A245769 -id:A245769 - cp's OEIS frontend

A268136 a(n) = (3/n)*Sum_{k=0..n-1} A245769(k)^2.

3, 3, 51, 507, 4947, 58243, 841443, 14240763, 269512483, 5524472451, 120183938835, 2738420763131, 64760819179635, 1579226738429187, 39515677808716739, 1010750709382934523, 26349289260686093379, 698387854199468231427, 18783213754115549685747, 511772677524431483886075

Offset: 1

Zhi-Wei Sun, Jan 26 2016

nonn

Conjecture: (i) All the terms are odd integers.

(ii) For n = 0,1,2,... let R_n(x) denote the polynomial sum_{k=0..n} binomial(n,k)*binomial(n+k,k)*x^k/(2k-1). Then, for each n = 1,2,3,.., all the coefficients of the polynomial (3/n)*Sum_{k=0..n-1} R_k(x)^2 are integral and the polynomial is irreducible over the field of rational numbers.

			a(3) = 51 since (3/3)*(A245769(0)^2 + A245769(1)^2 + A245769(2)^2) = (-1)^2 + 1^2 + 7^2 = 51.

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

Cf. A000290, A245769, A268137, A268138.

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

A268137 a(n) = (1/n)Sum_{k=0..n-1} A001850(k)A245769(k).

Original entry on oeis.org

-1, 1, 31, 417, 5919, 97217, 1828479, 38085249, 853450367, 20174707521, 496690317855, 12626836592289, 329476040177439, 8785359461936769, 238587766484265471, 6581966817521388033, 184067922884292651519, 5209333642085984431489, 148992465188631205367071, 4301514890878664802287777

Offset: 1

Zhi-Wei Sun, Jan 26 2016

sign

Conjecture: (i) All the terms are odd integers. For any prime p, if p == 3 (mod 4) then a(p) == -5 (mod p^2), otherwise a(p) == -1 (mod p).

(ii) For n = 0,1,2,... let D_n(x) = Sum_{k=0..n} binomial(n,k)*binomial(n+k,k)*x^k and R_n(x) = Sum_{k=0..n} binomial(n,k)*binomial(n+k,k)*x^k/(2k-1). For any positive integer n, all the coefficients of the polynomial (1/n)*Sum_{k=0..n-1} D_k(x)*R_k(x) are integral and the polynomial is irreducible over the field of rational numbers.

			a(3) = 31 since (A001850(0)*A245769(0) + A001850(1)*A245769(1) + A001850(2)*A245769(2))/3 = (1*(-1) + 3*1 + 13*7)/3 = 93/3 = 31.

Zhi-Wei Sun, Table of n, a(n) for n = 1..100
Zhi-Wei Sun, Two new kinds of numbers and related divisibility results, preprint, arXiv:1408.5381 [math.NT], 2014.
Zhi-Wei Sun, Arithmetic properties of Delannoy numbers and Schröder numbers, preprint, arXiv:1602.00574 [math.CO], 2016.

Cf. A001850, A245769, A268136, A268138.

d[n_]:=d[n]=Sum[Binomial[n,k]Binomial[n+k,k],{k,0,n}]
R[n_]:=R[n]=Sum[Binomial[n,k]Binomial[n+k,k]/(2k-1),{k,0,n}]
a[n_]:=a[n]=Sum[d[k]*R[k],{k,0,n-1}]/n
Do[Print[n," ",a[n]],{n,1,20}]

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

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

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

cp's OEIS Frontend

A268136 a(n) = (3/n)*Sum_{k=0..n-1} A245769(k)^2.

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A268137 a(n) = (1/n)*Sum_{k=0..n-1} A001850(k)*A245769(k).

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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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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)!).

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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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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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A268137 a(n) = (1/n)Sum_{k=0..n-1} A001850(k)A245769(k).

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)!).

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)!).