A014739 -id:A014739 - cp's OEIS frontend

A194050 Natural fractal sequence of A014739.

1, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 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, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

Offset: 1

Clark Kimberling, Aug 13 2011

nonn

See A194029 for definitions of natural fractal sequence and natural interspersion.

Cf. A194029, A194051, A194052.

z = 50;
c[k_] := LucasL[k + 1] - 2;
c = Table[c[k], {k, 1, z}]  (* A014739 *)
f[n_] := If[MemberQ[c, n], 1, 1 + f[n - 1]]
f = Table[f[n], {n, 1, 600}]  (* A194050 *)
r[n_] := Flatten[Position[f, n]]
t[n_, k_] := r[n][[k]]
TableForm[Table[t[n, k], {n, 1, 8}, {k, 1, 7}]]
p = Flatten[Table[t[k, n - k + 1], {n, 1, 12}, {k, 1, n}]] (* A194051 *)
q[n_] := Position[p, n]; Flatten[Table[q[n], {n, 1, 90}]] (* A194052 *)

A192744 Constant term of the reduction by x^2->x+1 of the polynomial p(n,x) defined below in Comments.

Original entry on oeis.org

1, 1, 3, 8, 29, 133, 762, 5215, 41257, 369032, 3676209, 40333241, 483094250, 6271446691, 87705811341, 1314473334832, 21017294666173, 357096406209005, 6424799978507178, 122024623087820183, 2439706330834135361, 51219771117454755544

Offset: 0

Clark Kimberling, Jul 09 2011

nonn

The titular polynomial is defined recursively by p(n,x)=x*p(n-1,x)+n! for n>0, where p(0,x)=1; see the Example.  For an introduction to polynomial reduction, see A192232.  The discussion at A192232 Comments continues here:
...
Let R(p,q,s) denote the "reduction of polynomial p by q->s" as defined at A192232.  Suppose that q(x)=x^k for some k>0 and that s(x)=s(k,0)*x^(k-1)+s(k,1)*x^(k-2)+...+s(k,k-2)x+s(k,k-1).
...
First, we shall take p(x)=x^n, where n>=0; the results will be used to formulate R(p,q,s) for general p.  Represent R(x^n,q,s) by
...
R(x^n)=s(n,0)*x^(k-1)+s(n,1)*x^(k-2)+...+s(n,k-2)*x+s(n,k-1).
...
Then each of the sequences u(n)=s(n,h), for h=0,1,...,k-1, satisfies this linear recurrence relation:
...
u(n)=s(k,0)*u(n-1)+s(k,1)*u(n-2)+...+s(k,k-2)*u(n-k-1)+s(k,k-1)*u(n-k), with initial values tabulated here:
...
n: ..s(n,0)...s(n,1)..s(n,2).......s(n,k-2)..s(n,k-1)
0: ....0........0.......0..............0.......1
1: ....0........0.......0..............1.......0
...
k-2: ..0........1.......0..............0.......0
k-1: ..0........0.......0..............0.......0
k: ..s(k,0)...s(k,1)..s(k,2).......s(k,k-2)..s(k,k-1)
...
That completes the formulation for p(x)=x^n.  Turning to the general case, suppose that
...
p(n,x)=p(n,0)*x^n+p(n,1)*x^(n-1)+...+p(n,n-1)*x+p(n,n)
...
is a polynomial of degree n>=0.  Then the reduction denoted by (R(p(n,x) by x^k -> s(x)) is the polynomial of degree k-1 given by the matrix product P*S*X, where P=(p(n,0)...p(n,1).........p(n-k)...p(n,n-k+1); X has all 0's except for main diagonal (x^(k-1), x^(k-2)...x,1); and S has
...
row 1: ... s(n,0) ... s(n,1) ...... s(n,k-2) . s(n,k-1)
row 2: ... s(n-1,0) . s(n-1,1) .... s(n-1,k-2) s(n-1,k-1)
...
row n-k+1: s(k,0).... s(k,1) ...... s(k,k-2) ..s(k,k-1)
row n-k+2: p(n,n-k+1) p(n,n-k+2) .. p(n,n-1) ..p(n,n)
*****
As a class of examples, suppose that (v(n)), for n>=0, is a sequence, that p(0,x)=1, and p(n,x)=v(n)+p(n-1,x) for n>0.  If q(x)=x^2 and s(x)=x+1, and we write the reduction R(p(n,x)) as u1(n)*x+u2(n), then the sequences u1 and u2 are convolutions with the Fibonacci sequence, viz., let F=(0,1,1,2,3,5,8,...)=A000045 and let G=(1,0,1,1,2,3,5,8...); then u1=G**v and u2=F**v, where ** denotes convolution.  Examples (with a few exceptions for initial terms):
...
If v(n)=n! then u1=A192744, u2=A192745.
If v(n)=n+1 then u1=A000071, u2=A001924.
If v(n)=2n then u1=A014739, u2=A027181.
If v(n)=2n+1 then u1=A001911, u2=A001891.
If v(n)=3n+1 then u1=A027961, u2=A023537.
If v(n)=3n+2 then u1=A192746, u2=A192747.
If v(n)=3n then u1=A154691, u2=A192748.
If v(n)=4n+1 then u1=A053311, u2=A192749.
If v(n)=4n+2 then u1=A192750, u2=A192751.
If v(n)=4n+3 then u1=A192752, u2=A192753.
If v(n)=4n then u1=A147728, u2=A023654.
If v(n)=5n+1 then u1=A192754, u2=A192755.
If v(n)=5n then u1=A166863, u2=A192756.
If v(n)=floor((n+1)tau) then u1=A192457, u2=A023611.
If v(n)=floor((n+2)/2) then u1=A052952, u2=A129696.
If v(n)=floor((n+3)/3) then u1=A004695, u2=A178982.
If v(n)=floor((n+4)/4) then u1=A080239, u2=A192758.
If v(n)=floor((n+5)/5) then u1=A124502, u2=A192759.
If v(n)=n+2 then u1=A001594, u2=A192760.
If v(n)=n+3 then u1=A022318, u2=A192761.
If v(n)=n+4 then u1=A022319, u2=A192762.
If v(n)=2^n then u1=A027934, u2=A008766.
If v(n)=3^n then u1=A106517, u2=A094688.

			The first five polynomials and their reductions:
1 -> 1
1+x -> 1+x
2+x+x^2 -> 3+2x
6+2x+x^2+x^3 -> 8+5x
24+6x+2x^2+x^3+x^4 -> 29+13x, so that
A192744=(1,1,3,8,29,...) and A192745=(0,1,2,5,13,...).

Cf. A192232.

A192744p := proc(n,x)
    option remember;
    if n = 0 then
        1;
    else
        x*procname(n-1,x)+n! ;
        expand(%) ;
    end if;
end proc:
A192744 := proc(n)
    local p;
    p := A192744p(n,x) ;
    while degree(p,x) > 1 do
        p := algsubs(x^2=x+1,p) ;
        p := expand(p) ;
    end do:
    coeftayl(p,x=0,0) ;
end proc: # R. J. Mathar, Dec 16 2015

q = x^2; s = x + 1; z = 40;
p[0, n_] := 1; p[n_, x_] := x*p[n - 1, x] + n!;
Table[Expand[p[n, x]], {n, 0, 7}]
reduce[{p1_, q_, s_, x_}] :=
FixedPoint[(s PolynomialQuotient @@ #1 +
       PolynomialRemainder @@ #1 &)[{#1, q, x}] &, p1]
t = Table[reduce[{p[n, x], q, s, x}], {n, 0, z}];
u1 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}]
  (* A192744 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}]
  (* A192745 *)

G.f.: (1-x)/(1-x-x^2)/Q(0), where Q(k)= 1 - x*(k+1)/(1 - x*(k+1)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, May 20 2013
Conjecture: a(n) +(-n-2)*a(n-1) +2*(n-1)*a(n-2) +3*a(n-3) +(-n+2)*a(n-4)=0. - R. J. Mathar, May 04 2014
Conjecture: (-n+2)*a(n) +(n^2-n-1)*a(n-1) +(-n^2+3*n-3)*a(n-2) -(n-1)^2*a(n-3)
=0. - R. J. Mathar, Dec 16 2015

A120562 Sum of binomial coefficients binomial(i+j, i) modulo 2 over all pairs (i,j) of positive integers satisfying 3i+j=n.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 2, 3, 1, 3, 2, 3, 2, 4, 3, 5, 1, 4, 3, 4, 2, 5, 3, 5, 2, 5, 4, 6, 3, 7, 5, 8, 1, 6, 4, 5, 3, 7, 4, 7, 2, 6, 5, 7, 3, 8, 5, 8, 2, 7, 5, 7, 4, 9, 6, 10, 3, 9, 7, 10, 5, 12, 8, 13, 1

Offset: 0

Sam Northshield (samuel.northshield(AT)plattsburgh.edu), Aug 07 2006

a(n) is the number of 'vectors' (..., e_k, e_{k-1}, ..., e_0) with e_k in {0,1,3} such that Sum_{k} e_k 2^k = n. a(2^n-1) = F(n+1)*a(2^{k+1}+j) + a(j) = a(2^k+j) + a(2^{k-1}+j) if 2^k > 4j. This sequence corresponds to the pair (3,1) as Stern's diatomic sequence [A002487] corresponds to (2,1) and Gould's sequence [A001316] corresponds to (1,1). There are many interesting similarities to A000119, the number of representations of n as a sum of distinct Fibonacci numbers.
A120562 can be generated from triangle A177444. Partial sums of A120562 = A177445. - Gary W. Adamson, May 08 2010
The Ca1 and Ca2 triangle sums, see A180662 for their definitions, of Sierpinski's triangle A047999 equal this sequence. Some A120562(2^n-p) sequences, 0 <= p <= 32, lead to known sequences, see the crossrefs. - Johannes W. Meijer, Jun 05 2011

			a(2^n)=1 since a(2n)=a(n).

Michael De Vlieger, Table of n, a(n) for n = 0..10000
K. Anders, Counting Non-Standard Binary Representations, JIS vol 19 (2016) #16.3.3 example 6.
Cristina Ballantine and George Beck, Partitions enumerated by self-similar sequences, arXiv:2303.11493 [math.CO], 2023.
S. Northshield, Sums across Pascal's triangle modulo 2, Congressus Numerantium, 200, pp. 35-52, 2010.

Cf. A001316 (1,1), A002487 (2,1), A120562 (3,1), A112970 (4,1), A191373 (5,1).
Cf. A177444, A177445. - Gary W. Adamson, May 08 2010
Cf. A000012 (p=0), A000045 (p=1, p=2, p=4, p=8, p=16, p=32), A000071 (p=3, p=6, p=12, p=13, p=24, p=26), A001610 (p=5, p=10, p=20), A001595 (p=7, p=14, p=28), A014739 (p=11, p=22, p=29), A111314 (p=15, p=30), A027961 (p=19), A154691 (p=21), A001911 (p=23). - Johannes W. Meijer, Jun 05 2011
Same recurrence for odd n as A000930.

p := product((1+x^(2^i)+x^(3*2^i)), i=0..25): s := series(p, x, 1000): for k from 0 to 250 do printf(`%d, `, coeff(s, x, k)) od:
A120562:=proc(n) option remember; if n <0 then A120562(n):=0 fi: if (n=0 or n=1) then 1 elif n mod 2 = 0 then A120562(n/2) else A120562((n-1)/2) + A120562((n-3)/2); fi; end: seq(A120562(n),n=0..64); # Johannes W. Meijer, Jun 05 2011

a[0] = a[1] = 1; a[n_?EvenQ] := a[n] = a[n/2]; a[n_?OddQ] := a[n] = a[(n-1)/2] + a[(n-1)/2 - 1]; Table[a[n], {n, 0, 64}] (* Jean-François Alcover, Sep 29 2011 *)
Nest[Append[#1, If[EvenQ@ #2, #1[[#2/2 + 1]], Total@ #1[[#2 ;; #2 + 1]] & @@ {#1, (#2 - 1)/2}]] & @@ {#, Length@ #} &, {1, 1}, 10^4 - 1] (* Michael De Vlieger, Feb 19 2019 *)

Recurrence; a(0)=a(1)=1, a(2*n)=a(n) and a(2*n+1)=a(n)+a(n-1).
G.f.: A(x) = Product_{i>=0} (1 + x^(2^i) + x^(3*2^i)) = (1 + x + x^3)*A(x^2).
a(n-1) << n^x with x = log_2(phi) = 0.69424... - Charles R Greathouse IV, Dec 27 2011

Reference edited and link added by Jason G. Wurtzel, Aug 22 2010

A166473 a(n) = 2^L(n+1) * 3^L(n)/12, where L(n) is the n-th Lucas number (A000032(n)).

Original entry on oeis.org

2, 36, 864, 373248, 3869835264, 17332899271409664, 804905577934332296851095552, 167416167663978753511691999938432197602574336

Offset: 1

Matthew Vandermast, Nov 05 2009

nonn

For m>1, A166469(A002110(m)*a(n)) = L(m+n).

A166469(a(n)) = L(n+2) - 2 = A014739(n).

G. C. Greubel, Table of n, a(n) for n = 1..14

Subsequence of A003586, A025487.

List([1..10], n-> 2^(Lucas(1,-1,n+1)[2]-2)*3^(Lucas(1,-1,n)[2]-1)); # G. C. Greubel, Jul 22 2019

[2^(Lucas(n+1)-2)*3^(Lucas(n)-1): n in [1..10]]; // G. C. Greubel, Jul 22 2019

Table[(2^LucasL[n+1] 3^LucasL[n])/12,{n,10}] (* Harvey P. Dale, Aug 17 2011 *)

lucas(n) = fibonacci(n+1) + fibonacci(n-1);
vector(10, n,  2^(lucas(n+1)-2)*3^(lucas(n)-1) ) \\ G. C. Greubel, Jul 22 2019

[2^(lucas_number2(n+1,1,-1)-2)*3^(lucas_number2(n,1,-1)-1) for n in (1..10)] # G. C. Greubel, Jul 22 2019

a(n) = A166471(n)/12.

For n>1, a(n) = 12*a(n-1) * a(n-2).

A171516 a(n) = a(n-1) + a(n-2) + k, n>1; with a(0) = 1, a(1) = 2, k = 3.

Original entry on oeis.org

1, 2, 6, 11, 20, 34, 57, 94, 154, 251, 408, 662, 1073, 1738, 2814, 4555, 7372, 11930, 19305, 31238, 50546, 81787, 132336, 214126, 346465, 560594, 907062, 1467659, 2374724, 3842386, 6217113, 10059502, 16276618, 26336123, 42612744, 68948870

Offset: 0

Gary W. Adamson, Dec 10 2009

In an infinite set of sequences converging to phi, H(n+1) = H(n) + H(n-1) + k.
The coincident formula = H(n) = (H(n+1) + H(n-2))/2, then proof of convergence to phi follows, [Gatta and D'Amico]: To get H(4) such that the average of H(4) and H(1) = H(3), the authors write H(4) = 2H(3) - H1 = 2H(1) + 2H(2) + 2k - H(1) = H(2) + (H(1) + H(2) + k) + k = H(2) + H(3) + k, then applying the iterative process to the latter, H(n+1) = H(n) + H(n-1) + k.
Cf. A014739 for a(0) = 1, a(1) = 2, k = 2, getting:
A014739 = (1, 2, 5,. 9, 16, 27, 45, 74, 121, 197,...)
A171516 = (1, 2, 6, 11, 20, 34, 57, 94, 154, 251,...), we obtain
A000071 = (0, 0, 1, .2,..4,..7,.12,.20,..33,..54,...).

			a(5) = a(4) + a(3) + 3 = 20 + 11 + 3 = 34.
a(4) = (a(5) + a(2)) / 2 = (34 + 6) / 2 = 20.

G. C. Greubel, Table of n, a(n) for n = 0..1000
F. Gatta and A. D'Amico, Sequences {H(n)} For Which H(n+1)/H(n) Approaches The Golden Ratio, The Fibonacci Quarterly, Vol. 46/47, Nov. 2008/2009, #4.
Index entries for linear recurrences with constant coefficients, signature (2,0,-1). [_R. J. Mathar_, Dec 12 2009]

Cf. A000045, A014739.

Cf. A000285 (first differences). [From R. J. Mathar, Dec 12 2009]

F:=Fibonacci;; List([0..40], n-> F(n+3)+2*F(n+1)-3); # G. C. Greubel, Jul 12 2019

F:=Fibonacci; [F(n+3)+2*F(n+1)-3: n in [0..40]]; // G. C. Greubel, Jul 12 2019

LinearRecurrence[{2,0,-1},{1,2,6},40] (* Harvey P. Dale, Apr 07 2012 *)
With[{F=Fibonacci}, Table[F[n+3]+2*F[n+1]-3, {n,0,40}]] (* G. C. Greubel, Jul 12 2019 *)

vector(40, n, n--; f=fibonacci; f(n+2)+2*f(n+1)-3) \\ G. C. Greubel, Jul 12 2019

f=fibonacci; [f(n+2)+2*f(n+1)-3 for n in (0..40)] # G. C. Greubel, Jul 12 2019

a(n) = a(n-1) + a(n-2) + 3, n>1; with a(0) = 1, a(1) = 2.
From R. J. Mathar, Dec 12 2009: (Start)
a(n) = 2*a(n-1) - a(n-3) = A000285(n+1) - 3.
G.f.: (1+2*x^2) / ((1-x)*(1-x-x^2)). (End)
a(n) = Fibonacci(n+3) + 2*Fibonacci(n+1) - 3. - G. C. Greubel, Jul 12 2019

More terms from R. J. Mathar, Dec 12 2009

Fixed typos in name, formula, crossrefs - Alex Ratushnyak, Apr 27 2012

A194051 Natural interspersion of A194050, a rectangular array, by antidiagonals.

Original entry on oeis.org

1, 2, 3, 5, 6, 4, 9, 10, 7, 8, 16, 17, 11, 12, 13, 27, 28, 18, 19, 20, 14, 45, 46, 29, 30, 31, 21, 15, 74, 75, 47, 48, 49, 32, 22, 23, 121, 122, 76, 77, 78, 50, 33, 34, 24, 197, 198, 123, 124, 125, 79, 51, 52, 35, 25, 320, 321, 199, 200, 201, 126, 80, 81, 53

Offset: 1

Clark Kimberling, Aug 13 2011

See A194029 for definitions of natural fractal sequence and natural interspersion. Every positive integer occurs exactly once (and every pair of rows intersperse), so that as a sequence, A194051 is a permutation of the positive integers; its inverse is A194052.

			Northwest corner:
1...2...5...9...16
3...6...10..17..28
4...7...11..18..29
8...12..19..30..48
13..20..31..49..78

Cf. A194029, A014739, A194050, A104052.

z = 50;
c[k_] := LucasL[k + 1] - 2;
c = Table[c[k], {k, 1, z}]  (* A014739 *)
f[n_] := If[MemberQ[c, n], 1, 1 + f[n - 1]]
f = Table[f[n], {n, 1, 600}]  (* A194050 *)
r[n_] := Flatten[Position[f, n]]
t[n_, k_] := r[n][[k]]
TableForm[Table[t[n, k], {n, 1, 8}, {k, 1, 7}]]
p = Flatten[Table[t[k, n - k + 1], {n, 1, 12}, {k, 1, n}]] (* A194051 *)
q[n_] := Position[p, n]; Flatten[Table[q[n], {n, 1, 90}]] (* A194052 *)

A207974 Triangle related to A152198.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 1, 1, 1, 4, 2, 2, 1, 1, 5, 2, 4, 1, 1, 1, 6, 3, 6, 3, 2, 1, 1, 7, 3, 9, 3, 5, 1, 1, 1, 8, 4, 12, 6, 8, 4, 2, 1, 1, 9, 4, 16, 6, 14, 4, 6, 1, 1, 1, 10, 5, 20, 10, 20, 10, 10, 5, 2, 1

Offset: 0

Philippe Deléham, Feb 22 2012

Row sums are A027383(n).
Diagonal sums are alternately A014739(n) and A001911(n+1).
The matrix inverse starts
1;
-1,1;
1,-2,1;
1,-1,-1,1;
-1,2,0,-2,1;
-1,1,2,-2,-1,1;
1,-2,-1,4,-1,-2,1;
1,-1,-3,3,3,-3,-1,1;
-1,2,2,-6,0,6,-2,-2,1;
-1,1,4,-4,-6,6,4,-4,-1,1;
1,-2,-3,8,2,-12,2,8,-3,-2,1;
apparently related to A158854. - R. J. Mathar, Apr 08 2013
From Gheorghe Coserea, Jun 11 2016: (Start)
T(n,k) is the number of terms of the sequence A057890 in the interval [2^n,2^(n+1)-1] having binary weight k+1.
T(n,k) = A007318(n,k) (mod 2) and the number of odd terms in row n of the triangle is 2^A000120(n).
(End)

			Triangle begins :
n\k  [0] [1] [2] [3] [4] [5] [6] [7] [8] [9]
[0]  1;
[1]  1,  1;
[2]  1,  2,  1;
[3]  1,  3,  1,  1;
[4]  1,  4,  2,  2,  1;
[5]  1,  5,  2,  4,  1,  1;
[6]  1,  6,  3,  6,  3,  2,  1;
[7]  1,  7,  3,  9,  3,  5,  1,  1;
[8]  1,  8,  4,  12, 6,  8,  4,  2,  1;
[9]  1,  9,  4,  16, 6,  14, 4,  6,  1,  1;
[10] ...

Gheorghe Coserea, Rows n = 0..200, flattened

Cf. Columns : A000012, A000027, A004526, A002620, A008805, A006918, A058187
Cf. Diagonals : A000012, A000034, A052938, A097362
Cf. A007318, A110813, A135278, A152201
Related to thickness: A000120, A027383, A057890, A274036.

A207974 := proc(n,k)
    if k = 0 then
        1;
    elif k < 0 or k > n then
        0 ;
    else
        procname(n-1,k-1)-(-1)^k*procname(n-1,k) ;
    end if;
end proc: # R. J. Mathar, Apr 08 2013

seq(N) = {
  my(t = vector(N+1, n, vector(n, k, k==1 || k == n)));
  for(n = 2, N+1, for (k = 2, n-1,
      t[n][k] = t[n-1][k-1] + (-1)^(k%2)*t[n-1][k]));
  return(t);
};
concat(seq(10))  \\ Gheorghe Coserea, Jun 09 2016

P(n) = ((2+x+(n%2)*x^2) * (1+x^2)^(n\2) - 2)/x;
concat(vector(11, n, Vecrev(P(n-1)))) \\ Gheorghe Coserea, Mar 14 2017

T(n,k) = T(n-1,k-1) - (-1)^k*T(n-1,k), k>0 ; T(n,0) = 1.
T(2n,2k) = T(2n+1,2k) = binomial(n,k) = A007318(n,k).
T(2n+1,2k+1) = A110813(n,k).
T(2n+2,2k+1) = 2*A135278(n,k).
T(n,2k) + T(n,2k+1) = A152201(n,k).
T(n,2k) = A152198(n,k).
T(n+1,2k+1) = A152201(n,k).
T(n,k) = T(n-2,k-2) + T(n-2,k).
T(2n,n) = A128014(n+1).
T(n,k) = card {p, 2^n <= A057890(p) <= 2^(n+1)-1 and A000120(A057890(p)) = k+1}. - Gheorghe Coserea, Jun 09 2016
P_n(x) = Sum_{k=0..n} T(n,k)*x^k = ((2+x+(n mod 2)*x^2)*(1+x^2)^(n\2) - 2)/x. - Gheorghe Coserea, Mar 14 2017

A372501 The 2-Zeckendorf array of the second kind, read by upward antidiagonals.

Original entry on oeis.org

0, 2, 1, 5, 4, 3, 7, 9, 8, 6, 10, 12, 16, 14, 11, 13, 17, 21, 27, 24, 19, 15, 22, 29, 35, 45, 40, 32, 18, 25, 37, 48, 58, 74, 66, 53, 20, 30, 42, 61, 79, 95, 121, 108, 87, 23, 33, 50, 69, 100, 129, 155, 197, 176, 142, 26, 38, 55, 82, 113, 163, 210, 252, 320, 286, 231

Offset: 1

A.H.M. Smeets, May 03 2024

The 2-Zeckendorf array of the second kind is based on the dual Zeckendorf representation of numbers (see A104326).
Column k contains the numbers whose dual Zeckendorf expansion ends "... 0 1^(k-1)" where ^ denotes repetition.
Rows satisfy this recurrence: T(n,k+1) = T(n,k) + T(n,k-1) + 2 for all n > 0 and k > 1.
As a sequence, the array is a permutation of the nonnegative integers.
As an array, T is an interspersion (hence also a dispersion). This holds as well for all Zeckendorf arrays of the second kind.
In general, for the m-Zeckendorf array of the second kind, the row recursion is given by T(n,k) = T(n,k-1) + T(n,k-m) + m, and the first column represent the "even" numbers.

			Array begins:
       k=1    2    3    4    5    6    7
      +---------------------------------
  n=1 |  0    1    3    6   11   19   32
  n=2 |  2    4    8   14   24   40   66
  n=3 |  5    9   16   27   45   74  121
  n=4 |  7   12   21   35   58   95  155
  n=5 | 10   17   29   48   79  129  210
  n=6 | 13   22   37   61  100  163  265
  n=7 | 15   25   42   69  113  184  299
The same in dual Zeckendorf form shows the pattern of digit suffixes, for example column k=3 is all numbers ending 011:
          k=1      2       3        4
      +------------------------------
  n=1 |     0      1      11      111
  n=2 |    10    101    1011    10111
  n=3 |   110   1101   11011   110111
  n=4 |  1010  10101  101011  1010111
  n=5 |  1110  11101  111011  1110111

Cf. A104326.
Rows n=1..3: A001911, A019274, A014739.
Columns k=1..3: A090909, A276885, A188012.
Cf. k-th prepended column: A022342 (k=1), A023444 (k=2).

T(n,1) = A090909(n+1).
T(1,k) = A001911(k-1).
T(2,k) = A019274(k-2).
T(3,k) = A014739(k-1).
T(n,1) = floor((n-1)*phi^2) and T(n,k+1) = floor((T(n,k)+1)*phi) for k > 0, where phi = (1+sqrt(5))/2. This can be considered as an alternative way to define the array.

A182415 a(0) = 1, a(1) = 2; for n>1, a(n) = a(n-1) + a(n-2) + 4.

Original entry on oeis.org

1, 2, 7, 13, 24, 41, 69, 114, 187, 305, 496, 805, 1305, 2114, 3423, 5541, 8968, 14513, 23485, 38002, 61491, 99497, 160992, 260493, 421489, 681986, 1103479, 1785469, 2888952, 4674425, 7563381, 12237810, 19801195, 32039009, 51840208, 83879221, 135719433, 219598658

Offset: 0

Alex Ratushnyak, Apr 28 2012

			a(3) = 7 + 2 + 4 = 13.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,0,-1).

Cf. A000045, A014739, A022095 (first differences), A171516.

F:=Fibonacci;; List([0..40], n-> F(n+3)+3*F(n+1)-4); # G. C. Greubel, Jul 22 2019

F:=Fibonacci; [F(n+3)+3*F(n+1)-4: n in [0..40]]; // G. C. Greubel, Jul 22 2019

With[{f=Fibonacci}, Table[F[n+3]+3*F[n+1]-4, {n,0,40}]] (* G. C. Greubel, Jul 22 2019 *)
RecurrenceTable[{a[0]==1,a[1]==2,a[n]==a[n-1]+a[n-2]+4},a,{n,40}] (* or *) LinearRecurrence[{2,0,-1},{1,2,7},40] (* Harvey P. Dale, Nov 24 2020 *)

vector(40, n, n--; f=fibonacci; f(n+3) +3*f(n+1) -4 ) \\ G. C. Greubel, Jul 22 2019

f=fibonacci; [f(n+3)+3*f(n+1)-4 for n in (0..40)] # G. C. Greubel, Jul 22 2019

From Colin Barker, May 07 2012: (Start)
a(n) = 2*a(n-1) - a(n-3).
G.f.: (1+3*x^2)/((1-x)*(1-x-x^2)). (End)
a(n) = Fibonacci(n+3) + 3*Fibonacci(n+1) - 4. - G. C. Greubel, Jul 22 2019

A114476 Triangle read by rows: inverse of triangle in A061554 with signs in each column +,+,-,-,+,+,-,-,...

Original entry on oeis.org

1, -1, 1, 3, -1, 1, -3, 4, -1, 1, 5, -4, 5, -1, 1, -5, 9, -5, 6, -1, 1, 7, -9, 14, -6, 7, -1, 1, -7, 16, -14, 20, -7, 8, -1, 1, 9, -16, 30, -20, 27, -8, 9, -1, 1, -9, 25, -30, 50, -27, 35, -9, 10, -1, 1, 11, -25, 55, -50, 77, -35, 44, -10, 11, -1, 1, -11, 36, -55, 105, -77, 112, -44, 54, -11, 12, -1, 1, 13, -36, 91, -105, 182, -112, 156, -54, 65, -12, 13, -1, 1

Offset: 0

Gary W. Adamson, Nov 27 2006

Unsigned row sums appear to be A014739.

			Start with a signed version of A061554:
1;
1, 1;
-2, 1, 1;
-3, -3, 1, 1;
6, -4, -4, 1, 1;
10, 10, -5, -5, 1;
...
and invert it, getting:
1
-1, 1;
3, -1, 1;
-3, 4, -1, 1;
5, -4, 5, -1, 1;
-5, 9, -5, 6, -1, 1;
...

A061554 := proc(n,k) binomial(n+k,floor(k/2)) ; end: nmax := 13 : A := array(1..nmax,1..nmax) : for r from 1 to nmax do for c from 1 to nmax do A[r,c] := A061554(c-1,r-c)*(-1)^floor((r-c)/2) ; od: od: A := linalg[inverse](A) : for r from 1 to nmax do for c from 1 to r do printf("%d, ", A[r,c]) ; od: od: # R. J. Mathar, Jan 31 2008

A061554[n_, k_] := Binomial[n+k, Floor[k/2]];
nmax = 13;
A = Table[A061554[c-1, r-c]*(-1)^Floor[(r-c)/2], {r, nmax}, {c, nmax}];
A = Inverse[A];
Table[A[[r, c]], {r, nmax}, {c, r}] // Flatten (* Jean-François Alcover, May 01 2023 *)

Edited by N. J. A. Sloane, Dec 01 2006

More terms from R. J. Mathar, Jan 31 2008

cp's OEIS Frontend

A194050 Natural fractal sequence of A014739.

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A192744 Constant term of the reduction by x^2->x+1 of the polynomial p(n,x) defined below in Comments.

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A120562 Sum of binomial coefficients binomial(i+j, i) modulo 2 over all pairs (i,j) of positive integers satisfying 3i+j=n.

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A166473 a(n) = 2^L(n+1) * 3^L(n)/12, where L(n) is the n-th Lucas number (A000032(n)).

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A171516 a(n) = a(n-1) + a(n-2) + k, n>1; with a(0) = 1, a(1) = 2, k = 3.

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A194051 Natural interspersion of A194050, a rectangular array, by antidiagonals.

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A207974 Triangle related to A152198.

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