A372972 -id:A372972 - cp's OEIS frontend

A372864 Numbers k such that A372720(k) = 0.

1, 500, 578, 722, 750, 1058, 1500, 1682, 1922, 2646, 2744, 3430, 3645, 4800, 5202, 5346, 5476, 5488, 5625, 6318, 6400, 6724, 7168, 7396, 8000, 8836, 10092, 10976, 11236, 11532, 11979, 12005, 13068, 13924, 14450, 14884, 15309, 16810, 16875, 16896, 18050, 18225

Offset: 1

Michael De Vlieger, Jun 02 2024

nonn

Let tau = A000005, let omega = A001221, let f = A008479, and let g = A372720.
For squarefree k, A372720(k) >= 0, since f(k) = 1 while tau(k) = 2^omega(k).
For prime power p^m, A372720(p^m) = 1, since f(p^m) = m while tau(k) = m+1.
Therefore, apart from a(1) = 1, this sequence is a proper subset of A126706.
In the sequence R = {k = m*s : rad(m) | s, s > 1 in A120944}, there is a smallest term k such that g(k) <= 0 and a largest term k such that g(k) is positive. For instance, in A033845 where s = 6, only {6, 12, 18, 24, 36, 48, 54, 72, 96, 108, 144, 192, 216, 288, 432, 576, 864} are such that g(k) > 0.
Apart from terms in this sequence, all the rest of the terms k in R are such that g(k) is negative.
There are no 3-smooth numbers k > 1 in this sequence, however there are 3 terms {500, 6400, 8000} in A033846 (with s = rad(k) = 10). For s = 2*3*23, there are 6 terms {19044, 25392, 38088, 70656, 536544, 953856}.
Conjecture: proper subset of A361098, hence of A360765 and A360768. This is to say that k = a(n) is such that A003557(k) >= A119288(k), i.e., k/rad(k) >= second smallest prime factor of k, and A003557(k) > A053669(k), where A053669(k) is the smallest prime q that does not divide k.

			a(1) = 1 since tau(1) - f(1) = 1 - 1 = 0.
a(2) = 500 = 2^2 * 5*3, since tau(500) - f(500)
     = (2+1)*(3+1) - card({10,20,40,50,80,100,160,200,250,320,400,500})
     = 12 - 12 = 0.
a(3) = 578 = 2*17^2, since tau(578) - f(578)
     = (1+1)*(2+1) - card({34,68,136,272,544,578})
     = 6 - 6 = 0, etc.

Michael De Vlieger, Table of n, a(n) for n = 1..2000

Cf. A000005, A001221, A007947, A008479, A010846, A372720, A372972.

rad[x_] := rad[x] = Times @@ FactorInteger[x][[All, 1]]; Position[Table[r = rad[n]; DivisorSigma[0, n] - Count[Range[n/r], ?(Divisible[r, rad[#]] &)], {n, 20000}], ?(# == 0 &)][[All, 1]]

A372868 Irregular triangle read by rows: T(n,k) is the number of flattened Catalan words of length n with exactly k runs of weak ascents, with 1 <= k <= ceiling(n/2).

Original entry on oeis.org

1, 2, 4, 1, 8, 6, 16, 24, 1, 32, 80, 10, 64, 240, 60, 1, 128, 672, 280, 14, 256, 1792, 1120, 112, 1, 512, 4608, 4032, 672, 18, 1024, 11520, 13440, 3360, 180, 1, 2048, 28160, 42240, 14784, 1320, 22, 4096, 67584, 126720, 59136, 7920, 264, 1, 8192, 159744, 366080, 219648, 41184, 2288, 26

Offset: 1

Stefano Spezia, May 15 2024

With offset 0 for the variable k, T(n,k) is the number of flattened Catalan words of length n with exactly k peaks. In such case, T(4,1) = 6 corresponds to 6 flattened Catalan words of length 4 with 1 peak: 0010, 0100, 0110, 0101, 0120, and 0121. See Baril et al. at page 20.

			The irregular triangle begins:
    1;
    2;
    4,    1;
    8,    6;
   16,   24,    1;
   32,   80,   10;
   64,  240,   60,   1;
  128,  672,  280,  14;
  256, 1792, 1120, 112, 1;
  ...
T(4,2) = 6 since there are 6 flattened Catalan words of length 4 with 2 runs of weak ascents: 0010, 0100, 0101, 0110, 0120, and 0121.

Jean-Luc Baril, Pamela E. Harris, and José L. Ramírez, Flattened Catalan Words, arXiv:2405.05357 [math.CO], 2024. See pp. 8-9.

Cf. A000079, A001788, A002409, A003472, A007051 (row sums), A110654 (row lengths), A140325, A172242.

Cf. A372852, A372972.

T[n_,k_]:=SeriesCoefficient[(1-2x)*x*y/(1-4*x+4*x^2-x^2*y),{x,0,n},{y,0,k}]; Table[T[n,k],{n,14},{k,Ceiling[n/2]}] //Flatten (* or *)
T[n_,k_]:=2^(n-2k+1)Binomial[n-1,2k-2]; Table[T[n,k],{n,14},{k,Ceiling[n/2]}]

G.f.: (1-2*x)*x*y/(1 - 4*x + 4*x^2 - x^2*y).
T(n,k) = 2^(n-2*k+1)*binomial(n-1, 2*k-2).
T(n,1) = A000079(n-1).
T(n,2) = A001788(n-2).
T(n,3) = A003472(n-1).
T(n,4) = A002409(n-7).
T(n,5) = A140325(n-9).
T(n,6) = A172242(n-1).
Sum_{k>=0} T(n,k) = A007051(n-1).

cp's OEIS Frontend

A372864 Numbers k such that A372720(k) = 0.

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