(2n+1,n), array T as in A054110.

Original entry on oeis.org

1, 8, 28, 98, 350, 1274, 4706, 17576, 66196, 250952, 956384, 3660540, 14061140, 54177740, 209295260, 810375650, 3143981870, 12219117170, 47564380970, 185410909790, 723668784230, 2827767747950, 11061198475550, 43308802158650, 169719408596402, 665637941544506
Offset: 0

Views

Author

Keywords

Comments

Apart from the initial term, identical to A066796(n+1).

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(-1+3*x+5*x^2-4*x^3+Sqrt[1-4*x])/(x*(1-x)(1-4*x)), {x,0,30}],x] (* Georg Fischer, Apr 09 2020 *)

Formula

G.f.: (-1+3*x+5*x^2-4*x^3+sqrt(1-4*x))/(x*(1-x)*(1-4*x)). - Ralf Stephan, Apr 03 2004; corrected by Georg Fischer, Apr 09 2020
For n>0, a(n) = sum(k=1, n+1, C(2k, k)). - Ralf Stephan, Apr 03 2004

Extensions

Corrected by Franklin T. Adams-Watters, Oct 25 2006
a(23)-a(25) corrected by Georg Fischer, Apr 09 2020
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