A246138 -id:A246138 - cp's OEIS frontend

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

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

Offset: 1

Zhi-Wei Sun, Aug 26 2014

nonn

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

			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.

Zhi-Wei Sun, Table of n, a(n) for n = 1..100
Zhi-Wei Sun, Two new kinds of numbers and related divisibility results, arXiv:1408.5381, 2014.

Cf. A245769, A246065, A246138, A246459.

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

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}]

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

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

A246065 a(n) = Sum_{k=0..n}C(n,k)^2C(2k,k)/(2k-1), where C(n,k) denotes the binomial coefficient n!/(k!(n-k)!).

Original entry on oeis.org

-1, 1, 9, 39, 177, 927, 5463, 34857, 234657, 1641471, 11820135, 87080265, 653499135, 4979882385, 38441107305, 300027646647, 2364113123073, 18784242756927, 150351698420247, 1211310469545081, 9816017765368671, 79963826730913809, 654504197331971961, 5380270242617370951

Offset: 0

Zhi-Wei Sun, Aug 24 2014

sign

a(n) is always an integer since (2k-1)|C(2k,k) for any nonnegative integer k.
Conjecture: (i) The sequence a(n+1)/a(n) (n = 2,3,...) is strictly increasing to the limit 9, and the sequence a(n+1)^(1/(n+1))/a(n)^(1/n) (n = 1,2,3,...) is strictly decreasing to the limit 1.
(ii) sum_{k=0}^{n-1}a(k) == 0 (mod n^2) for all n > 0. Moreover, for any prime p we have sum_{k=0}^{p-1}a(k) == -p^2*(1+9*(p/3))/2 (mod p^3), where (p/3) is the Legendre symbol.
We are able to prove n | sum_{k=0}^{n-1}a(k). Note also that sum_{k=0}^{n-1}a(k)*9^(n-1-k) = -n^2*A086618(n-1) for all n > 0 since both sides satisfy the same recurrence via the Zeilberger algorithm.
The congruence (0 mod n^2) in (ii) is true, see the formula for A246138 in terms of A005802. - Mark van Hoeij, Nov 07 2023

			a(2) = 9 since Sum_{k=0,1,2}C(2,k)^2*C(2k,k)/(2k-1) = -1 + 8 + 6/3 = 9.

Zhi-Wei Sun, Table of n, a(n) for n = 0..150
Zhi-Wei Sun, A new kind of numbers and their arithmetic properties, preprint, arXiv:1408.5381 [math.NT], 2014-2018.

Cf. A086618, A245769.

a := n -> -hypergeom([-1/2, -n, -n], [1, 1], 4):
seq(simplify(a(n)), n=0..23); # Peter Luschny, Nov 07 2023
ogf := -(1-9*x)^(1/4)*hypergeom([-1/4, 3/4],[1],64*x^3/((1-9*x)*(x-1)^3))/(1-x)^(5/4);
series(ogf, x=0, 25); # Mark van Hoeij, Nov 12 2023

a[n_]:=Sum[Binomial[n,k]^2*Binomial[2k,k]/(2k-1),{k,0,n}]
Table[a[n],{n,0,20}]

Recurrence (obtained via the Zeilberger algorithm):
9*(n+1)^2*a(n) -(19n^2+58n+63)*a(n+1) + (11n^2+46n+47)*a(n+2)-(n+3)^2*a(n+3) = 0.
a(n) ~ A086618(n)/2 ~ 3^(2*n + 5/2)/(16*Pi*n^2) as n tends to the infinity.
a(n) = (9*(2*n+1)^2*A002893(n) - 4*(n+1)^2*A002893(n+1))/3. - Mark van Hoeij, Nov 07 2023
a(n) = -hypergeom([-1/2, -n, -n], [1, 1], 4). - Peter Luschny, Nov 07 2023

A246459 a(n) = Sum_{k=0..n} C(n,k)^2C(2k,k)(2k+1), where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!).

Original entry on oeis.org

1, 7, 55, 465, 4047, 35673, 316521, 2819295, 25173855, 225157881, 2016242265, 18070920255, 162071863425, 1454320387575, 13055422263255, 117237213829953, 1053070838993151, 9461217421304505, 85019389336077225, 764113545253570191, 6868417199986308129

Offset: 0

Zhi-Wei Sun, Aug 26 2014

nonn

Zhi-Wei Sun proved that for any n > 0 we have Sum_{k=0..n-1} a(k) = n^2*A086618(n-1), and (Sum_{k=0..n-1}a(k,x))/n is a polynomial with integer coefficients, where a(k,x) = sum_{j=0..k}C(k,j)^2*C(2j,j)*(2j+1)*x^j.

			a(2) = 55 since Sum_{k=0,1,2} C(2,k)^2*C(2k,k)(2k+1) = 1 + 8*3 + 6*5 = 55.

Zhi-Wei Sun, Table of n, a(n) for n = 0..160
Zhi-Wei Sun, Two new kinds of numbers and their arithmetic properties, arXiv:1408.5381 [math.NT], 2014-2018.

Cf. A086618, A245769, A246065, A246138.

A246459:=n->add(binomial(n,k)^2*binomial(2*k,k)*(2*k+1), k=0..n): seq(A246459(n), n=0..20); # Wesley Ivan Hurt, Aug 26 2014

a[n_]:=Sum[Binomial[n,k]^2*Binomial[2k,k](2k+1),{k,0,n}]
Table[a[n],{n,0,20}]

Recurrence (obtained via the Zeilberger algorithm): 9*(n+1)^2*a(n) - (19*n^2+74*n+87)*a(n+1) + (n+3)*(11*n+29)*a(n+2) - (n+3)^2*a(n+3) = 0.
a(n) ~ 3^(2*n+1/2) / Pi. - Vaclav Kotesovec, Aug 27 2014
a(n) = (4*n+3)*A002893(n)/3. - Mark van Hoeij, Nov 12 2023

A246511 a(n) = (Sum_{k=0..n-1} (-1)^k(2k+1)C(n-1,k)^2C(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

Zhi-Wei Sun, Aug 27 2014

sign

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.

			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).

Zhi-Wei Sun, Table of n, a(n) for n = 1..100
Zhi-Wei Sun, Two new kinds of numbers and their arithmetic properties, arXiv:1408.5381 [math.NT], 2014-2018.

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

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

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}]

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

Zhi-Wei Sun, Aug 26 2014

nonn

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).

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

Zhi-Wei Sun, Table of n, a(n) for n = 0..100
Zhi-Wei Sun, Two new kinds of numbers and their arithmetic properties, arXiv:1408.5381 [math.NT], 2014-2018.

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

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

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

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)^2C(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

Zhi-Wei Sun, Aug 26 2014

nonn

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

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

Zhi-Wei Sun, Table of n, a(n) for n = 0..100
Zhi-Wei Sun, Two new kinds of numbers and their arithmetic properties, arXiv:1408.5381 [math.NT], 2014-2018.

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

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

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

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)^2C(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

Zhi-Wei Sun, Aug 28 2014

nonn

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,....

			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.

Zhi-Wei Sun, Table of n, a(n) for n = 1..100
Zhi-Wei Sun, Two new kinds of numbers and related divisibility results, arXiv:1408.5381 [math.NT], 2014-2018.
Zuo-Ru Zhang, Proof of two conjectures of Z.-W. Sun on combinatorial sequences, arXiv:2112.12427 [math.CO], 2021.

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

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}]

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

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.

cp's OEIS Frontend

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)!).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions

A246065 a(n) = Sum_{k=0..n}C(n,k)^2*C(2k,k)/(2k-1), where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A246459 a(n) = Sum_{k=0..n} C(n,k)^2*C(2k,k)*(2k+1), where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

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)!).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

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)!).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

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)!).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

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)!).

Views

Author

Keywords

Comments

Examples

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

A246065 a(n) = Sum_{k=0..n}C(n,k)^2C(2k,k)/(2k-1), where C(n,k) denotes the binomial coefficient n!/(k!(n-k)!).

A246459 a(n) = Sum_{k=0..n} C(n,k)^2C(2k,k)(2k+1), where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!).

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

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)!).

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

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