A246542 -id:A246542 - cp's OEIS frontend

A246543 a(n) = (2/n^3)( Sum_{k=0..n-1} (-1)^k(3k^2+3k+1)binomial(n-1,k)^3binomial(n+k,k)^3 ).

2, -47, 1142, 3793, -4094806, 371557891, -13021558306, -1374157073639, 281067953420114, -22220280272696387, -51611579093593498, 257837341935815261683, -35155217354672369625958, 1761633462267526777842223, 202464167122130621896038062

Offset: 1

Zhi-Wei Sun, Aug 29 2014

sign

Conjecture: Let n be any positive integer. For m = 0, 2, 4, ..., we have Sum_{k=0..n-1} (3k^2+3k+1)*(binomial(n-1,k)*binomial(n+k,k))^m == 0 (mod n^3); for m = 1, 3, 5, ... we have 2*Sum_{k=0..n-1} (-1)^k*(3k^2+3k+1)*(binomial(n-1,k)*binomial(n+k,k))^m == 0 (mod n^3).
The Zeilberger algorithm could yield a complicated fifth-order recurrence for a(n).
The author proved the conjecture in the latest version of arXiv:1408.5381. - Zhi-Wei Sun, Sep 14 2014

			a(2) = -47 since (2/2^3)*( Sum_{k=0..1} (-1)^k*(3k^2+3k+1)*binomial(1,k)^3*binomial(2+k,k)^3 ) = (1/4)*(1-7*3^3) = -47.

Zhi-Wei Sun, Table of n, a(n) for n = 1..70
Zhi-Wei Sun, Two new kinds of numbers and related divisibility results, arXiv:1408.5381 [math.NT], 2014-2018.

Cf. A246512, A246542.

a[n_] := Sum[(3 k^2 + 3 k + 1) (-1)^k (Binomial[n - 1, k] Binomial[n + k, k])^3, {k, 0, n - 1}] 2/n^3
Table[a[n], {n, 1, 14}]

A243101 a(n) = (sum_{k=0}^{n-1}(4k^3-1)C(n-1,k)C(n+k,k))/n^2, where C(m,k) denotes the binomial coefficient m!/(k!(m-k)!).

Original entry on oeis.org

-1, 2, 37, 324, 2403, 16582, 109961, 711176, 4521607, 28394442, 176648877, 1090974156, 6698429931, 40930511502, 249105886737, 1510954029072, 9138180665871, 55128765109906, 331851698655797, 1993747216968788

Offset: 1

Zhi-Wei Sun, Aug 29 2014

sign

Conjecture: Let n be any positive integer.
(i) For any m = 0,1,2,..., we have sum_{k=0}^{n-1}(-1)^k*(4*k^3-1)*C(n-1,k)^m*C(-n-1,k)^m == 0 (mod n^2). Also, we may replace 4*k^3-1  by 2*k^2+2*k+1 or 2*k^3+k^2+k.
(ii) For any s,t = 0,1,2,..., both sum_{k=0}^{n-1}(4*k^3-1)*C(n-1,k)^s*C(-n-1,k)^t and sum_{k=0}^{n-1}(-1)^k*(4*k^3-1)*C(n-1,k)^s*C(-n-1,k)^t are multiples of n.
The author proved the conjecture in the latest version of arXiv:1408.5381, thus a(n) is always an integer. - Zhi-Wei Sun, Sep 01 2014

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

Zhi-Wei Sun, Table of n, a(n) for n = 1..200
Zhi-Wei Sun, Two new kinds of numbers and related divisibility results, arXiv:1408.5381, 2014.

Cf. A246542, A246543.

A243101:=n->add((4*k^3-1)*binomial(n-1,k)*binomial(n+k,k), k=0..n-1)/n^2: seq(A243101(n), n=1..20); # Wesley Ivan Hurt, Sep 01 2014

a[n_]:=Sum[(4k^3-1)*Binomial[n-1,k]Binomial[n+k,k],{k,0,n-1}]/n^2
Table[a[n],{n,1,20}]

Recurrence (obtained via the Zeilberger algorithm):
n*(2*n+3)*(n^2+3*n+1)*a(n) - 2*(2*n^2+4*n+1)*(3*n^2+6*n+2)*a(n+1) + (n+2)*(2*n+1)*(n^2+n-1)*a(n+2) = 0.
a(n) ~ (3+2*sqrt(2))^n * sqrt(n) / (2^(5/4) * sqrt(Pi)). - Vaclav Kotesovec, Sep 02 2014

A180364 a(n) = sum_{k=0..n} C(n,k)C(n+k,k)(2k+1)^2, where C(m,k) denotes the binomial coefficient m!/(k!(m-k)!).

Original entry on oeis.org

1, 19, 205, 1839, 14961, 114483, 839917, 5975455, 41524897, 283272723, 1903686093, 12636115407, 83007985425, 540484102707, 3492471392493, 22418010385983, 143062290575937, 908253002030355, 5739641232682957, 36121371405797743, 226475167518421681

Offset: 0

Zhi-Wei Sun, Aug 29 2014

nonn

For any n > 0, we have a(0)+ ... + a(n-1) = n*sum_{k=0}^{n-1} (2*k+1)*C(n-1,k)*C(n+k,k) = n^2*A002002(n). The first equality can be easily deduced, and the second equality holds since both sides satisfy the same recurrence by the Zeilberger algorithm.

Conjecture: The sequence a(n+1)/a(n) (n = 0,1,...) is strictly decreasing to the limit 3+2*sqrt(2), and the sequence a(n+1)^(1/(n+1))/a(n)^(1/n) (n = 1,2,3,..) is strictly increasing to the limit 1.

			a(1) = 19 since sum_{k=0,1} C(1,k)*C(1+k,k)*(2k+1)^2 = 1 + 2*3^2 = 19.

Zhi-Wei Sun, Table of n, a(n) for n = 0..200
Zhi-Wei Sun, Two kinds of numbers and their arithmetic properties, arXiv:1408.5381, 2014.

Cf. A002002, A246460, A246542, A246543.

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

Recurrence (obtained via the Zeilberger algorithm):
-(n+1)*(2*n^2+10*n+11)*a(n) + (2*n+3)*(6*n^2+18*n-7)*a(n+1) - (n+2)*(2*n^2+2*n-1)*a(n+2) = 0.
a(n) ~ sqrt(8+6*sqrt(2)) * (3+2*sqrt(2))^n * n^(3/2) / (2*sqrt(Pi)). - Vaclav Kotesovec, Sep 02 2014

A246567 a(n) = (sum_{k=0}^{n-1}C(n-1,k)^2C(-n-1,k)^2/(4k^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

Zhi-Wei Sun, Aug 30 2014

sign

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

			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.

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. A243101, A246460, A246511, A246542, A246543.

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);

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}]

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

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

cp's OEIS Frontend

A246543 a(n) = (2/n^3)*( Sum_{k=0..n-1} (-1)^k*(3*k^2+3*k+1)*binomial(n-1,k)^3*binomial(n+k,k)^3 ).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A243101 a(n) = (sum_{k=0}^{n-1}(4*k^3-1)*C(n-1,k)*C(n+k,k))/n^2, where C(m,k) denotes the binomial coefficient m!/(k!*(m-k)!).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A180364 a(n) = sum_{k=0..n} C(n,k)*C(n+k,k)*(2*k+1)^2, where C(m,k) denotes the binomial coefficient m!/(k!*(m-k)!).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

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).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A246543 a(n) = (2/n^3)( Sum_{k=0..n-1} (-1)^k(3k^2+3k+1)binomial(n-1,k)^3binomial(n+k,k)^3 ).

A243101 a(n) = (sum_{k=0}^{n-1}(4k^3-1)C(n-1,k)C(n+k,k))/n^2, where C(m,k) denotes the binomial coefficient m!/(k!(m-k)!).

A180364 a(n) = sum_{k=0..n} C(n,k)C(n+k,k)(2k+1)^2, where C(m,k) denotes the binomial coefficient m!/(k!(m-k)!).

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