A246461 -id:A246461

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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)!).

+10
5

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

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.

LINKS

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.

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

EXAMPLE

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).

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}]

CROSSREFS

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

KEYWORD

sign

AUTHOR

Zhi-Wei Sun, Aug 27 2014

STATUS

approved

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)!).

+10
3

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,2

COMMENTS

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

LINKS

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.

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

EXAMPLE

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

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}]

CROSSREFS

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

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Aug 26 2014

STATUS

approved

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)!).

+10
2

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

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,....

LINKS

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.

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.

EXAMPLE

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.

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}]

PROG

(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

CROSSREFS

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

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Aug 28 2014

STATUS

approved

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