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 1-7 of 7 results.

A246513 a(n) = (4/n^2)*( Sum_{k=0..n-1} k*A246459(k) ).

Original entry on oeis.org

0, 7, 52, 378, 2832, 21785, 171036, 1364391, 11023264, 89985681, 740894700, 6144227430, 51267563280, 430045297695, 3623966778180, 30662599042530, 260367332354496, 2217928838577641, 18947382204700044, 162281586037920126
Offset: 1

Views

Author

Zhi-Wei Sun, Aug 28 2014

Keywords

Comments

Conjecture: a(n) is always an integer.
Note: the formula for a(n) in terms of A005802 proves that a(n) is an integer, divisible by n-1. - Mark van Hoeij, Nov 06 2023

Examples

			a(2) = 7 since (4/2^2)*( Sum_{k=0..1} k*A246459(k) ) = A246459(1) = 7.

Links

Crossrefs

Cf. A246459, A246460.

Programs

  • Maple
    h := n -> hypergeom([1/2, 1 - n, -n], [2, 2], 4):
a := n -> (n - 1) * ((n + 1)^2 * h(n) / n - n * h(n - 1)):
seq(simplify(a(n)), n = 1..20);  # Peter Luschny, Nov 06 2023
ogf := (((-54*x^4+18*x^3+30*x^2+6*x)*hypergeom([4/3, 4/3],[2],-27*x*(x-1)^2/(9*x-1)^2)+(-1701*x^3+783*x^2-111*x+5)*hypergeom([1/3, 1/3],[1],-27*x*(x-1)^2/(9*x-1)^2))/(1-9*x)^(8/3) - 5)/6;
series(ogf, x=0, 25); # Mark van Hoeij, Nov 12 2023
  • Mathematica
    s[n_] := Sum[Binomial[n, k]^2 Binomial[2 k, k] (2 k + 1), {k, 0, n}]
a[n_] := Sum[k s[k], {k, 0, n-1}] 4/n^2
Table[a[n], {n, 1, 20}]

Formula

Recurrence: (n-2)*n^2*(2*n-7)*(4*n-5)*a(n) = (n-1)*(80*n^4 - 532*n^3 + 1126*n^2 - 893*n + 195)*a(n-1) - 9*(n-2)^2*(n-1)*(2*n-5)*(4*n-1)*a(n-2). - Vaclav Kotesovec, Aug 28 2014
a(n) ~ 3^(2*n+1/2) / (2*Pi*n). - Vaclav Kotesovec, Aug 28 2014
a(n) = (n-1) * ((n+1)^2 * A005802(n-1) - (n-1)*n * A005802(n-2)). - Mark van Hoeij, Nov 06 2023

A246460 a(n) = (sum_{k=0}^{n-1} (2k+1)*C(n-1,k)^2*C(n+k,k)^2)/n^2, where C(n,k) denotes the binomial coefficient n!/(k!(n-k)!).

Original entry on oeis.org

1, 7, 77, 1211, 23009, 489035, 11203765, 270937315, 6825612185, 177559028087, 4739821161173, 129244697791951, 3587524535220001, 101099089948850323, 2886373390151379397, 83343790441133767475, 2430567530705659113545, 71508611747063572974095, 2120357936904537499679125, 63315310358625743871987019
Offset: 1

Views

Author

Zhi-Wei Sun, Aug 26 2014

Keywords

Comments

Conjecture: a(n) is always an integer.
The author proved this in the latest version of arXiv:1408.5381. - Zhi-Wei Sun, Sep 01 2014

Examples

			a(2) = 7 since sum_{k=0,1} (2k+1)C(1,k)^2*C(2+k,k)^2 = 1 + 3*3^2 = 28 = 2^2*7.

Links

Crossrefs

Cf. A245769, A246065, A246138, A246459.

Programs

  • Maple
    A246460:=n->add((2*k+1)*binomial(n-1,k)^2*binomial(n+k,k)^2/n^2, k=0..n-1): seq(A246460(n), n=1..20); # Wesley Ivan Hurt, Aug 26 2014
  • Mathematica
    a[n_]:=Sum[(2k+1)*Binomial[n-1,k]^2*Binomial[n+k,k]^2,{k,0,n-1}]/n^2
Table[a[n],{n,1,20}]

Formula

Recurrence (obtained via the Zeilberger algorithm):
-n^3*(2*n + 5)*(3*n^2 + 12*n + 11)*a(n) + (2*n + 5)*(105*n^5 + 675*n^4 + 1579*n^3 + 1663*n^2 + 768*n+126)*a(n+1) - (2*n + 1)*(105*n^5 + 900*n^4 + 2929*n^3 + 4448*n^2 + 3048*n + 684)*a(n+2) + (n + 3)^3*(2*n + 1)*(3*n^2 + 6*n + 2)*a(n+3) = 0.
a(n) ~ 2^(1/4) * (17+12*sqrt(2))^n / (8 * Pi^(3/2) * n^(5/2)). - Vaclav Kotesovec, Aug 27 2014

Extensions

Typo in cross-reference corrected by Vaclav Kotesovec, Aug 27 2014

A246511 a(n) = (Sum_{k=0..n-1} (-1)^k*(2k+1)*C(n-1,k)^2*C(n+k,k)^2)/n, where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!).

Original entry on oeis.org

1, -13, 103, 219, -26139, 503957, -4066061, -54914149, 2550230113, -43157232273, 192777017511, 10118180981037, -318814450789587, 4344955121014089, 6807591584551563, -1781238363905009253, 42912636577174295769, -425791821468024981709, -5452095049517604924017, 305524943325956601071159
Offset: 1

Views

Author

Zhi-Wei Sun, Aug 27 2014

Keywords

Comments

Zhi-Wei Sun proved that a(n) is always an integer, and that Sum_{k=0..n-1}(2k+1)*A(k) = n^3*a(n), where A(k) = Sum_{j=0..k} (-1)^j*(2j+1)^2*C(k,j)^2*C(k+j,j)^2.
The Zeilberger algorithm could yield a complicated fourth-order recurrence for this sequence.

Examples

			a(2) = -13 since Sum_{k=0,1}(-1)^k*(2k+1)C(1,k)^2*C(2+k,k)^2 = 1 - 3*3^2 = 2*(-13).

Links

Crossrefs

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

Programs

  • Maple
    a:= n -> add((-1)^k*(2*k+1)*binomial(n-1,k)^2*binomial(n+k,k)^2,k=0..n-1)/n:
seq(a(n),n=1..40); # Robert Israel, Aug 28 2014
  • Mathematica
    a[n_]:=Sum[(-1)^k*(2k+1)*Binomial[n-1,k]^2*Binomial[n+k,k]^2,{k,0,n-1}]/n
Table[a[n],{n,1,20}]

Formula

a(n) = hypergeom([3/2, 1-n, 1-n, n+1, n+1], [1/2, 1, 1, 1], -1)/n. - Robert Israel, Aug 28 2014
Recurrence: (n-1)^2*n^3*(2*n-7)*(2*n-5)*(40*n^6 - 600*n^5 + 3612*n^4 - 11120*n^3 + 18354*n^2 - 15270*n + 4949)*a(n) = -2*(n-1)^2*(2*n-7)*(1120*n^10 - 21280*n^9 + 173136*n^8 - 789528*n^7 + 2217244*n^6 - 3965700*n^5 + 4511984*n^4 - 3162198*n^3 + 1267357*n^2 - 247675*n + 14910)*a(n-1) - 2*(n-2)*(2*n-7)*(2*n-1)*(9080*n^10 - 181600*n^9 + 1569004*n^8 - 7670464*n^7 + 23311258*n^6 - 45445432*n^5 + 56332869*n^4 - 42029480*n^3 + 16243359*n^2 - 1773884*n - 347928)*a(n-2) - 2*(n-3)^2*(2*n-1)*(1120*n^10 - 23520*n^9 + 213456*n^8 - 1095144*n^7 + 3485308*n^6 - 7092252*n^5 + 9139424*n^4 - 7057450*n^3 + 2811541*n^2 - 317773*n - 61278)*a(n-3) - (n-4)^3*(n-3)^2*(2*n-3)*(2*n-1)*(40*n^6 - 360*n^5 + 1212*n^4 - 1872*n^3 + 1266*n^2 - 234*n - 35)*a(n-4). - Vaclav Kotesovec, Sep 07 2014

A246461 a(n) = Sum_{k=0..n} ((2k+1)*C(n,k)*C(n+k,k))^2, where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!).

Original entry on oeis.org

1, 37, 1225, 43397, 1563401, 56309885, 2020496185, 72190600165, 2569004841385, 91095128385485, 3220006254279233, 113505318773615741, 3991330807880182105, 140050346341652428141, 4904787249549605102233, 171480516047539645266725
Offset: 0

Views

Author

Zhi-Wei Sun, Aug 26 2014

Keywords

Comments

Zhi-Wei Sun noted that for any positive integer n we have Sum_{k=0..n-1} (2k+1)*a(k) = n^4*A246460(n).

Examples

			a(1) = 37 since Sum_{k=0..1} ((2k+1)*C(1,k)*C(1+k,k))^2 = 1^2 + (3*2)^2 = 37.

Links

Crossrefs

Cf. A245769, A246065, A246138, A246459, A246460.

Programs

  • Maple
    A246461:=n->add(((2*k+1)*binomial(n,k)*binomial(n+k,k))^2, k=0..n): seq(A246461(n), n=0..20); # Wesley Ivan Hurt, Aug 26 2014
  • Mathematica
    a[n_]:=Sum[((2k+1)*Binomial[n,k]*Binomial[n+k,k])^2,{k,0,n}]
Table[a[n],{n,0,15}]

Formula

Recurrence: n^3*(2*n-3)*(4*n^4 - 24*n^3 + 50*n^2 - 42*n + 11)*(6*n^4 - 36*n^3 + 67*n^2 - 39*n - 4)*a(n) = (2*n-1)*(840*n^11 - 10464*n^10 + 53192*n^9 - 137864*n^8 + 172296*n^7 - 19912*n^6 - 226019*n^5 + 271559*n^4 - 92324*n^3 - 42188*n^2 + 39128*n - 8466)*a(n-1) - (2*n-3)*(840*n^11 - 8016*n^10 + 28712*n^9 - 44872*n^8 + 15880*n^7 + 43992*n^6 - 64675*n^5 + 32567*n^4 + 1692*n^3 - 9364*n^2 + 4072*n - 606)*a(n-2) + (n-2)^3*(2*n-1)*(4*n^4 - 8*n^3 + 2*n^2 + 2*n - 1)*(6*n^4 - 12*n^3 - 5*n^2 + 11*n - 6)*a(n-3). - Vaclav Kotesovec, Aug 27 2014
a(n) ~ sqrt(24+17*sqrt(2)) * (17+12*sqrt(2))^n * sqrt(n) / (2*sqrt(2)*Pi^(3/2)). - Vaclav Kotesovec, Aug 27 2014

A246462 a(n) = Sum_{k=0..n} (2k+1)*C(n,k)^2*C(n+k,k)^2, where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!).

Original entry on oeis.org

1, 13, 289, 7733, 223001, 6689045, 205569505, 6422252485, 203029535305, 6476057609045, 208013166524153, 6718923443380109, 218021269879802377, 7101635058978727909, 232072490781790669153, 7604916953685880646885
Offset: 0

Views

Author

Zhi-Wei Sun, Aug 26 2014

Keywords

Comments

For any n > 0, n^3 divides Sum_{k=0..n-1} (2k+1)*a(k).

Examples

			a(1) = 13 since Sum_{k=0..1} (2k+1)*C(1,k)^2*C(1+k,k)^2 = 1 + 3*2^2 = 13.

Links

Crossrefs

Cf. A245769, A246065, A246138, A246459, A246460, A246461.

Programs

  • Maple
    A246462:=n->add((2*k+1)*binomial(n,k)^2*binomial(n+k,k)^2, k=0..n): seq(A246462(n), n=0..20); # Wesley Ivan Hurt, Aug 27 2014
  • Mathematica
    a[n_]:=Sum[(2k+1)*Binomial[n,k]^2*Binomial[n+k,k]^2,{k,0,n}]
Table[a[n],{n,0,15}]

Formula

Recurrence: n^3*(2*n-3)*(8*n^4 - 48*n^3 + 96*n^2 - 72*n + 13)*a(n) = (2*n-1)*(280*n^7 - 2096*n^6 + 5728*n^5 - 6536*n^4 + 1383*n^3 + 3160*n^2 - 2432*n + 552)*a(n-1) - (2*n-3)*(280*n^7 - 1824*n^6 + 4096*n^5 - 3384*n^4 - 345*n^3 + 2046*n^2 - 1100*n + 192)*a(n-2) + (n-2)^3*(2*n-1)*(8*n^4 - 16*n^3 + 8*n - 3)*a(n-3). - Vaclav Kotesovec, Aug 27 2014
a(n) ~ sqrt(24+17*sqrt(2)) * (17+12*sqrt(2))^n / (4*Pi^(3/2)*sqrt(n)). - Vaclav Kotesovec, Aug 27 2014

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.

A246761 a(n) = Sum_{k=0..n} C(n,k)^3 * C(2k,k) * C(3k,k) * (3k + 1).

Original entry on oeis.org

1, 25, 823, 34459, 1663267, 85847347, 4598058505, 252738855901, 14170006731643, 806810379495379, 46503528950782309, 2707097765891635585, 158884136607368717797, 9389663462839346537221, 558176792747732603265463, 33349982885530909490561203
Offset: 0

Views

Author

Zhi-Wei Sun, Sep 02 2014

Keywords

Comments

Conjecture 1: Let m be any positive integer, and set S(m,n,x) = Sum_{k=0..n} C(n,k)^m*((k*m+1)!/(k!)^m)*x^k for n = 0, 1, 2, .... Then, for any n > 0, all the coefficients of the polynomial Sum_{k=0..n-1} S(m,k,x) are multiples of n, i.e., ((k*m+1)!/(k!)^m)*Sum_{h=k..n-1} C(h,k)^m == 0 (mod n) for all k = 0, ..., n-1.
In the case m = 3, this implies that Sum_{k=0..n-1} a(k) == 0 (mod n). Conjecture 1 with m = 2 was proved by the author in arXiv:1408.5381. Conjecture 1 with m = 1 is easy since (k+1)*Sum_{h=k..n-1} C(h,k) = (k+1)*C(n,k+1) = n*C(n-1,k) for all k = 0, ..., n-1. By using Bernoulli numbers, we obtain that (m+1)!*Sum_{h=1..n-1} C(h,1)^m == 0 (mod n).
We also formulate the following q-analog of Conjecture 1.
Conjecture 2: Let m be any positive integer. For n = 0,1,..., define S(q;m,n,x) = Sum_{k=0..n} C(q;n,k)^m*(Factorial(q,k*m+1)/Factorial(q,k)^m)*x^k, where C(q;n,k) is the q-analog of C(n,k) and Factorial(q,k) = Product_{j=1..k} [j]q is the q-analog of the factorial k! with [j]_q=(1-q^j)/(1-q). Then, for any n > 0, all the coefficients of the polynomial Sum{k=0..n-1} q^k*S(q;m,k,x) in x are divisible by the polynomial [n]q (the q-analog of n), i.e., (Factorial(q,k*m+1)/Factorial(q,k)^m)*Sum{h=k..n-1} q^h*C(q;h,k)^m == 0 (mod [n]_q) for all k = 0, ..., n-1.
We are able to prove Conjecture 2 for m = 1, 2, 3.
The Zeilberger algorithm could yield a complicated 5th-order recurrence for a(n).

Examples

			a(1) = 25 since Sum_{k=0..1} C(1,k)^3 * C(2k,k) * C(3k,k)*(3k+1) = 1 + 2*3*4 = 25.

Links

Crossrefs

Cf. A246459.

Programs

  • Mathematica
    a[n_]:=Sum[Binomial[n,k]^3*Binomial[2k,k]Binomial[3k,k](3k+1),{k,0,n}]
Table[a[n],{n,0,15}]

Formula

a(n) ~ 2^(6*n+2) / (Pi^2 * n). - Vaclav Kotesovec, Nov 27 2017
Showing 1-7 of 7 results.

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