A027961 -id:A027961 - cp's OEIS frontend

A263225 Positive values of n such that A027961(n) is divisible by A000217(n).

1, 60, 240, 600, 660, 768, 1008, 1200, 1320, 1800, 1860, 2160, 2688, 2736, 3000, 3300, 3360, 3888, 4620, 4800, 5280, 5520, 5568, 6120, 6480, 6720, 6840, 7320, 7680, 8208, 8640, 9000, 9600, 9720, 10368, 11160, 12240, 12288, 13200, 13248, 13440, 13680, 13868, 14400, 15120, 15360

Offset: 1

Altug Alkan, Oct 12 2015

Is there a maximum value of a(n) - a(n-1)?
A263161 is not a subsequence although they have many common terms.
Terms that are not congruent to 0 (mod 6): 1, 13868, 16016, 34988, 158252, 196412, 313988, 1287788, 2056748, 2212412, 2542028, 2847260, 2951708, 6117548, 7538108, 7756988, 9056732, 9865628, ... . - Robert G. Wilson v, Oct 15 2015

			For n = 1, A027961(1) = 1 is divisible by A000217(1) = 1.
For n = 60, A027961(60) = 9062201101800 = 1830*4952022460, therefore it is divisible by A000217(60) = 1830.

Cf. A000204, A000217, A027961, A263161.

[n: n in [1..20000] | IsDivisibleBy(Lucas(n+2)-3,n*(n+1) div 2)]; // Bruno Berselli, Oct 19 2015

fQ[n_] := Mod[ Fibonacci[n + 1] + Fibonacci[n + 3] - 3, n (n + 1)/2] == 0; Select[ Range@ 16000, fQ] (* Robert G. Wilson v, Oct 15 2015 *)

for(n=1, 20000, if((fibonacci(n+3) + fibonacci(n+1)-3) % (n*(n+1)/2) == 0, print1(n", ")));

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

A101220 a(n) = Sum_{k=0..n} Fibonacci(n-k)*n^k.

Original entry on oeis.org

0, 1, 3, 14, 91, 820, 9650, 140601, 2440317, 49109632, 1123595495, 28792920872, 816742025772, 25402428294801, 859492240650847, 31427791175659690, 1234928473553777403, 51893300561135516404, 2322083099525697299278

Offset: 0

Ross La Haye, Dec 14 2004

In what follows a(i,j,k) denotes a three-dimensional array, the terms a(n) are defined as a(n,n,n) in that array. - Joerg Arndt, Jan 03 2021
Previous name was: Three-dimensional array: a(i,j,k) = expansion of x*(1 + (i-j)*x)/((1-j*x)*(1-x-x^2)), read by a(n,n,n).
a(i,j,k) = the k-th value of the convolution of the Fibonacci numbers (A000045) with the powers of i = Sum_{m=0..k} a(i-1,j,m), both for i = j and i > 0; a(i,j,k) = a(i-1,j,k) + a(j,j,k-1), for i,k > 0; a(i,1,k) = Sum_{m=0..k} a(i-1,0,m), for i > 0. With F = Fibonacci and L = Lucas, then a(1,1,k) = F(k+2) - 1; a(2,1,k) = F(k+3) - 2; a(3,1,k) = L(k+2) - 3; a(4,1,k) = 4*F(k+1) + F(k) - 4; a(1,2,k) = 2^k - F(k+1); a(2,2,k) = 2^(k+1) - F(k+3); a(3,2,k) = 3(2^k - F(k+2)) + F(k); a(4,2,k) = 2^(k+2) - F(k+4) - F(k+2); a(1,3,k) = (3^k + L(k-1))/5, for k > 0; a(2,3,k) = (2 * 3^k - L(k)) /5, for k > 0; a(3,3,k) = (3^(k+1) - L(k+2))/5; a(4,3,k) = (4 * 3^k - L(k+2) - L(k+1))/5, etc..

			a(1,3,3) = 6 because a(1,3,0) = 0, a(1,3,1) = 1, a(1,3,2) = 2 and 4*2 - 2*1 - 3*0 = 6.

G. C. Greubel, Table of n, a(n) for n = 0..385
Eric Weisstein's World of Mathematics, Fibonacci Number
Eric Weisstein's World of Mathematics, Lucas Number

a(0, j, k) = A000045(k).
a(1, 2, k+1) - a(1, 2, k) = A099036(k).
a(3, 2, k+1) - a(3, 2, k) = A104004(k).
a(4, 2, k+1) - a(4, 2, k) = A027973(k).
a(1, 3, k+1) - a(1, 3, k) = A099159(k).
a(i, 0, k) = A109754(i, k).
a(i, i+1, 3) = A002522(i+1).
a(i, i+1, 4) = A071568(i+1).
a(2^i-2, 0, k+1) = A118654(i, k), for i > 0.
Sequences of the form a(n, 0, k): A000045(k+1) (n=1), A000032(k) (n=2), A000285(k-1) (n=3), A022095(k-1) (n=4), A022096(k-1) (n=5), A022097(k-1) (n=6), A022098(k-1) (n=7), A022099(k-1) (n=8), A022100(k-1) (n=9), A022101(k-1) (n=10), A022102(k-1) (n=11), A022103(k-1) (n=12), A022104(k-1) (n=13), A022105(k-1) (n=14), A022106(k-1) (n=15), A022107(k-1) (n=16), A022108(k-1) (n=17), A022109(k-1) (n=18), A022110(k-1) (n=19), A088209(k-2) (n=k-2), A007502(k) (n=k-1), A094588(k) (n=k).
Sequences of the form a(1, n, k): A000071(k+2) (n=1), A027934(k-1) (n=2), A098703(k) (n=3).
Sequences of the form a(2, n, k): A001911(k) (n=1), A008466(k+1) (n=2), A106517(k-1) (n=3).
Sequences of the form a(3, n, k): A027961(k) (n=1), A094688(k) (n=3).
Sequences of the form a(4, n, k): A053311(k-1) (n=1), A027974(k-1) (n=2).
Cf. A001622, A001891, A001924, A023537, A104161, A106517.

A101220:= func< n | (&+[n^k*Fibonacci(n-k): k in [0..n]]) >;
[A101220(n): n in [0..30]]; // G. C. Greubel, Jun 01 2025

Join[{0}, Table[Sum[Fibonacci[n-k]*n^k, {k, 0, n}], {n, 1, 20}]] (* Vaclav Kotesovec, Jan 03 2021 *)

a(n)=sum(k=0,n,fibonacci(n-k)*n^k) \\ Joerg Arndt, Jan 03 2021

def A101220(n): return sum(n^k*fibonacci(n-k) for k in range(n+1))
print([A101220(n) for n in range(31)]) # G. C. Greubel, Jun 01 2025

a(i, j, 0) = 0, a(i, j, 1) = 1, a(i, j, 2) = i+1; a(i, j, k) = ((j+1)*a(i, j, k-1)) - ((j-1)*a(i, j, k-2)) - (j*a(i, j, k-3)), for k > 2.
a(i, j, k) = Fibonacci(k) + i*a(j, j, k-1), for i, k > 0.
a(i, j, k) = (Phi^k - (-Phi)^-k + i(((j^k - Phi^k) / (j - Phi)) - ((j^k - (-Phi)^-k) / (j - (-Phi)^-1)))) / sqrt(5), where Phi denotes the golden mean/ratio (A001622).
i^k = a(i-1, i, k) + a(i-2, i, k+1).
A104161(k) = Sum_{m=0..k} a(k-m, 0, m).
a(i, j, 0) = 0, a(i, j, 1) = 1, a(i, j, 2) = i+1, a(i, j, 3) = i*(j+1) + 2; a(i, j, k) = (j+2)*a(i, j, k-1) - 2*j*a(i, j, k-2) - a(i, j, k-3) + j*a(i, j, k-4), for k > 3. a(i, j, 0) = 0, a(i, j, 1) = 1; a(i, j, k) = a(i, j, k-1) + a(i, j, k-2) + i * j^(k-2), for k > 1.
G.f.: x*(1 + (i-j)*x)/((1-j*x)*(1-x-x^2)).
a(n, n, n) = Sum_{k=0..n} Fibonacci(n-k) * n^k. - Ross La Haye, Jan 14 2006
Sum_{m=0..k} binomial(k,m)*(i-1)^m = a(i-1,i,k) + a(i-2,i,k+1), for i > 1. - Ross La Haye, May 29 2006
From Ross La Haye, Jun 03 2006: (Start)
a(3, 3, k+1) - a(3, 3, k) = A106517(k).
a(1, 1, k) = A001924(k) - A001924(k-1), for k > 0.
a(2, 1, k) = A001891(k) - A001891(k-1), for k > 0.
a(3, 1, k) = A023537(k) - A023537(k-1), for k > 0.
Sum_{j=0..i+1} a(i-j+1, 0, j) - Sum_{j=0..i} a(i-j, 0, j) = A001595(i). (End)
a(i,j,k) = a(j,j,k) + (i-j)*a(j,j,k-1), for k > 0.
a(n) ~ n^(n-1). - Vaclav Kotesovec, Jan 03 2021

New name from Joerg Arndt, Jan 03 2021

A027960 'Lucas array': triangular array T read by rows.

Original entry on oeis.org

1, 1, 3, 1, 1, 3, 4, 4, 1, 1, 3, 4, 7, 8, 5, 1, 1, 3, 4, 7, 11, 15, 13, 6, 1, 1, 3, 4, 7, 11, 18, 26, 28, 19, 7, 1, 1, 3, 4, 7, 11, 18, 29, 44, 54, 47, 26, 8, 1, 1, 3, 4, 7, 11, 18, 29, 47, 73, 98, 101, 73, 34, 9, 1, 1, 3, 4, 7, 11, 18, 29, 47, 76, 120, 171, 199, 174, 107, 43, 10, 1

Offset: 0

Clark Kimberling

The k-th row contains 2k+1 numbers.
Columns in the right half consist of convolutions of the Lucas numbers with the natural numbers.
T(n,k) = number of strings s(0),...,s(n) such that s(n)=n-k. s(0) in {0,1,2}, s(1)=1 if s(0) in {1,2}, s(1) in {0,1,2} if s(0)=0 and for 1 <= i <= n, s(i) = s(i-1)+d, with d in {0,2} if s(i)=2i, in {0,1,2} if s(i)=2i-1, in {0,1} if 0 <= s(i) <= 2i-2.

			                           1
                       1,  3,  1
                   1,  3,  4,  4,  1
               1,  3,  4,  7,  8,  5,   1
           1,  3,  4,  7, 11, 15, 13,   6,  1
        1, 3,  4,  7, 11, 18, 26, 28,  19,  7,  1
     1, 3, 4,  7, 11, 18, 29, 44, 54,  47, 26,  8, 1
  1, 3, 4, 7, 11, 18, 29, 47, 73, 98, 101, 73, 34, 9, 1

Nathaniel Johnston, Table of n, a(n) for n = 0..10000

Central column is the Lucas numbers without initial 2: A000204.
Columns in the right half include A027961, A027962, A027963, A027964, A053298.
Bisection triangles are in A026998 and A027011.
Row sums: A036563, A153881 (alternating sign).
Sums include: A023537, A027973, A027974, A027975, A027976, A027978, A027979, A027980, A027981, A027982, A027984, A027985, A027986, A027987, A053298.
Diagonals of the form T(n, 2*n-p): A000012 (p=0), A000027 (p=1), A034856 (p=2), A027965 (p=3), A027966 (p=4), A027967 (p=5), A027968 (p=6), A027969 (p=7), A027970 (p=8), A027971 (p=9), A027972 (p=10).
Diagonals of the form T(n, n+p): A000032 (p=0), A027961 (p=1), A023537 (p=2), A027963 (p=3), A027964 (p=4), A053298 (p=5), A027002 U A027018 (p=6), A027007 U A027014 (p=7), A027003 U A027019 (p=8).
Cf. A000032, A004396, A026998, A027011, A027977, A075193, A168616.

function T(n,k) // T = A027960
      if k le n then return Lucas(k+1);
      elif k gt 2*n then return 0;
      else return T(n-1, k-2) + T(n-1, k-1);
      end if;
end function;
[T(n,k): k in [0..2*n], n in [0..12]]; // G. C. Greubel, Jun 08 2025

T:=proc(n,k)option remember:if(k=0 or k=2*n)then return 1:elif(k=1)then return 3:else return T(n-1,k-2) + T(n-1,k-1):fi:end:
for n from 0 to 6 do for k from 0 to 2*n do print(T(n,k));od:od: # Nathaniel Johnston, Apr 18 2011

(* First program *)
t[, 0] = 1; t[, 1] = 3; t[n_, k_] /; (k == 2*n) = 1; t[n_, k_] := t[n, k] = t[n-1, k-2] + t[n-1, k-1]; Table[t[n, k], {n, 0, 8}, {k, 0, 2*n}] // Flatten (* Jean-François Alcover, Dec 27 2013 *)
(* Second program *)
f[n_, k_]:= f[n,k]= Sum[Binomial[2*n-k+j,j]*LucasL[2*(k-n-j)], {j,0,k-n-1}];
A027960[n_, k_]:= LucasL[k+1] - f[n,k]*Boole[k>n];
Table[A027960[n,k], {n,0,12}, {k,0,2*n}]//Flatten (* G. C. Greubel, Jun 08 2025 *)

T(r,n)=if(r<0||n>2*r,return(0)); if(n==0||n==2*r,return(1)); if(n==1,3,T(r-1,n-1)+T(r-1,n-2)) /* Ralf Stephan, May 04 2005 */

@CachedFunction
def T(n, k): # T = A027960
    if (k>2*n): return 0
    elif (kG. C. Greubel, Jun 01 2019; Jun 08 2025

T(n, k) = Lucas(k+1) for k <= n, otherwise the (2n-k)th coefficient of the power series for (1+2*x)/{(1-x-x^2)*(1-x)^(k-n)}.
Recurrence: T(n, 0)=T(n, 2n)=1 for n >= 0; T(n, 1)=3 for n >= 1; and for n >= 2, T(n, k) = T(n-1, k-2) + T(n-1, k-1) for 2 <= k <= 2*n-1.
From G. C. Greubel, Jun 08 2025: (Start)
T(n, k) = A000032(k+1) - f(n, k)*[k > n], where f(n, k) = Sum_{j=0..k-n-1} binomial(2*n -k+j, j)*A000032(2*(k-n-j)).
Sum_{k=0..A004396(n+1)} T(n-k, k) = A027975(n).
Sum_{k=0..n} T(n, k) = A027961(n).
Sum_{k=0..2*n} T(n, k) = A168616(n+2) + 2.
Sum_{k=n+1..2*n} (-1)^k*T(n, k) = A075193(n-1), n >= 1. (End)

Edited by Ralf Stephan, May 04 2005

A023537 a(n) = Lucas(n+4) - (3*n+7).

Original entry on oeis.org

1, 5, 13, 28, 54, 98, 171, 291, 487, 806, 1324, 2164, 3525, 5729, 9297, 15072, 24418, 39542, 64015, 103615, 167691, 271370, 439128, 710568, 1149769, 1860413, 3010261, 4870756, 7881102, 12751946, 20633139, 33385179, 54018415, 87403694, 141422212, 228826012

Offset: 1

Clark Kimberling

Define a triangle with T(n, 1) = n*(n-1) + 1 and T(n, n) = n for n = 1, 2, 3, ... The interior terms T(r, c) = T(r - 1, c) + T(r - 2, c - 1); this triangle will give the sum of terms in row(n) = a(n). The rows begin 1; 3 2; 7 3 3; 13 6 5 4; 21 13 8 7 5 having row(n) sums 1, 5, 13, 28, 54. - J. M. Bergot, Feb 17 2013

Wolfdieter Lang in "Applications of Fibonacci Numbers", Vol. 7, p. 235, eds.: G. E. Bergum et al, Kluwer, 1998.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (3,-2,-1,1).

Cf. A000032, A000045, A027960, A027961, A101220.

List([1..40], n-> Lucas(1, -1, n+4)[2] -(3*n+7) ); # G. C. Greubel, Jun 01 2019

[Lucas(n+4) -(3*n+7): n in [1..40]]; // Vincenzo Librandi, Apr 16 2011

with(combinat): L:=n->fibonacci(n+2)-fibonacci(n-2): seq(L(n),n=0..12): seq(L(n+4)-3*n-7,n=1..40); # Emeric Deutsch, Aug 08 2005

Table[LucasL[n + 4] - (3n + 7), {n, 40}] (* Alonso del Arte, Feb 17 2013 *)

Vec(x*(1+2*x)/((1-x-x^2)*(1-x)^2) + O(x^40)) \\ Colin Barker, Mar 11 2017

[lucas_number2(n+4, 1, -1) -(3*n+7) for n in (1..40)] # G. C. Greubel, Jun 01 2019

def lucas(n: BigInt): BigInt = {
  val zero = BigInt(0)
  def fibTail(n: BigInt, a: BigInt, b: BigInt): BigInt = n match {
    case `zero` => a
    case _ => fibTail(n - 1, b, a + b)
  }
  fibTail(n, 2, 1)
}
(1 to 40).map(n => lucas(n + 4) - (3 * n + 7)) // Alonso del Arte, Oct 20 2019

Convolution of natural numbers with Lucas numbers A000204.
a(n) = A027960(n+1, n+3).
From Wolfdieter Lang: (Start)
a(n) = 7*(F(n+1) - 1) + 4*F(n) - 3*n; F(n) = A000045 (Fibonacci);
G.f.: x*(1 + 2*x)/((1-x)^2*(1 - x - x^2)). (End)
a(n) - a(n-1) = A101220(3, 1, n). - Ross La Haye, May 31 2006
a(n+1) - a(n) = A027961(n+1). - R. J. Mathar, Feb 21 2013
From Colin Barker, Mar 11 2017: (Start)
a(n) = -4 + (2^(-1 - n)*((1 - sqrt(5))^n*(-15 + 7*sqrt(5)) + (1 + sqrt(5))^n*(15 + 7*sqrt(5)))) / sqrt(5) - 3*(1+n).
a(n) = 3*a(n-1) - 2*a(n-2) - a(n-3) + a(n-4) for n > 4. (End)
From G. C. Greubel, Jun 08 2025: (Start)
a(n) = a(n-1) + a(n-2) + 3*n - 2.
a(n) = Sum_{i=0..n-1} Sum_{j=0..i} A027960(i,j).
E.g.f.: exp(x/2)*( 7*cosh(sqrt(5)*x/2) + 3*sqrt(5)*sinh(sqrt(5)*x/2) ) - (3*x+7)*exp(x). (End)

More terms from Emeric Deutsch, Aug 08 2005

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

A104765 Triangle T(n,k) read by rows: row n contains the first n Lucas numbers A000204.

Original entry on oeis.org

1, 1, 3, 1, 3, 4, 1, 3, 4, 7, 1, 3, 4, 7, 11, 1, 3, 4, 7, 11, 18, 1, 3, 4, 7, 11, 18, 29, 1, 3, 4, 7, 11, 18, 29, 47, 1, 3, 4, 7, 11, 18, 29, 47, 76, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 1, 3, 4, 7, 11

Offset: 1

Gary W. Adamson, Mar 24 2005

Reading rows from the right to the left yields A104764.

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 A104765 is the reluctant sequence of A000204. - Boris Putievskiy, Dec 14 2012

			First few rows of the triangle are:
  1;
  1, 3;
  1, 3, 4;
  1, 3, 4, 7;
  1, 3, 4, 7, 11;
  1, 3, 4, 7, 11, 18;
  ...

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Boris Putievskiy, Transformations Integer Sequences And Pairing Functions, arXiv:1212.2732 [math.CO], 2012.

Cf. A000204, A104765, A104762, A104763.

Cf. A027961 (row sums).

Table[LucasL[k], {n, 1, 10}, {k, 1, n}] // Flatten (* G. C. Greubel, Dec 21 2017 *)
Module[{nn=20,luc},luc=LucasL[Range[nn]];Table[Take[luc,n],{n,nn}]]//Flatten (* Harvey P. Dale, Jul 10 2024 *)

for(n=1,10, for(k=1,n, print1(fibonacci(k+1) + fibonacci(k-1), ", "))) \\ G. C. Greubel, Dec 21 2017

T(n,k) = A000204(k), 1<=k<=n.
T(n,k) = A104764(n,n-k+1).
a(n) = A000204(m), where m = n-t(t+1)/2, t = floor((-1+sqrt(8*n-7))/2). - Boris Putievskiy, Dec 14 2012
G.f.: (x*y*(2*x*y+1))/((x-1)*(x^2*y^2+x*y-1)). - Vladimir Kruchinin, Jun 21 2025

Edited and extended by R. J. Mathar, Jul 23 2008

A137176 Hyperlucas number array T(r,n) = L(n)^(r), read by ascending antidiagonals (r >= 0, n >= 0).

Original entry on oeis.org

0, 0, 1, 0, 1, 3, 0, 1, 4, 4, 0, 1, 5, 8, 7, 0, 1, 6, 13, 15, 11, 0, 1, 7, 19, 28, 26, 18, 0, 1, 8, 26, 47, 54, 44, 29, 0, 1, 9, 34, 73, 101, 98, 73, 47, 0, 1, 10, 43, 107, 174, 199, 171, 120, 76, 0, 1, 11, 53, 150, 281, 373, 370, 291, 196, 123

Offset: 0

Jonathan Vos Post, Apr 04 2008

In Theorem 17, Dil and Mezo (2008) connect the hyperlucas numbers (this array) with the incomplete Lucas numbers (A324242). - Petros Hadjicostas, Sep 03 2019

			The array T(r,n) = L(n)^(r) begins:
.....|n=0|n=1|.n=2|.n=3|.n=4.|.n=5.|..n=6.|.n=7..|..n=8..|..n=9..|.n=10..|.in.OEIS
r=0..|.0.|.1.|..3.|..4.|...7.|..11.|...18.|...29.|....47.|....76.|...123.|.A000204
r=1..|.0.|.1.|..4.|..8.|..15.|..26.|...44.|...73.|...120.|...196.|...319.|.A027961
r=2..|.0.|.1.|..5.|.13.|..28.|..54.|...98.|..171.|...291.|...487.|...806.|.A023537
r=3..|.0.|.1.|..6.|.19.|..47.|.101.|..199.|..370.|...661.|..1148.|..1954.|.A027963
r=4..|.0.|.1.|..7.|.26.|..73.|.174.|..373.|..743.|..1404.|..2552.|..4506.|.A027964
r=5..|.0.|.1.|..8.|.34.|.107.|.281.|..654.|.1397.|..2801.|..5353.|..9859.|.A053298
r=6..|.0.|.1.|..9.|.43.|.150.|.431.|.1085.|.2482.|..5283.|.10636.|.20495.|.new
r=7..|.0.|.1.|.10.|.53.|.203.|.634.|.1719.|.4201.|..9484.|.20120.|.40615.|.new
r=8..|.0.|.1.|.11.|.64.|.267.|.901.|.2620.|.6821.|.16305.|.36425.|.77040.|.new
r=9..|.0.|.1.|.12.|.76.|.343.|1244.|.3864.|10685.|.26990.|.63415.|140455.|.new
For example, T(4,5) = L(5)^(4) = L(0)^(3) + L(1)^(3) + L(2)^(3) + L(3)^(3) + L(4)^(3) + L(5)^(3) = 0 + 1 + 6 + 19 + 47 + 101 = 174. - _Petros Hadjicostas_, Sep 03 2019

Ayhan Dil and Istvan Mezo, A Symmetric Algorithm for Hyperharmonic and Fibonacci Numbers, arXiv:0803.4388 [math.NT], 2008.
Ayhan Dil and Istvan Mezo, A Symmetric Algorithm for Hyperharmonic and Fibonacci Numbers, Applied Mathematics and Computation 206(2) (2008), 942-951.
Eric W. Weisstein, Steenrod Algebra.

Cf. A000204, A027961, A023537, A027963, A027964, A053298, A123736, A136431.

Cf. A038730, A038792, and A134511 for incomplete Fibonacci sequences, and A324242 for incomplete Lucas sequences.

L:= proc(r, n) option remember; `if`(n=0, 0, `if`(r=0,
      `if`(n<3, 2*n-1, L(0, n-2)+L(0, n-1)), L(r-1, n)+L(r, n-1)))
    end:
seq(seq(L(d-n, n), n=0..d), d=0..12);  # Alois P. Heinz, Sep 03 2019

L[r_, n_] := L[r, n] = If[n == 0, 0, If[r == 0, If[n < 3, 2n-1, L[0, n-2] + L[0, n-1]], L[r-1, n] + L[r, n-1]]];
Table[L[d-n, n], {d, 0, 12}, {n, 0, d}] // Flatten (* Jean-François Alcover, Sep 26 2019, from Maple *)

T(r,n) = L(n)^(r) = Apply partial sum operator r times to Lucas numbers A000204.
From Petros Hadjicostas, Sep 03 2019: (Start)
T(r, n) = L(n)^(r) = Sum_{k = 0..n} L(k)^(r-1) for r >= 1, with T(0,n) = L(n)^(0) = L(n) = A000204(n), T(r,0) = L(0)^(r) = 0, and T(r,1) = L(1)^(r) = 1. (See Definition 13 in Dil and Mezo (2008).)
G.f. for row r: Sum_{n >= 0} L(n)^(r)*t^n = t * (1+2*t)/((1-t-t^2) * (1-t)^r). (Corrected from Proposition 14 in Dil and Mezo (2008).)
(End)

A111058 Numbers k such that the average of the first k Lucas numbers is an integer.

Original entry on oeis.org

1, 2, 8, 12, 20, 24, 48, 60, 68, 72, 92, 96, 120, 140, 144, 188, 192, 200, 212, 216, 240, 288, 300, 332, 336, 360, 384, 428, 432, 440, 452, 480, 500, 548, 576, 600, 648, 660, 668, 672, 680, 692, 696, 720, 768, 780, 788, 812, 864, 908, 932, 960, 1008, 1028, 1052

Offset: 1

Jonathan Vos Post, Oct 07 2005

A111035 is the equivalent for Fibonacci numbers and has many elements in common with this sequence. T. D. Noe, who extended this sequence, noticed that, for some reason, 24 divides many of those k.

All terms are even except for the first term. - Harvey P. Dale, Apr 22 2024

Alois P. Heinz, Table of n, a(n) for n = 1..10000 (first 1000 terms from Harvey P. Dale)

Cf. A000032, A000204, A027961, A111035, A160665.

Lucas[n_] := Fibonacci[n+1]+Fibonacci[n-1]; lst={}; s=0; Do[s=s+Lucas[n]; If[Mod[s, n]==0, AppendTo[lst, n]], {n, 1000}]; lst (* T. D. Noe *)
Module[{nn=1000,ln},ln=LucasL[Range[nn]];Table[If[IntegerQ[Mean[Take[ln,n]]],n,Nothing],{n,nn}]] (* Harvey P. Dale, Apr 22 2024 *)

k such that (Sum_{i=1..k} A000204(i))/k is an integer.

{ k : A027961(k) == 0 (mod k) }. - Alois P. Heinz, Apr 23 2024

A202871 Symmetric matrix based on the Lucas sequence, A000032, by antidiagonals.

Original entry on oeis.org

1, 3, 3, 4, 10, 4, 7, 15, 15, 7, 11, 25, 26, 25, 11, 18, 40, 43, 43, 40, 18, 29, 65, 69, 75, 69, 65, 29, 47, 105, 112, 120, 120, 112, 105, 47, 76, 170, 181, 195, 196, 195, 181, 170, 76, 123, 275, 293, 315, 318, 318, 315, 293, 275, 123, 199, 445, 474, 510, 514

Offset: 1

Clark Kimberling, Dec 26 2011

Let s=(1,3,4,7,11,...)=A000201 and let T be the infinite square matrix whose n-th row is formed by putting n-1 zeros before the terms of s. Let T' be the transpose of T. Then A202871 represents the matrix product M=T'*T. M is the self-fusion matrix of s, as defined at A193722. See A202872 for characteristic polynomials of principal submatrices of M, with interlacing zeros.

			Northwest corner:
1....3....4....7....11...18
3....10...15...25...40...65
4....15...26...43...69...112
7....25...43...75...120..195
11...40...69...120..196..318

Cf. A202872.

s[k_] := LucasL[k];
U = NestList[Most[Prepend[#, 0]] &, #, Length[#] - 1] &[Table[s[k], {k, 1, 15}]];
L = Transpose[U]; M = L.U; TableForm[M]
m[i_, j_] := M[[i]][[j]];
Flatten[Table[m[i, n + 1 - i], {n, 1, 12}, {i, 1, n}]]
f[n_] := Sum[m[i, n], {i, 1, n}] + Sum[m[n, j], {j, 1, n - 1}]
Table[f[n], {n, 1, 12}]
Table[Sqrt[f[n]], {n, 1, 12}] (* A027961 *)
Table[m[1, j], {j, 1, 12}]    (* A000032 *)

cp's OEIS Frontend

A263225 Positive values of n such that A027961(n) is divisible by A000217(n).

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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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A101220 a(n) = Sum_{k=0..n} Fibonacci(n-k)*n^k.

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A027960 'Lucas array': triangular array T read by rows.

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A023537 a(n) = Lucas(n+4) - (3*n+7).

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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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