|
| |
|
|
A004737
|
|
Concatenation of sequences (1,2,...,n-1,n,n-1,...,1) for n >= 1.
|
|
36
|
|
|
|
1, 1, 2, 1, 1, 2, 3, 2, 1, 1, 2, 3, 4, 3, 2, 1, 1, 2, 3, 4, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 7, 6, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 7, 8, 7, 6, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 8, 7, 6, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,3
|
|
|
COMMENTS
|
The ordinal transform of a sequence b_0, b_1, b_2, ... is the sequence a_0, a_1, a_2, ... where a_n is the number of times b_n has occurred in {b_0 ... b_n}.
This sequence is the even subset of A003983 for odd p=2,4,6,8,....
Given the triangle rows: (1; 1,2,1; 1,2,3,2,1; ...) as polcoeff with offset 0:
q = (1 + 2x + x^2), r = (1 + 2x + 3x^2 + 2x^3 +x^4), etc.; then
(1 + 2x + 3x^2 + ...) = q(x) * q(x^2) * q(x^4) * q(x^8) * ...
..................... = r(x) * r(x^3) * r(x^9) * r(x^27) * ...
..................... = s(x) * s(x^4) * s(x^16)* s(x^64) * ...
... (End)
Let U_1(t)=1, U_2(t)=2*t, and U_r(t)=2*t*U_(r-1)(t)-U(r-2)(t), r>2, be Chebyshev polynomials of the second kind. For q>1 an integer, let N=2*q and x_k=cos((2*k-1)*Pi/N), and define the ordered column vectors V_k=[U_k(x_1), U_k(x_2), ..., U_k(x_q)]^T, k=1,...,q, where A^T denotes the transpose of matrix A. Let E_N=[V_1, V_2, ..., V_q] be the q X q matrix formed from the ordered components of the V_k. E_N contains the joint spectra of the Danzer basis (see [Jeffery]) associated with N. Let M_N=(1/q)*[E_N]^T*E_N. For the trivial case q=1, let M_2=[1]. CONJECTURE: E_N and M_N are always integral and symmetric, with M_N having diagonal entries {1,2,...} beginning at entries 1,j (j odd) in the first row and i,1 (i odd) in the first column and with zeros elsewhere. If N is allowed to increase without bound, and assuming the conjecture is true, then triangle A004737 emerges in its entirety from the successive antidiagonals containing those entries [M_N]_(i,j) such that i+j=2*v, for each v in {1,2,...,floor((q+1)/2)}. For example, for N=18 and q=9 (omitting the zeros for clarity),
M_18=[
(1 1 1 1 1);
( 2 2 2 2 );
(1 3 3 3 3);
( 2 4 4 4 );
(1 3 5 5 5);
( 2 4 6 6 );
(1 3 5 7 7);
( 2 4 6 8 );
(1 3 5 7 9)],
from which the first five rows of the sequence can be read off in succession. (End)
T(n,k) = min(n,k). The order of the list T(n,k) is by sides of squares from T(1,n) to T(n,n), then from T(n,n) to T(n,1). - Boris Putievskiy, Jan 13 2013
Expanded form of T(2,k) k=0,1,...,2m for ascending m-nomial triangles. - Bob Selcoe, Feb 07 2014
Terms in the first nine rows of the triangle can be duplicated by performing (111...)^2 with <= nine ones. By way of example, (11111)^2 = 123454321. - Gary W. Adamson, Mar 27 2015
|
|
|
REFERENCES
|
Miklós Laczkovich, Conjecture and Proof, TypoTex, Budapest, 1998. See Chapter 10.
F. Smarandache, "Numerical Sequences", University of Craiova, 1975.
|
|
|
LINKS
|
Jerry Brown et al., Problem 4619, School Science and Mathematics, USA, Vol. 97 (4), 1997, pp. 221-222.
|
|
|
FORMULA
|
a(n) = if n<3 then 1 else (if a(n-1)=1 then 1 + 0^(a(n-2)-1) else a(n-1) - 0^X + (a(n-1)-a(n-2))*(1 - 0^X)), where X = A003059(n-1)-a(n-1). - Reinhard Zumkeller, Mar 10 2006
If the sequence is read as a triangular array, beginning [1]; [1,2,1]; [1,2,3,2,1]; ..., then the o.g.f. is (1+qx)/((1-x)(1-qx)(1-q^2x)) = 1 + x(1 + 2q + q^2) + x^2(1 + 2q + 3q^2 + 2q^3 +q^4) + .... The row polynomials for this triangle are (1 + q + ... + q^n)^2 =[n,2]_q + q[n-1,2]_q, where [n,2]_q are Gaussian polynomials (see A008967). - Peter Bala, Sep 23 2007
a(n) = floor(sqrt(n-1)) - |n - floor(sqrt(n-1))^2 - floor(sqrt(n-1)) - 1| + 1. - Boris Putievskiy, Jan 13 2013
Read as a triangular array, then T(n,k) = n - |n-k-1|; T(n,0) = 1; T(n,n-1) = n. - Juan Pablo Herrera P., Oct 17 2016
|
|
|
EXAMPLE
|
The start of the sequence as a table:
1 1 1 1 1 1 ...
1 2 2 2 2 2 ...
1 2 3 3 3 3 ...
1 2 3 4 4 4 ...
1 2 3 4 5 5 ...
1 2 3 4 5 6 ...
...
The start of the sequence as an irregular triangle array read by rows:
1;
1,2,1;
1,2,3,2,1;
1,2,3,4,3,2,1;
1,2,3,4,5,4,3,2,1;
1,2,3,4,5,6,5,4,3,2,1;
...
Row number k contains 2*k-1 numbers: 1,2,...,k-1,k,k-1,...,1. (End)
The sequence of fractions A196199/A004737 = 0/1, -1/1, 0/2, 1/1, -2/1, -1/2, 0/3, 1/2, 2/1, -3/1, -2/2, -1/3, 0/4, 1/3, 2/2, 3/1, -4/4. -3/2, ... contains every rational number (infinitely often) [Laczkovich]. - N. J. A. Sloane, Oct 09 2013
|
|
|
MATHEMATICA
|
Table[Min[n - #^2, (# + 1)^2 - n + 1] &@ Floor[Sqrt[n - 1]], {n, 105}] (* or *)
Table[Floor@ # - Abs[n - Floor[#]^2 - Floor@ # - 1] + 1 &@ Sqrt[n - 1], {n, 105}] (* Michael De Vlieger, Oct 21 2016 *)
Table[Join[Range[n], Range[n-1, 1, -1]], {n, 20}]//Flatten (* Harvey P. Dale, Dec 27 2019 *)
|
|
|
PROG
|
(Haskell)
import Data.List (inits)
a004737 n = a004737_list !! (n-1)
a004737_list = concatMap f $ tail $ inits [1..]
where f xs = xs ++ tail (reverse xs)
(PARI) a(n) = n--; my(m=sqrtint(n)); m+1-abs(n-m^2-m) \\ David A. Corneth, Oct 18 2016
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,frac,easy,tabf
|
|
|
AUTHOR
|
R. Muller
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
