A361717 -id:A361717 - cp's OEIS frontend

A060542 a(n) = (1/6)multinomial(3n;n,n,n).

1, 15, 280, 5775, 126126, 2858856, 66512160, 1577585295, 37978905250, 925166131890, 22754499243840, 564121960420200, 14079683012144400, 353428777651788000, 8915829964229105280, 225890910734335847055, 5744976449471863238250, 146603287914300510042750

Offset: 1

Henry Bottomley, Apr 02 2001

Number of ways of dividing 3n labeled items into 3 unlabeled boxes with n items in each box.
From Antonio Campello (campello(AT)ime.unicamp.br), Nov 11 2009: (Start)
A060542(t) is the number of optimal [n,2,d] binary codes that correct at most t errors, i.e., having Hamming distance 2*t + 1 (achieved on length n = 3*t + 2). These codes are all isometric.
It is also the number of optimal [n,2,d] binary codes that detect 2*t + 1 errors, i.e., having Hamming distance 2t+2 (obtained by adding an overall parity check to the n = 3*t + 2 optimal codes). These codes are also all isometric.
For t = 0, we have the famous MDS, cyclic, simplex code {(000), (101), (110), (011)}. (End)
Also the number of distinct adjacency matrices of the complete tripartite graph K_{n,n,n}. - Eric W. Weisstein, Apr 21 2017

Harry J. Smith, Table of n, a(n) for n = 1..100
Eric Weisstein's World of Mathematics, Adjacency Matrix
Eric Weisstein's World of Mathematics, Complete Tripartite Graph

Row 3 of A060540.

Cf. A006480, A025035, A361717.

a:= n-> combinat[multinomial](3*n,n$3)/3!:
seq(a(n), n=1..18);  # Alois P. Heinz, Jul 29 2023

Table[(3*n)!/(n!^3*6),{n,1,20}] (* Vaclav Kotesovec, Sep 23 2013 *)
Table[Multinomial[n, n, n], {n, 20}]/6 (* Eric W. Weisstein, Apr 21 2017 *)

{ a=1/6; for (n=1, 100, write("b060542.txt", n, " ", a=a*3*(3*n - 1)*(3*n - 2)/n^2); ) } \\ Harry J. Smith, Jul 06 2009

a(n) = (3*n)!/((n!)^3*6) = a(n-1)*3*(3*n - 1)*(3*n - 2)/n^2 = A060540(3,n) = A006480(n)/6. - corrected by Vaclav Kotesovec, Sep 23 2013
a(n) ~ 3^(3*n-1/2)/(4*Pi*n). - Vaclav Kotesovec, Sep 23 2013
a(n) = 1/(8*n^3) * Sum_{k = 0..2*n} (-1)^(n+k) * k*(2*n-k)^2 * binomial(2*n, k)^3. - Peter Bala, Oct 11 2024
a(n) = 1/(n^2) * Sum_{k = 0..n} (-1)^(n+k+1) * (n-k)^2 * binomial(2*n, k)^3. - Peter Bala, Nov 03 2024

Definition revised by N. J. A. Sloane, Feb 02 2009

A361712 a(n) = Sum_{k = 0..n-1} binomial(n,k)^2binomial(n+k,k)binomial(n+k-1,k).

Original entry on oeis.org

0, 1, 25, 649, 16921, 448751, 12160177, 336745053, 9513822745, 273585035755, 7988828082775, 236367018090017, 7072779699975601, 213701611408357567, 6511338458568750853, 199850727914988936149, 6173376842290368719385, 191776434791965521115235, 5987554996434696230487955

Offset: 0

Peter Bala, Mar 21 2023

Conjecture 1: the supercongruence a(p) == a(1) (mod p^5) holds for all primes p >= 7 (checked up to p = 199).
Conjecture 2: for r >= 2, the supercongruence a(p^r) == a(p^(r-1)) (mod p^(3*r+3)) holds for all primes p >= 5.
Compare with the Apéry numbers A005259(n) = Sum_{k = 0..n} binomial(n,k)^2 * binomial(n+k,k)^2, which satisfy the weaker supercongruences A005259(p^r) == A005259(p^(r-1)) (mod p^(3*r)) for all primes p >= 5.

			a(7) - a(1) = (2^2)*(7^5)*5009 == 0 (mod 7^5)
a(11) - a(1) = (2^5)*(11^5)*45864163 == 0 (mod 11^5)
a(7^2) - a(7) = (2*3)*(7^9)*377052719*240136524699189343838527* 17965610580703155723668147409587 == 0 (mod 7^9)

Paolo Xausa, Table of n, a(n) for n = 0..650
Peter Bala, Recurrence equation for A361712

Cf. A005259, A212334, A361713, A361714, A361715, A361717.

Cf. A361877, A361878.

seq(add(binomial(n,k)^2*binomial(n+k,k)*binomial(n+k-1,k), k = 0..n-1), n = 0..25);
# Alternative:
A361712 := n -> hypergeom([-n, -n, n, n + 1], [1, 1, 1], 1) - binomial(2*n, n)*binomial(2*n-1, n): seq(simplify(A361712(n)), n = 0..18); # Peter Luschny, Mar 27 2023

A361712[n_] := HypergeometricPFQ[{-n, -n, n, n+1}, {1, 1, 1}, 1] - Binomial[2*n, n]*Binomial[2*n-1, n]; Array[A361712, 20, 0] (* Paolo Xausa, Jul 10 2024 *)

a(n) = (1/12)*(7*A005259(n) + A005259(n-1)) - (1/2)*binomial(2*n,n)^2.
a(n) ~ 2^(1/4)*(1 + sqrt(2))^(4*n+1)/(4*Pi^(3/2)*n^(3/2)).
a(n) = hypergeom([-n, -n, n, n + 1], [1, 1, 1], 1) - binomial(2*n, n)*binomial(2*n - 1, n) = A361878(n) - A361877(n). - Peter Luschny, Mar 27 2023

A361713 a(n) = Sum_{k = 0..n-1} binomial(n,k)^2 * binomial(n+k-1,k)^2.

Original entry on oeis.org

0, 1, 17, 406, 10257, 268126, 7213166, 198978074, 5609330705, 161095277710, 4700175389142, 138986764820410, 4157185583199534, 125568602682092818, 3825026187780837266, 117376010145070696906, 3625095243230562818065, 112596592142021739522670, 3514965607470183733302470

Offset: 0

Peter Bala, Mar 21 2023

Conjecture 1: the supercongruence a(p) == a(1) (mod p^5) holds for all primes p >= 7 (checked up to p = 199).
Conjecture 2: for r >= 2, the supercongruence a(p^r) == a(p^(r-1)) (mod p^(4*r+1)) holds for all primes p >= 7.
Compare with the Apéry numbers A005259(n) = Sum_{k = 0..n} binomial(n,k)^2 * binomial(n+k,k)^2, which satisfy the weaker supercongruences A005259(p^r) == A005259(p^(r-1)) (mod p^(3*r)) for all positive integers r and all primes p >= 5.

Paolo Xausa, Table of n, a(n) for n = 0..650
Peter Bala, Recurrence equation for A361713

Cf. A005259, A060150, A177316, A212334, A361712, A361714, A361715, A361717.

seq(add(binomial(n,k)^2*binomial(n+k-1,k)^2, k = 0..n-1), n = 0..25);
# Alternative:
A361713 := n -> hypergeom([-n, -n, n, n], [1, 1, 1], 1) - binomial(2*n - 1, n)^2:
seq(simplify(A361713(n)), n = 0..18); # Peter Luschny, Mar 27 2023

A361713[n_] := HypergeometricPFQ[{-n, -n, n, n}, {1, 1, 1}, 1] - Binomial[2*n-1, n]^2; Array[A361713, 20, 0] (* Paolo Xausa, Jul 11 2024 *)

a(n) = (1/3)*(A005259(n) + A005259(n-1)) - (1/4)*binomial(2*n,n)^2 = A177316(n) - A060150(n).
a(n) ~ C*(12*sqrt(2) + 17)^n/n^(3/2), where C = 1/(2^(5/4)*Pi^(3/2)).
a(n) = hypergeom([-n, -n, n, n], [1, 1, 1], 1) - binomial(2*n-1, n)^2. This is another way to write the first formula. - Peter Luschny, Mar 27 2023

A361714 a(n) = Sum_{k = 0..n-1} (-1)^(n+k+1)binomial(n,k)binomial(n+k-1,k)^2.

Original entry on oeis.org

0, 1, 7, 82, 1063, 14376, 199204, 2806770, 40053031, 577468684, 8397778882, 123029274666, 1814016998116, 26898142793068, 400836647993292, 5999796281063082, 90162110212198695, 1359731143731297396, 20571691450059355174, 312134224830052880826, 4748435338386591995938

Offset: 0

Peter Bala, Mar 21 2023

Conjecture 1: the supercongruence a(p) == a(1) (mod p^5) holds for all primes p >= 7 (checked up to p = 199).
Conjecture 2: for r >= 2, the supercongruence a(p^r) == a(p^(r-1)) (mod p^(3*r+3)) holds for all primes p >= 7.
Compare with the Apéry numbers A005258(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n,k) * binomial(n+k,k)^2, which satisfy the weaker supercongruences A005258(p^r) == A005258(p^(r-1)) (mod p^(3*r)) for positive integer r and all primes p >= 5.

			Examples of supercongruence:
a(11) - a(1) = 23029274666 - 1 = 5*(11^5)*152783 == 0 (mod 11^5).
a(13) - a(1) = 26898142793068 - 1 = (3^2)*7*(13^5)*1149913 == 0 (mod 13^5).
a(5^2) - a(5) = 3994642669575050040375014376 - 14376 = (2^6)*(3^6)*(5^9)*103* 425601520324429 == 0 (mod 5^9).

Paolo Xausa, Table of n, a(n) for n = 0..800

Cf. A005258, A352654, A361712, A361713, A361715, A361717.

seq(add((-1)^(n+k+1)*binomial(n,k)*binomial(n+k-1,k)^2, k = 0..n-1), n = 0..20);
# Alternative:
A361714 := n -> binomial(2*n-1, n)^2 - (-1)^n*hypergeom([-n, n, n], [1, 1], 1):
seq(simplify(A361714(n)), n = 0..20); # Peter Luschny, Mar 27 2023

A361714[n_] := Binomial[2*n-1, n]^2 - (-1)^n*HypergeometricPFQ[{-n, n, n}, {1, 1}, 1]; Array[A361714, 20, 0] (* Paolo Xausa, Jul 11 2024 *)

a(n) = binomial(2*n-1,n)^2 - (1/5)*(A005258(n) - 3*A005258(n-1)) for n >= 1.
P-recursive:
(395*n^10 - 6083*n^9 + 39816*n^8 - 144606*n^7 + 318639*n^6 - 436307*n^5 + 362870*n^4 - 167820*n^3 + 33096*n^2)*a(n) = (10665*n^10 - 174906*n^9 + 1243697*n^8 - 5033114*n^7 + 12789951*n^6 - 21235254*n^5 + 23221451*n^4 - 16437246*n^3 + 7182940*n^2 - 1753656*n + 185472)*a(n-1) - (69125*n^10 - 1202775*n^9 + 9159576*n^8 - 40005738*n^7 + 110271201*n^6 - 198723383*n^5 + 234346978*n^4 - 175661976*n^3 + 78402944*n^2 - 18529392*n + 1901088)*a(n-2) -  4*(n - 3)^2*(1580*n^8 - 19592*n^7 + 101515*n^6 - 284307*n^5 + 464411*n^4 - 444309*n^3 + 236490*n^2 - 62500*n + 7000)*a(n-3) with a(0) = 0, a(1) = 1 and a(2) = 7.
a(n) = binomial(2*n-1, n)^2 - (-1)^n*hypergeom([-n, n, n], [1, 1], 1). This is another way to write the first formula. - Peter Luschny, Mar 27 2023

A361715 a(n) = Sum_{k = 0..n-1} binomial(n,k)^2*binomial(n+k-1,k).

Original entry on oeis.org

0, 1, 9, 82, 745, 6876, 64764, 621860, 6070761, 60085720, 601493134, 6078225792, 61907445340, 634751002718, 6545478537810, 67830084149832, 705950951578089, 7375212511115184, 77310175072063914, 812839577957617640, 8569327793354169870, 90562666708303706642, 959212007563384494522, 10180245921386807485152

Offset: 0

Peter Bala, Mar 23 2023

Conjecture 1: the supercongruence a(p) == a(1) (mod p^5) holds for all primes p >= 7 (checked up to p = 199).
Conjecture 2: for r >= 2, the supercongruence a(p^r) == a(p^(r-1)) (mod p^(3*r+3)) holds for all primes p >= 5.
Compare with the Apéry numbers A005258(n) = Sum_{k = 0..n} binomial(n,k)^2 * binomial(n+k,k), which satisfy the weaker supercongruences A005258(p^r) == A005258(p^(r-1)) (mod p^(3*r)) for all primes p >= 5.

Cf. A005258, A103882, A361712, A361713, A361714, A361717.

seq( add( binomial(n,k)^2*binomial(n+k-1,k), k = 0..n-1), n = 0..25);
#faster alternative program
P(n) := 145*n^4 - 1217*n^3 + 3763*n^2 - 5079*n + 2532:
Q(n) := (n - 1)*(n - 2)*(2175*n^6 - 20140*n^5 + 73132*n^4 - 131786*n^3 + 122789*n^2 - 55626*n + 9936):
R(n) := (n - 2)*(6235*n^7 - 67846*n^6 + 304860*n^5 - 731294*n^4 + 1008701*n^3 - 798060*n^2 + 335340*n - 58320):
a := proc(n) option remember; if n = 0 then 0 elif n = 1 then 1 elif n = 2 then 9 else (Q(n)*a(n-1) - R(n)*a(n-2) - 2*(n - 1)*(n - 3)^2*(2*n - 5)*P(n+1)*a(n-3))/((n - 1)*(n - 2)*n^2*P(n)) end if; end:
seq(a(n), n = 0..25);
# Alternative:
A361715 := n -> hypergeom([-n, -n, n], [1, 1], 1) - binomial(2*n-1, n):
seq(simplify(A361715(n)), n = 0..23); # Peter Luschny, Mar 27 2023

Table[Sum[Binomial[n,k]^2 Binomial[n+k-1,k],{k,0,n-1}],{n,0,30}] (* Harvey P. Dale, Nov 01 2023 *)

a(n) = A103882(n) - binomial(2*n-1,n) = (3*A005258(n) + A005258(n-1))/5 - binomial(2*n-1,n) for n >= 1.
a(n) ~ sqrt(sqrt(5)/10 + 1/4)*(5*sqrt(5)/2 + 11/2)^n/(Pi*n)
P-recursive:
(n - 1)*(n - 2)*n^2*P(n)*a(n) = Q(n)*a(n - 1) - R(n)*a(n-2) - 2*(n - 1)*(n - 3)^2*(2*n - 5)*P(n+1)*a(n-3) with a(0) = 0, a(1) = 1 and a(2) = 9 and where
P(n) = 145*n^4 - 1217*n^3 + 3763*n^2 - 5079*n + 2532,
Q(n) = (n - 1)*(n - 2)*(2175*n^6 - 20140*n^5 + 73132*n^4 - 131786*n^3 + 122789*n^2 - 55626*n + 9936) and
R(n) = (n - 2)*(6235*n^7 - 67846*n^6 + 304860*n^5 - 731294*n^4 + 1008701*n^3 - 798060*n^2 + 335340*n - 58320).
a(n) = hypergeom([-n, -n, n], [1, 1], 1) - binomial(2*n-1, n). This is another way to write the first formula. - Peter Luschny, Mar 27 2023

cp's OEIS Frontend

A060542 a(n) = (1/6)*multinomial(3*n;n,n,n).

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A361712 a(n) = Sum_{k = 0..n-1} binomial(n,k)^2*binomial(n+k,k)*binomial(n+k-1,k).

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A361713 a(n) = Sum_{k = 0..n-1} binomial(n,k)^2 * binomial(n+k-1,k)^2.

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A361714 a(n) = Sum_{k = 0..n-1} (-1)^(n+k+1)*binomial(n,k)*binomial(n+k-1,k)^2.

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A361715 a(n) = Sum_{k = 0..n-1} binomial(n,k)^2*binomial(n+k-1,k).

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A060542 a(n) = (1/6)multinomial(3n;n,n,n).

A361712 a(n) = Sum_{k = 0..n-1} binomial(n,k)^2binomial(n+k,k)binomial(n+k-1,k).

A361714 a(n) = Sum_{k = 0..n-1} (-1)^(n+k+1)binomial(n,k)binomial(n+k-1,k)^2.