A001891 -id:A001891 - cp's OEIS frontend

A053808 Partial sums of A001891.

1, 5, 15, 36, 76, 148, 273, 485, 839, 1424, 2384, 3952, 6505, 10653, 17383, 28292, 45964, 74580, 120905, 195885, 317231, 513600, 831360, 1345536, 2177521, 3523733, 5701983, 9226500, 14929324, 24156724, 39087009, 63244757, 102332855, 165578768, 267912848

Offset: 0

Barry E. Williams, Mar 27 2000

Antidiagonal sums of the convolution array A213579 and row 1 of the convolution array A213590. - Clark Kimberling, Jun 18 2012

Also number CG(n,2) of complete games with n players of 2 types. - N. J. A. Sloane, Dec 29 2012

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

G. C. Greubel, Table of n, a(n) for n = 0..1000
J. Freixas and S. Kurz, The golden number and Fibonacci sequences in the design of voting structures, 2012. - From _N. J. A. Sloane_, Dec 29 2012
W. Lang, Problem B-858, Fibonacci Quarterly, 36,3 (1998) 373-374; Solution, ibid. 37,2 (1999) 183-184.
Index entries for linear recurrences with constant coefficients, signature (4,-5,1,2,-1).

Convolution of A000290 (squares) with A000045, n >= 1. (Fibonacci) - Wolfdieter Lang, Apr 10 2000

Right-hand column 7 of triangle A011794.

List([0..40], n-> Fibonacci(n+8) - (n^2 +8*n+20)); # G. C. Greubel, Jul 06 2019

[Fibonacci(n+8) - (n^2+8*n+20): n in [0..40]]; // G. C. Greubel, Jul 06 2019

Table[Fibonacci[n+8] -(n^2 +8*n+20), {n,0,40}] (* G. C. Greubel, Jul 06 2019 *)
LinearRecurrence[{4,-5,1,2,-1},{1,5,15,36,76},40] (* Harvey P. Dale, Apr 14 2022 *)

vector(40, n, n--; fibonacci(n+8) - (n^2 +8*n+20)) \\ G. C. Greubel, Jul 06 2019

[fibonacci(n+8) - (n^2 +8*n+20) for n in (0..20)] # G. C. Greubel, Jul 06 2019

a(n) = a(n-1) + a(n-2) + (n+1)^2, a(-n)=0.
G.f.: (1+x)/((1-x-x^2)*(1-x)^3).
a(n) = Fibonacci(n+6) - (n^2 + 4*n + 8), n >= 2 (see p. 184 of FQ reference).
a(n-2) = Sum_{i=0..n} Fibonacci(i)*(n-i)^2. - Benoit Cloitre, Mar 06 2004

A053809 Second partial sums of A001891.

Original entry on oeis.org

1, 6, 21, 57, 133, 281, 554, 1039, 1878, 3302, 5686, 9638, 16143, 26796, 44179, 72471, 118435, 193015, 313920, 509805, 827036, 1340636, 2171996, 3517532, 5695053, 9218786, 14920769, 24147269, 39076593, 63233317, 102320326

Offset: 0

Barry E. Williams, Mar 27 2000

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-9,6,1,-3,1).

Cf. A001911, A001891, A053808.

Right-hand column 9 of triangle A011794. Pairwise sums of A014166.

List([0..40], n-> Fibonacci(n+10) - (2*n^3 + 27*n^2 + 145*n + 324)/6) # G. C. Greubel, Jul 06 2019

[Fibonacci(n+10) - (2*n^3 + 27*n^2 + 145*n + 324)/6: n in [0..40]]; // G. C. Greubel, Jul 06 2019

Table[Fibonacci[n+10] - (2*n^3+27*n^2+145*n+324)/6, {n,0,40}] (* G. C. Greubel, Jul 06 2019 *)

vector(40, n, n--; fibonacci(n+10) - (2*n^3 + 27*n^2 + 145*n + 324)/6) \\ G. C. Greubel, Jul 06 2019

[fibonacci(n+10) - (2*n^3 + 27*n^2 + 145*n + 324)/6 for n in (0..40)] # G. C. Greubel, Jul 06 2019

a(n) = a(n-1) + a(n-2) + (2*n+3)*C(n+2, 2)/3; a(-x)=0.
a(n) = Fibonacci(n+10) - (2*n^3 + 27*n^2 + 145*n + 324)/6.
G.f.: (1+x)/((1-x)^4*(1-x-x^2)).
a(n) = 5*a(n-1) - 9*a(n-2) + 6*a(n-3) + a(n-4) - 3*a(n-5) + a(n-6). - Wesley Ivan Hurt, Apr 21 2021

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

A001924 Apply partial sum operator twice to Fibonacci numbers.

Original entry on oeis.org

0, 1, 3, 7, 14, 26, 46, 79, 133, 221, 364, 596, 972, 1581, 2567, 4163, 6746, 10926, 17690, 28635, 46345, 75001, 121368, 196392, 317784, 514201, 832011, 1346239, 2178278, 3524546, 5702854, 9227431, 14930317, 24157781, 39088132, 63245948, 102334116, 165580101

Offset: 0

N. J. A. Sloane

Leading coefficients in certain rook polynomials (for n>=2; see p. 18 of the Riordan paper). - Emeric Deutsch, Mar 08 2004
(1, 3, 7, 14, ...) = row sums of triangle A141289. - Gary W. Adamson, Jun 22 2008
a(n) is the number of nonempty subsets of {1,2,...,n} such that the difference of successive elements is at most 2. See example below. Generally, the o.g.f. for the number of nonempty subsets of {1,2,...,n} such that the difference of successive elements is <= k is: x/((1-x)*(1-2*x+x^(k+1))). Cf. A000217 the case for k=1, A001477 the case for k=0 (counts singleton subsets). - Geoffrey Critzer, Feb 17 2012
-Fibonacci(n-2) = p(-1) where p(x) is the unique degree-n polynomial such that p(k) = a(k) for k = 0, 1, ..., n. - Michael Somos, Dec 31 2012
a(n) is the number of bit strings of length n+1 with the pattern 00 and without the pattern 011, see example. - John M. Campbell, Feb 10 2013
From Jianing Song, Apr 28 2025: (Start)
For n >= 2, a(n-2) is the number of subsets of {1,2,...,n} with 2 or more elements that contain no consecutive elements (i.e., such that the difference of successive elements is at least 2). Note that the number of such subsets with k elements is binomial(n+1-k,k), and Sum_{k=2..floor((n+1)/2)} binomial(n+1-k,k) = F(n+2) - binomial(n+1,0) - binomial(n,1) = F(n+2) - (n+1).
If subsets of {1,2,...,n} are required to contain no consecutive elements module n, then the result is A023548(n-3). (End)

			a(5) = 26 because there are 31 nonempty subsets of {1,2,3,4,5} but 5 of these have successive elements that differ by 3 or more: {1,4}, {1,5}, {2,5}, {1,2,5}, {1,4,5}. - _Geoffrey Critzer_, Feb 17 2012
From _John M. Campbell_, Feb 10 2013: (Start)
There are a(5) = 26 bit strings with the pattern 00 and without the pattern 011 of length 5+1:
   000000, 000001, 000010, 000100, 000101, 001000,
   001001, 001010, 010000, 010001, 010010, 010100,
   100000, 100001, 100010, 100100, 100101, 101000, 101001,
   110000, 110001, 110010, 110100, 111000, 111001, 111100.
(End)

J. Riordan, Discordant permutations, Scripta Math., 20 (1954), 14-23.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..500
Bader AlBdaiwi, On the Number of Cycles in a Graph, arXiv preprint arXiv:1603.01807 [cs.DM], 2016.
Jean-Luc Baril and Jean-Marcel Pallo, Motzkin subposet and Motzkin geodesics in Tamari lattices, 2013.
Ning-Ning Cao and Feng-Zhen Zhao, Some Properties of Hyperfibonacci and Hyperlucas Numbers, J. Int. Seq. 13 (2010) # 10.8.8.
Hung Viet Chu, Various Sequences from Counting Subsets, arXiv:2005.10081 [math.CO], 2020.
Hung Viet Chu, Partial Sums of the Fibonacci Sequence, arXiv:2106.03659 [math.CO], 2021.
Ligia Loretta Cristea, Ivica Martinjak, and Igor Urbiha, Hyperfibonacci Sequences and Polytopic Numbers, arXiv:1606.06228 [math.CO], 2016.
Emrah Kiliç and Pantelimon Stănică, Generating matrices for weighted sums of second order linear recurrences, JIS 12 (2009) 09.2.7.
Wolfdieter Lang, Problem B-858, Fibonacci Quarterly, 36 (1998), 373-374, Solution, ibid. 37 (1999) 183-184.
Candice A. Marshall, Construction of Pseudo-Involutions in the Riordan Group, Dissertation, Morgan State University, 2017.
Igor Pak, Boris Shapiro, Ilya Smirnov, and Ken-ichi Yoshida, Hilbert-Kunz multiplicity of quadrics via the Ehrhart theory, Stockholm Univ. (Sweden, 2025). See p. 6.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
J. Riordan, Discordant permutations, Scripta Math., 20 (1954), 14-23. [Annotated scanned copy]
Stacey Wagner, Enumerating Alternating Permutations with One Alternating Descent, DePaul Discoveries: Vol. 2: Iss. 1, Article 2.
Index entries for linear recurrences with constant coefficients, signature (3,-2,-1,1).

Cf. A000045, A023548, A001891, A133640, A141289.
Right-hand column 4 of triangle A011794.
Cf. A014162, A039834, A077880, A122595, A129696.
Cf. A065220.

List([0..40], n-> Fibonacci(n+4) -n-3); # G. C. Greubel, Jul 08 2019

a001924 n = a001924_list !! n
a001924_list = drop 3 $ zipWith (-) (tail a000045_list) [0..]
-- Reinhard Zumkeller, Nov 17 2013

[Fibonacci(n+4)-(n+3): n in [0..40]]; // Vincenzo Librandi, Jun 23 2016

A001924:=-1/(z**2+z-1)/(z-1)**2; # Conjectured by Simon Plouffe in his 1992 dissertation.
##
a:= n-> (<<0|1|0|0>, <0|0|1|0>, <0|0|0|1>, <1|-1|-2|3>>^n.
         <<0, 1, 3, 7>>)[1, 1]:
seq(a(n), n=0..40);  # Alois P. Heinz, Oct 05 2012

a[n_]:= Fibonacci[n+4] -3-n; Array[a, 40, 0]  (* Robert G. Wilson v *)
LinearRecurrence[{3,-2,-1,1},{0,1,3,7},40] (* Harvey P. Dale, Jan 24 2015 *)
Nest[Accumulate,Fibonacci[Range[0,40]],2] (* Harvey P. Dale, Jun 15 2016 *)

a(n)=fibonacci(n+4)-n-3 \\ Charles R Greathouse IV, Feb 24 2011

[fibonacci(n+4) -n-3 for n in (0..40)] # G. C. Greubel, Jul 08 2019

From Wolfdieter Lang: (Start)
G.f.: x/((1-x-x^2)*(1-x)^2).
Convolution of natural numbers n >= 1 with Fibonacci numbers F(k).
a(n) = Fibonacci(n+4) - (3+n). (End)
From Henry Bottomley, Jan 03 2003: (Start)
a(n) = a(n-1) + a(n-2) + n = a(n-1) + A000071(n+2).
a(n) = A001891(n) - a(n-1) = n + A001891(n-1).
a(n) = A065220(n+4) + 1 = A000126(n+1) - 1. (End)
a(n) = Sum_{k=0..n} Sum_{i=0..k} Fibonacci(i). - Benoit Cloitre, Jan 26 2003
a(n) = (sqrt(5)/2 + 1/2)^n*(7*sqrt(5)/10 + 3/2) + (3/2 - 7*sqrt(5)/10)*(sqrt(5)/2 - 1/2)^n*(-1)^n - n - 3. - Paul Barry, Mar 26 2003
a(n) = Sum_{k=0..n} Fibonacci(k)*(n-k). - Benoit Cloitre, Jun 07 2004
A107909(a(n)) = A000225(n) = 2^n - 1. - Reinhard Zumkeller, May 28 2005
a(n) - a(n-1) = A101220(1,1,n). - Ross La Haye, May 31 2006
F(n) + a(n-3) = A133640(n). - Gary W. Adamson, Sep 19 2007
a(n) = A077880(-3-n) = 2*a(n-1) - a(n-3) + 1. - Michael Somos, Dec 31 2012
INVERT transform is A122595. PSUM transform is A014162. PSUMSIGN transform is A129696. BINOMIAL transform of A039834 with 0,1 prepended is this sequence. - Michael Somos, Dec 31 2012
a(n) = A228074(n+1,3) for n > 1. - Reinhard Zumkeller, Aug 15 2013
a(n) = Sum_{k=0..n} Sum_{i=0..n} i * C(n-k,k-i). - Wesley Ivan Hurt, Sep 21 2017
E.g.f.: exp(x/2)*(15*cosh(sqrt(5)*x/2) + 7*sqrt(5)*sinh(sqrt(5)*x/2))/5 - exp(x)*(3 + x). - Stefano Spezia, Jun 25 2022

Description improved by N. J. A. Sloane, Jan 01 1997

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

A192951 Coefficient of x in the reduction by x^2 -> x+1 of the polynomial p(n,x) defined at Comments.

Original entry on oeis.org

0, 1, 3, 9, 20, 40, 74, 131, 225, 379, 630, 1038, 1700, 2773, 4511, 7325, 11880, 19252, 31182, 50487, 81725, 132271, 214058, 346394, 560520, 906985, 1467579, 2374641, 3842300, 6217024, 10059410, 16276523, 26336025, 42612643, 68948766

Offset: 0

Clark Kimberling, Jul 13 2011

The titular polynomials are defined recursively:  p(n,x) = x*p(n-1,x) + 3n - 1, with p(0,x)=1.  For an introduction to reductions of polynomials by substitutions such as x^2 -> x+1, see A192232 and A192744.
...
The list of examples at A192744 is extended here; the recurrence is given by p(n,x) = x*p(n-1,x) + v(n), with p(0,x)=1, and the reduction of p(n,x) by x^2 -> x+1 is represented by u1 + u2*x:
...
If v(n)=        n, then u1=A001595, u2=A104161.
If v(n)=      n-1, then u1=A001610, u2=A066982.
If v(n)=     3n-1, then u1=A171516, u2=A192951.
If v(n)=     3n-2, then u1=A192746, u2=A192952.
If v(n)=     2n-1, then u1=A111314, u2=A192953.
If v(n)=      n^2, then u1=A192954, u2=A192955.
If v(n)=   -1+n^2, then u1=A192956, u2=A192957.
If v(n)=    1+n^2, then u1=A192953, u2=A192389.
If v(n)=   -2+n^2, then u1=A192958, u2=A192959.
If v(n)=    2+n^2, then u1=A192960, u2=A192961.
If v(n)=    n+n^2, then u1=A192962, u2=A192963.
If v(n)=   -n+n^2, then u1=A192964, u2=A192965.
If v(n)= n(n+1)/2, then u1=A030119, u2=A192966.
If v(n)= n(n-1)/2, then u1=A192967, u2=A192968.
If v(n)= n(n+3)/2, then u1=A192969, u2=A192970.
If v(n)=     2n^2, then u1=A192971, u2=A192972.
If v(n)=   1+2n^2, then u1=A192973, u2=A192974.
If v(n)=  -1+2n^2, then u1=A192975, u2=A192976.
If v(n)=  1+n+n^2, then u1=A027181, u2=A192978.
If v(n)=  1-n+n^2, then u1=A192979, u2=A192980.
If v(n)=  (n+1)^2, then u1=A001891, u2=A053808.
If v(n)=  (n-1)^2, then u1=A192981, u2=A192982.

Vincenzo Librandi, Table of n, a(n) for n = 0..400
Index entries for linear recurrences with constant coefficients, signature (3, -2, -1, 1).

Cf. A000045, A192232, A192744.

F:=Fibonacci;; List([0..40], n-> F(n+4)+2*F(n+2)-(3*n+5)); # G. C. Greubel, Jul 12 2019

I:=[0, 1, 3, 9]; [n le 4 select I[n] else 3*Self(n-1)-2*Self(n-2)-1*Self(n-3)+Self(n-4): n in [1..40]]; // Vincenzo Librandi, Nov 16 2011

F:=Fibonacci; [F(n+4)+2*F(n+2)-(3*n+5): n in [0..40]]; // G. C. Greubel, Jul 12 2019

(* First program *)
q = x^2; s = x + 1; z = 40;
p[0, x]:= 1;
p[n_, x_]:= x*p[n-1, x] + 3n - 1;
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}] (* A171516 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192951 *)
(* Additional programs *)
LinearRecurrence[{3,-2,-1,1},{0,1,3,9},40] (* Vincenzo Librandi, Nov 16 2011 *)
With[{F=Fibonacci}, Table[F[n+4]+2*F[n+2]-(3*n+5), {n,0,40}]] (* G. C. Greubel, Jul 12 2019 *)

a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; 1,-1,-2,3]^n*[0;1;3;9])[1,1] \\ Charles R Greathouse IV, Mar 22 2016

vector(40, n, n--; f=fibonacci; f(n+4)+2*f(n+2)-(3*n+5)) \\ G. C. Greubel, Jul 12 2019

f=fibonacci; [f(n+4)+2*f(n+2)-(3*n+5) for n in (0..40)] # G. C. Greubel, Jul 12 2019

a(n) = 3*a(n-1) - 2*a(n-2) - a(n-3) + a(n-4).
From Bruno Berselli, Nov 16 2011: (Start)
G.f.: x*(1+2*x^2)/((1-x)^2*(1 - x - x^2)).
a(n) = ((25+13*t)*(1+t)^n + (25-13*t)*(1-t)^n)/(10*2^n) - 3*n - 5 = A000285(n+2) - 3*n - 5  where t=sqrt(5). (End)
a(n) = Fibonacci(n+4) + 2*Fibonacci(n+2) - (3*n+5). - G. C. Greubel, Jul 12 2019

A001883 Number of permutations s of {1,2,...,n} such that |s(i)-i|>1 for each i=1,2,...,n.

Original entry on oeis.org

1, 0, 0, 0, 1, 4, 29, 206, 1708, 15702, 159737, 1780696, 21599745, 283294740, 3995630216, 60312696452, 970234088153, 16571597074140, 299518677455165, 5711583170669554, 114601867572247060, 2413623459384988298, 53238503492701261201, 1227382998752177970288, 29520591675204638641249

Offset: 0

N. J. A. Sloane

nonn

Permanent of the (0,1)-matrix having (i,j)-th entry equal to 0 iff this is on the first lower-diagonal, diagonal or the first upper-diagonal. - Simone Severini, Oct 14 2004

J. Riordan, "The enumeration of permutations with three-ply staircase restrictions," unpublished memorandum, Bell Telephone Laboratories, Murray Hill, NJ, Oct 1963.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Eric M. Schmidt, Table of n, a(n) for n = 0..200
Dmitry Efimov, Determinants of generalized binary band matrices, arXiv:1702.05655 [math.RA], 2017.
J. Riordan, Letter, Oct 31 1977
N. J. A. Sloane, Annotated copy of Riordan's Three-Ply Staircase paper
Donovan Young, The Number of Domino Matchings in the Game of Memory, J. Int. Seq., Vol. 21 (2018), Article 18.8.1.
Donovan Young, Generating Functions for Domino Matchings in the 2 X k Game of Memory, J. Int. Seq., Vol. 22 (2019), Article 19.8.7.
D. Zeilberger, Automatic Enumeration of Generalized Menage Numbers, arXiv preprint arXiv:1401.1089 [math.CO], 2014.

Cf. A001884-A001891, A075851, A075852.
Also a diagonal of A080018.
Column k=0 of A323671.

b:= proc(n, s) option remember; `if`(n=0, 1, add(
      `if`(abs(n-i)<=1, 0, b(n-1, s minus {i})), i=s))
    end:
a:= n-> b(n, {$1..n}):
seq(a(n), n=0..15);  # Alois P. Heinz, Jul 04 2015

b[n_, s_List] := b[n, s] = If[n == 0, 1, Sum[If[Abs[n-i] <= 1, 0, b[n-1, s ~Complement~ {i}]], {i, s}]]; a[n_] := b[n, Range[n]]; Table[Print[a[n]]; a[n], {n, 4, 24}] (* Jean-François Alcover, Nov 10 2015, after Alois P. Heinz *)

permRWNb(a)=n=matsize(a)[1]; if(n==1,return(a[1,1])); sg=1; in=vectorv(n); x=in; x=a[,n]-sum(j=1,n,a[,j])/2; p=prod(i=1,n,x[i]); for(k=1,2^(n-1)-1,sg=-sg; j=valuation(k,2)+1; z=1-2*in[j]; in[j]+=z; x+=z*a[,j]; p+=prod(i=1,n,x[i],sg)); return(2*(2*(n%2)-1)*p)
for(n=1,23,a=matrix(n,n,i,j,abs(i-j)>1);print1(permRWNb(a)",")) \\ Herman Jamke (hermanjamke(AT)fastmail.fm), May 16 2007

a(n) = (n+1)*a(n-1) - (n-3)*a(n-2) - (n-4)*a(n-3) + (n-4)*a(n-4) + a(n-5) + (-1)^n * Lucas(n-3), n > 4. [Riordan] (Note: There is a slight mistake in Riordan's paper. On p. 3 it should say that a_5 = 3.) - Eric M. Schmidt, Oct 09 2017
From Vaclav Kotesovec, Oct 10 2017: (Start)
a(n) = n*a(n-1) + 4*a(n-2) - 3*(n-3)*a(n-3) + (n-4)*a(n-4) + 2*(n-5)*a(n-5) - (n-7)*a(n-6) - a(n-7).
a(n) ~ exp(-3) * n!.
(End)

More terms and better description from Reiner Martin, Oct 14 2002
More terms from Vladimir Baltic, Vladeta Jovovic, Jan 04 2003
More terms from Herman Jamke (hermanjamke(AT)fastmail.fm), May 16 2007
a(22)-a(24) from Alois P. Heinz, Jul 04 2015
a(0)-a(3) from Eric M. Schmidt, Oct 09 2017

A011794 Triangle defined by T(n+1, k) = T(n, k-1) + T(n-1, k), T(n,1) = 1, T(1,k) = 1, T(2,k) = min(2,k).

Original entry on oeis.org

1, 1, 2, 1, 2, 3, 1, 3, 4, 5, 1, 3, 6, 7, 8, 1, 4, 7, 11, 12, 13, 1, 4, 10, 14, 19, 20, 21, 1, 5, 11, 21, 26, 32, 33, 34, 1, 5, 15, 25, 40, 46, 53, 54, 55, 1, 6, 16, 36, 51, 72, 79, 87, 88, 89, 1, 6, 21, 41, 76, 97, 125, 133, 142, 143, 144, 1, 7, 22, 57, 92, 148, 176, 212, 221, 231, 232, 233

Offset: 1

N. J. A. Sloane and David Broadhurst

			matrix(10,10,n,k,a(n-1,k-1))
  [ 0 0 0 0 0 0 0 0 0 0 ]
  [ 0 1 1 1 1 1 1 1 1 1 ]
  [ 0 1 2 2 2 2 2 2 2 2 ]
  [ 0 1 2 3 3 3 3 3 3 3 ]
  [ 0 1 3 4 5 5 5 5 5 5 ]
  [ 0 1 3 6 7 8 8 8 8 8 ]
Triangle begins as:
  1;
  1, 2;
  1, 2,  3;
  1, 3,  4,  5;
  1, 3,  6,  7,  8;
  1, 4,  7, 11, 12, 13;
  1, 4, 10, 14, 19, 20, 21;
  1, 5, 11, 21, 26, 32, 33, 34;
  1, 5, 15, 25, 40, 46, 53, 54, 55;
  1, 6, 16, 36, 51, 72, 79, 87, 88, 89;

G. C. Greubel, Rows n = 1..100 of the triangle, flattened
D. J. Broadhurst, On the enumeration of irreducible k-fold Euler sums and their roles in knot theory and field theory, arXiv:hep-th/9604128, 1996.

Columns include A008619 and (essentially) A055802, A055803, A055804, A055805, A055806.
Right-hand columns 1-14 are A000045, A000071, A001911, A001924, A001891, A014162, A053808, A014166, A053809, A053739, A054469, A053295, A054470, A053296.
Essentially a reflected version of A055801.
Sums include: A039834 (signed row), A131913 (row).
Cf. A008466, A074878, A103609.

function T(n,k) // T = A011794(n,k)
  if k eq 1 or n eq 1 then return 1;
  elif n eq 2 then return Min(2, k);
  else return T(n-1,k-1) + T(n-2,k);
  end if;
end function;
[T(n,k): k in [1..n], n in [1..15]]; // G. C. Greubel, Oct 21 2024

T[n_, k_]:= T[n, k]= T[n-1, k-1] + T[n-2, k]; T[n_, 1] = 1; T[1, k_] = 1; T[2, k_] := Min[2, k]; Table[T[n, k], {n,15}, {k,n}]//Flatten (* Jean-François Alcover, Feb 26 2013 *)

T(n,k)=if(n<=0 || k<=0,0, if(n<=2 || k==1, min(n,k), T(n-1,k-1)+T(n-2,k)))

def T(n, k): # T = A011794
    if (k==1 or n==1): return 1
    elif (n==2): return min(2,k)
    else: return T(n-1, k-1) + T(n-2, k)
flatten([[T(n, k) for k in range(1,n+1)] for n in range(1,16)]) # G. C. Greubel, Oct 21 2024

T(n,n) = Fibonacci(n+1). - Jean-François Alcover, Feb 26 2013
From G. C. Greubel, Oct 21 2024: (Start)
Sum_{k=1..n} T(n, k) = A131913(n-1).
Sum_{k=1..n} (-1)^(k-1)*T(n, k) = A039834(n).
Sum_{k=1..floor((n+1)/2)} T(n-k+1,k) = (1/2)*((1-(-1)^n)*A074878((n+3)/2) + (1+(-1)^n)*A008466((n+6)/2)) (diagonal row sums).
Sum_{k=1..floor((n+1)/2)} (-1)^(k-1)*T(n-k+1,k) = (-1)^floor((n-1)/2)*A103609(n) + [n=1] (signed diagonal row sums). (End)

Entry improved by comments from Michael Somos

More terms added by G. C. Greubel, Oct 21 2024

A261019 Irregular triangle read by rows: T(n,k) (0 <= k <= A261017(n)) = number of binary strings of length n such that the smallest number whose binary representation is not visible in the string is k.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 1, 1, 1, 3, 6, 4, 1, 1, 1, 4, 11, 10, 5, 1, 1, 5, 19, 21, 15, 0, 2, 1, 1, 6, 32, 40, 35, 2, 9, 2, 1, 1, 7, 53, 72, 73, 6, 31, 10, 2, 1, 1, 8, 87, 125, 144, 15, 79, 40, 12, 1, 1, 9, 142, 212, 274, 32, 185, 116, 52, 1, 1, 10, 231, 354, 509, 64, 408, 296, 168, 2, 4

Offset: 1

Alois P. Heinz and N. J. A. Sloane, Aug 17 2015

This is a more compact version of the triangle in A261015, ending each row at the last nonzero entry.

			The first 16 rows are:
1, 1,
1, 1,  1,    1,
1, 1,  2,    3,    1,
1, 1,  3,    6,    4,    1,
1, 1,  4,   11,   10,    5,
1, 1,  5,   19,   21,   15,    0,     2,
1, 1,  6,   32,   40,   35,    2,     9,     2,
1, 1,  7,   53,   72,   73,    6,    31,    10,     2,
1, 1,  8,   87,  125,  144,   15,    79,    40,    12,
1, 1,  9,  142,  212,  274,   32,   185,   116,    52,
1, 1, 10,  231,  354,  509,   64,   408,   296,   168,    2,    4,
1, 1, 11,  375,  585,  931,  120,   864,   699,   461,   24,   24,
1, 1, 12,  608,  960, 1685,  218,  1771,  1557,  1133,  130,  110,   2,   4,
1, 1, 13,  985, 1568, 3027,  385,  3555,  3325,  2612,  471,  387,  14,  24,  0,  16,
1, 1, 14, 1595, 2553, 5409,  668,  7021,  6893,  5759, 1401, 1135,  92, 120,  0,  90, 16,
1, 1, 15, 2582, 4148, 9628, 1142, 13696, 13964, 12309, 3734, 2972, 373, 439, 28, 390, 98, 16,
...

Alois P. Heinz, N. J. A. Sloane, R. Zumkeller and Hiroaki Yamanouchi, Rows n = 1..58, flattened (rows 17..25 from R. Zumkeller, rows 26..36 from Alois P. Heinz)
R. Zumkeller, The first 25 rows of the triangle, displayed as a triangle (similar to the way the rows are shown in the Example section, but showing 25 rows).

Cf. A261015, A261016.
The row lengths are given by A261017.
Columns k=3-10 give: A001911, A001891, A261441, A261442, A261443, A261473, A261474, A261475.
Cf. A076478, A030308, A000079 (row sums), A261392 (max per row).

import Data.List (isInfixOf, sort, group)
a261019 n k = a261019_tabf !! (n-1) !! k
a261019_row n = a261019_tabf !! (n-1)
a261019_tabf = map (i 0 . group . sort . map f) a076478_tabf
   where f bs = g a030308_tabf where
           g (cs:css) | isInfixOf cs bs = g css
                      | otherwise = foldr (\d v -> 2 * v + d) 0 cs
         i _ [] = []
         i r gss'@(gs:gss) | head gs == r = (length gs) : i (r + 1) gss
                           | otherwise    = 0 : i (r + 1) gss'
-- Reinhard Zumkeller, Aug 18 2015

A119457 Triangle read by rows: T(n, 1) = n, T(n, 2) = 2*(n-1) for n>1 and T(n, k) = T(n-1, k-1) + T(n-2, k-2) for 2 < k <= n.

Original entry on oeis.org

1, 2, 2, 3, 4, 3, 4, 6, 6, 5, 5, 8, 9, 10, 8, 6, 10, 12, 15, 16, 13, 7, 12, 15, 20, 24, 26, 21, 8, 14, 18, 25, 32, 39, 42, 34, 9, 16, 21, 30, 40, 52, 63, 68, 55, 10, 18, 24, 35, 48, 65, 84, 102, 110, 89, 11, 20, 27, 40, 56, 78, 105, 136, 165, 178, 144, 12, 22, 30, 45, 64, 91, 126, 170, 220, 267, 288, 233

Offset: 1

Reinhard Zumkeller, May 20 2006

			Triangle begins as:
   1;
   2,  2;
   3,  4,  3;
   4,  6,  6,  5;
   5,  8,  9, 10,  8;
   6, 10, 12, 15, 16, 13;
   7, 12, 15, 20, 24, 26,  21;
   8, 14, 18, 25, 32, 39,  42,  34;
   9, 16, 21, 30, 40, 52,  63,  68,  55;
  10, 18, 24, 35, 48, 65,  84, 102, 110,  89;
  11, 20, 27, 40, 56, 78, 105, 136, 165, 178, 144;
  12, 22, 30, 45, 64, 91, 126, 170, 220, 267, 288, 233;

G. C. Greubel, Rows n = 1..50 of the triangle, flattened
Eric Weisstein's World of Mathematics, Fibonacci Number

Main diagonal: A023607(n).
Sums: A001891 (row), A355020 (signed row).
Columns: A000027(n) (k=1), A005843(n-1) (k=2), A008585(n-2) (k=3), A008587(n-3) (k=4), A008590(n-4) (k=5), A008595(n-5) (k=6), A008603(n-6) (k=7).
Diagonals: A000045(n+1) (k=n), A006355(n+1) (k=n-1), A022086(n-1) (k=n-2), A022087(n-2) (k=n-3), A022088(n-3) (k=n-4), A022089(n-4) (k=n-5), A022090(n-5) (k=n-6).
Cf. A023652, A112469.

A119457:= func< n,k | (n-k+1)*Fibonacci(k+1) >;
[A119457(n,k): k in [1..n], n in [1..12]]; // G. C. Greubel, Apr 16 2025

(* First program *)
T[n_, 1] := n;
T[n_ /; n > 1, 2] := 2 n - 2;
T[n_, k_] /; 2 < k <= n := T[n, k] = T[n - 1, k - 1] + T[n - 2, k - 2];
Table[T[n, k], {n, 1, 12}, {k, 1, n}] // Flatten (* Jean-François Alcover, Dec 01 2021 *)
(* Second program *)
A119457[n_,k_]:= (n-k+1)*Fibonacci[k+1];
Table[A119457[n,k], {n,13}, {k,n}]//Flatten (* G. C. Greubel, Apr 16 2025 *)

def A119457(n,k): return (n-k+1)*fibonacci(k+1)
print(flatten([[A119457(n,k) for k in range(1,n+1)] for n in range(1,13)])) # G. C. Greubel, Apr 16 2025

T(n, k) = (n-k+1)*T(k,k) for 1 <= k < n, with T(n, n) = A000045(n+1).
From G. C. Greubel, Apr 15 2025: (Start)
T(n, k) = (n-k+1)*Fibonacci(k+1).
Sum_{k=1..floor((n+1)/2)} T(n-k+1, k) = (1/2)*(1-(-1)^n)*A023652(floor((n+1)/2)) + (1+(-1)^n)*A001891(floor(n/2)).
Sum_{k=1..floor((n+1)/2)} (-1)^(k-1)*T(n-k+1, k) = (1/2)*(1-(-1)^n)*A112469(floor((n-1)/2)) + (1+(-1)^n)*A355020(floor((n-2)/2)). (End)

cp's OEIS Frontend

A053808 Partial sums of A001891.

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A053809 Second partial sums of A001891.

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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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A001924 Apply partial sum operator twice to Fibonacci numbers.

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

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A192951 Coefficient of x in the reduction by x^2 -> x+1 of the polynomial p(n,x) defined at Comments.

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