A067049 -id:A067049 - cp's OEIS frontend

A061297 a(n) = Sum_{ r = 0 to n} L(n,r), where L(n,r) (A067049) = lcm(n, n-1, n-2, ..., n-r+1)/lcm(1, 2, 3, ..., r).

1, 2, 4, 8, 14, 32, 39, 114, 166, 266, 421, 1608, 1005, 3980, 6894, 4206, 8666, 40904, 49559, 315478, 162321, 79180, 269878, 1647124, 937553, 1810092, 8445654, 7791356, 3978238, 33071544, 19578861, 283536170, 327438714, 117635956, 742042967, 154748984, 88779589, 1532487536, 10514107742, 3761632498

Offset: 0

Amarnath Murthy, Apr 26 2001

The following sequences all appear to have the same parity: A003071, A029886, A061297, A092524, A093431, A102393, A104258, A122248, A128975. - Jeremy Gardiner, Dec 28 2008

			a(0) = 1, a(4) = 14: we have L(4,0) = 1, L(4,1) = 4, L(4,2) = 6, L(4,3) = 2, L(4,4) = 1. For r = 0 to 4, sigma {L(4,r)}= 1 + 4 + 6 + 2 + 1 = 14.

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

Tanya Khovanova, There are no coincidences, arXiv preprint 1410.2193 [math.CO], 2014.

Row sums of A067049.

tlcm(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);}
a(n) = sum(r = 0, n , tlcm(n, r)); \\ Michel Marcus, Sep 14 2013

A093430 Triangle read by rows: T(n,k) = lcm(n, n-1, ..., n-k+2, n-k+1)/lcm(1, 2, ..., k) (1 <= k <= n).

Original entry on oeis.org

1, 2, 1, 3, 3, 1, 4, 6, 2, 1, 5, 10, 10, 5, 1, 6, 15, 10, 5, 1, 1, 7, 21, 35, 35, 7, 7, 1, 8, 28, 28, 70, 14, 14, 2, 1, 9, 36, 84, 42, 42, 42, 6, 3, 1, 10, 45, 60, 210, 42, 42, 6, 3, 1, 1, 11, 55, 165, 330, 462, 462, 66, 33, 11, 11, 1, 12, 66, 110, 165, 66, 462, 66, 33, 11, 11, 1, 1, 13

Offset: 1

Amarnath Murthy, Mar 31 2004

An LCM-analog of the binomial coefficients. - N. J. A. Sloane, Aug 26 2015

			T(7,3) = lcm(7,6,5)/lcm(1,2,3) = 210/6 = 35.
Triangle starts:
  1;
  2,  1;
  3,  3,  1;
  4,  6,  2,  1;
  5, 10, 10,  5,  1;
  6, 15, 10,  5,  1,  1;
  ...

Bakir Farhi, An analog of the arithmetic triangle obtained by replacing the products by the least common multiples, arXiv:1002.1383 [math.NT], 2010.
Siao Hong and Guoyou Qian, On the lcm-analog of binomial coefficient, Asian-European Journal of Mathematics, Volume 07, Issue 04, December 2014; DOI: 10.1142/S1793557114500569.

Cf. A067049 (same triangle with an additional leading column of ones).
Cf. A093431, A093432, A093433.
Row sums yield A093431.

T:=(n,k)->lcm(seq(i,i=n-k+1..n))/lcm(seq(j,j=1..k)): for n from 1 to 13 do seq(T(n,k),k=1..n) od; # yields sequence in triangular form # Emeric Deutsch, Jan 30 2006

t[n_, k_] :=  LCM @@ Table[j, {j, n-k+1, n}] / LCM @@ Table[j, {j, 1, k}]; t[, 0] = 1; Table[t[n, k], {n, 1, 13}, {k, 1, n}] // Flatten (* _Jean-François Alcover, Apr 23 2014 *)

More terms from Emeric Deutsch, Jan 30 2006

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

A241475 Triangle t(n,r) = s(n,r)*s(n,r+1), where s(n,r) = lcm(n,n-1,...,n-r+1)/lcm(1,2,...,r-1,r), n >= 1 and 0 <= r < n.

Original entry on oeis.org

1, 2, 2, 3, 9, 3, 4, 24, 12, 2, 5, 50, 100, 50, 5, 6, 90, 150, 50, 5, 1, 7, 147, 735, 1225, 245, 49, 7, 8, 224, 784, 1960, 980, 196, 28, 2, 9, 324, 3024, 3528, 1764, 1764, 252, 18, 3, 10, 450, 2700, 12600, 8820, 1764, 252, 18, 3, 1, 11, 605, 9075, 54450, 152460, 213444, 30492, 2178, 363, 121, 11

Offset: 1

Jean-François Alcover, Apr 23 2014

The first eight terms and the first two terms of every row are identical to those of A132812.

			Triangle begins:
  1;
  2,  2;
  3,  9,   3;
  4, 24,  12,  2;
  5, 50, 100, 50, 5;
  6, 90, 150, 50, 5, 1;
  ...

S. M. Khairnar, Anant W. Vyawahare and J. N. Salunkhe, On Smarandache least common multiple ratio, Scientia Magna Vol. 5 (2009), No. 1, 29-36.
Amarnath Murthy, Some Notions on Least Common Multiples, Smarandache Notions Journal, Vol. 12, No. 1-2-3, Spring 2001.

Cf. A067049, A093430, A132812.

s[, 0] = 1; s[n, r_?NumericQ] := LCM @@ Table[n-k+1, {k, 1, r}] / LCM @@ Table[k, {k, 1, r}]; t[n_, r_] := s[n, r]*s[n, r+1]; Table[t[n, r] , {n, 1, 12}, {r, 0, n-1}] // Flatten

cp's OEIS Frontend

A061297 a(n) = Sum_{ r = 0 to n} L(n,r), where L(n,r) (A067049) = lcm(n, n-1, n-2, ..., n-r+1)/lcm(1, 2, 3, ..., r).

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A093430 Triangle read by rows: T(n,k) = lcm(n, n-1, ..., n-k+2, n-k+1)/lcm(1, 2, ..., k) (1 <= k <= n).

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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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A241475 Triangle t(n,r) = s(n,r)*s(n,r+1), where s(n,r) = lcm(n,n-1,...,n-r+1)/lcm(1,2,...,r-1,r), n >= 1 and 0 <= r < n.

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Programs

Mathematica