A000071 -id:A000071 - cp's OEIS frontend

A053606 a(n) = (Fibonacci(6*n+3) - 2)/4.

0, 8, 152, 2736, 49104, 881144, 15811496, 283725792, 5091252768, 91358824040, 1639367579960, 29417257615248, 527871269494512, 9472265593285976, 169972909409653064, 3050040103780469184, 54730748958638792256, 982103441151717791432

Offset: 0

N. J. A. Sloane, James Sellers, Jan 20 2000

Define a(1)=0, a(2)=8 with 5*(a(1)^2) + 5*a(1) + 1 = j(1)^2 = 1^2 and 5*(a(2)^2) + 5*a(2) + 1 = j(2)^2 = 19^2. Then a(n) = a(n-2) + 8*sqrt(5*(a(n-1)^2) + 5*a(n-1)+1). Another definition: a(n) such that 5*(a(n)^2) + 5*a(n) + 1 = j(n)^2. - Pierre CAMI, Mar 30 2005
It appears this sequence gives all nonnegative m such that 5*m^2 + 5*m + 1 is a square. - Gerald McGarvey, Apr 03 2005
sqrt(5*a(n)^2+5*a(n)+1) = A049629(n). - Gerald McGarvey, Apr 19 2005
a(n) is such that 5*a(n)^2 + 5*a(n) + 1 = j^2 = the square of A049629(n). Also A049629(n)/a(n) tends to sqrt(5) as n increases. - Pierre CAMI, Apr 21 2005
From Russell Jay Hendel, Apr 25 2015: (Start)
We prove the two McGarvey-CAMI conjectures mentioned at the beginning of the Comments section. Let, as usual, F(n) = A000045(n), the Fibonacci numbers. In the sequel we indicate equations with upper case letters ((A), (B), (C), (D)) for ease of reference.
Then we must prove (A), 5*((F(6*n+3)-2)/4)^2 + 5*((F(6*n+3)-2)/4) + 1 = ((F(6*n+5)-F(6*n+1))/4)^2. Let m = 3*n+1 so that 6*n+1, 6*n+3, and 6*n+5 are 2*m-1, 2*m+1, and 2*m+3 respectively. Define G(m) = F(6*n+3) = F(2*m+1) = A001519(m+1), the bisected Fibonacci numbers. We can now simplify equation (A) by i) multiplying the LHS and RHS by 16, ii) expanding squares, and iii) gathering like terms. This shows proof of (A) equivalent to proving (B), 5*G(m)^2-4 = (G(m+1)-G(m-1))^2.
By Jarden's theorem (D. Jarden, Recurring sequences, 2nd ed. Jerusalem, Riveon Lematematika, (1966)), if {H(n)}{n >=1} is any recursive sequence satisfying (C), H(n)=3H(n-1)-H(n-2), then {H(n)}^2{n >=1} is also a recursive sequence satisfying (D), H(n)^2=8H(n-1)^2-8H(n-2)^2+H(n-3)^2. As noted in the Formula section of A001519, {G(m)}_{m >= 1} satisfies (C).
Proof of (B) is now straightforward. Since {G(m)}{m >=1} satisfies (C), it follows that {G(m)^2}{m >=1} satisfies (D), and therefore, {5G(m)^2-4}_{m >=1} also satisfies (D).
Similarly, since {G(m)}{m >=1} satisfies (C), it follows that both {G(m+1)}{m >=1}, {G(m-1)}{m >=1} and their difference {G(m+1)-G(m-1)}{m >=1} satisfy (C), and therefore {G(m+1)-G(m-1)}^2_{m >=1} satisfies (D).
But then the LHS and RHS of (B) are equal for m=1,2,3 and satisfy the same recursion, (D). Hence the LHS and RHS of (B) are equal for all m. This completes the proof. (End)

G. C. Greubel, Table of n, a(n) for n = 0..790
F. Ellermann, Illustration of binomial transforms
Index entries for linear recurrences with constant coefficients, signature (19,-19,1).

Cf. A049629.

Related to sum of Fibonacci(kn) over n.. A000071, A027941, A099919, A058038, A138134.

List([0..30], n-> (Fibonacci(6*n+3)-2)/4); # G. C. Greubel, May 16 2019

[(Fibonacci(6*n+3)-2)/4: n in [0..30]]; // Vincenzo Librandi, Apr 20 2011

A053606 := proc(n) add(combinat[fibonacci](6*k),k=0..n) ;end proc:
seq(A053606(n),n=0..30) ;

Table[(Fibonacci[6n+3] -2)/4, {n,0,30}] (* Vladimir Joseph Stephan Orlovsky, Jul 01 2011 *)

a(n)=fibonacci(6*n+3)\4 \\ Charles R Greathouse IV, Jul 02 2013

[(fibonacci(6*n+3)-2)/4 for n in (0..30)] # G. C. Greubel, May 16 2019

a(n) = 8*A049664(n).
a(n+1) = 9*a(n) + 2*sqrt(5*(2*a(n)+1)^2-1) + 4. - Richard Choulet, Aug 30 2007
G.f.: 8*x/((1-x)*(1-18*x+x^2)). - Richard Choulet, Oct 09 2007
a(n) = 18*a(n-1) - a(n-2) + 8, n > 1. - Gary Detlefs, Dec 07 2010
a(n) = Sum_{k=0..n} A134492(k). - Gary Detlefs, Dec 07 2010
a(n) = (Fibonacci(6*n+6) - Fibonacci(6*n) - 8)/16. - Gary Detlefs, Dec 08 2010

A099919 a(n) = F(3) + F(6) + F(9) + ... + F(3n), F(n) = Fibonacci numbers A000045.

Original entry on oeis.org

0, 2, 10, 44, 188, 798, 3382, 14328, 60696, 257114, 1089154, 4613732, 19544084, 82790070, 350704366, 1485607536, 6293134512, 26658145586, 112925716858, 478361013020, 2026369768940, 8583840088782, 36361730124070, 154030760585064, 652484772464328, 2763969850442378

Offset: 0

Ralf Stephan, Oct 30 2004

Partial sum of the even Fibonacci numbers. - Vladimir Joseph Stephan Orlovsky, Nov 28 2010

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 25.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Project Euler, Problem 2: Even Fibonacci Numbers.
Index entries for linear recurrences with constant coefficients, signature (5,-3,-1).

Partial sums of A014445.
Cf. A000045, A004794, A015448, A049652.
Cf. A087635.
Case k = 3 of partial sums of fibonacci(k*n): A000071, A027941, A058038, A138134, A053606.

[(Fibonacci(3*n+2) - 1)/2: n in [0..30]]; // G. C. Greubel, Jan 17 2018

CoefficientList[Series[2 x/((1 - x) (1 - 4 x - x^2)), {x, 0, 40}], x] (* Vincenzo Librandi, Mar 15 2014 *)
LinearRecurrence[{5, -3, -1}, {0, 2, 10}, 30] (* G. C. Greubel, Jan 17 2018 *)
Accumulate[Fibonacci[3Range[0, 19]]] (* Alonso del Arte, Dec 23 2018 *)

a(n) = sum(i=1, n, fibonacci(3*i)); \\ Michel Marcus, Mar 15 2014

a(n) = fibonacci(3*n+2)\2 \\ Charles R Greathouse IV, Jun 11 2015

a(n) = (Fibonacci(3*n + 2) - 1)/2 = (A015448(n+1)-1)/2.
G.f.: 2*x/((1 - x)*(1 - 4*x - x^2)).
a(n) = (F(3n + 2) - 1)/2 = 2 * A049652(n).
a(n) = Sum_{0 <= j <= i <= n} binomial(i, j)*F(i + j). - Benoit Cloitre, May 21 2005
From Gary Detlefs, Dec 08 2010: (Start)
a(n) = 4*a(n - 1) + a(n - 2) + 2, n > 1.
a(n) = 5*a(n - 1) - 3*a(n - 2) - a(n - 3), n > 2.
a(n) = (Fibonacci(3*n + 3) + Fibonacci(3*n) - 2)/4. (End)
a(n) = (-10 + (5 - 3*sqrt(5))*(2 - sqrt(5))^n + (2 + sqrt(5))^n*(5 + 3*sqrt(5)))/20. - Colin Barker, Nov 26 2016
E.g.f.: exp(x)*(exp(x)*(5*cosh(sqrt(5)*x) + 3*sqrt(5)*sinh(sqrt(5)*x)) - 5)/10. - Stefano Spezia, Jun 03 2024

a(0) = 0 prepended by Joerg Arndt, Mar 13 2014

A130152 Triangle read by rows: T(n,k) = number of permutations p of [n] such that max(|p(i)-i|)=k (n>=1, 0<=k<=n-1).

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 4, 9, 10, 1, 7, 23, 47, 42, 1, 12, 60, 157, 274, 216, 1, 20, 151, 503, 1227, 1818, 1320, 1, 33, 366, 1669, 4833, 10402, 13656, 9360, 1, 54, 877, 5472, 18827, 50879, 96090, 115080, 75600, 1, 88, 2088, 17531, 75693, 234061, 569602, 966456, 1077840, 685440, 1, 143, 4937, 55135, 304900, 1076807, 3111243, 6791994, 10553640, 11123280, 6894720

Offset: 1

Emeric Deutsch, May 27 2007

Row sums are the factorials. T(n,n) = (n-2)!*(2n-3) = A007680(n-2) (for n>=2). T(n,1) = Fibonacci(n+1)-1 = A000071(n+1). Sum_{k=0..n-1} k*T(n,k) = A130153(n). For the statistic max(p(i)-i) see A056151.

			T(4,1) = 4 because we have 1243, 1324, 2134 and 2143.
Triangle starts:
  1;
  1,  1;
  1,  2,  3;
  1,  4,  9,  10;
  1,  7, 23,  47,  42;
  1, 12, 60, 157, 274, 216;
  ...

Alois P. Heinz, Rows n = 1..23, flattened
Torleiv Kløve, Spheres of Permutations under the Infinity Norm - Permutations with limited displacement, Reports in Informatics, Department of Informatics, University of Bergen, Norway, no. 376, November 2008.

Columns k=0-10 give: A000012, A000071(n+1), A323798, A323799, A323800, A323801, A323802, A323803, A323804, A323805, A323806.
Row sums give A000142.
T(n,floor(n/2)) gives A323807.
Cf. A007680, A130153, A056151, A299789, A306209.

with(combinat): for n from 1 to 7 do P:=permute(n): for i from 0 to n-1 do ct[i]:=0 od: for j from 1 to n! do if max(seq(abs(P[j][i]-i),i=1..n))=0 then ct[0]:=ct[0]+1 elif max(seq(abs(P[j][i]-i),i=1..n))=1 then ct[1]:=ct[1]+1 elif max(seq(abs(P[j][i]-i),i=1..n))=2 then ct[2]:=ct[2]+1 elif max(seq(abs(P[j][i]-i),i=1..n))=3 then ct[3]:=ct[3]+1 elif max(seq(abs(P[j][i]-i),i=1..n))=4 then ct[4]:=ct[4]+1 elif max(seq(abs(P[j][i]-i),i=1..n))=5 then ct[5]:=ct[5]+1 elif max(seq(abs(P[j][i]-i),i=1..n))=6 then ct[6]:=ct[6]+1 else fi od: a[n]:=seq(ct[i],i=0..n-1): od: for n from 1 to 7 do a[n] od; # a cumbersome program to obtain, by straightforward counting, the first 7 rows of the triangle
n := 8: st := proc (p) max(seq(abs(p[j]-j), j = 1 .. nops(p))) end proc: with(combinat): P := permute(n): f := sort(add(t^st(P[i]), i = 1 .. factorial(n))); # program gives the row generating polynomial for the specified n - Emeric Deutsch, Aug 13 2009
# second Maple program:
b:= proc(s) option remember; (n-> `if`(n=0, 1, add((p-> add(
      coeff(p, x, i)*x^max(i, abs(n-j)), i=0..degree(p)))(
        b(s minus {j})), j=s)))(nops(s))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n-1))(b({$1..n})):
seq(T(n), n=1..10);  # Alois P. Heinz, Jan 21 2019
# third Maple program:
A:= proc(n, k) option remember; LinearAlgebra[Permanent](
      Matrix(n, (i, j)-> `if`(abs(i-j)<=k, 1, 0)))
    end:
T:= (n, k)-> A(n, k)-A(n, k-1):
seq(seq(T(n, k), k=0..n-1), n=1..10);  # Alois P. Heinz, Jan 22 2019

(* from second Maple program: *)
b[s_List] := b[s] = Function[n, If[n == 0, 1, Sum[Function[p, Sum[ Coefficient[p, x, i]*x^Max[i, Abs[n - j]], {i, 0, Exponent[p, x]}]][b[s ~Complement~ {j}]], {j, s}]]][Length[s]];
T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 0, n-1}]][b[Range[n]] ];
Table[T[n], {n, 1, 11}] // Flatten
(* from third Maple program: *)
A[n_, k_] := A[n, k] = Permanent[Table[If[Abs[i-j] <= k, 1, 0], {i, 1, n}, {j, 1, n}]];
T[n_, k_] := A[n, k] - A[n, k - 1];
Table[Table[T[n, k], {k, 0, n - 1}], {n, 1, 11}] // Flatten (* Jean-François Alcover, Dec 06 2019, after Alois P. Heinz *)

T(n,k) = A306209(n,k) - A306209(n,k-1) for k > 0, T(n,0) = 1. - Alois P. Heinz, Jan 29 2019

More terms from R. J. Mathar, Oct 15 2007

A172119 Sum the k preceding elements in the same column and add 1 every time.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 2, 1, 1, 4, 4, 2, 1, 1, 5, 7, 4, 2, 1, 1, 6, 12, 8, 4, 2, 1, 1, 7, 20, 15, 8, 4, 2, 1, 1, 8, 33, 28, 16, 8, 4, 2, 1, 1, 9, 54, 52, 31, 16, 8, 4, 2, 1, 1, 10, 88, 96, 60, 32, 16, 8, 4, 2, 1, 1, 11, 143, 177, 116, 63, 32, 16, 8, 4, 2, 1, 1, 12, 232, 326, 224, 124, 64, 32, 16

Offset: 0

Mats Granvik, Jan 26 2010

Columns are related to Fibonacci n-step numbers. Are there closed forms for the sequences in the columns?
We denote by a(n,k) the number which is in the (n+1)-th row and (k+1)-th-column. With help of the definition, we also have the recurrence relation: a(n+k+1, k) = 2*a(n+k, k) - a(n, k). We see on the main diagonal the numbers 1,2,4, 8, ..., which is clear from the formula for the general term d(n)=2^n. - Richard Choulet, Jan 31 2010
Most of the paper by Dunkel (1925) is a study of the columns of this table. - Petros Hadjicostas, Jun 14 2019

			Triangle begins:
n\k|....0....1....2....3....4....5....6....7....8....9...10
---|-------------------------------------------------------
0..|....1
1..|....1....1
2..|....1....2....1
3..|....1....3....2....1
4..|....1....4....4....2....1
5..|....1....5....7....4....2....1
6..|....1....6...12....8....4....2....1
7..|....1....7...20...15....8....4....2....1
8..|....1....8...33...28...16....8....4....2....1
9..|....1....9...54...52...31...16....8....4....2....1
10.|....1...10...88...96...60...32...16....8....4....2....1

Erik Bates, Blan Morrison, Mason Rogers, Arianna Serafini, and Anav Sood, A new combinatorial interpretation of partial sums of m-step Fibonacci numbers, arXiv:2503.11055 [math.CO], 2025. See p. 3.
Otto Dunkel, Solutions of a probability difference equation, Amer. Math. Monthly, 32 (1925), 354-370; see p. 356.
Thomas Langley, Jeffrey Liese, and Jeffrey Remmel, Generating Functions for Wilf Equivalence Under Generalized Factor Order, J. Int. Seq. 14 (2011), # 11.4.2.
Eric Weisstein's World of Mathematics, Fibonacci n-Step Number.
Wikipedia, Fibonacci number.

Cf. A000071 (col. 3), A008937 (col. 4), A107066 (col. 5), A001949 (col. 6), A172316 (col. 7), A172317 (col. 8), A172318 (col. 9), A172319 (col. 10), A172320 (col. 11), A144428.

Cf. (1-((-1)^T(n, k)))/2 = A051731, see formula by Hieronymus Fischer in A022003.

T:= function(n,k)
    if k=0 and k=n then return 1;
    elif k<0 or k>n then return 0;
    else return 1 + Sum([1..k], j-> T(n-j,k));
    fi;
  end;
Flat(List([0..12], n-> List([0..n], k-> T(n,k) ))); # G. C. Greubel, Jul 27 2019

T:= func< n,k | (&+[(-1)^j*2^(n-k-(k+1)*j)*Binomial(n-k-k*j, n-k-(k+1)*j): j in [0..Floor((n-k)/(k+1))]]) >;
[[T(n,k): k in [0..n]]: n in [0..12]]; // G. C. Greubel, Jul 27 2019

for k from 0 to 20 do for n from 0 to 20 do b(n):=sum((-1)^j*binomial(n-k*j,n-(k+1)*j)*2^(n-(k+1)*j),j=0..floor(n/(k+1))):od: seq(b(n),n=0..20):od; # Richard Choulet, Jan 31 2010
A172119 := proc(n,k)
    option remember;
    if k = 0 then
        1;
    elif k > n then
        0;
    else
        1+add(procname(n-k+i,k),i=0..k-1) ;
    end if;
end proc:
seq(seq(A172119(n,k),k=0..n),n=0..12) ; # R. J. Mathar, Sep 16 2017

T[, 0] = 1; T[n, n_] = 1; T[n_, k_] /; k>n = 0; T[n_, k_] := T[n, k] = Sum[T[n-k+i, k], {i, 0, k-1}] + 1;
Table[T[n, k], {n, 0, 12}, {k, 0, n}]//Flatten
Table[Sum[(-1)^j*2^(n-k-(k+1)*j)*Binomial[n-k-k*j, n-k-(k+1)*j], {j, 0, Floor[(n-k)/(k+1)]}], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jul 27 2019 *)

T(n,k) = if(k<0 || k>n, 0, k==1 && k==n, 1, 1 + sum(j=1,k, T(n-j,k)));
for(n=1,12, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Jul 27 2019

@CachedFunction
def T(n, k):
    if (k==0 and k==n): return 1
    elif (k<0 or k>n): return 0
    else: return 1 + sum(T(n-j, k) for j in (1..k))
[[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Jul 27 2019

T(n,0) = 1.
T(n,1) = n.
T(n,2) = A000071(n+1).
T(n,3) = A008937(n-2).
The general term in the n-th row and k-th column is given by: a(n, k) = Sum_{j=0..floor(n/(k+1))} ((-1)^j binomial(n-k*j,n-(k+1)*j)*2^(n-(k+1)*j)). For example: a(5,3) = binomial(5,5)*2^5 - binomial(2,1)*2^1 = 28. The generating function of the (k+1)-th column satisfies: psi(k)(z)=1/(1-2*z+z^(k+1)) (for k=0 we have the known result psi(0)(z)=1/(1-z)). - Richard Choulet, Jan 31 2010 [By saying "(k+1)-th column" the author actually means "k-th column" for k = 0, 1, 2, ... - Petros Hadjicostas, Jul 26 2019]

A058038 a(n) = Fibonacci(2n)Fibonacci(2*n+2).

Original entry on oeis.org

0, 3, 24, 168, 1155, 7920, 54288, 372099, 2550408, 17480760, 119814915, 821223648, 5628750624, 38580030723, 264431464440, 1812440220360, 12422650078083, 85146110326224, 583600122205488, 4000054745112195, 27416783093579880, 187917426909946968

Offset: 0

N. J. A. Sloane, Jun 09 2002

Partial sums of A033888, i.e., a(n) = Sum_{k=0..n} Fibonacci(4*k). - Vladeta Jovovic, Jun 09 2002
From Paul Weisenhorn, May 17 2009: (Start)
a(n) is the solution of the 2 equations a(n)+1=A^2 and 5*a(n)+1=B^2
which are equivalent to the Pell equation (10*a(n)+3)^2-5*(A*B)^2=4.
(End)
Numbers a(n) such as a(n)+1 and 5*a(n)+1 are perfect squares. - Sture Sjöstedt, Nov 03 2011

			G.f. = 3*x + 24*x^2 + 168*x^3 + 1155*x^4 + 7920*x^5 + 54288*x^6 + ... - _Michael Somos_, Jan 23 2025

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 29.
H. J. H. Tuenter, Fibonacci summation identities arising from Catalan's identity, Fib. Q., 60:4 (2022), 312-319.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A033888, A004187, A094874.
Bisection of A059929, A064831 and A080097.
Related to sum of fibonacci(kn) over n; cf. A000071, A099919, A027941, A138134, A053606.

[Fibonacci(2*n)*Fibonacci(2*n+2): n in [0..30]]; // Vincenzo Librandi, Apr 18 2011

fs4:=n->sum(fibonacci(4*k),k=0..n):seq(fs4(n),n=0..21); # Gary Detlefs, Dec 07 2010

Table[Fibonacci[2 n]*Fibonacci[2 n + 2], {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Jul 01 2011 *)
Accumulate[Fibonacci[4*Range[0,30]]] (* or *) LinearRecurrence[{8,-8,1},{0,3,24},30] (* Harvey P. Dale, Jul 25 2013 *)

a(n)=fibonacci(2*n)*fibonacci(2*n+2) \\ Charles R Greathouse IV, Jul 02 2013

a(n) = -3/5 + (1/5*sqrt(5)+3/5)*(2*1/(7+3*sqrt(5)))^n/(7+3*sqrt(5)) + (1/5*sqrt(5)-3/5)*(-2*1/(-7+3*sqrt(5)))^n/(-7+3*sqrt(5)). Recurrence: a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3). G.f.: 3*x/(1-7*x+x^2)/(1-x). - Vladeta Jovovic, Jun 09 2002
a(n) = A081068(n) - 1.
a(n) is the next integer from ((3+sqrt(5))*((7+3*sqrt(5))/2)^(n-1)-6)/10. - Paul Weisenhorn, May 17 2009
a(n) = 7*a(n-1) - a(n-2) + 3, n>1. - Gary Detlefs, Dec 07 2010
a(n) = sum_{k=0..n} Fibonacci(4k). - Gary Detlefs, Dec 07 2010
a(n) = (Lucas(4n+2)-3)/5, where Lucas(n)= A000032(n). - Gary Detlefs, Dec 07 2010
a(n) = (1/5)*(Fibonacci(4n+4) - Fibonacci(4n)-3). - Gary Detlefs, Dec 08 2010
a(n) = 3*A092521(n). - R. J. Mathar, Nov 03 2011
a(0)=0, a(1)=3, a(2)=24, a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3). - Harvey P. Dale, Jul 25 2013
a(n) = A001906(n)*A001906(n+1). - R. J. Mathar, Jul 09 2019
Sum_{n>=1} 1/a(n) = 2/(3 + sqrt(5)) = A094874 - 1. - Amiram Eldar, Oct 05 2020
a(n) = a(-1-n) for all n in Z. - Michael Somos, Jan 23 2025

A061667 a(n) = Fibonacci(2*n+1) - 2^(n-1).

Original entry on oeis.org

1, 3, 9, 26, 73, 201, 546, 1469, 3925, 10434, 27633, 72977, 192322, 506037, 1329885, 3491810, 9161929, 24026745, 62983842, 165055853, 432445861, 1132806018, 2967020769, 7770353441, 20348233858, 53282736741, 139516753581, 365301078434, 956453590585

Offset: 1

Emeric Deutsch, Jun 16 2001

Number of cells in the bottom row of all directed column-convex polyominoes of area n+1.
Also the binomial transform of A000071 (after removing its 2 leading zeros). - R. J. Mathar, Nov 04 2008
Equals row sums of triangle A147293. - Gary W. Adamson, Nov 05 2008

Harry J. Smith, Table of n, a(n) for n = 1..200
Elena Barcucci, Renzo Pinzani, and Renzo Sprugnoli, Directed column-convex polyominoes by recurrence relations, Lecture Notes in Computer Science, No. 668, Springer, Berlin (1993), pp. 282-298.
Alexander Burstein and Toufik Mansour, Words restricted by 3-letter generalized multipermutation patterns, arXiv:math/0112281 [math.CO], 2001.
Alexander Burstein and Toufik Mansour, Words restricted by 3-letter generalized multipermutation patterns, Annals. Combin., 7 (2003), 1-14; see Th. 3.8.
Manosij Ghosh Dastidar and Michael Wallner, Bijections and congruences involving lattice paths and integer compositions, arXiv:2402.17849 [math.CO], 2024. See p. 22.
Index entries for linear recurrences with constant coefficients, signature (5,-7,2).

Cf. A000045.

Cf. A147293. - Gary W. Adamson, Nov 05 2008

Table[Fibonacci[2 n + 1] - 2^(n - 1), {n, 1, 100}] (* Vladimir Joseph Stephan Orlovsky, Jul 01 2011 *)

a(n) = { fibonacci(2*n + 1) - 2^(n - 1) } \\ Harry J. Smith, Jul 26 2009

Vec(x*(1-x)^2/((1-2*x)*(1-3*x+x^2)) + O(x^50)) \\ Michel Marcus, Nov 29 2014

G.f.: x*(1-x)^2/((1-2*x)*(1-3*x+x^2)). - corrected by Philip B. Zhang, Nov 28 2014
a(n) = Sum_{k=0..n+1} C(n+1, k)*sum{j=0..floor(k/2), Fibonacci(k-2j)}. - Paul Barry, Apr 17 2005
a(n) = 2*A001906(n+1)-A001906(n)-A000079(n). - R. J. Mathar, Nov 16 2007
From Colin Barker, Jun 05 2017: (Start)
a(n) = 2^(-1-n)*(-5*4^n - (3-sqrt(5))^n*(-5+sqrt(5)) + (3+sqrt(5))^n*(5+sqrt(5))) / 5.
a(n) = 5*a(n-1) - 7*a(n-2) + 2*a(n-3) for n>3. (End)

Offset changed from 0 to 1 by Harry J. Smith, Jul 26 2009

A065220 a(n) = Fibonacci(n) - n.

Original entry on oeis.org

0, 0, -1, -1, -1, 0, 2, 6, 13, 25, 45, 78, 132, 220, 363, 595, 971, 1580, 2566, 4162, 6745, 10925, 17689, 28634, 46344, 75000, 121367, 196391, 317783, 514200, 832010, 1346238, 2178277, 3524545, 5702853, 9227430, 14930316, 24157780, 39088131, 63245947, 102334115, 165580100, 267914254

Offset: 0

Henry Bottomley, Oct 22 2001

E(n) = Fib(n+4)-(n+4): cost of maximum height Huffman tree of size n for Fibonacci sequence (Fibonacci sequence is minimizing absolutely ordered sequence of Huffman tree). - Alex Vinokur (alexvn(AT)barak-online.net), Oct 26 2004

Vinokur A.B, Huffman trees and Fibonacci numbers, Kibernetika Issue 6 (1986) 9-12 (in Russian); English translation in Cybernetics 21, Issue 6 (1986), 692-696.

Harry J. Smith, Table of n, a(n) for n = 0..300
Gregory Dresden, On the Brousseau sums Sum_{i=1..n} i^p*Fibonacci(i), arxiv.org:2206.00115 [math.NT], 2022.
Albert Frank, International Contest Of Logical Sequences, 2002 - 2003. Item 7
Albert Frank, Solutions of International Contest Of Logical Sequences, 2002 - 2003.
A. B. Vinokur, Huffman trees and Fibonacci numbers, Cybernetics 21, Issue 6 (1986), 692-696; also at Research Gate.
Alex Vinokur, Fibonacci connection between Huffman codes and Wythoff array, arXiv:cs/0410013 [cs.DM], 2004-2005.
Index entries for linear recurrences with constant coefficients, signature (3,-2,-1,1).

Cf. A002062, A045925.

List([0..50], n-> Fibonacci(n) - n); # G. C. Greubel, Jul 09 2019

a065220 n = a065220_list !! n
a065220_list = zipWith (-) a000045_list [0..]
-- Reinhard Zumkeller, Nov 06 2012

[Fibonacci(n) - n: n in [0..50]]; // G. C. Greubel, Jul 09 2019

a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=a[n-1]+a[n-2] od: seq(a[n]-n, n=0..42); # Zerinvary Lajos, Mar 20 2008

lst={};Do[f=Fibonacci[n]-n;AppendTo[lst,f],{n,0,5!}];lst (* Vladimir Joseph Stephan Orlovsky, Mar 21 2009 *)
Table[Fibonacci[n]-n,{n,0,50}] (* or *) LinearRecurrence[{3,-2,-1,1},{0,0,-1,-1},50] (* Harvey P. Dale, May 29 2017 *)

a(n) = { fibonacci(n) - n } \\ Harry J. Smith, Oct 14 2009

[fibonacci(n) - n for n in (0..50)] # G. C. Greubel, Jul 09 2019

a(n) = A000045(n) - A001477(n) = A000126(n-3) - 2 = A001924(n-4) - 1.
a(n) = a(n-1) + a(n-2) + n - 3 = a(n-1) + A000071(n-2).
G.f.: x^2*(2x-1)/((1-x-x^2)*(1-x)^2).
a(n) = Sum_{i=0..n} (i - 2)*F(n-i) for F(n) the Fibonacci sequence A000045. - Greg Dresden, Jun 01 2022

A104763 Triangle read by rows: Fibonacci(1), Fibonacci(2), ..., Fibonacci(n) in row n.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 2, 3, 5, 1, 1, 2, 3, 5, 8, 1, 1, 2, 3, 5, 8, 13, 1, 1, 2, 3, 5, 8, 13, 21, 1, 1, 2, 3, 5, 8, 13, 21, 34, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233

Offset: 1

Gary W. Adamson, Mar 23 2005

Triangle of A104762, Fibonacci sequence in each row starts from the right.
The triangle or chess sums, see A180662 for their definitions, link the Fibonacci(n) triangle to sixteen different sequences, see the crossrefs. The knight sums Kn14 - Kn18 have been added. As could be expected all sums are related to the Fibonacci numbers. - Johannes W. Meijer, Sep 22 2010
Sequence B is called a reluctant sequence of sequence A, if B is triangle array read by rows: row number k coincides with first k elements of the sequence A. Sequence A104763 is reluctant sequence of Fibonacci numbers (A000045), except 0. - Boris Putievskiy, Dec 13 2012

			First few rows of the triangle are:
  1;
  1, 1;
  1, 1, 2;
  1, 1, 2, 3;
  1, 1, 2, 3, 5;
  1, 1, 2, 3, 5, 8;
  1, 1, 2, 3, 5, 8, 13; ...

Reinhard Zumkeller, Rows n = 1..100 of table, flattened
Boris Putievskiy, Transformations Integer Sequences And Pairing Functions, arXiv:1212.2732 [math.CO], 2012.

Cf. A104762, A000045.
Cf. A000071 (row sums). - R. J. Mathar, Jul 22 2009
Triangle sums (see the comments): A000071 (Row1; Kn4 & Ca1 & Ca4 & Gi1 & Gi4); A008346 (Row2); A131524 (Kn11); A001911 (Kn12); A006327 (Kn13); A167616 (Kn14); A180671 (Kn15); A180672 (Kn16); A180673 (Kn17); A180674 (Kn18); A052952 (Kn21 & Kn22 & Kn23 & Fi2 & Ze2); A001906 (Kn3 &Fi1 & Ze3); A004695 (Ca2 & Ze4); A001076 (Ca3 & Ze1); A080239 (Gi2); A081016 (Gi3). - Johannes W. Meijer, Sep 22 2010

Flat(List([1..15], n -> List([1..n], k -> Fibonacci(k)))); # G. C. Greubel, Jul 13 2019

a104763 n k = a104763_tabl !! (n-1) !! (k-1)
a104763_row n = a104763_tabl !! (n-1)
a104763_tabl = map (flip take $ tail a000045_list) [1..]
-- Reinhard Zumkeller, Aug 15 2013

[Fibonacci(k): k in [1..n], n in [1..15]]; // G. C. Greubel, Jul 13 2019

Table[Fibonacci[k], {n,15}, {k,n}]//Flatten (* G. C. Greubel, Jul 13 2019 *)

for(n=1,15, for(k=1,n, print1(fibonacci(k), ", "))) \\ G. C. Greubel, Jul 13 2019

[[fibonacci(k) for k in (1..n)] for n in (1..15)] # G. C. Greubel, Jul 13 2019

F(1) through F(n) starting from the left in n-th row.
T(n,k) = A000045(k), 1<=k<=n. - R. J. Mathar, May 02 2008
a(n) = A000045(m), where m= n-t(t+1)/2, t=floor((-1+sqrt(8*n-7))/2). - Boris Putievskiy, Dec 13 2012
G.f.: (x*y)/((x-1)*(x^2*y^2+x*y-1)). - Vladimir Kruchinin, Jun 21 2025

Edited by R. J. Mathar, May 02 2008

Extended by R. J. Mathar, Aug 27 2008

A129696 Antidiagonal sums of triangular array T defined in A014430: T(j,k) = binomial(j+1, k) - 1 for 1 <= k <= j.

Original entry on oeis.org

1, 2, 5, 9, 17, 29, 50, 83, 138, 226, 370, 602, 979, 1588, 2575, 4171, 6755, 10935, 17700, 28645, 46356, 75012, 121380, 196404, 317797, 514214, 832025, 1346253, 2178293, 3524561, 5702870, 9227447, 14930334, 24157798, 39088150, 63245966

Offset: 1

Paul Curtz, Jun 01 2007

If T is construed as a lower triangular matrix M over the rational field, the inverse M^-1 is a lower triangular matrix containing fractions. Its row sums are the Bernoulli numbers. First column of M^-1 is 1, -1, 2/3, -1/4, -1/30, 1/12, 1/42, -1/12, ... . Multiplied by j! this gives 1, -2, 4, -6, -4, 60, 120, -3660, ... .
The Kn22 sums, see A180662 for the definition of these sums, of the 'Races with Ties' triangle A035317 lead to this sequence. - Johannes W. Meijer, Jul 20 2011
This sequence is the convolution of (1,1,2,3,5,8,13,21,...) and (1,1,2,2,3,3,4,4,5,5,...), i.e., the Fibonacci numbers (A000045) and A008619. - Clark Kimberling, May 28 2012
a(n) is the sum of the first summands over all Arndt compositions of n (see the Checa link). - Daniel Checa, Jan 01 2024

Paul Curtz, Intégration numérique des systèmes différentiels à conditions initiales. Note no. 12 du Centre de Calcul Scientifique de l'Armement, 1969.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Daniel F. Checa, Arndt Compositions: Connections with Fibonacci Numbers, Statistics, and Generalizations, 2023, p. 31.
Larry Ericksen, Problem 7, Problem Section, The Fibonacci Quarterly, Vol. 57, No. 5 (2019), pp. 172, 175-183.
Index entries for linear recurrences with constant coefficients, signature (2,1,-3,0,1).

Cf. A000045, A000071, A001924, A008619, A014430, A027641, A035317, A052952 (first differences), A180662.

Cf. A000120, A002487.

m:=36; M:=ZeroMatrix(IntegerRing(), m, m); for j:=1 to m do for k:=1 to j do M[j, k]:=Binomial(j+1, k)-1; end for; end for; [ &+[ M[j-k+1, k]: k in [1..(j+1) div 2] ]: j in [1..m] ]; // Klaus Brockhaus, Jun 11 2007

[Fibonacci(n+3)-2-Floor(n/2): n in [1..40]]; // Vincenzo Librandi, Nov 23 2014

with(combinat): a := proc (n) options operator, arrow: fibonacci(n+3)-2-floor((1/2)*n) end proc: seq(a(n), n = 1 .. 34); # Emeric Deutsch, Nov 22 2014

a[n_]:= a[n]= If[n<3, n, a[n-1] + a[n-2] + (n + Mod[n, 2])/2];
Table[a[n], {n,40}] (* Jean-François Alcover, Mar 04 2013 *)
Table[Fibonacci[n+3] -2 -Floor[n/2], {n, 100}] (* Vincenzo Librandi, Nov 23 2014 *)

prpr = 1
prev = 2
for n in range(2,100):
    print(prpr, end=", ")
    curr = prpr+prev + 1 + n//2
    prpr = prev
    prev = curr
# Alex Ratushnyak, Jul 30 2012

[fibonacci(n+3) -2 -(n//2) for n in range(1,41)] # G. C. Greubel, Mar 17 2023

From Paul Barry, Jan 18 2009: (Start)
a(n) = Sum_{k=0..floor(n/2)} A000071(n-2*k+3).
a(n) = Sum_{k=0..floor(n/2)} (Sum_{j=0..n-2*k} Fibonacci(j+1)). (End)
a(n+1) = a(n-1) + a(n) + 1 + floor(n/2) for n>1, a(0)=1, a(1)=2. - Alex Ratushnyak, Jul 30 2012
From R. J. Mathar, Jul 25 2013: (Start)
G.f.: x/((1 + x)*(1 - x)^2*(1 - x - x^2)).
a(n) + a(n+1) = A001924(n+1). (End)
a(n) = Fibonacci(n+3) - 2 - floor(n/2). - Emeric Deutsch, Nov 22 2014
a(n) = (-5/4 - (-1)^n/4 + (2^(-n)*((1 - t)^n*(-2 + t) + (1 + t)^n*(2 + t)))/t + (-1 - n)/2), where t=sqrt(5). - Colin Barker, Feb 09 2017
E.g.f.: (4*exp(x/2)*(5*cosh(sqrt(5)*x/2) + 2*sqrt(5)*sinh(sqrt(5)*x/2)) - 5*(4 + x)*cosh(x) - 5*(3 + x)*sinh(x))/10. - Stefano Spezia, Apr 06 2024
a(n) = max_{k = 2^(n+1)..2^(n+2)-1} (A002487(k) - A000120(k)) (Ericksen, 2019). - Amiram Eldar, Jan 30 2025

Edited and extended by Klaus Brockhaus, Jun 11 2007

A188607 T(n,k)=Number of nXk binary arrays without the pattern 1 1 0 diagonally, vertically, antidiagonally or horizontally.

Original entry on oeis.org

2, 4, 4, 7, 16, 7, 12, 49, 49, 12, 20, 144, 211, 144, 20, 33, 400, 847, 883, 400, 33, 54, 1089, 3282, 4914, 3354, 1089, 54, 88, 2916, 12405, 26723, 24711, 12529, 2916, 88, 143, 7744, 45860, 140617, 181626, 123984, 45705, 7744, 143, 232, 20449, 167525, 721597

Offset: 1

R. H. Hardin Apr 05 2011

Table starts
...2.....4.......7.......12.........20..........33............54.............88
...4....16......49......144........400........1089..........2916...........7744
...7....49.....211......847.......3282.......12405.........45860.........167525
..12...144.....883.....4914......26723......140617........721597........3648942
..20...400....3354....24711.....181626.....1288734.......8951764.......61078115
..33..1089...12529...123984....1238275....11895711.....111888105.....1031943536
..54..2916...45705...602049....8077216...104225541....1318303126....16339159241
..88..7744..165506..2925040...52884231...919272280...15651179908...260970703443
.143.20449..595370.14100232..342253692..7996186202..182869043451..4096182114761
.232.53824.2135861.67998376.2219051847.69774651726.2144672066828.64567134345340

			Some solutions for 5X3
..0..0..1....0..0..0....0..1..0....0..1..1....1..0..0....0..0..0....0..0..0
..0..1..0....0..0..0....0..0..1....0..0..0....0..0..0....0..1..0....1..0..1
..1..0..0....1..0..0....0..0..0....1..0..1....1..1..1....1..0..0....0..0..0
..0..0..1....1..1..1....1..1..1....1..0..0....0..0..0....0..0..1....1..1..1
..0..1..0....1..1..1....1..0..1....1..0..0....0..1..1....0..0..1....0..0..0

R. H. Hardin, Table of n, a(n) for n = 1..312

Row 1 and Column 1 are A000071(n+3)

Row 2 and Column 2 are A188516

cp's OEIS Frontend

A053606 a(n) = (Fibonacci(6*n+3) - 2)/4.

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A099919 a(n) = F(3) + F(6) + F(9) + ... + F(3n), F(n) = Fibonacci numbers A000045.

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A130152 Triangle read by rows: T(n,k) = number of permutations p of [n] such that max(|p(i)-i|)=k (n>=1, 0<=k<=n-1).

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A172119 Sum the k preceding elements in the same column and add 1 every time.

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A058038 a(n) = Fibonacci(2*n)*Fibonacci(2*n+2).

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A061667 a(n) = Fibonacci(2*n+1) - 2^(n-1).

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A058038 a(n) = Fibonacci(2n)Fibonacci(2*n+2).