A067047 -id:A067047 - cp's OEIS frontend

A067049 Triangle T(n,r) = lcm(n,n-1,n-2,...,n-r+1)/lcm(1,2,3,...,r-1,r), 0 <= r < n.

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 2, 1, 1, 5, 10, 10, 5, 1, 1, 6, 15, 10, 5, 1, 1, 1, 7, 21, 35, 35, 7, 7, 1, 1, 8, 28, 28, 70, 14, 14, 2, 1, 1, 9, 36, 84, 42, 42, 42, 6, 3, 1, 1, 10, 45, 60, 210, 42, 42, 6, 3, 1, 1, 1, 11, 55, 165, 330, 462, 462, 66, 33, 11, 11, 1, 1, 12, 66, 110

Offset: 0

Amarnath Murthy, Dec 30 2001

			Triangle begins:
  1;
  1, 1;
  1, 2, 1;
  1, 3, 3, 1;
  1, 4, 6, 2, 1; ...

Amarnath Murthy, Some Notions on Least Common Multiples, Smarandache Notions Journal, Vol. 12, No. 1-2-3, Spring 2001.

Diagonals give A067046, A067047, A067048. Row sums give A061297.

t[n_, r_] :=  LCM @@ Table[n-k+1, {k, 1, r}] / LCM @@ Table[k, {k, 1, r}]; t[, 0] = 1; Table[t[n, r], {n, 0, 12}, {r, 0, n}] // Flatten (* _Jean-François Alcover, Apr 22 2014 *)

t(n, r) = {nt = 1; for (k = n-r+1, n, nt = lcm(nt, k);); dt = 1; for (k = 1, r, dt = lcm(dt, k);); return (nt/dt);} \\ Michel Marcus, Sep 14 2013

More terms from Vladeta Jovovic, Dec 31 2001

A067048 a(n) = lcm(n, n+1, n+2, n+3, n+4) / 60.

Original entry on oeis.org

1, 1, 7, 14, 42, 42, 462, 66, 429, 1001, 1001, 364, 6188, 1428, 3876, 3876, 6783, 4389, 33649, 3542, 17710, 32890, 26910, 8190, 118755, 23751, 56637, 50344, 79112, 46376, 324632, 31416, 145299, 250971, 191919, 54834, 749398, 141778, 320866, 271502, 407253

Offset: 1

Amarnath Murthy, Dec 30 2001

			a(6) = 42 as lcm(6,7,8,9,10)/60 = 2520/60 = 42.

Harry J. Smith, Table of n, a(n) for n = 1..1000
Amarnath Murthy, Some Notions on Least Common Multiples, Smarandache Notions Journal, Vol. 12, No. 1-2-3 (Spring 2001), pp. 307-308.
Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -15, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 20, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -15, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1).
Index entries for sequences related to lcm's.

Cf. A067046, A067047.

seq(ilcm(n,n+1,n+2,n+3,n+4)/60,n=1..100); # Robert Israel, Feb 07 2016

Table[LCM @@ Range[n, n + 4]/60, {n, 1, 50}] (* Amiram Eldar, Sep 29 2022 *)

a(n)={lcm([n, n+1, n+2, n+3, n+4])/60} \\ Harry J. Smith, May 01 2010

From Gary Detlefs Apr 14 2011 and Apr 18 2011: (Start)
a(n) = (n+4)!*gcd(n-1,3)/(360*(n-1)!*gcd(n,4))
a(n) = (n+4)!*(5-4*cos((2*n+1)*Pi/3))/(1080*(n-1)!*(2+(-1)^n+cos(n*Pi/2)))
a(n) = (n+4)!*gcd(n-1,6)/(180*(n-1)!*2^((2*cos(n*Pi/2)+9+(-1)^n)/4)), n>1. (End)
120 <= n*(n+1)*(n+2)*(n+3)*(n+4)/a(n) <= 1440. - Charles R Greathouse IV, Sep 19 2012
Sum_{n>=1} 1/a(n) = 80 - 40*log(sqrt(3)+2)/sqrt(3) - 490*log(2)/3 + 60*log(3). - Amiram Eldar, Sep 29 2022

A189046 a(n) = lcm(n,n+1,n+2,n+3,n+4,n+5)/60.

Original entry on oeis.org

0, 1, 7, 14, 42, 42, 462, 462, 858, 3003, 1001, 4004, 6188, 18564, 27132, 3876, 27132, 74613, 100947, 67298, 17710, 230230, 296010, 188370, 237510, 118755, 736281, 453096, 553784, 1344904, 324632

Offset: 0

Gary Detlefs, Apr 15 2011

a(n) mod 2 has a period of 8, repeating [0,1,1,0,0,0,0,0].

Nathaniel Johnston, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 7, 0, 0, 0, 0, -28, 0, 0, 0, 0, 84, 0, 0, 0, 0, -203, 0, 0, 0, 0, 413, 0, 0, 0, 0, -728, 0, 0, 0, 0, 1128, 0, 0, 0, 0, -1554, 0, 0, 0, 0, 1918, 0, 0, 0, 0, -2128, 0, 0, 0, 0, 2128, 0, 0, 0, 0, -1918, 0, 0, 0, 0, 1554, 0, 0, 0, 0, -1128, 0, 0, 0, 0, 728, 0, 0, 0, 0, -413, 0, 0, 0, 0, 203, 0, 0, 0, 0, -84, 0, 0, 0, 0, 28, 0, 0, 0, 0, -7, 0, 0, 0, 0, 1).
Index entries for sequences related to lcm's.

Cf. A000217 ( = lcm(n,n+1)/2), A021913, A067046, A067047, A067048.

seq(lcm(n,n+1,n+2,n+3,n+4,n+5)/60,n=0..30)

Table[(LCM@@(n+Range[0,5]))/60,{n,0,40}]  (* Harvey P. Dale, Apr 17 2011 *)

a(n)=lcm([n..n+5])/60 \\ Charles R Greathouse IV, Sep 30 2016

a(n) = n*(n+1)*(n+2)*(n+3)*(n+4)*(n+5)*(4*(n^4 mod 5)+1)/(1800*((n^3 mod 4)+((n-1)^3 mod 4)+1)).
a(n) = binomial(n+5,6)/(gcd(n,5)*(A021913(n-1)+1)).
a(n) = binomial(n+5,6)/(gcd(n,5)*floor(((n-1) mod 4)/2+1)). - Gary Detlefs, Apr 22 2011
Sum_{n>=1} 1/a(n) = 92 + (54/5-18*sqrt(5)+6*sqrt(178-398/sqrt(5)))*Pi. - Amiram Eldar, Sep 29 2022

A241262 Array t(n,k) = binomial(n*k, n+1)/n, where n >= 1 and k >= 2, read by ascending antidiagonals.

Original entry on oeis.org

1, 2, 3, 5, 10, 6, 14, 42, 28, 10, 42, 198, 165, 60, 15, 132, 1001, 1092, 455, 110, 21, 429, 5304, 7752, 3876, 1020, 182, 28, 1430, 29070, 57684, 35420, 10626, 1995, 280, 36, 4862, 163438, 444015, 339300, 118755, 24570, 3542, 408, 45, 16796, 937365, 3506100, 3362260, 1391280, 324632, 50344, 5850, 570, 55

Offset: 1

Jean-François Alcover, Apr 18 2014

About the "root estimation" question asked in MathOverflow, one can check (at least numerically) that, for instance with k = 4 and a = 1/11, the series a^-1 + (k - 1) + Sum_{n>=} (-1)^n*binomial(n*k, n+1)/n*a^n evaluates to the positive solution of x^k = (x+1)^(k-1).
Row 1 is A000217 (triangular numbers),
Row 2 is A006331 (twice the square pyramidal numbers),
Row 3 is A067047(3n) = lcm(3n, 3n+1, 3n+2, 3n+3)/12 (from column r=4 of A067049),
Row 4 is A222715(2n) = (n-1)*n*(2n-1)*(4n-3)*(4n-1)/15,
Row 5 is not in the OEIS.
Column 1 is A000108 (Catalan numbers),
Column 2 is A007226 left shifted 1 place,
Column 4 is A007228 left shifted 1 place,
Column 5 is A124724 left shifted 1 place,
Column 6 is not in the OEIS.

			Array begins:
    1,    3,     6,     10,      15,      21, ...
    2,   10,    28,     60,     110,     182, ...
    5,   42,   165,    455,    1020,    1995, ...
   14,  198,  1092,   3876,   10626,   24570, ...
   42, 1001,  7752,  35420,  118755,  324632, ...
  132, 5304, 57684, 339300, 1391280, 4496388, ...
  etc.

N. S. S. Gu, H. Prodinger, S. Wagner, Bijections for a class of labeled plane trees, Eur. J. Combinat. 31 (2010) 720-732, doi|10.1016/j.ejc.2009.10.007, Theorem 2

MathOverflow, Root estimation

Cf. A000217, A006331, A000108, A007226, A007228, A067047, A067049, A124724, A222715.

t[n_, k_] := Binomial[n*k, n+1]/n; Table[t[n-k+2, k], {n, 1, 10}, {k, 2, n+1}] // Flatten

cp's OEIS Frontend

A067049 Triangle T(n,r) = lcm(n,n-1,n-2,...,n-r+1)/lcm(1,2,3,...,r-1,r), 0 <= r < n.

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A067048 a(n) = lcm(n, n+1, n+2, n+3, n+4) / 60.

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A189046 a(n) = lcm(n,n+1,n+2,n+3,n+4,n+5)/60.

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Formula

A241262 Array t(n,k) = binomial(n*k, n+1)/n, where n >= 1 and k >= 2, read by ascending antidiagonals.

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