A246065 -id:A246065 - cp's OEIS frontend

A246138 a(n) = (Sum_{k=0..n-1} A246065(k)) / n^2.

-1, 0, 1, 3, 9, 32, 135, 648, 3409, 19176, 113535, 700125, 4463415, 29256120, 196334697, 1344542787, 9371335905, 66335058128, 476022873279, 3457886816997, 25394948961831, 188353304179920, 1409578821465129, 10635308054118792, 80845157085234975

Offset: 1

Zhi-Wei Sun, Aug 25 2014

sign

Part (ii) of the conjecture in A246065 implies that all the terms in the current sequence are integers.

Conjecture: The sequence a(n+1)/a(n) (n = 4,5,...) is strictly increasing to the limit 9, and the sequence a(n+1)^(1/(n+1))/a(n)^(1/n) (n = 3,4,...) is strictly decreasing to the limit 1.

			a(5) = 9 since sum_{k=0}^{5-1}A246065(k) = -1 + 1 + 9 + 39 + 177 = 225 = 5^2*9.

Zhi-Wei Sun, Table of n, a(n) for n = 1..170

Cf. A005802, A246065.

ogf := (1-((9*x-1)/(x-1))^(3/4)*hypergeom([-1/4, 3/4],[1],-64*x/(9*x-1)^3/(x-1)))/6;
series(ogf, x=0, 25); # Mark van Hoeij, Nov 12 2023

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

Recurrence: n^2*a(n) = 2*(n-2)*(5*n-8)*a(n-1) - 9*(n-2)^2*a(n-2). - Vaclav Kotesovec, Aug 27 2014
a(n) ~ 3^(2*n+5/2) / (128*Pi*n^4). - Vaclav Kotesovec, Aug 27 2014
a(n) = ((3*n+2)*(3*n-2)*A005802(n-1) - (n+2)^2*A005802(n))/4. - Mark van Hoeij, Nov 06 2023

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)!).

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

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

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.

A367024 Triangle read by rows, T(n, k) = [x^k] -hypergeom([-1/2, -n, -n], [1, 1], 4*x).

Original entry on oeis.org

-1, -1, 2, -1, 8, 2, -1, 18, 18, 4, -1, 32, 72, 64, 10, -1, 50, 200, 400, 250, 28, -1, 72, 450, 1600, 2250, 1008, 84, -1, 98, 882, 4900, 12250, 12348, 4116, 264, -1, 128, 1568, 12544, 49000, 87808, 65856, 16896, 858, -1, 162, 2592, 28224, 158760, 444528, 592704, 342144, 69498, 2860

Offset: 0

Peter Luschny, Nov 07 2023

			Triangle T(n, k) starts:
  [0] -1;
  [1] -1,   2;
  [2] -1,   8,    2;
  [3] -1,  18,   18,     4;
  [4] -1,  32,   72,    64,     10;
  [5] -1,  50,  200,   400,    250,     28;
  [6] -1,  72,  450,  1600,   2250,   1008,     84;
  [7] -1,  98,  882,  4900,  12250,  12348,   4116,    264;
  [8] -1, 128, 1568, 12544,  49000,  87808,  65856,  16896,   858;
  [9] -1, 162, 2592, 28224, 158760, 444528, 592704, 342144, 69498, 2860;

Cf. A246065 (row sums), -A002420 and A284016 (main diagonal).

Cf. A367022, A367023, A387025.

p := n -> -hypergeom([-1/2, -n, -n], [1, 1], 4*x):
T := (n, k) -> coeff(simplify(p(n)), x, k):
seq(seq(T(n, k), k = 0..n), n = 0..9);

T(n, k) = binomial(n, k)^2 * binomial(2*k, k) / (2*k - 1).

cp's OEIS Frontend

A246138 a(n) = (Sum_{k=0..n-1} A246065(k)) / n^2.

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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)!).

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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)!).

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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)!).

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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)!).

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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)!).

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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)!).

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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)!).

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)!).