cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A160161 First differences of the 3D toothpick numbers A160160.

Original entry on oeis.org

0, 1, 2, 4, 8, 8, 8, 8, 16, 32, 56, 32, 16, 8, 16, 32, 56, 56, 64, 80, 152, 232, 352, 144, 48, 32, 24, 40, 56, 56, 64, 80, 152, 232, 352, 216, 168, 176, 272, 360, 496, 448, 536, 664, 1168, 1488, 2000, 768, 304, 336, 264, 192, 112, 120, 128, 112, 168, 240, 352, 216, 168, 176, 272, 360, 496
Offset: 0

Views

Author

Omar E. Pol, May 03 2009

Keywords

Comments

Number of toothpicks added at n-th stage to the three-dimensional toothpick structure of A160160.
The sequence should start with a(1) = 1 = A160160(1) - A160160(0), the initial a(0) = 0 seems purely conventional and not given in terms of A160160. The sequence can be written as a table with rows r >= 0 of length 1, 1, 1, 3, 9, 18, 36, ... = 9*2^(r-4) for row r >= 4. In that case, rows 0..3 are filled with 2^r, and all rows r >= 3 have the form (x_r, y_r, x_r) where x_r and y_r have 3*2^(r-4) elements, all multiples of 8. Moreover, y_r[1] = a(A033484(r-2)) = x_{r+1}[1] = a(A176449(r-3)) is the largest element of row r and thus a record value of the sequence. - M. F. Hasler, Dec 11 2018

Examples

			Array begins:
===================
    x     y     z
===================
          0     1
    2     4     8
    8     8     8
   16    32    56
   32    16     8
   16    32    56
   56    64    80
  152   232   352
  144    48    32
...
From _Omar E. Pol_, Feb 28 2018: (Start)
Also, starting with 1, the sequence can be written as an irregular triangle in which the row lengths are the terms of A011782 multiplied by 3, as shown below:
   1,  2,  4;
   8,  8,  8;
   8, 16, 32, 56, 32, 16;
   8, 16, 32, 56, 56, 64, 80, 152, 232, 352, 144, 48;
  32, 24, 40, 56, 56, 64, 80, 152, 232, 352, 216, 168, 176, 272, 360, 496, 448, ...
(End)
If one starts rows with a(A176449(k) = 9*2^k-2), they are of the form A_k, B_k, A_k where A_k and B_k have 3*2^k elements and the first element of A_k is the first element of B_{k-1} and the largest of that (previous) row:
   k | a(9*2^k-2, ...) = A_k ; B_k ; A_k
  ---+-------------------------------------
     | a( 1 .. 6) = (1, 2, 4, 8, 8, 8)   (One might consider a row (8 ; 8 ; 8).)
   0 | a( 7, ...) = (8, 16, 32 ; 56, 32, 16 ; 8, 16, 32)
   1 | a(16, ...) = (56, 56, 64, 80, 152, 232 ; 352, 144, 48, 32, 24, 40 ;
     |               56, 56, 64, 80, 152, 232)
   2 | a(34, ...) = (352, 216, 168, 176, 272, 360, 496, 448, 536, 664, 1168, 1488 ;
     |               2000, 768, 304, 336, 264, 192, 112, 120, 128, 112, 168, 240 ;
     |               352, 216, 168, 176, 272, 360, 496, 448, 536, 664, 1168, 1488)
   3 | a(70, ...) = (2000, 984, ... ; 10576, 4304, ... ; 2000, 984, ...)
   4 | a(142, ...) = (10576, 5016, ... ; 54328, 24120, ...; 10576, 5016, ...)
  etc. - _M. F. Hasler_, Dec 11 2018
		

Crossrefs

Programs

  • PARI
    A160161_vec(n)=(n=A160160_vec(n))-concat(0,n[^-1]) \\ M. F. Hasler, Dec 11 2018
    
  • PARI
    A160161_vec(n)={local(E=[Vecsmall([1,1,1])], s(U)=[Vecsmall(Vec(V)+U)|V<-E], J=[], M, A, B, U); [if(i>4,8*#E=setminus(setunion(A=s(U=matid(3)[i%3+1,]), B=select(vecmin,s(-U))), J=setunion(setunion(setintersect(A, B), E), J)),2^(i-1))|i<-[1..n]]} \\ Returns the vector a(1..n). (A160160 is actually given as partial sums of this sequence, rather than the converse.) - M. F. Hasler, Dec 12 2018

Formula

a(9*2^k - m) = a(6*2^k - m) for all k >= 0 and 2 <= m <= 3*2^(k-1) + 2. - M. F. Hasler, Dec 12 2018

Extensions

Extended to 78 terms with C++ program by R. J. Mathar, Jan 09 2010
Edited and extended by M. F. Hasler, Dec 11 2018

A250656 T(n,k)=Number of (n+1)X(k+1) 0..1 arrays with nondecreasing x(i,j)-x(i,j-1) in the i direction and nondecreasing min(x(i,j),x(i-1,j)) in the j direction.

Original entry on oeis.org

9, 16, 19, 25, 34, 39, 36, 53, 70, 79, 49, 76, 109, 142, 159, 64, 103, 156, 221, 286, 319, 81, 134, 211, 316, 445, 574, 639, 100, 169, 274, 427, 636, 893, 1150, 1279, 121, 208, 345, 554, 859, 1276, 1789, 2302, 2559, 144, 251, 424, 697, 1114, 1723, 2556, 3581
Offset: 1

Views

Author

R. H. Hardin, Nov 26 2014

Keywords

Comments

Table starts
....9...16....25....36....49....64....81...100...121...144...169....196....225
...19...34....53....76...103...134...169...208...251...298...349....404....463
...39...70...109...156...211...274...345...424...511...606...709....820....939
...79..142...221...316...427...554...697...856..1031..1222..1429...1652...1891
..159..286...445...636...859..1114..1401..1720..2071..2454..2869...3316...3795
..319..574...893..1276..1723..2234..2809..3448..4151..4918..5749...6644...7603
..639.1150..1789..2556..3451..4474..5625..6904..8311..9846.11509..13300..15219
.1279.2302..3581..5116..6907..8954.11257.13816.16631.19702.23029..26612..30451
.2559.4606..7165.10236.13819.17914.22521.27640.33271.39414.46069..53236..60915
.5119.9214.14333.20476.27643.35834.45049.55288.66551.78838.92149.106484.121843

Examples

			Some solutions for n=4 k=4
..1..1..0..1..1....0..0..0..0..0....0..0..0..0..0....1..1..1..0..0
..0..0..0..1..1....1..1..1..1..1....1..1..1..1..1....0..0..0..0..0
..0..0..0..1..1....1..1..1..1..1....0..0..0..0..0....0..0..0..0..0
..0..0..0..1..1....0..0..0..0..0....1..1..1..1..1....1..1..1..1..1
..0..0..0..1..1....0..1..1..1..1....1..1..1..1..1....0..0..0..1..1
		

Crossrefs

Column 1 is A052549(n+1)
Column 2 is A176449
Column 3 is A156127(n+1)
Column 4 is A048487(n+2)
Row 1 is A000290(n+2)
Row 2 is A168244(n+3)

Formula

Empirical: T(n,k) = 2^(n-1)*k^2 + (5*2^(n-1)-1)*k + 2^(n+1)
Empirical for column k:
k=1: a(n) = 3*a(n-1) -2*a(n-2); also a(n) = 2^(n-1) +(5*2^(n-1) -1) +2^(n+1)
k=2: a(n) = 3*a(n-1) -2*a(n-2); also a(n) = 2^(n-1)*4 +(5*2^(n-1) -1)*2 +2^(n+1)
k=3: a(n) = 3*a(n-1) -2*a(n-2); also a(n) = 2^(n-1)*9 +(5*2^(n-1) -1)*3 +2^(n+1)
k=4: a(n) = 3*a(n-1) -2*a(n-2); also a(n) = 2^(n-1)*16 +(5*2^(n-1) -1)*4 +2^(n+1)
k=5: a(n) = 3*a(n-1) -2*a(n-2); also a(n) = 2^(n-1)*25 +(5*2^(n-1) -1)*5 +2^(n+1)
k=6: a(n) = 3*a(n-1) -2*a(n-2); also a(n) = 2^(n-1)*36 +(5*2^(n-1) -1)*6 +2^(n+1)
k=7: a(n) = 3*a(n-1) -2*a(n-2); also a(n) = 2^(n-1)*49 +(5*2^(n-1) -1)*7 +2^(n+1)
Empirical for row n:
n=1: a(n) = 1*n^2 + 4*n + 4
n=2: a(n) = 2*n^2 + 9*n + 8
n=3: a(n) = 4*n^2 + 19*n + 16
n=4: a(n) = 8*n^2 + 39*n + 32
n=5: a(n) = 16*n^2 + 79*n + 64
n=6: a(n) = 32*n^2 + 159*n + 128
n=7: a(n) = 64*n^2 + 319*n + 256

A360692 a(0) = 0. Thereafter a(n+1) = a(a(n)) if a(n) has not occurred previously, otherwise a(n+1) = n - 1 - a(n-1).

Original entry on oeis.org

0, 0, 0, 1, 0, 2, 0, 3, 1, 4, 0, 5, 2, 6, 0, 7, 3, 8, 1, 9, 4, 10, 0, 11, 5, 12, 2, 13, 6, 14, 0, 15, 7, 16, 3, 17, 8, 18, 1, 19, 9, 20, 4, 21, 10, 22, 0, 23, 11, 24, 5, 25, 12, 26, 2, 27, 13, 28, 6, 29, 14, 30, 0, 31, 15, 32, 7, 33, 16, 34, 3, 35, 17, 36, 8
Offset: 0

Views

Author

David James Sycamore, Feb 16 2023

Keywords

Comments

An inductive argument shows that a(n) <= n for all n, with equality iff n = 0. It follows that a(n) is well defined, and the sequence is infinite.
Apart from a(1) = 0 every repeat term is followed by a novel term, and vice versa.
Every nonnegative integer appears infinitely many times.
The proper subsequence given by a(2*k) for k >= 2 is the sequence itself, which is therefore fractal.
Starting from a(1) = 0 the sequence is the nonnegative integers interleaved with itself.

Examples

			a(0) = 0 is a novel term so a(1) = a(a(0)) = 0. Since a(1) is a repeat term a(2) = 0 - a(0) = 0 - 0 = 0. a(1,2) = 0,0 is the only case of consecutive repeat terms.
Since a(2) = 0 is a repeat term, a(3) = 1 - a(1) = 1 - 0 = 1, a novel term so a(4) = a(a(1)) = 0, and so on.
a(16) = 3, a repeat term (last seen at a(7)), so a(17) = 15 - a(15) = 15 - 7 = 8.
		

Crossrefs

Programs

Formula

a(2*n + 1) = n for all n >= 0.
A027383(n) = 0. (n >= 0) gives the positions of all zeros after a(0) = 0.
a((2*k + 3)*2^n - 2) = k (n >= 0) gives the positions of all k > 0.
The number of nonnegative terms occurring between consecutive zeros is 0,0,1,1,3,3,7,7,15,15,... (A000225(n), repeat).
a(n) = A101279(n+2) - 1. - Rémy Sigrist, Feb 18 2023

Extensions

More terms from Rémy Sigrist, Feb 18 2023

A370882 Square array T(n,k) = 9*2^k - n read by ascending antidiagonals.

Original entry on oeis.org

9, 8, 18, 7, 17, 36, 6, 16, 35, 72, 5, 15, 34, 71, 144, 4, 14, 33, 70, 143, 288, 3, 13, 32, 69, 142, 287, 576, 2, 12, 31, 68, 141, 286, 575, 1152, 1, 11, 30, 67, 140, 285, 574, 1151, 2304, 0, 10, 29, 66, 139, 284, 573, 1150, 2303, 4608, -1, 9, 28, 65, 138, 283, 572, 1149, 2302, 4607, 9216
Offset: 0

Views

Author

Paul Curtz, Mar 05 2024

Keywords

Comments

Just after A367559 and A368826.

Examples

			Table begins:
       k=0  1  2  3   4   5
  n=0:   9 18 36 72 144 288 ...
  n=1:   8 17 35 71 143 287 ...
  n=2:   7 16 34 70 142 286 ...
  n=3:   6 15 33 69 141 285 ...
  n=4:   5 14 32 68 140 284 ...
  n=5:   4 13 31 67 139 283 ...
Every line has the signature (3,-2). For n=1: 3*17 - 2*8 = 35.
Main diagonal's difference table:
  9   17   34   69   140   283   570  1145  ...  =  b(n)
  8   17   35   71   143   287   575  1151  ...  =  A052996(n+2)
  9   18   36   72   144   288   576  1152  ...  =  A005010(n)
  ...
b(n+1) - 2*b(n) = A023443(n).
		

Crossrefs

Programs

  • Mathematica
    T[n_, k_] := 9*2^k - n; Table[T[n - k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Mar 06 2024 *)

Formula

T(0,k) = 9*2^k = A005010(k);
T(1,k) = 9*2^k - 1 = A052996(k+2);
T(2,k) = 9*2^k - 2 = A176449(k);
T(3,k) = 9*2^k - 3 = 3*A083329(k);
T(4,k) = 9*2^k - 4 = A053209(k);
T(5,k) = 9*2^k - 5 = A304383(k+3);
T(6,k) = 9*2^k - 6 = 3*A033484(k);
T(7,k) = 9*2^k - 7 = A154251(k+1);
T(8,k) = 9*2^k - 8 = A048491(k);
T(9,k) = 9*2^k - 9 = 3*A000225(k).
G.f.: (9 - 9*y + x*(11*y - 10))/((1 - x)^2*(1 - y)*(1 - 2*y)). - Stefano Spezia, Mar 17 2024
Showing 1-4 of 4 results.