A015457 -id:A015457 - cp's OEIS frontend

A055830 Triangle T read by rows: diagonal differences of triangle A037027.

1, 1, 0, 2, 1, 0, 3, 3, 1, 0, 5, 7, 4, 1, 0, 8, 15, 12, 5, 1, 0, 13, 30, 31, 18, 6, 1, 0, 21, 58, 73, 54, 25, 7, 1, 0, 34, 109, 162, 145, 85, 33, 8, 1, 0, 55, 201, 344, 361, 255, 125, 42, 9, 1, 0, 89, 365, 707, 850, 701, 413, 175, 52, 10, 1, 0, 144, 655, 1416, 1918, 1806, 1239, 630, 236, 63, 11, 1, 0

Offset: 0

Clark Kimberling, May 28 2000

Or, coefficients of a generalized Lucas-Pell polynomial read by rows. - Philippe Deléham, Nov 05 2006

Equals A046854(shifted) * Pascal's triangle; where A046854 is shifted down one row and "1" inserted at (0,0). - Gary W. Adamson, Dec 24 2008

			Triangle begins:
   1
   1,   0
   2,   1,   0
   3,   3,   1,   0
   5,   7,   4,   1,   0
   8,  15,  12,   5,   1,   0
  13,  30,  31,  18,   6,   1,  0
  21,  58,  73,  54,  25,   7,  1, 0
  34, 109, 162, 145,  85,  33,  8, 1, 0
  55, 201, 344, 361, 255, 125, 42, 9, 1, 0
  ...

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Clark Kimberling, Path-counting and Fibonacci numbers, Fib. Quart. 40 (4) (2002) 328-338, Example 3D.

Left-hand columns include A000045, A023610.
Right-hand columns include A055831, A055832, A055833, A055834, A055835, A055836, A055837, A055838, A055839, A055840.
Row sums: A001333 (numerators of continued fraction convergents to sqrt(2)).
Cf. A122075 (another version).
Cf. A046854. - Gary W. Adamson, Dec 24 2008

function T(n,k)
  if k lt 0 or k gt n then return 0;
  elif k eq 0 then return Fibonacci(n+1);
  elif n eq 1 and k eq 1 then return 0;
  else return T(n-1,k-1) + T(n-1,k) + T(n-2,k);
  end if; return T; end function;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 21 2020

with(combinat);
T:= proc(n, k) option remember;
      if k<0 or k>n then 0
    elif k=0 then fibonacci(n+1)
    elif n=1 and k=1 then 0
    else T(n-1, k-1) + T(n-1, k) + T(n-2, k)
      fi; end:
seq(seq(T(n, k), k=0..n), n=0..12); # G. C. Greubel, Jan 21 2020

T[n_, k_]:= T[n, k]= If[k<0 || k>n, 0, If[k==0, Fibonacci[n+1], If[n==1 && k==1, 0, T[n-1, k-1] + T[n-1, k] + T[n-2, k]]]]; Table[T[n, k], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Dec 19 2017 *)

T(n,k) = if(k<0 || k>n, 0, if(k==0, fibonacci(n+1), if(n==1 && k==1, 0, T(n-1, k-1) + T(n-1, k) + T(n-2, k) )));
for(n=0,12, for(k=0, n, print1(T(n,k), ", "))) \\ G. C. Greubel, Jan 21 2020

@CachedFunction
def T(n, k):
    if (k<0 or k>n): return 0
    elif (k==0): return fibonacci(n+1)
    elif (n==1 and k==1): return 0
    else: return T(n-1, k-1) + T(n-1, k) + T(n-2, k)
[[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Jan 21 2020

G.f.: (1-y*z) / (1-y*(1+y+z)).
T(i, j) = R(i-j, j), where R(0, 0)=1, R(0, j)=0 for j >= 1, R(1, j)=1 for j >= 0, R(i, j) = Sum_{k=0..j} (R(i-2, k) + R(i-1, k)) for i >= 1, j >= 1.
Sum_{k=0..n} x^k*T(n,k) = A039834(n-2), A000012(n), A000045(n+1), A001333(n), A003688(n), A015448(n), A015449(n), A015451(n), A015453(n), A015454(n), A015455(n), A015456(n), A015457(n) for x= -2,-1,0,1,2,3,4,5,6,7,8,9,10. - Philippe Deléham, Oct 22 2006
Sum_{k=0..floor(n/2)} T(n-k,k) = A011782(n). - Philippe Deléham, Oct 22 2006
Triangle T(n,k), 0 <= k <= n, given by [1, 1, -1, 0, 0, 0, 0, 0, ...] DELTA [0, 1, 0, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 05 2006
T(n,0) = Fibonacci(n+1) = A000045(n+1). Sum_{k=0..n} T(n,k) = A001333(n). T(n,k)=0 if k > n or if k < 0, T(0,0)=1, T(1,1)=0, T(n,k) = T(n-1,k-1) + T(n-1,k) + T(n-2,k). - Philippe Deléham, Nov 05 2006

Edited by Ralf Stephan, Jan 12 2005

A213900 The minimum number of 11's in the relation n*[n,11,11,...,11,n] = [x,...,x] between simple terminating continued fractions.

Original entry on oeis.org

2, 3, 5, 4, 11, 7, 5, 11, 14, 1, 11, 6, 23, 19, 11, 8, 11, 17, 29, 7, 5, 23, 11, 24, 20, 35, 23, 13, 59, 5, 23, 3, 8, 39, 11, 18, 17, 27, 29, 3, 23, 43, 5, 59, 23, 15, 11, 55, 74, 35, 41, 26, 35, 9, 23, 35, 41, 57, 59, 2, 5, 23, 47, 34, 11, 67, 17, 23, 119, 13

Offset: 2

Art DuPre, Jun 24 2012

nonn

In a variant of A213891, multiply n by a number with simple continued fraction [n,11,11,..,11,n] and increase the number of 11's until the continued fraction of the product has the same first and last entry (called x in the NAME). Examples are
2 * [2, 11, 11, 2] = [4, 5, 1, 1, 5, 4],
3 * [3, 11, 11, 11, 3] = [9, 3, 1, 2, 3, 2, 1, 3, 9],
4 * [4, 11, 11, 11, 11, 11, 4] = [16, 2, 1, 3, 2, 1, 1, 10, 1, 1, 2, 3, 1, 2, 16],
5 * [5, 11, 11, 11, 11, 5] = [25, 2, 4, 1, 1, 2, 2, 1, 1, 4, 2, 25] ,
6 * [6, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 6] = [36, 1, 1, 5, 1, 1, 2, 7, 16, 1, 1, 1, 2, 1, 6, 1, 2, 1, 1, 1, 16, 7, 2, 1, 1, 5, 1, 1, 36].
The number of 11's needed defines the sequence a(n).
If we consider the fixed points such that a(n)=n, we conjecture to obtain the sequence A000057. This sequence consists of prime numbers. We conjecture that this sequence of prime numbers, in addition to its well-known relation to the collection of Fibonacci sequences (sequences satisfying f(n)=f(n-1)+f(n-2) with arbitrary positive integer values for f(1) and f(2)) it also refers to the sequences satisfying f(n)=11*f(n-1)+f(n-2), A049666, A015457, etc. This would mean that a prime is in the sequence A000057 if and only if it divides some term in each of the sequences satisfying f(n)=11*f(n-1)+f(n-2).
It is surprising that the fixed points of this sequence seem to be the same as for the variant A213648 where 11 is replaced by 1, while for the other variants A262212 - A262220 (where the repeated term is 2, ..., 10) the fixed points are different, see A213891 - A213899. - M. F. Hasler, Sep 15 2015

Cf. A000057, A213358, A213891 - A213899.

Cf. A213648, A262212 - A262220, A213900.

f[m_, n_] := Block[{c, k = 1}, c[x_, y_] := ContinuedFraction[x FromContinuedFraction[Join[{x}, Table[m, {y}], {x}]]]; While[First@ c[n, k] != Last@ c[n, k], k++]; k]; f[11, #] & /@ Range[2, 120] (* Michael De Vlieger, Sep 16 2015 *)

\\ This PARI program will generate sequence A000057
{a(n) = local(t, m=1); if( n<2, 0, while( 1,
   t = contfracpnqn( concat([n, vector(m,i,11), n]));
   t = contfrac(n*t[1,1]/t[2,1]);
   if(t[1]

A214992 Power ceiling-floor sequence of (golden ratio)^4.

Original entry on oeis.org

7, 47, 323, 2213, 15169, 103969, 712615, 4884335, 33477731, 229459781, 1572740737, 10779725377, 73885336903, 506417632943, 3471038093699, 23790849022949, 163064905066945, 1117663486445665, 7660579500052711

Offset: 0

Clark Kimberling, Nov 08 2012, Jan 24 2013

Let f = floor and c = ceiling.  For x > 1, define four sequences as functions of x, as follows:
p1(0) = f(x), p1(n) = f(x*p1(n-1));
p2(0) = f(x), p2(n) = c(x*p2(n-1)) if n is odd and p2(n) = f(x*p1(n-1)) if n is even;
p3(0) = c(x), p3(n) = f(x*p3(n-1)) if n is odd and p3(n) = c(x*p3(n-1)) if n is even;
p4(0) = c(x), p4(n) = c(x*p4(n-1)).
The present sequence is given by a(n) = p3(n).
Following the terminology at A214986, call the four sequences power floor, power floor-ceiling, power ceiling-floor, and power ceiling sequences.  In the table below, a sequence is identified with an A-numbered sequence if they appear to agree except possibly for initial terms.  Notation: S(t)=sqrt(t), r = (1+S(5))/2 = golden ratio, and Limit = limit of p3(n)/p2(n).
x ......p1..... p2..... p3..... p4.......Limit
r.......A000045 A000045 A000045 A000045..r
r^2.....A001519 A001654 A061646 A001906..-1+S(5)
r^3.....A024551 A001076 A015448 A049652..-1+S(5)
r^4.....A049685 A157335 A214992 A004187..-19+9*S(5)
r^5.....A214993 A049666 A015457 A214994...(-9+5*S(5))/2
r^6.....A007805 A156085 A214995 A049660..-151+68*S(5)
1+S(2)..A024537 A000129 A001333 A048739..S(2)
2+S(2)..A007052 A214996 A214997 A007070..(1+S(2))/2
1+S(3)..A057960 A002605 A028859 A077846..(1+S(3))/2
2+S(3)..A001835 A109437 A214998 A001353..-4+3*S(3)
S(5)....A214999 A215091 A218982 A218983..1.26879683...
2+S(5)..A024551 A001076 A015448 A049652..-1+S(5)
2+S(6)..A218984 A090017 A123347 A218985..S(3/2)
2+S(7)..A218986 A015530 A126473 A218987..(1+S(7))/3
2+S(8)..A218988 A057087 A086347 A218989..(1+S(2))/2
3+S(8)..A001653 A084158 A218990 A001109..-13+10*S(2)
3+S(10).A218991 A005668 A015451 A218992..-2+S(10)
...
Properties of p1, p2, p3, p4:
(1) If x > 2, the terms of p2 and p3 interlace: p2(0) < p3(0) < p2(1) < p3(1) < p2(2) < p3(2)... Also, p1(n) <= p2(n) <= p3(n) <= p4(n) <= p1(n+1) for all x>0 and n>=0.
(2) If x > 2, the limits L(x) = limit(p/x^n) exist for the four functions p(x), and L1(x) <= L2(x) <= L3(x) <= L4 (x). See the Mathematica programs for plots of the four functions; one of them also occurs in the Odlyzko and Wilf article, along with a discussion of the special case x = 3/2.
(3) Suppose that x = u + sqrt(v) where v is a nonsquare positive integer.  If u = f(x) or u = c(x), then p1, p2, p3, p4 are linear recurrence sequences.  Is this true for sequences p1, p2, p3, p4 obtained from x = (u + sqrt(v))^q for every positive integer q?
(4) Suppose that x is a Pisot-Vijayaraghavan number. Must p1, p2, p3, p4 then be linearly recurrent?  If x is also a quadratic irrational b + c*sqrt(d), must the four limits L(x) be in the field Q(sqrt(d))?
(5) The Odlyzko and Wilf article (page 239) raises three interesting questions about the power ceiling function; it appears that they remain open.

			a(0) = ceiling(r) = 7, where r = ((1+sqrt(5))/2)^4 = 6.8...; a(1) = floor(7*r) = 47; a(2) = ceiling(47) = 323.

Clark Kimberling, Table of n, a(n) for n = 0..250
A. M. Odlyzko and H. S. Wilf, Functional iteration and the Josephus problem, Glasgow Math. J. 33, 235-240, 1991.
Index entries for linear recurrences with constant coefficients, signature (6,6,-1).
Index entries for sequences related to the Josephus Problem.

Cf. A001622, A214986, A049685, A157335, A004187.

(* Program 1.  A214992 and related sequences *)
x = GoldenRatio^4; z = 30; (* z = # terms in sequences *)
z1 = 100; (* z1 = # digits in approximations *)
f[x_] := Floor[x]; c[x_] := Ceiling[x];
p1[0] = f[x]; p2[0] = f[x]; p3[0] = c[x]; p4[0] = c[x];
p1[n_] := f[x*p1[n - 1]]
p2[n_] := If[Mod[n, 2] == 1, c[x*p2[n - 1]], f[x*p2[n - 1]]]
p3[n_] := If[Mod[n, 2] == 1, f[x*p3[n - 1]], c[x*p3[n - 1]]]
p4[n_] := c[x*p4[n - 1]]
Table[p1[n], {n, 0, z}]  (* A049685 *)
Table[p2[n], {n, 0, z}]  (* A157335 *)
Table[p3[n], {n, 0, z}]  (* A214992 *)
Table[p4[n], {n, 0, z}]  (* A004187 *)
Table[p4[n] - p1[n], {n, 0, z}]  (* A004187 *)
Table[p3[n] - p2[n], {n, 0, z}]  (* A098305 *)
(* Program 2.  Plot of power floor and power ceiling functions, p1(x) and p4(x) *)
f[x_] := f[x] = Floor[x]; c[x_] := c[x] = Ceiling[x];
p1[x_, 0] := f[x]; p1[x_, n_] := f[x*p1[x, n - 1]];
p4[x_, 0] := c[x]; p4[x_, n_] := c[x*p4[x, n - 1]];
Plot[Evaluate[{p1[x, 10]/x^10, p4[x, 10]/x^10}], {x, 2, 3}, PlotRange -> {0, 4}]
(* Program 3. Plot of power floor-ceiling and power ceiling-floor functions, p2(x) and p3(x) *)
f[x_] := f[x] = Floor[x]; c[x_] := c[x] = Ceiling[x];
p2[x_, 0] := f[x]; p3[x_, 0] := c[x];
p2[x_, n_] := If[Mod[n, 2] == 1, c[x*p2[x, n - 1]], f[x*p2[x, n - 1]]]
p3[x_, n_] := If[Mod[n, 2] == 1, f[x*p3[x, n - 1]], c[x*p3[x, n - 1]]]
Plot[Evaluate[{p2[x, 10]/x^10, p3[x, 10]/x^10}], {x, 2, 3}, PlotRange -> {0, 4}]

a(n) = floor(r*a(n-1)) if n is odd and a(n) = ceiling(r*a(n-1)) if n is even, where a(0) = ceiling(r), r = (golden ratio)^4 = (7 + sqrt(5))/2.
a(n) = 6*a(n-1) + 6*a(n-2) - a(n-3).
G.f.: (7 + 5*x - x^2)/((1 + x)*(1 - 7*x + x^2)).
a(n) = (10*(-2)^n+(10+3*sqrt(5))*(7-3*sqrt(5))^(n+2)+(10-3*sqrt(5))*(7+3*sqrt(5))^(n+2))/(90*2^n). - Bruno Berselli, Nov 14 2012
a(n) = 7*A157335(n) + 5*A157335(n-1) - A157335(n-2). - R. J. Mathar, Feb 05 2020
E.g.f.: exp(-x)*(5 + 2*exp(9*x/2)*(155*cosh(3*sqrt(5)*x/2) + 69*sqrt(5)*sinh(3*sqrt(5)*x/2)))/45. - Stefano Spezia, Oct 28 2024

A135597 Square array read by antidiagonals: row m (m >= 1) satisfies b(0) = b(1) = 1; b(n) = m*b(n-1) + b(n-2).

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 7, 5, 1, 1, 5, 13, 17, 8, 1, 1, 6, 21, 43, 41, 13, 1, 1, 7, 31, 89, 142, 99, 21, 1, 1, 8, 43, 161, 377, 469, 239, 34, 1, 1, 9, 57, 265, 836, 1597, 1549, 577, 55, 1, 1, 10, 73, 407, 1633, 4341, 6765, 5116, 1393, 89, 1, 1, 11, 91, 593, 2906

Offset: 1

N. J. A. Sloane, Mar 02 2008

For n > 1, the number of independent vertex sets in the graph K_m X P_{n-1}. For example, in K_3 X P_1 there are 4 independent vertex sets. - Andrew Howroyd, May 23 2017

			Array begins:
========================================================
m\n| 0 1 2  3   4    5     6      7       8        9
---|----------------------------------------------------
1  | 1 1 2  3   5    8    13     21      34       55 ...
2  | 1 1 3  7  17   41    99    239     577     1393 ...
3  | 1 1 4 13  43  142   469   1549    5116    16897 ...
4  | 1 1 5 21  89  377  1597   6765   28657   121393 ...
5  | 1 1 6 31 161  836  4341  22541  117046   607771 ...
6  | 1 1 7 43 265 1633 10063  62011  382129  2354785 ...
7  | 1 1 8 57 407 2906 20749 148149 1057792  7552693 ...
8  | 1 1 9 73 593 4817 39129 317849 2581921 20973217 ...
...

Andrew Howroyd, Table of n, a(n) for n = 1..1275

Cf. A121875, A000045, A287376.

Rows 2-11 are A001333, A003688, A015448, A015449, A015451, A015453-A015457.

A135597 := proc(m,c) coeftayl( (m*x-x-1)/(x^2+m*x-1),x=0,c) ; end: for d from 1 to 15 do for c from 0 to d-1 do printf("%d,",A135597(d-c,c)) ; od: od: # R. J. Mathar, Apr 21 2008

a[, 0] = a[, 1] = 1; a[m_, n_] := m*a[m, n-1] + a[m, n-2]; Table[a[m-n+1, n], {m, 0, 11}, {n, 0, m}] // Flatten (* Jean-François Alcover, Jan 20 2014 *)

O.g.f. row m: (mx-x-1)/(x^2+mx-1). - R. J. Mathar, Apr 21 2008

More terms from R. J. Mathar, Apr 21 2008

A214993 Power floor sequence of (golden ratio)^5.

Original entry on oeis.org

11, 121, 1341, 14871, 164921, 1829001, 20283931, 224952241, 2494758581, 27667296631, 306835021521, 3402852533361, 37738212888491, 418523194306761, 4641493350262861, 51474950047198231, 570865943869443401, 6331000332611075641, 70211869602591275451

Offset: 0

Clark Kimberling, Nov 09 2012

See A214992 for a discussion of power floor sequence and also the power floor function, p1(x) = limit of a(n,x)/x^n. The present sequence is a(n,r), where r = (golden ratio)^5, and the limit p1(r) = (3/22)*(3+2*sqrt(5)).

			a(0) = [r] = [11.0902] = 11, where r = (1+sqrt(5))^5.
a(1) = [11*r] = 121; a(2) = [121*r] = 1341.

Clark Kimberling, Table of n, a(n) for n = 0..250
Yaohui Zhu, Kaiming Sun, Zhengdong Luo, and Lingfeng Wang, Progressive Self-Learning for Domain Adaptation on Symbolic Regression of Integer Sequences, Proc. 39th AAAI Conf. Artif. Intel. (2025) Vol. 39, No. 1, 1692-1699. See p. 1698.
Index entries for linear recurrences with constant coefficients, signature (12,-10,-1).

Cf. A214992, A214994, A049666, A015457.

I:=[11,121,1341]; [n le 3 select I[n] else 12*Self(n-1)-10*Self(n-2)-Self(n-3): n in [1..30]]; // G. C. Greubel, Feb 01 2018

x = GoldenRatio^5; z = 30; (* z = # terms in sequences *)
z1 = 100; (* z1 = # digits in approximations *)
f[x_] := Floor[x]; c[x_] := Ceiling[x];
p1[0] = f[x]; p2[0] = f[x]; p3[0] = c[x]; p4[0] = c[x];
p1[n_] := f[x*p1[n - 1]]
p2[n_] := If[Mod[n, 2] == 1, c[x*p2[n - 1]], f[x*p2[n - 1]]]
p3[n_] := If[Mod[n, 2] == 1, f[x*p3[n - 1]], c[x*p3[n - 1]]]
p4[n_] := c[x*p4[n - 1]]
Table[p1[n], {n, 0, z}]  (* A214993 *)
Table[p2[n], {n, 0, z}]  (* A049666 *)
Table[p3[n], {n, 0, z}]  (* A015457 *)
Table[p4[n], {n, 0, z}]  (* A214994 *)
LinearRecurrence[{12,-10,-1}, {11,121,1341}, 30] (* G. C. Greubel, Feb 01 2018 *)

Vec((11 - 11*x - x^2) / ((1 - x)*(1 - 11*x - x^2)) + O(x^20)) \\ Colin Barker, Nov 13 2017

a(n) = [x*a(n-1)], where x=((1+sqrt(5))/2)^5, a(0) = [x].
a(n) = 1 (mod 10).
a(n) = 12*a(n-1) - 10*a(n-2) - a(n-3).
G.f.: (11 - 11*x - x^2)/(1 - 12*x + 10*x^2 + x^3).
a(n) = (1/55)*(5 + (300-134*sqrt(5))*((11-5*sqrt(5))/2)^n + 2*(11/2+(5*sqrt(5))/2)^n*(150+67*sqrt(5))). - Colin Barker, Nov 13 2017

A214994 Power ceiling sequence of (golden ratio)^5.

Original entry on oeis.org

12, 134, 1487, 16492, 182900, 2028393, 22495224, 249475858, 2766729663, 30683502152, 340285253336, 3773821288849, 41852319430676, 464149335026286, 5147495004719823, 57086594386944340, 633100033261107564, 7021186960259127545, 77866156596111510560

Offset: 0

Clark Kimberling, Nov 09 2012

See A214992 for a discussion of power ceiling sequence and the power ceiling function, p4(x) = limit of a(n,x)/x^n. The present sequence is a(n,r), where r = (golden ratio)^5, and the limit p4(r) = (1/30)*(105+47*sqrt(5)).

See A214993 for the power floor sequence and power floor function, p1. For comparison with p4, we have p4(r)/p1(r) = (5 + 3*sqrt(5))/10.

			a(0) = ceiling(r) = [11.0902]=12, where r=(1+sqrt(5))^5.
a(1) = ceiling(12) = 134; a(2) = ceiling(134 ) = 1487.

Clark Kimberling, Table of n, a(n) for n = 0..250
Index entries for linear recurrences with constant coefficients, signature (12,-10,-1).

Cf. A214992, A214993, A049666, A015457.

I:=[12,134,1487]; [n le 3 select I[n] else 12*Self(n-1) - 10*Self(n-2) - Self(n-3): n in [1..30]]; // G. C. Greubel, Feb 01 2018

(See A214993.)
LinearRecurrence[{12,-10,-1}, {12,134,1487}, 30] (* G. C. Greubel, Feb 01 2018 *)

Vec((12 - 10*x - x^2) / ((1 - x)*(1 - 11*x - x^2)) + O(x^40)) \\ Colin Barker, Nov 13 2017

a(n) = ceiling(x*a(n-1)), x=((1+sqrt(5))/2)^5, a(0) = ceiling(x).
a(n) = 12*a(n-1) - 10*a(n-2) - a(n-3).
G.f.: (12 - 10*x - x^2)/(1 - 12*x + 10*x^2 + x^3).
a(n) = (1/550)*(-50 + (3325-1487*sqrt(5))*((11-5*sqrt(5))/2)^n + ((11+5*sqrt(5))/2)^n*(3325+1487*sqrt(5))). - Colin Barker, Nov 13 2017

A153764 Triangle T(n,k), 0 <= k <= n, read by rows, given by [1,0,-1,0,0,0,0,0,0,0,0,...] DELTA [0,1,0,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 2, 1, 1, 0, 1, 2, 3, 1, 1, 0, 1, 3, 3, 4, 1, 1, 0, 1, 3, 6, 4, 5, 1, 1, 0, 1, 4, 6, 10, 5, 6, 1, 1, 0, 1, 4, 10, 10, 15, 6, 7, 1, 1, 0, 1, 5, 10, 20, 15, 21, 7, 8, 1, 1, 0, 1, 5, 15, 20, 35, 21, 28, 8, 9, 1, 1, 0, 1, 6, 15, 35, 35, 56, 28, 36, 9, 10, 1, 1, 0

Offset: 0

Philippe Deléham, Jan 01 2009

A130595*A153342 as infinite lower triangular matrices. Reflected version of A103631. Another version of A046854. Row sums are Fibonacci numbers (A000045).

A055830*A130595 as infinite lower triangular matrices.

			Triangle begins:
  1;
  1, 0;
  1, 1, 0;
  1, 1, 1, 0;
  1, 2, 1, 1, 0;
  1, 2, 3, 1, 1, 0;
  1, 3, 3, 4, 1, 1, 0;
  ...

G. C. Greubel, Table of n, a(n) for the first 45 rows

Cf. A046854, A103631.

/* As triangle */ [[Binomial(Floor((n+k-1)/2),k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Aug 28 2016

Table[Binomial[Floor[(n + k - 1)/2], k], {n, 0, 45}, {k, 0, n}] // Flatten (* G. C. Greubel, Aug 27 2016 *)

T(n,k) = binomial(floor((n+k-1)/2),k).
Sum_{k=0..n} T(n,k)*x^k = A122335(n-1), A039834(n-2), A000012(n), A000045(n+1), A001333(n), A003688(n), A015448(n), A015449(n), A015451(n), A015453(n), A015454(n), A015455(n), A015456(n), A015457(n) for x = -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 respectively. - Philippe Deléham, Dec 17 2011
Sum_{k=0..n} T(n,k)*x^(n-k) = A152163(n), A000007(n), A000045(n+1), A026597(n), A122994(n+1), A158608(n), A122995(n+1), A158797(n), A122996(n+1), A158798(n), A158609(n) for x = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively. - Philippe Deléham, Dec 17 2011
G.f.: (1+(1-y)*x)/(1-y*x-x^2). - Philippe Deléham, Dec 17 2011
T(n,k) = T(n-1,k-1) + T(n-2,k), T(0,0) = T(1,0) = T(2,0) = T(2,1) = 1, T(1,1) = T(2,2) = 0, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Nov 09 2013

cp's OEIS Frontend

A055830 Triangle T read by rows: diagonal differences of triangle A037027.

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A213900 The minimum number of 11's in the relation n*[n,11,11,...,11,n] = [x,...,x] between simple terminating continued fractions.

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A214992 Power ceiling-floor sequence of (golden ratio)^4.

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A135597 Square array read by antidiagonals: row m (m >= 1) satisfies b(0) = b(1) = 1; b(n) = m*b(n-1) + b(n-2).

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A214993 Power floor sequence of (golden ratio)^5.

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A214994 Power ceiling sequence of (golden ratio)^5.

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A153764 Triangle T(n,k), 0 <= k <= n, read by rows, given by [1,0,-1,0,0,0,0,0,0,0,0,...] DELTA [0,1,0,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.

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