A153893 -id:A153893 - cp's OEIS frontend

A330530 Lexicographically earliest sequence of distinct positive integers such that the product of two consecutive terms is always divisible by 4.

1, 4, 2, 6, 8, 3, 12, 5, 16, 7, 20, 9, 24, 10, 14, 18, 22, 26, 28, 11, 32, 13, 36, 15, 40, 17, 44, 19, 48, 21, 52, 23, 56, 25, 60, 27, 64, 29, 68, 30, 34, 38, 42, 46, 50, 54, 58, 62, 66, 70, 72, 31, 76, 33, 80, 35, 84, 37, 88, 39, 92, 41, 96, 43, 100, 45, 104

Offset: 1

Rémy Sigrist, Dec 17 2019

For any k > 0, let f_k be the lexicographically earliest sequence of distinct positive integers such that the product of two consecutive terms is always divisible by k:
- in particular:
     f_1 = f_2 = A000027,
     f_3 = A006368,
     f_4 = a (this sequence),
     f_6 = A330531,
- f_k is a permutation of the natural numbers,
- f_k(1) = 1, f_k(2) = max(2, k),
- if k is prime, then f_k corresponds to the integers that are not multiple of k interspersed with the integers that are multiple of k.
Apparently:
- for m > 0, the m-th run of consecutive terms such that gcd(a(n), 4) = 2 has A153893(m+1) terms,
- for m > 1, the m-th run of consecutive terms such that gcd(a(n), 4) = 1 or 4 has A068156(m+1) terms.

			The first terms, alongside their product with the next term, are:
  n   a(n)  a(n)*a(n+1)
  --  ----  -----------
   1     1            4
   2     4            8
   3     2           12
   4     6           48
   5     8           24
   6     3           36
   7    12           60
   8     5           80
   9    16          112
  10     7          140

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Rémy Sigrist, Colored scatterplot of the first 10000 terms
Index entries for sequences that are permutations of the natural numbers

Cf. A006368, A068156, A153893, A330531 (f_6), A330576 (inverse).

s=0; v=1; for (n=1, 10 000, print (n " " v); s+=2^v; for (w=1, oo, if (!bittest(s,w) && (v*w)%4==0, v=w; break)))

A204202 Triangle based on (0,2/3,1) averaging array.

Original entry on oeis.org

2, 2, 5, 2, 7, 11, 2, 9, 18, 23, 2, 11, 27, 41, 47, 2, 13, 38, 68, 88, 95, 2, 15, 51, 106, 156, 183, 191, 2, 17, 66, 157, 262, 339, 374, 383, 2, 19, 83, 223, 419, 601, 713, 757, 767, 2, 21, 102, 306, 642, 1020, 1314, 1470, 1524, 1535, 2, 23, 123, 408, 948

Offset: 1

Clark Kimberling, Jan 12 2012

See A204201 for a discussion of averaging arrays and related triangles

			First six rows:
2
2...5
2...7....11
2...9....18...23
2...11...27...41...47
2...13...38...68...88..95

Cf. A204201.

a = 0; r = 2/3; b = 1;
t[1, 1] = r;
t[n_, 1] := (a + t[n - 1, 1])/2;
t[n_, n_] := (b + t[n - 1, n - 1])/2;
t[n_, k_] := (t[n - 1, k - 1] + t[n - 1, k])/2;
u[n_] := Table[t[n, k], {k, 1, n}]
Table[u[n], {n, 1, 5}]   (* averaging array *)
u = Table[(1/r) 2^n*u[n], {n, 1, 12}];
TableForm[u]  (* A204202 triangle *)
Flatten[u]    (* A204202 sequence *)

From Philippe Deléham, Dec 24 2013: (Start)
T(n,n) = A055010(n) = A083329(n) = A153893(n-1).
Sum_{k=1..n} T(n,k) = A066373(n+1).
T(n,k) = T(n-1,k)+3*T(n-1,k-1)-2*T(n-2,k-1)-2*T(n-2,k-2), T(1,1)=2, T(2,1)=2, T(2,2)=5, T(n,k)=0 if k<1 or if k>n. (End)

A182460 a(n) = (3/5)*2^(4n+1) - (1/5).

Original entry on oeis.org

1, 19, 307, 4915, 78643, 1258291, 20132659, 322122547, 5153960755, 82463372083, 1319413953331, 21110623253299, 337769972052787, 5404319552844595, 86469112845513523, 1383505805528216371, 22136092888451461939, 354177486215223391027, 5666839779443574256435, 90669436471097188102963, 1450710983537555009647411

Offset: 0

Brad Clardy, Apr 30 2012

Bisection of A112627.

Index entries for linear recurrences with constant coefficients, signature (17,-16).

Cf. A112627, A016029, A153893.

[(3/5)*2^(4*n+1) - (1/5): n in [0..20]];

(3*2^(4*Range[0,20]+1)-1)/5 (* or *) LinearRecurrence[{17,-16},{1,19},30] (* Harvey P. Dale, Jul 21 2021 *)

a(n) = (3/5)*2^(4*n+1) - (1/5).
a(n) = 16*a(n-1) + 3 for n > 0.
a(n) = (1/5)*A153893(4*n+1).
a(n) = A016029(4*n+2).
a(n) = A112627(2*n+1).
G.f.: (1+2*x)/((1-x)*(1-16*x)). - Colin Barker, May 06 2012

A213668 Irregular triangle read by rows: T(n,k) is the number of dominating subsets with k vertices of the graph G(n) consisting of a pair of endvertices joined by n internally disjoint paths of length 2 (the n-ary generalized theta graph THETA_{2,2,...2}; n>=1, 1<=k<=n+2).

Original entry on oeis.org

1, 3, 1, 0, 6, 4, 1, 0, 7, 10, 5, 1, 0, 9, 16, 15, 6, 1, 0, 11, 25, 30, 21, 7, 1, 0, 13, 36, 55, 50, 28, 8, 1, 0, 15, 49, 91, 105, 77, 36, 9, 1, 0, 17, 64, 140, 196, 182, 112, 45, 10, 1, 0, 19, 81, 204, 336, 378, 294, 156, 55, 11, 1

Offset: 1

Emeric Deutsch, Jul 06 2012

Row n>=2 contains n+1 entries.
Sum of entries in row n=3*2^n-1 = A052940(n) = A153893(n) = A055010(n+1) = A083329(n+1).
The graph G(n) is the join of the graph consisting of 2 isolated vertices and the graph consisting of n isolated vertices. Then the expression of the domination polynomial follows from Theorem 12 of the Akbari et al. reference.

			Row 1 is 1,3,1 because the graph G(1) is the path abc; there are 1 dominating subset of size 1 ({b}), 3 dominating subsets of size 2 ({a,b}, {a,c}, {b,c}), and 1 dominating subset of size 3 ({a,b,c}).
Row 2 is 0,6,4,1 because the graph G(2) is the cycle a-b-c-d-a and has dominating subsets ab, ac, ad, bc, bd, cd, abc, abd, acd, bcd, and abcd (see A212634).
Triangle starts:
1,3,1;
0,6,4,1;
0,7,10,5,1;
0,9,16,15,6,1;

S. Akbari, S. Alikhani, and Y. H. Peng, Characterization of graphs using domination polynomials, European J. Comb., 31, 2010, 1714-1724.

S. Alikhani and Y. H. Peng, Introduction to domination polynomial of a graph, arXiv:0905.2251.
T. Kotek, J. Preen, F. Simon, P. Tittmann, and M. Trinks, Recurrence relations and splitting formulas for the domination polynomial, arXiv:1206.5926.

Cf. A052940, A153893, A055010, A083329, A212634

p := proc (n) options operator, arrow: ((1+x)^n-1)*((1+x)^2-1)+x^n+x^2 end proc: for n to 12 do seq(coeff(p(n), x, k), k = 1 .. n+2) end do; # yields sequence in triangular form

T[n_, k_] := SeriesCoefficient[((1+x)^n-1) ((1+x)^2-1)+x^n+x^2, {x, 0, k}];
Table[T[n, k], {n, 1, 9}, {k, 1, n+2}] // Flatten (* Jean-François Alcover, Dec 06 2017 *)

The generating polynomial of row n is p(n)=((1+x)^n-1)*((1+x)^2-1)+x^n+x^2; by definition, p(n) is the domination polynomial of the graph G(n).
Bivariate g.f.: x*z/(1-x*z)-2*x*z/(1-z)+x*z*(1+x)*(2+x)/(1-z-x*z).
T(n,3)=n^2 for n!=3.

A345401 a(n) is the unique odd number h such that BCR(h*2^m-1) = 2n (except for BCR(0) = 1) where BCR is bit complement and reverse per A036044.

Original entry on oeis.org

1, 3, 7, 5, 15, 11, 13, 9, 31, 23, 27, 19, 29, 21, 25, 17, 63, 47, 55, 39, 59, 43, 51, 35, 61, 45, 53, 37, 57, 41, 49, 33, 127, 95, 111, 79, 119, 87, 103, 71, 123, 91, 107, 75, 115, 83, 99, 67, 125, 93, 109, 77, 117, 85, 101, 69, 121, 89, 105, 73, 113, 81, 97, 65, 255, 191

Offset: 0

Bernard Schott, Jun 18 2021

This sequence is a permutation of the odd numbers.
We have BCR(a(n)*2^m-1) = 2n when n = 0 for m >= 1, and BCR(a(n)*2^m-1) = 2n when n >= 1 for m >= 0.
Why this exception when n = 0? As a(0) = 1, we have BCR(1*2^m-1) = 2*0 = 0 only for m >= 1, because, for m = 0, we have BCR(1*2^0-1) = BCR(0) = 1 <> 2*0 = 0.

			a(0) = 1 because BCR(1*2^m-1) = 2*0 = 0 for m >= 1 (A000225).
a(1) = 3 because BCR(3*2^m-1) = 2*1 = 2 for m >= 0 (A153893).
a(2) = 7 because BCR(7*2^m-1) = 2*2 = 4 for m >= 0 (A086224).
Indeed, a(1) = 3 because 3*2^m-1 = 1011..11_2 (i.e., 10 followed by m 1's), whose bit complement is 0100..00, which reverses to 10_2 = 2 = 2*1.
Also, a(43) = 75 because 75*2^m-1 = 100101011..11_2 (i.e., 1001010 followed by m 1's), whose bit complement is 011010100..00, which reverses to 1010110_2 = 86 = 2*43.

Cf. A036044 (BCR), A059894.

When BCR(n) = 0, 2, 4, 6, 8, 10, 12, then corresponding a(n) = h = 1, 3, 7, 5, 15, 11, 13 and numbers h*2^m-1 are respectively in A000225, A153893, A086224, A153894, A196305, A086225, A198274.

a(n) = BCR(2*n) + 1 for n >= 1.

a(n) = 2*A059894(n) + 1 for n >= 1. - Hugo Pfoertner, Jun 18 2021

cp's OEIS Frontend

A330530 Lexicographically earliest sequence of distinct positive integers such that the product of two consecutive terms is always divisible by 4.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

A204202 Triangle based on (0,2/3,1) averaging array.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Formula

A182460 a(n) = (3/5)*2^(4n+1) - (1/5).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

Formula

A213668 Irregular triangle read by rows: T(n,k) is the number of dominating subsets with k vertices of the graph G(n) consisting of a pair of endvertices joined by n internally disjoint paths of length 2 (the n-ary generalized theta graph THETA_{2,2,...2}; n>=1, 1<=k<=n+2).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A345401 a(n) is the unique odd number h such that BCR(h*2^m-1) = 2n (except for BCR(0) = 1) where BCR is bit complement and reverse per A036044.

Views

Author

Keywords

Comments

Examples

Crossrefs

Formula