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 1-4 of 4 results.

A246542 a(n) = 2/n^2*( sum_{k=0..n-1} (2*k+1)*C(n-1,k)^2*C(n+k,k) ), where C(m,k) denotes the binomial coefficient m!/(k!*(m-k)!).

Original entry on oeis.org

2, 5, 22, 132, 918, 6981, 56390, 475796, 4149286, 37133043, 339307098, 3154030050, 29741815998, 283896719073, 2738445478758, 26656533873204, 261561469613190, 2584718580416919, 25703179602581234, 257046296680889600, 2583719988283365322, 26090463844931102715, 264570229302222957162, 2693187696469413499902, 27511970457139362253598
Offset: 1

Views

Author

Zhi-Wei Sun, Aug 29 2014

Keywords

Comments

The following general conjecture implies that a(n) is always an integer.
Conjecture: Let a and b be nonnegative integers, and let n be a positive integer. If a == b (mod 2), then sum_{k=0}^{n-1}(2k+1)*C(n-1,k)^a*C(-n-1,k)^b == 0 (mod n^2); if a - b is odd, then 2*sum_{k=0}^{n-1}(-1)^k*(2k+1)C(n-1,k)^a*C(-n-1,k)^b == 0 (mod n^2).
An extension of the conjecture was proved in the latest version of arXiv:1408.5381. - Zhi-Wei Sun, Sep 14 2014

Examples

			a(2) = 5 since 2/2^2*( sum_{k=0,1} (2k+1)*C(1,k)^2*C(2+k,k) ) = 1/2*(1+3*3) = 5.

Links

Crossrefs

Cf. A180364, A246460, A246511, A246543.

Programs

  • Mathematica
    a[n_] := Sum[(2 k + 1)Binomial[n - 1, k]^2 Binomial[n + k, k], {k, 0, n - 1}] 2/n^2
Table[a[n], {n, 1, 25}]

Formula

Recurrence (obtained via the Zeilberger algorithm):
-n*(n-2)*(20*n^2+50*n+29)*a(n) - (220*n^4+550*n^3+369*n^2+59*n-10)*a(n+1)
+ (n+2)^2*(20*n^2+10*n-1)*a(n+2) = 0.

A246512 a(n) = (sum_{k=0}^{n-1}(3k^2+3k+1)*C(n-1,k)^2*C(n+k,k)^2)/n^3, where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!).

Original entry on oeis.org

1, 8, 87, 1334, 25045, 529080, 12076435, 291307490, 7325385345, 190294925864, 5074233846583, 138240914882394, 3834434331534781, 107990908896551192, 3081524055740420811, 88938694296657330170, 2592715751635344852505, 76252823735941187830920, 2260342454730542009915455, 67476975730679069406101870
Offset: 1

Views

Author

Zhi-Wei Sun, Aug 28 2014

Keywords

Comments

In the latest version of arXiv:1408:5381, the author proved that a(n) is always an integer. Notice that a(65) is relatively prime to 65. - Zhi-Wei Sun, Sep 14 2014
Conjecture: The sequence a(n+1)/a(n) (n = 1,2,3,...) is strictly increasing to the limit 17+12*sqrt(2), and the sequence a(n+1)^(1/(n+1))/a(n)^(1/n) (n = 1,2,3,...) is strictly decreasing to the limit 1.
Note that sum_{k=0}^{n-1}(2k+1)*A(k) = n^5*a(n) for all n > 0, where A(n) = sum_{k=0..n}C(n,k)^2*C(n+k,k)^2*(6k^3+9k^2+5k+1) for n = 0,1,2,....

Examples

			a(2) = 8 since sum_{k=0,1} (3k^2+3k+1)C(1,k)^2*C(2+k,k)^2 = 1 + 7*3^2 = 64 = 2^3*8.

Links

Crossrefs

Cf. A020639, A246065, A246138, A246459, A246460, A246461, A246462, A246511.

Programs

  • Mathematica
    a[n_]:=Sum[(3k^2+3k+1)*(Binomial[n-1,k]Binomial[n+k,k])^2,{k,0,n-1}]/(n^3)
Table[a[n],{n,1,20}]
  • PARI
    a(n) = sum(k=0, n-1, (3*k^2+3*k+1)*binomial(n-1,k)^2*binomial(n+k,k)^2) /n^3; \\ Michel Marcus, Dec 24 2021

Formula

Recurrence (obtained via the Zeilberger algorithm):
n^3*(n + 1)*(2n + 5)*(3n^2 + 12n + 11)*(6n^2 + 24n + 25)*a(n) - (n + 1)*(2n + 5)*(630n^7 + 6552n^6 + 28137n^5 + 64134n^4 + 82777n^3 + 59512n^2 + 21646n + 3076)*a(n+1) + (n + 2)*(2n + 1)*(630n^7 + 6678n^6 + 29271n^5 + 68751n^4 + 93469n^3 + 73445n^2 + 30640n + 5072)*a(n+2) - (n + 2)*(n + 3)^3*(2n + 1)*(3n^2 + 6n + 2)*(6n^2 + 12n + 7)*a(n+3) = 0.

A246567 a(n) = (sum_{k=0}^{n-1}C(n-1,k)^2*C(-n-1,k)^2/(4*k^2-1))/n, where C(x,k) refers to binomial(x,k).

Original entry on oeis.org

-1, 1, 9, 61, 587, 7575, 117485, 2057365, 39314175, 802816213, 17275712297, 387886408443, 9020881956707, 216101556811603, 5309497149531957, 133334756362738885, 3412887111988377575, 88838285028658754625, 2347236720247792005665, 62849602943515066525633
Offset: 1

Views

Author

Zhi-Wei Sun, Aug 30 2014

Keywords

Comments

The following conjecture implies that a(n) is an integer.
Conjecture: (i) For any positive integers m and n, the sum sum_{k=0}^{n-1}C(n-1,k)^m*C(-n-1,k)^m/(4k^2-1) is always an integer divisible by n.
(ii) The sequence a(n+1)/a(n) (n > 2) is strictly increasing to the limit 17+12*sqrt(2), and the sequence a(n+1)^(1/(n+1))/a(n)^(1/n) (n > 1) is strictly decreasing to the limit 1.
For any positive integer n, we have sum_{k=0}^{n-1}C(n-1,k)*C(-n-1,k)/(4k^2-1) = -n, and n^3*a(n) = sum_{k=0}^{n-1}(2*k+1)*sum_{j=0..k}C(k,j)^2*C(k+j,j)^2/(2*j-1).
In the latest version of arXiv:1408.5381, the author proved part (i) of the conjecture, thus a(n) is indeed integral. - Zhi-Wei Sun, Sep 04 2014

Examples

			a(2) = 1 since 1/2*sum_{k=0,1}C(1,k)^2*C(-3,k)^2/(4*k^2-1) = 1/2*(-1+9/3) = 1.

Links

Crossrefs

Cf. A243101, A246460, A246511, A246542, A246543.

Programs

  • Maple
    A246567:=n->add((binomial(n-1,k)*binomial(-n-1,k))^2/(4*k^2-1), k=0..n-1)/n: seq(A246567(n), n=1..20);
  • Mathematica
    a[n_]:=Sum[(Binomial[n-1,k]*Binomial[-n-1,k])^2/(4*k^2-1),{k,0,n-1}]/n
Table[a[n],{n,1,20}]
  • PARI
    a(n) = sum(k=0, n-1, binomial(n-1,k)^2*binomial(n+k,k)^2/(4*k^2-1))/n; \\ Michel Marcus, Dec 24 2021

Formula

Recurrence (obtained via the Zeilberger algorithm): n^3*(n+1)*(2*n+5)*a(n) - (n+1)*(2*n+5)*(35*n^3+152*n^2+191*n+62)*a(n+1) + (n+2)*(2*n+1)*(35*n^3+163*n^2+224*n+88)*a(n+2) - (n+2)*(n+3)^3*(2*n+1)*a(n+3) = 0.
a(n) ~ (17+12*sqrt(2))^n / (2^(17/4) * Pi^(3/2) * n^(9/2)). - Vaclav Kotesovec, Sep 07 2014

A246875 a(n) = (Sum_{k=0..n-1} C(n-1,k)^2*C(-n-1,k)^2/C(k+2,2))/n.

Original entry on oeis.org

1, 2, 13, 134, 1783, 27950, 491335, 9401390, 192033565, 4131488426, 92723165533, 2155279960586, 51602299168639, 1267128734047142, 31803430252162579, 813628992468938750, 21168533016938471665, 559044288633621863810, 14962460440143262653685, 405299365266569619086462
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 07 2014

Keywords

Comments

The author proved in arXiv:1408.5381 that a(n) is always an integer.
Note that Sum_{k=0..n-1} C(n-1,k)*C(-n-1,k)/C(k+2,2) = 0 for n > 1.
Conjecture: The sequence a(n+1)/a(n) (n > 0) is strictly increasing to the limit 17+12*sqrt(2), and the sequence a(n+1)^(1/(n+1))/a(n)^(1/n) (n > 1) is strictly decreasing to the limit 1.

Examples

			a(2) = 2 since (Sum_{k=0..1} C(2-1,k)^2*C(-2-1,k)^2/C(2+k,2))/2 = (1 + (-3)^2/3)/2 = 2.

Links

Crossrefs

Cf. A243101, A246460, A246511, A246543, A246567.

Programs

  • Mathematica
    a[n_]:=Sum[(Binomial[n-1,k]*Binomial[-n-1,k])^2/Binomial[k+2,2],{k,0,n-1}]/n
Table[a[n],{n,1,20}]
Table[HypergeometricPFQ[{1-n,1-n,1+n,1+n},{1,1,3},1]/n,{n,1,10}] (* Benedict W. J. Irwin, Apr 04 2017 *)
  • PARI
    a(n) = sum(k=0, n - 1, (binomial(n - 1, k) * binomial(-n - 1, k))^2/binomial(k + 2, 2))/n; \\ Indranil Ghosh, Apr 04 2017

Formula

Recurrence (obtained via the Zeilberger algorithm): -n*(n-1)^2*(2*n+3)*a(n) + 4*(17*n^4+68*n^3+92*n^2+48*n+9)*a(n+1) - (n+2)*(n+3)^2*(2*n+1)*a(n+2) = 0.
a(n) ~ (17+12*sqrt(2))^n / (2^(5/4) * Pi^(3/2) * n^(9/2)). - Vaclav Kotesovec, Sep 07 2014
a(n) = 4F3(1-n,1-n,1+n,1+n;1,1,3;1)/n. - Benedict W. J. Irwin, Apr 04 2017
Showing 1-4 of 4 results.

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