A023610 -id:A023610 - cp's OEIS frontend

A067418 Triangle A067330 with rows read backwards.

1, 2, 1, 5, 3, 2, 10, 7, 5, 3, 20, 15, 12, 8, 5, 38, 30, 25, 19, 13, 8, 71, 58, 50, 40, 31, 21, 13, 130, 109, 96, 80, 65, 50, 34, 21, 235, 201, 180, 154, 130, 105, 81, 55, 34, 420, 365, 331, 289, 250, 210, 170, 131, 89, 55, 744, 655, 600, 532, 469, 404, 340, 275, 212, 144, 89, 1308, 1164, 1075, 965

Offset: 0

Wolfdieter Lang, Feb 15 2002

The column m (without leading 0's) gives the convolution of Fibonacci numbers F(n+1) := A000045(n+1), n>=0, with those with m-shifted index: a(n+m,m)=sum(F(k+1)*F(m+n+1-k),k=0..n), n>=0, m=0,1,...
The row polynomials p(n,x) := sum(a(n,m)*x^m,m=0..n) are generated by A(z)*(A(z)-x*A(x*z))/(1-x), with A(x) := 1/(1-x-x^2) (g.f. for Fibonacci F(n+1), n>=0).
The columns give A001629(n+2), A023610, A067331-4, A067430-1, A067977-8 for m= 0..9, respectively. Row sums give A067988.

			{1}; {2,1}; {5,3,2}; {10,7,5,3}; ...; p(2,n)=5+3*x+2*x^2.

Michael De Vlieger, Table of n, a(n) for n = 0..10000

Reverse /@ Table[Sum[Fibonacci[k + 1] Fibonacci[n - k + 1], {k, 0, m}], {n, 0, 11}, {m, 0, n}] // Flatten (* Michael De Vlieger, Apr 11 2016 *)

a(n, m)=A067330(n, n-m), n>=m>=0, else 0.
a(n, m)= (((3*(n-m)+5)*F(n-m+1)+(n-m+1)*F(n-m))*F(m+1)+((n-m)*F(n-m+1)+2*(n-m+1)*F(n-m))*F(m))/5.
G.f. for column m=0, 1, ...: (x^m)*(F(m+1)+F(m)*x)/(1-x-x^2)^2, with F(m) := A000045(m) (Fibonacci).
a(n, m) = ((-1)^m*F(n-2*m+1)-m*L(n+2)+n*L(n+2)+5*F(n)+4*F(n-1))/5, with F(-n) = (-1)^(n+1)*F(n), hence a(n, m) = (2*(n-m+1)*L(n+2)-A067990(n, m))/5, n>=m>=0. - Ehren Metcalfe, Apr 11 2016

A091533 Triangle read by rows, related to Pascal's triangle, starting with rows 1; 1,1.

Original entry on oeis.org

1, 1, 1, 2, 3, 2, 3, 7, 7, 3, 5, 15, 21, 15, 5, 8, 30, 53, 53, 30, 8, 13, 58, 124, 157, 124, 58, 13, 21, 109, 273, 417, 417, 273, 109, 21, 34, 201, 577, 1029, 1239, 1029, 577, 201, 34, 55, 365, 1181, 2405, 3375, 3375, 2405, 1181, 365, 55, 89, 655, 2358, 5393, 8625, 10047, 8625, 5393, 2358, 655, 89

Offset: 0

Christian G. Bower, Jan 19 2004

T(n,k) is the number of lattice paths from (0,0) to (k,n-k) using steps (1,0),(2,0),(0,1),(0,2),(1,1). - Seiichi Manyama, Apr 26 2025.

			This triangle begins:
   1;
   1,   1;
   2,   3,    2;
   3,   7,    7,    3;
   5,  15,   21,   15,    5;
   8,  30,   53,   53,   30,     8;
  13,  58,  124,  157,  124,    58,   13;
  21, 109,  273,  417,  417,   273,  109,   21;
  34, 201,  577, 1029, 1239,  1029,  577,  201,   34;
  55, 365, 1181, 2405, 3375,  3375, 2405, 1181,  365,  55;
  89, 655, 2358, 5393, 8625, 10047, 8625, 5393, 2358, 655, 89;
  ...

Seiichi Manyama, Rows n = 0..139, flattened

Row sums: A015518(n+1). Columns 0-1: A000045(n+1), A023610(n-1).
Cf. A090174, A212338 (column 2), A192364 (central terms).
Cf. A036355.

T:= proc(n, k) option remember; `if`(k<0 or k>n, 0,
     `if`(n<1, 1, add(add(T(n-i, k-j), j=0..i), i=1..2)))
    end:
seq(seq(T(n, k), k=0..n), n=0..10);  # Alois P. Heinz, Jan 14 2022

A091533[-2, n2_] = 0; A091533[n1_, -2] = 0; A091533[-1, n2_] = 0; A091533[n1_, -1] = 0; A091533[0, 0] = 1; A091533[n1_, n2_] := A091533[n1, n2] = A091533[n1 - 1, n2] + A091533[n1, n2 - 1] + A091533[n1 - 1, n2 - 1] + A091533[n1 - 2, n2] + A091533[n1, n2 - 2]; Table[A091533[x - y, y], {x, 0, 9}, {y, 0, x}] // Flatten (* Robert P. P. McKone, Jan 14 2022 *)

T(n, k) = T(n-1, k) + T(n-1, k-1) + T(n-2, k) + T(n-2, k-1) + T(n-2, k-2) for n >= 2, k >= 0, with initial conditions specified by first two rows.
G.f.: A(x, y) = 1/(1-x-x*y-x^2-x^2*y-x^2*y^2).
Sum_{k = 0..n} T(n,k)*x^k = A000045(n+1), A015518(n+1), A015524(n+1), A200069(n+1) for x = 0, 1, 2, 3 respectively. - Philippe Deléham, Oct 30 2013
Sum_{k = 0..floor(n/2)} T(n-k,k) = (-1)^n*A079926(n). - Philippe Deléham, Oct 30 2013

A074083 Coefficient of q^3 in nu(n), where nu(0)=1, nu(1)=b and, for n>=2, nu(n)=bnu(n-1)+lambda(1+q+q^2+...+q^(n-2))*nu(n-2) with (b,lambda)=(1,1).

Original entry on oeis.org

0, 0, 0, 0, 0, 4, 14, 39, 97, 224, 494, 1051, 2177, 4412, 8784, 17228, 33360, 63886, 121164, 227833, 425147, 787916, 1451198, 2657821, 4842727, 8782230, 15857426, 28517864, 51095760, 91232520, 162372682, 288115147, 509790277, 899630376

Offset: 0

Y. Kelly Itakura (yitkr(AT)mta.ca), Aug 19 2002

nonn

The coefficient of q^0 in nu(n) is the Fibonacci number F(n+1). The coefficient of q^1 is A023610(n-3).

			The first 6 nu polynomials are nu(0)=1, nu(1)=1, nu(2)=2, nu(3)=3+q, nu(4)=5+3q+2q^2, nu(5)=8+7q+6q^2+4q^3+q^4, so the coefficients of q^3 are 0,0,0,0,0,4.

M. Beattie, S. Dăscălescu and S. Raianu, Lifting of Nichols Algebras of Type B_2, arXiv:math/0204075 [math.QA], 2002.
Index entries for linear recurrences with constant coefficients, signature (4, -2, -8, 5, 8, -2, -4, -1).

Coefficients of q^0, q^1 and q^2 are in A000045, A023610 and A074082. Related sequences with different values of b and lambda are in A074084-A074089.

b=1; lambda=1; expon=3; nu[0]=1; nu[1]=b; nu[n_] := nu[n]=Together[b*nu[n-1]+lambda(1-q^(n-1))/(1-q)nu[n-2]]; a[n_] := Coefficient[nu[n], q, expon]
(* Second program: *)
Join[{0, 0, 0}, LinearRecurrence[{4, -2, -8, 5, 8, -2, -4, -1}, {0, 0, 4, 14, 39, 97, 224, 494}, 31]] (* Jean-François Alcover, Jan 27 2019 *)

G.f.: (4x^5-2x^6-9x^7+x^8+6x^9+2x^10)/(1-x-x^2)^4.

a(n) = 4a(n-1)-2a(n-2)-8a(n-3)+5a(n-4)+8a(n-5)-2a(n-6)-4a(n-7)-a(n-8) for n>=11.

Edited by Dean Hickerson, Aug 21 2002

A134400 M * A007318, where M = triangle with (1, 1, 2, 3, ...) in the main diagonal and the rest zeros.

Original entry on oeis.org

1, 1, 1, 2, 4, 2, 3, 9, 9, 3, 4, 16, 24, 16, 4, 5, 25, 50, 50, 25, 5, 6, 36, 90, 120, 90, 36, 6, 7, 49, 147, 245, 245, 147, 49, 7, 8, 64, 224, 448, 560, 448, 224, 64, 8, 9, 81, 324, 756, 1134, 1134, 756, 324, 81, 9, 10, 100, 450, 1200, 2100, 2520, 2100, 1200, 450, 100, 10

Offset: 0

Gary W. Adamson, Oct 23 2007

Row sums = A134401: (1, 2, 8, 24, 64, 160, 384, ...).
Triangle T(n,k), read by rows, given by [1,1,-1,1,0,0,0,0,0,...] DELTA [1,1,-1,1,0,0,0,0,0,...] where DELTA is the operator defined in A084938. A134402*A007318 as infinite lower triangular matrices. - Philippe Deléham, Oct 26 2007
For n > 0, from n athletes, select a team of k players and then choose a coach who is allowed to be on the team or not. - Geoffrey Critzer, Mar 13 2010
Row sums are A036289 if first term changed to zero. Diagonal sums are A023610, starting with the 2nd diagonal. Partial sums of diagonals are A002940 if first term changed to zero. - John Molokach, Jul 06 2013
For n > 0, T(n,k) is the number of states in Sokoban puzzle with n non-obstacles cells and k boxes (see Russell and Norvig at page 157). - Stefano Spezia, Dec 03 2023

			First few rows of the triangle:
  1;
  1,  1;
  2,  4,   2;
  3,  9,   9,   3;
  4, 16,  24,  16,   4;
  5, 25,  50,  50,  25,   5;
  6, 36,  90, 120,  90,  36,  6;
  7, 49, 147, 245, 245, 147, 49, 7;
  ...

Stuart Russell and Peter Norvig, Artificial Intelligence: A Modern Approach, Fourth Edition, Hoboken: Pearson, 2021.

Wikipedia, Sokoban

Cf. A002940, A003506, A007318, A036289, A134401, A134402.

T(2n,n) give A002011(n-1) for n>=1.

with(combstruct): for n from 0 to 10 do seq(`if`(n=0, 1, n)* count( Combination(n), size=m), m=0..n) od; # Zerinvary Lajos, Apr 09 2008

Join[{1},Table[Table[n*Binomial[n, k], {k,0, n}], {n, 10}]] //Flatten (* Geoffrey Critzer, Mar 13 2010 adapted by Stefano Spezia, Dec 03 2023 *)

From Geoffrey Critzer, Mar 13 2010: (Start)
T(0,0) = 1 and T(n,k) = n * binomial(n,k) for n > 0.
E.g.f. for column k is: (x^k/k!)*exp(x)*(x+k). (End)
T(n,k) = A003506(n,k) + A003506(n,k-1). - Geoffrey Critzer, Mar 13 2010
G.f.: (1-x-x*y+x^2+x^2*y+x^2*y^2)/(1-2*x-2*x*y+x^2+2*x^2*y+x^2*y^2). - Philippe Deléham, Nov 14 2013
T(n,k) = 2*T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k) - 2*T(n-2,k-1) - T(n-2,k-2), T(0,0)=T(1,0)=T(1,1)=1, T(2,0)=T(2,2)=2, T(2,1)=4, T(n,k)=0 if k < 0 or if k > n. - Philippe Deléham, Nov 14 2013
E.g.f.: 1 + exp(y*x)*exp(x)*(y*x + x). - Geoffrey Critzer, Mar 15 2015

a(55)-a(65) from Stefano Spezia, Dec 03 2023

A207610 Triangle of coefficients of polynomials u(n,x) jointly generated with A207611; see the Formula section.

Original entry on oeis.org

1, 2, 4, 1, 7, 3, 1, 12, 7, 3, 1, 20, 15, 8, 3, 1, 33, 30, 19, 9, 3, 1, 54, 58, 42, 23, 10, 3, 1, 88, 109, 89, 55, 27, 11, 3, 1, 143, 201, 182, 125, 69, 31, 12, 3, 1, 232, 365, 363, 273, 166, 84, 35, 13, 3, 1, 376, 655, 709, 579, 383, 212, 100, 39, 14, 3, 1, 609

Offset: 1

Clark Kimberling, Feb 19 2012

			First five rows:
1
2
4...1
7...3...1
12...7...3...1

Cf. A207611.

Cf. A000071 (column 1), A023610 (column 2).

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + v[n - 1, x]
v[n_, x_] := u[n - 1, x] + x*v[n - 1, x] + 1
Table[Factor[u[n, x]], {n, 1, z}]
Table[Factor[v[n, x]], {n, 1, z}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]    (* A207610 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A207611 *)

from sympy import Poly
from sympy.abc import x
def u(n, x): return 1 if n==1 else u(n - 1, x) + v(n - 1, x)
def v(n, x): return 1 if n==1 else u(n - 1, x) + x*v(n - 1, x) + 1
def a(n): return Poly(u(n, x), x).all_coeffs()[::-1]
for n in range(1, 13): print(a(n)) # Indranil Ghosh, May 28 2017

u(n,x)=u(n-1,x)+v(n-1,x), v(n,x)=u(n-1,x)+x*v(n-1,x)+1, where u(1,x)=1, v(1,x)=1.

A207612 Triangle of coefficients of polynomials u(n,x) jointly generated with A207613; see the Formula section.

Original entry on oeis.org

1, 2, 4, 2, 7, 6, 4, 12, 14, 12, 8, 20, 30, 32, 24, 16, 33, 60, 76, 72, 48, 32, 54, 116, 168, 184, 160, 96, 64, 88, 218, 356, 440, 432, 352, 192, 128, 143, 402, 728, 1000, 1104, 992, 768, 384, 256, 232, 730, 1452, 2184, 2656, 2688, 2240, 1664, 768, 512

Offset: 1

Clark Kimberling, Feb 19 2012

Column 1: A000071

Column 2: 2*A023610

			First five rows:
1
2
4....2
7....6....4
12...14...12...8

Cf. A207613.

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + v[n - 1, x]
v[n_, x_] := u[n - 1, x] + 2 x*v[n - 1, x] + 1
Table[Factor[u[n, x]], {n, 1, z}]
Table[Factor[v[n, x]], {n, 1, z}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]    (* A207612 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A207613 *)

from sympy import Poly
from sympy.abc import x
def u(n, x): return 1 if n==1 else u(n - 1, x) + v(n - 1, x)
def v(n, x): return 1 if n==1 else u(n - 1, x) + 2*x*v(n - 1, x) + 1
def a(n): return Poly(u(n, x), x).all_coeffs()[::-1]
for n in range(1, 13): print(a(n)) # Indranil Ghosh, May 28 2017

u(n,x)=u(n-1,x)+v(n-1,x), v(n,x)=u(n-1,x)+2x*v(n-1,x)+1,

where u(1,x)=1, v(1,x)=1.

A213777 Rectangular array: (row n) = bc, where b(h) = F(h), c(h) = F(h+1), F=A000045 (Fibonacci numbers), n>=1, h>=1, and = convolution.

Original entry on oeis.org

1, 3, 2, 7, 5, 3, 15, 12, 8, 5, 30, 25, 19, 13, 8, 58, 50, 40, 31, 21, 13, 109, 96, 80, 65, 50, 34, 21, 201, 180, 154, 130, 105, 81, 55, 34, 365, 331, 289, 250, 210, 170, 131, 89, 55, 655, 600, 532, 469, 404, 340, 275, 212, 144, 89, 1164, 1075, 965, 863

Offset: 1

Clark Kimberling, Jun 21 2012

Principal diagonal:  A001870
Antidiagonal sums:  A152881
row 1,  (1,1,2,3,5,8,...)**(1,2,3,5,8,13,...): A023610(k-1)
row 2,  (1,1,2,3,5,8,...)**(2,3,5,8,13,21,...): A067331(k-1)
row 3,  (1,1,2,3,5,8,...)**(3,5,8,13,21,34,...)
For a guide to related arrays, see A213500.

			Northwest corner (the array is read by falling antidiagonals):
1....3....7....15....30....58
2....5....12...25....50....96
3....8....19...40....80....154
5....13...31...65....130...250
8....21...50...105...210...404
13...34...81...170...340...654

Clark Kimberling, Antidiagonals n=1..80, flattened

Cf. A213500.

b[n_] := Fibonacci[n]; c[n_] := Fibonacci[n + 1];
t[n_, k_] := Sum[b[k - i] c[n + i], {i, 0, k - 1}]
TableForm[Table[t[n, k], {n, 1, 10}, {k, 1, 10}]]
Flatten[Table[t[n - k + 1, k], {n, 12}, {k, n, 1, -1}]]
r[n_] := Table[t[n, k], {k, 1, 60}]  (* A213777 *)
Table[t[n, n], {n, 1, 40}] (* A001870 *)
s[n_] := Sum[t[i, n + 1 - i], {i, 1, n}]
Table[s[n], {n, 1, 50}] (* A152881 *)

T(n,k) = 2*T(n,k-1) + T(n,k-2) - 2*T(n,k-3) - T(n,k-4).
G.f. for row n:  f(x)/g(x), where f(x) = F(n-1) + F(n-2)*x and g(x) = (1 - x - x^2)^2.
T(n,k) = (k*Lucas(n+k+1) + Lucas(n)*Fibonacci(k))/5. - Ehren Metcalfe, Jul 10 2019

A378707 Array read by ascending antidiagonals: A(n,k) is the total number of inner points of n-Fibonacci polyominoes with k columns, where k > 0.

Original entry on oeis.org

0, 0, 1, 0, 3, 3, 0, 5, 10, 7, 0, 7, 18, 26, 15, 0, 9, 26, 50, 63, 30, 0, 11, 34, 74, 124, 143, 58, 0, 13, 42, 98, 190, 296, 313, 109, 0, 15, 50, 122, 254, 457, 679, 668, 201, 0, 17, 58, 146, 318, 622, 1070, 1517, 1398, 365, 0, 19, 66, 170, 382, 782, 1461, 2439, 3325, 2883, 655

Offset: 2

Stefano Spezia, Dec 05 2024

			The array begins as:
  0,  1,  3,   7,  15,  30,   58,  109,   201,   365, ...
  0,  3, 10,  26,  63, 143,  313,  668,  1398,  2883, ...
  0,  5, 18,  50, 124, 296,  679, 1517,  3325,  7184, ...
  0,  7, 26,  74, 190, 457, 1070, 2439,  5453, 12013, ...
  0,  9, 34,  98, 254, 622, 1461, 3361,  7583, 16857, ...
  0, 11, 42, 122, 318, 782, 1854, 4272,  9681, 21615, ...
  0, 13, 50, 146, 382, 942, 2238, 5182, 11754, 26302, ...
  ...

Juan F. Pulido, José L. Ramírez, and Andrés R. Vindas-Meléndez, Generating Trees and Fibonacci Polyominoes, arXiv:2411.17812 [math.CO], 2024. See page 12.

Cf. A023610, A378704, A378706, A378716.

A[n_, k_]:=SeriesCoefficient[y((6-4n)y-(2-4n)y^2-(3-n)n y^n-2(2-n)^2y^(n+1)+(2-5n+n^2)y^(n+2)+2y^(2n+1))/(2(-1+y)(1-2y+y^(n+1))^2), {y, 0, k}]; Table[A[n-k+1, k], {n, 2, 12}, {k, n-1}]//Flatten

A(n, k) = [y^k] y*((6 - 4*n)*y - (2 - 4*n)*y^2 - (3 - n)*n*y^n -2*(2 - n)^2*y^(n+1) + (2 - 5*n + n^2)*y^(n+2) + 2*y^(2n+1))/(2*(-1 + y)*(1 - 2*y + y^(n+1))^2).

A(2, n) = A023610(n-2) for n > 1.

A129710 Triangle read by rows: T(n,k) is the number of Fibonacci binary words of length n and having k 01 subwords (0 <= k <= floor(n/2)). A Fibonacci binary word is a binary word having no 00 subword.

Original entry on oeis.org

1, 2, 2, 1, 2, 3, 2, 5, 1, 2, 7, 4, 2, 9, 9, 1, 2, 11, 16, 5, 2, 13, 25, 14, 1, 2, 15, 36, 30, 6, 2, 17, 49, 55, 20, 1, 2, 19, 64, 91, 50, 7, 2, 21, 81, 140, 105, 27, 1, 2, 23, 100, 204, 196, 77, 8, 2, 25, 121, 285, 336, 182, 35, 1, 2, 27, 144, 385, 540, 378, 112, 9, 2, 29, 169, 506

Offset: 0

Emeric Deutsch, May 12 2007

Also number of Fibonacci binary words of length n and having k 10 subwords.
Row n has 1+floor(n/2) terms.
Row sums are the Fibonacci numbers (A000045).
T(n,0)=2 for n >= 1.
Sum_{k>=0} k*T(n,k) = A023610(n-2).
Triangle, with zeros omitted, given by (2, -1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1/2, -1/2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Jan 14 2012
Riordan array ((1+x)/(1-x), x^2/(1-x)), zeros omitted. - Philippe Deléham, Jan 14 2012

			T(5,2)=4 because we have 10101, 01101, 01010 and 01011.
Triangle starts:
  1;
  2;
  2, 1;
  2, 3;
  2, 5, 1;
  2, 7, 4;
  2, 9, 9, 1;
Triangle (2, -1, 0, 0, 0, 0, 0, ...) DELTA (0, 1/2, -1/2, 0, 0, 0, 0, 0, 0, ...) begins:
  1;
  2, 0;
  2, 1, 0;
  2, 3, 0, 0;
  2, 5, 1, 0, 0;
  2, 7, 4, 0, 0, 0;
  2, 9, 9, 1, 0, 0, 0;

Michael De Vlieger, Table of n, a(n) for n = 0..10200 (rows 0 <= n <= 200, flattened.)
Jean-Luc Baril and José Luis Ramírez, Descent distribution on Catalan words avoiding ordered pairs of Relations, arXiv:2302.12741 [math.CO], 2023.
Thomas Grubb and Frederick Rajasekaran, Set Partition Patterns and the Dimension Index, arXiv:2009.00650 [math.CO], 2020. Mentions this sequence.

Cf. A000045, A023610.
Cf. A029635, A029653.
Columns: A040000, A005408, A000290, A000330, A002415, A005585, A040977, A050486, A053347, A054333, A054334, A057788.

T:=proc(n,k) if n=0 and k=0 then 1 elif k<=floor(n/2) then binomial(n-k,k)+binomial(n-k-1,k) else 0 fi end: for n from 0 to 18 do seq(T(n,k),k=0..floor(n/2)) od; # yields sequence in triangular form

MapAt[# - 1 &, #, 1] &@ Table[Binomial[n - k, k] + Binomial[n - k - 1, k], {n, 0, 16}, {k, 0, Floor[n/2]}] // Flatten (* Michael De Vlieger, Nov 15 2019 *)

T(n,k) = binomial(n-k,k) + binomial(n-k-1,k) for n >= 1 and 0 <= k <= floor(n/2).
G.f. = G(t,z) = (1+z)/(1-z-tz^2).
Sum_{k=0..n} T(n,k)*x^k = (-1)^n*A078050(n), A057079(n), A040000(n), A000045(n+2), A000079(n), A006138(n), A026597(n), A133407(n), A133467(n), A133469(n), A133479(n), A133558(n), A133577(n), A063092(n) for x = -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 respectively. - Philippe Deléham, Jan 14 2012
T(n,k) = T(n-1,k) + T(n-2,k-1) with T(0,0)=1, T(1,0)=2, T(1,1)=0 and T(n,k) = 0 if k > n or if k < 0. - Philippe Deléham, Jan 14 2012

A181630 Duplicate of A112310.

Original entry on oeis.org

1, 1, 2, 2, 2, 3, 2, 3, 3, 3, 4, 3, 3, 4, 3, 4, 4, 4, 5

Offset: 2

Gary W. Adamson, Nov 02 2010

dead

A181631 = number of leading 1's in Fibonacci Maximal notation.
This notation uses headings: (...5, 3, 2, 1); and fills in entries starting from the right, as opposed to the left. For example n=8 = 1011 = (5 + 2 + 1) as opposed to Minimal notation which would be (1100) = (5 + 3).
Conjectured row sums = A023610: (1, 3, 7, 15, 30, 58, ...).
Is this the same as A112310? - R. J. Mathar, Nov 03 2010
Contribution from Gary W. Adamson, Nov 03 2010: (Start)
The next row of 13 terms = (3, 4, 4, 5, 6, 4, 4, 5, 4, 5, 5, 5, 6), sum = 58. (End)

			First few terms of the triangle:
  1;
  1, 2;
  2, 2, 3;
  2, 3, 3, 3, 4;
  3, 3, 4, 3, 4, 4, 4, 5;
  ...
Example: 10 in Fibonacci Maximal = 1110, having three 1's, so a(10) = 3.

Cf. A023610, A181631.

Count numbers of 1's in Fibonacci Maximal notation for n.

cp's OEIS Frontend

A067418 Triangle A067330 with rows read backwards.

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A091533 Triangle read by rows, related to Pascal's triangle, starting with rows 1; 1,1.

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A074083 Coefficient of q^3 in nu(n), where nu(0)=1, nu(1)=b and, for n>=2, nu(n)=b*nu(n-1)+lambda*(1+q+q^2+...+q^(n-2))*nu(n-2) with (b,lambda)=(1,1).

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A134400 M * A007318, where M = triangle with (1, 1, 2, 3, ...) in the main diagonal and the rest zeros.

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A207610 Triangle of coefficients of polynomials u(n,x) jointly generated with A207611; see the Formula section.

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A207612 Triangle of coefficients of polynomials u(n,x) jointly generated with A207613; see the Formula section.

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A213777 Rectangular array: (row n) = b**c, where b(h) = F(h), c(h) = F(h+1), F=A000045 (Fibonacci numbers), n>=1, h>=1, and ** = convolution.

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A074083 Coefficient of q^3 in nu(n), where nu(0)=1, nu(1)=b and, for n>=2, nu(n)=bnu(n-1)+lambda(1+q+q^2+...+q^(n-2))*nu(n-2) with (b,lambda)=(1,1).

A213777 Rectangular array: (row n) = bc, where b(h) = F(h), c(h) = F(h+1), F=A000045 (Fibonacci numbers), n>=1, h>=1, and = convolution.