Displaying 1-10 of 11 results found.
Recursive sequence based on the central polygonal numbers ( A000124) and A004738.
+20
2
1, 2, 4, 3, 6, 9, 7, 8, 10, 13, 5, 15, 12, 14, 16, 19, 11, 23, 20, 17, 22, 18, 24, 27, 21, 31, 35, 28, 32, 34, 26, 33, 29, 37, 25, 41, 45, 39, 47, 30, 44, 46, 42, 40, 36, 49, 43, 53, 57, 51, 58, 50, 61, 54, 52, 60, 55, 59, 38, 63, 56, 67, 71, 65, 72, 75, 70
COMMENTS
Conjectured to be a permutation of the natural numbers.
The central polygonal numbers can be constructed by starting with the natural numbers, setting A000124(0)=1 and obtaining A000124(n+1) by reversing the order of the next A000124(n) numbers after A000124(n). This procedure doesn't produce a permutation of the natural numbers for A000124 because the sequence is strictly increasing. The present sequence is constructed by the same procedure, except that a(n+1) is obtained by reversing the next a( A004738(n)) numbers.
EXAMPLE
Start with the natural numbers:
1, 2, 3, 4, 5, 6, 7, 8, 9...
a( A004738(1))=1, so reverse the order of the next term, leaving the sequence unchanged: (1)
1, (2), 3, 4, 5, 6, 7, 8, 9...
a( A004738(2))=2, so reverse the order of the next 2 terms: (2)
1, 2, (4, 3), 5, 6, 7, 8, 9...
a( A004738(3))=1, so reverse the order of the next term, leaving the sequence unchanged: (1)
1, 2, 4, (3), 5, 6, 7, 8, 9...
a( A004738(4))=2, so reverse the order of the next 2 terms: (2)
1, 2, 4, 3, (6, 5), 7, 8, 9...
a( A004738(5))=4, so reverse the order of the next 4 terms: (4)
1, 2, 4, 3, 6, (9, 8, 7, 5)...
a( A004738(6))=2, so reverse the order of the next 2 terms: (2)
1, 2, 4, 3, 6, 9, (7, 8), 5...
a( A004738(7))=1, so reverse the order of the next term, leaving the sequence unchanged: (1)
1, 2, 4, 3, 6, 9, 7, (8), 5...
Sums of rows of the sequence of triangles with nonnegative integers and row widths defined by A004738.
+20
1
0, 3, 3, 9, 21, 19, 11, 25, 45, 74, 66, 49, 26, 55, 90, 134, 190, 170, 138, 97, 50, 103, 162, 230, 310, 405, 365, 310, 243, 167, 85, 173, 267, 370, 485, 615, 763, 693, 605, 502, 387, 263, 133, 269, 411, 562, 725, 903, 1099, 1316, 1204, 1071, 920, 754, 576, 389
COMMENTS
Row widths: A004738(n): 1, 2, 1, 2, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 4, 5... Pits: A051925(n+1): 0, 3, 11, 26, 50, 85, 133, 196, 276, 375, 495, 638... Peak tops: A007290(n+3): 2, 8, 20, 40, 70, 112, 168, 240, 330, 440, 572... Peak bases: A084990(n+1): 1, 6, 17, 36, 65, 106, 161, 232, 321, 430, 561...
EXAMPLE
The sequence of triangles begins:
0
1 2
3
4 5
6 7 8
9 10
11
12 13
14 15 16
17 18 19 20
21 22 23
24 25
26
27 28
29 30 31
32 33 34 35
36 37 38 39 40
41 42 43 44
45 46 47
48 49
50
51 52
PROG
(Python)
curSign=-1
curLength=sum=0
rowLength=topLength=1
for n in range(1232):
sum += n
curLength += 1
if curLength==rowLength:
print(sum, end=', ')
curLength = sum = 0
if rowLength==1 or rowLength==topLength:
curSign = -curSign
topLength += (rowLength==1)
rowLength += curSign
CROSSREFS
Cf. A027480: sums of rows of a triangle with increasing row widths: 0; 1,2; 3,4,5; 6,7,8,9; ...
Indices of occurrences of 2 in A004738.
+20
0
2, 4, 6, 8, 12, 14, 20, 22, 30, 32, 42, 44, 56, 58, 72, 74, 90, 92, 110, 112, 132, 134, 156, 158, 182, 184, 210, 212, 240, 242, 272, 274, 306, 308, 342, 344, 380, 382, 420, 422, 462, 464, 506, 508, 552, 554, 600, 602, 650, 652, 702, 704, 756, 758, 812, 814, 870, 872, 930, 932, 992, 994, 1056, 1058, 1122, 1124, 1190, 1192, 1260, 1262, 1332, 1334, 1406, 1408, 1482, 1484, 1560, 1562
COMMENTS
Indices of occurrences of 1 in A004738 are given by A002061, b(n) = n^2 - n + 1 (the central polygonal numbers). All entries are even.
MAPLE
local f ;
f := floor(sqrt(n)+1/2) ;
f+1-abs(n-1-f^2) ;
end proc:
for n from 1 to 1600 do
printf("%d, ", n) ;
end if;
MATHEMATICA
LinearRecurrence[{1, 2, -2, -1, 1}, {2, 4, 6, 8, 12}, 80] (* Harvey P. Dale, Jun 16 2017 *)
Partial sums of A004738, leftmost column of the sequence of triangles defined in A206492.
+20
0
1, 3, 4, 6, 9, 11, 12, 14, 17, 21, 24, 26, 27, 29, 32, 36, 41, 45, 48, 50, 51, 53, 56, 60, 65, 71, 76, 80, 83, 85, 86, 88, 91, 95, 100, 106, 113, 119, 124, 128, 131, 133, 134, 136, 139, 143, 148, 154, 161, 169, 176, 182, 187, 191, 194, 196, 197, 199, 202
FORMULA
For n>0, a(n) = a(n-1) + A004738(n).
Concatenation of sequences (1,2,...,n-1,n,n-1,...,1) for n >= 1.
+10
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
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
Pyramidal sequence: distance to nearest product of two consecutive integers (promic or heteromecic numbers).
+10
11
0, 1, 0, 1, 2, 1, 0, 1, 2, 3, 2, 1, 0, 1, 2, 3, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 9, 8, 7
COMMENTS
Table A049581 T(n,k) = |n-k| read by sides of squares from T(1,n) to T(n,n), then from T(n,n) to T(n,1). - Boris Putievskiy, Jan 29 2013
FORMULA
Let u(1)=1, u(n) = n - u(n-sqrtint(n)) (cf. A037458); then a(0)=0 and for n > 0 a(n) = 2*u(n) - n. - Benoit Cloitre, Dec 22 2002 a(0)=0 then a(n) = floor(sqrt(n)) - a(n - floor(sqrt(n))). - Benoit Cloitre, May 03 2004
EXAMPLE
a(10) = |10 - 3*4| = 2.
The start of the sequence as table:
0, 1, 2, 3, 4, 5, 6, 7, ...
1, 0, 1, 2, 3, 4, 5, 6, ...
2, 1, 0, 1, 2, 3, 4, 5, ...
3, 2, 1, 0, 1, 2, 3, 4, ...
4, 3, 2, 1, 0, 1, 2, 3, ...
5, 4, 3, 2, 1, 0, 1, 2, ...
6, 5, 4, 3, 2, 1, 0, 1, ...
...
The start of the sequence as triangle array read by rows:
0;
1, 0, 1;
2, 1, 0, 1, 2;
3, 2, 1, 0, 1, 2, 3;
4, 3, 2, 1, 0, 1, 2, 3, 4;
5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5;
6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6;
7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7;
...
Row number r contains 2*r-1 numbers: r-1, r-2, ..., 0, 1, 2, ..., r-1. (End)
MATHEMATICA
Join[{0}, Module[{nn=150, ptci}, ptci=Times@@@Partition[Range[nn/2+1], 2, 1]; Table[Abs[n-Nearest[ptci, n]], {n, nn}][[All, 1]]]] (* Harvey P. Dale, Aug 29 2020 *)
PROG
(PARI) a(n)=if(n<1, 0, sqrtint(n)-a(n-sqrtint(n)))
Concatenation of sequences (1,2,2,...,n-1,n-1,n,n,n-1,n-1,...,2,2,1) for n >= 1.
+10
5
1, 1, 1, 2, 2, 1, 1, 2, 3, 3, 2, 1, 1, 2, 3, 4, 4, 3, 2, 1, 1, 2, 3, 4, 5, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 6, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 7, 7, 6, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 7, 8, 8, 7, 6, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 9, 8, 7, 6, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 10, 9, 8, 7
COMMENTS
Zeta(2, k/p) + Zeta(2, (p-k)/p) = (Pi/sin((Pi*a(n))/p))*2, where p=2,3,4, k=1..p-1.
This sequence is the odd subset of A003983 for odd p=3,5,7,9,.... Table T(n,k) n, k > 0, T(n,k) = n-k+1, if n >= k, T(n,k) = k-n, if n < k. Table read by sides of squares from T(1,n) to T(n,n), then from T(n,n) to T(n,1). General case A209301. Let m be a natural number. The first column of the table T(n,1) is the sequence of the natural numbers A000027. In all columns with number k (k > 1) the segment with the length of (k-1): {m+k-2, m+k-3, ..., m} shifts the sequence A000027. For m=1 the result is A004739, for m=2 the result is A004738, for m=3 the result is A209301. - Boris Putievskiy, Jan 24 2013
FORMULA
For the general case,
a(n) = m*v + (2*v-1)*(t*t-n) + t, where t = floor((sqrt(n) - 1/2) + 1 and v = floor((n-1)/t) - t + 1.
For m=1,
a(n) = v + (2*v-1)*(t*t-n) + t, where t = floor((sqrt(n) - 1/2) + 1 and v = floor((n-1)/t) - t + 1. (End)
EXAMPLE
The start of the sequence as table:
1, 1, 2, 3, 4, 5, 6, ...
2, 1, 1, 2, 3, 4, 5, ...
3, 2, 1, 1, 2, 3, 4, ...
4, 3, 2, 1, 1, 2, 3, ...
5, 4, 3, 2, 1, 1, 2, ...
6, 5, 4, 3, 2, 1, 1, ...
7, 6, 5, 4, 3, 2, 1, ...
...
The start of the sequence as triangle array read by rows:
1;
1, 1, 2;
2, 1, 1, 2, 3;
3, 2, 1, 1, 2, 3, 4;
4, 3, 2, 1, 1, 2, 3, 4, 5;
5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6;
6, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 7;
...
Row number r contains 2*r - 1 numbers: r-1, r-2, ..., 1, 1, 2, ..., r. (End)
MATHEMATICA
aa = {}; Do[Do[AppendTo[aa, (p/Pi) ArcSin[Sqrt[1/((1/Pi^2) (Zeta[2, k/p] + Zeta[2, (p - k)/p]))]]], {k, 1, p - 1}], {p, 3, 50, 2}]; Round[N[aa, 50]] (* Artur Jasinski, Mar 07 2010 *)
PROG
(Haskell)
a004739 n = a004739_list !! (n-1)
a004739_list = concat $ map (\n -> [1..n] ++ [n, n-1..1]) [1..]
Pyramidal sequence built with powers of 2.
+10
3
1, 2, 1, 2, 4, 2, 1, 2, 4, 8, 4, 2, 1, 2, 4, 8, 16, 8, 4, 2, 1, 2, 4, 8, 16, 32, 16, 8, 4, 2, 1, 2, 4, 8, 16, 32, 64, 32, 16, 8, 4, 2, 1, 2, 4, 8, 16, 32, 64, 128, 64, 32, 16, 8, 4, 2, 1, 2, 4, 8, 16, 32, 64, 128, 256, 128, 64, 32, 16, 8, 4, 2, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 256, 128
EXAMPLE
Triangle begins:
1
2 1 2
4 2 1 2 4
8 4 2 1 2 4 8
16 8 4 2 1 2 4 8 16
32 16 8 4 2 1 2 4 8 16 32
64 32 16 8 4 2 1 2 4 8 16 32 64
128 64 32 16 8 4 2 1 2 4 8 16 32 64 128
256 128 64 32 16 8 4 2 1 2 4 8 16 32 64 128 256
512 256 128 64 32 16 8 4 2 1 2 4 8 16 32 64 128 256 512
1024 512 256 128 64 32 16 8 4 2 1 2 4 8 16 32 64 128 256 512 1024
MATHEMATICA
pow2Pyram[row_] := Module[{st = 2^Range[0, row]}, Join[st, Reverse[Most[Rest[st]]]]]; Flatten[Array[pow2Pyram, 10]] (* Harvey P. Dale, May 09 2012 *) Flatten[Table[Table[2^Abs[col], {col, -row, row}], {row, 0, 7}]] (* Alonso del Arte, Apr 15 2017 *)
PROG
(PARI) for(i=0, 9, forstep(j=i, 0, -1, print1(1<<j", ")); for(j=1, i, print1(1<<j", "))) \\ Charles R Greathouse IV, Mar 20 2013
Table T(n,k) n, k > 0, T(n,k)=n-k+1, if n>=k, T(n,k)=k-n+2, if n < k. Table read by sides of squares from T(1,n) to T(n,n), then from T(n,n) to T(n,1).
+10
3
1, 3, 1, 2, 4, 3, 1, 2, 3, 5, 4, 3, 1, 2, 3, 4, 6, 5, 4, 3, 1, 2, 3, 4, 5, 7, 6, 5, 4, 3, 1, 2, 3, 4, 5, 6, 8, 7, 6, 5, 4, 3, 1, 2, 3, 4, 5, 6, 7, 9, 8, 7, 6, 5, 4, 3, 1, 2, 3, 4, 5, 6, 7, 8, 10, 9, 8, 7, 6, 5, 4, 3, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 10, 9, 8, 7
COMMENTS
In general, let m be natural number. The first column of the table T(n,1) is the sequence of the natural numbers A000027. In all columns with number k (k > 1) the segment with the length of (k-1): {m+k-2, m+k-3, ..., m} shifts the sequence A000027. For m=1 the result is A004739, for m=2 the result is A004738. This sequence is result for m=3.
FORMULA
For the general case
a(n ) = m*v+(2*v-1)*(t*t-n)+t,
where
t = floor((sqrt(n)-1/2)+1,
v = floor((n-1)/t)-t+1.
For m=3
a(n ) = 3*v+(2*v-1)*(t*t-n)+t,
where
t = floor((sqrt(n)-1/2)+1,
v = floor((n-1)/t)-t+1.
EXAMPLE
The start of the sequence as table for the general case:
1....m..m+1..m+2..m+3..m+4..m+5...
2....1....m..m+1..m+2..m+3..m+4...
3....2....1....m..m+1..m+2..m+3...
4....3....2....1....m..m+1..m+2...
5....4....3....2....1....m..m+1...
6....5....4....3....2....1....m...
7....6....5....4....3....2....1...
...
The start of the sequence as triangle array read by rows for the general case:
1;
m,1,2;
m+1,m,1,2,3;
m+2,m+1,m,1,2,3,4;
m+3,m+2,m+1,m,1,2,3,4,5;
m+4, m+3,m+2,m+1,m,1,2,3,4,5,6;
m+5, m+4, m+3,m+2,m+1,m,1,2,3,4,5,6,7;
...
Row number r contains 2*r -1 numbers: m+r-2, m+r-1,...m,1,2,...r.
The start of the sequence as triangle array read by rows for m=3:
1;
3,1,2;
4,3,1,2,3;
5,4,3,1,2,3,4;
6,5,4,3,1,2,3,4,5;
7,6,5,4,3,1,2,3,4,5,6;
8,7,6,5,4,3,1,2,3,4,5,6,7;
...
PROG
(Python)
t=int((math.sqrt(n))-0.5)+1
v=int((n-1)/t)-t+1
result=k*v+(2*v-1)*(t**2-n)+t
a(1)=0,a(2)=0,a(3)=1 then a(n)=abs(a(n-1)-a(n-2))-a(n-3).
+10
1
0, 0, 1, 1, 0, 0, -1, 1, 2, 2, -1, 1, 0, 2, 1, 1, -2, 2, 3, 3, -2, 2, 1, 3, 0, 2, -1, 3, 2, 2, -3, 3, 4, 4, -3, 3, 2, 4, -1, 3, 0, 4, 1, 3, -2, 4, 3, 3, -4, 4, 5, 5, -4, 4, 3, 5, -2, 4, 1, 5, 0, 4, -1, 5, 2, 4, -3, 5, 4, 4, -5, 5, 6, 6, -5, 5, 4, 6, -3, 5, 2, 6, -1, 5, 0, 6, 1, 5, -2, 6, 3, 5, -4, 6, 5, 5, -6, 6, 7, 7
COMMENTS
See A104156 for an order 2 example. b(n) = a(2n)-floor(sqrt(n))+1 is an infinite binary word consisting of a sequence of block (0,1) and single 0's where 0's occur when n is of form k^2-1, k>=2 i.e. b(n) begins for n>=1 : (0,1),0,(0,1),(0,1),0,(0,1),(0,1),(0,1),0,(0,1),... and single 0's occur at n=3,8,15,...
REFERENCES
B. Cloitre, On strange predictible recursions, preprint 2006
FORMULA
abs(a(2n-1)) = A004738(n)-1 where sign(a(2*n-1)) alternates between 2 consecutive zeros.
PROG
(PARI) an=vector(10000); an[1]=0; an[2]=0; an[3]=1; a(n)=if(n<0, 0, an[n]);
for(n=4, 10000, an[n]=abs(a(n-1)-a(n-2))-a(n-3))
an
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