cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 36 results. Next

A072337 Inverse EULER transform of A064831 (with its initial 1 omitted).

Original entry on oeis.org

1, 3, 3, 5, 10, 24, 50, 120, 270, 640, 1500, 3600, 8610, 20880, 50700, 124024, 304290, 750120, 1854400, 4600200, 11440548, 28527320, 71289000, 178526880, 447910470, 1125750120, 2833885800, 7144449920, 18036373140, 45591631800, 115381697740, 292329067800
Offset: 0

Views

Author

N. J. A. Sloane, Jul 16 2002

Keywords

Comments

Also, dimensions of the graded Yang-Mills algebra ym(3) [Herscovich and Solotar]. - N. J. A. Sloane, Jan 02 2013
Apparently the same as A032170 for n > 2. - Ralf Stephan, Feb 01 2004
From Petros Hadjicostas, Dec 03 2017: (Start)
We clarify the name of the sequence and provide a sketch of a proof of R. Stephan's statement above.
For n>=0, let b(n) = A064831(n+1). Then B(x) = Sum_{n>=0} b(n)*x^n = Sum_{n>=0} A064831(n+1)*x^n = 1/((1-x^2)*(1-3*x+x^2)) (see the formula section for sequence A064831). Define a(0) = 1, and (a(n): n>=1) = invEULER(b(n): n>=0), i.e., we give a new (more correct) definition of the current sequence. Using Bernstein and Sloane (1995), we define the sequence (d(n): n>=1) via Sum_{n>=1} (d(n)/n)*x^n = log B(x). Since EULER(a(n): n>=1) = (b(n): n>=0), we have a(n) = (1/n)*Sum_{s|n} mu(s)*d(n/s). Using the change of indexes m=n/s, we get A(x) = 1 + Sum_{n>=1} a(n)*x^n = 1 + Sum_{s>=1} (mu(s)/s)*Sum_{m>=1} (d(m)/m)*(x^s)^m = 1 + Sum_{s>=1} (mu(s)/s)*log B(x^s).
From the documentation of sequence A032170, we see that its g.f. is Sum_{s>=1} (mu(s)/s)*log C(x^s), where C(x) = (1-x)^2/(1-3*x+x^2). To prove R. Stephan's claim above, we need to prove that Sum_{s>=1} (mu(s)/s)*log C(x^s) - x - 2*x^2 = Sum_{s>=1} (mu(s)/s)*log B(x^s) - 3*x - 3*x^2. The last equality is equivalent to Sum_{s>=1} (mu(s)/s)*log(B(x^s)/C(x^s)) = 2*x + x^2, which in turn is equivalent to -Sum_{s>=1} (mu(s)/s) log((1-x^s)^2*(1-x^{2*s})) = 2*x + x^2. The last equality follows from the identity -Sum_{s>=1} (mu(s)/s)*log(1-y^s) = y, which follows from the Lambert series Sum_{s>=1} mu(s)*y^s/(1-y^s) = y.
(End)

Links

Crossrefs

Cf. A032170, A064831.

Programs

  • Mathematica
    mob[m_, n_] := If[Mod[m, n] == 0, MoebiusMu[m/n], 0];
EULERi[b_] := Module[{a, c, i, d}, c = {}; For[i = 1, i <= Length[b], i++, c = Append[c, i*b[[i]] - Sum[c[[d]]*b[[i - d]], {d, 1, i - 1}]]]; a = {}; For[i = 1, i <= Length[b], i++, a = Append[a, (1/i)*Sum[mob[i, d]*c[[d]], {d, 1, i}]]]; Return[a]];
Join[{1}, EULERi[LinearRecurrence[{3, 0, -3, 1}, {3, 9, 24, 64}, 31]]] (* Jean-François Alcover, Aug 19 2018 *)

Formula

G.f.: 1 + Sum_{s>=1} (mu(s)/s)*log B(x^s), where B(x) = 1/((1-x^2) * (1-3*x+x^2)). - Petros Hadjicostas, Dec 03 2017
a(n) ~ ((1 + sqrt(5))/2)^(2*n) / n. - Vaclav Kotesovec, Aug 19 2018

A072336 EULER transform of A064831 (with its initial 1 omitted).

Original entry on oeis.org

1, 3, 15, 61, 250, 966, 3691, 13708, 50198, 180756, 642883, 2259194, 7859454, 27086808, 92579267, 314011481, 1057679231, 3539727315, 11776328715, 38963355007, 128255099092, 420155031109, 1370225125574, 4449779889260, 14393128980829, 46380744221539
Offset: 0

Views

Author

N. J. A. Sloane, Jul 16 2002

Keywords

Links

A192232 Constant term of the reduction of n-th Fibonacci polynomial by x^2 -> x+1. (See Comments.)

Original entry on oeis.org

1, 0, 2, 1, 6, 7, 22, 36, 89, 168, 377, 756, 1630, 3353, 7110, 14783, 31130, 65016, 136513, 285648, 599041, 1254456, 2629418, 5508097, 11542854, 24183271, 50674318, 106173180, 222470009, 466131960, 976694489, 2046447180, 4287928678, 8984443769, 18825088134
Offset: 1

Views

Author

Clark Kimberling, Jun 26 2011

Keywords

Comments

Polynomial reduction: an introduction
...
We begin with an example. Suppose that p(x) is a polynomial, so that p(x)=(x^2)t(x)+r(x) for some polynomials t(x) and r(x), where r(x) has degree 0 or 1. Replace x^2 by x+1 to get (x+1)t(x)+r(x), which is (x^2)u(x)+v(x) for some u(x) and v(x), where v(x) has degree 0 or 1. Continuing in this manner results in a fixed polynomial w(x) of degree 0 or 1. If p(x)=x^n, then w(x)=x*F(n)+F(n-1), where F=A000045, the sequence of Fibonacci numbers.
In order to generalize, write d(g) for the degree of an arbitrary polynomial g(x), and suppose that p, q, s are polynomials satisfying d(s)s in this manner until reaching w such that d(w)s.
The coefficients of (reduction of p by q->s) comprise a vector of length d(q)-1, so that a sequence p(n,x) of polynomials begets a sequence of vectors, such as (F(n), F(n-1)) in the above example. We are interested in the component sequences (e.g., F(n-1) and F(n)) for various choices of p(n,x).
Following are examples of reduction by x^2->x+1:
n-th Fibonacci p(x) -> A192232+x*A112576
n-th cyclotomic p(x) -> A192233+x*A051258
n-th 1st-kind Chebyshev p(x) -> A192234+x*A071101
n-th 2nd-kind Chebyshev p(x) -> A192235+x*A192236
x(x+1)(x+2)...(x+n-1) -> A192238+x*A192239
(x+1)^n -> A001519+x*A001906
(x^2+x+1)^n -> A154626+x*A087635
(x+2)^n -> A020876+x*A030191
(x+3)^n -> A192240+x*A099453
...
Suppose that b=(b(0), b(1),...) is a sequence, and let p(n,x)=b(0)+b(1)x+b(2)x^2+...+b(n)x^n. We define (reduction of sequence b by q->s) to be the vector given by (reduction of p(n,x) by q->s), with components in the order of powers, from 0 up to d(q)-1. For k=0,1,...,d(q)-1, we then have the "k-sequence of (reduction of sequence b by q->s)". Continuing the example, if b is the sequence given by b(k)=1 if k=n and b(k)=0 otherwise, then the 0-sequence of (reduction of b by x^2->x+1) is (F(n-1)), and the 1-sequence is (F(n)).
...
For selected sequences b, here are the 0-sequences and 1-sequences of (reduction of b by x^2->x+1):
b=A000045, Fibonacci sequence (1,1,2,3,5,8,...) yields
0-sequence A166536 and 1-sequence A064831.
b=(1,A000045)=(1,1,1,2,3,5,8,...) yields
0-sequence A166516 and 1-sequence A001654.
b=A000027, natural number sequence (1,2,3,4,...) yields
0-sequence A190062 and 1-sequence A122491.
b=A000032, Lucas sequence (1,3,4,7,11,...) yields
0-sequence A192243 and 1-sequence A192068.
b=A000217, triangular sequence (1,3,6,10,...) yields
0-sequence A192244 and 1-sequence A192245.
b=A000290, squares sequence (1,4,9,16,...) yields
0-sequence A192254 and 1-sequence A192255.
More examples: A192245-A192257.
...
More comments:
(1) If s(n,x)=(reduction of x^n by q->s) and
p(x)=p(0)x^n+p(1)x^(n-1)+...+p(n)x^0, then
(reduction of p by q->s)=p(0)s(n,x)+p(1)s(n-1,x)
+...+p(n-1)s(1,x)+p(n)s(0,x). See A192744.
(2) For any polynomial p(x), let P(x)=(reduction of p(x)
by q->s). Then P(r)=p(r) for each zero r of
q(x)-s(x). In particular, if q(x)=x^2 and s(x)=x+1,
then P(r)=p(r) if r=(1+sqrt(5))/2 (golden ratio) or
r=(1-sqrt(5))/2.

Examples

			The first four Fibonacci polynomials and their reductions by x^2->x+1 are shown here:
F1(x)=1 -> 1 + 0x
F2(x)=x -> 0 + 1x
F3(x)=x^2+1 -> 2+1x
F4(x)=x^3+2x -> 1+4x
F5(x)=x^4+3x^2+1 -> (x+1)^2+3(x+1)+1 -> 6+6x.
From these, read A192232=(1,0,1,1,6,...) and A112576=(0,1,1,4,6,...).

Links

Crossrefs

Cf. A168561, A192233 - A192240, A192744.
Cf. A206296, A265398, A265399, A265752, A265753.

Programs

  • Mathematica
    q[x_] := x + 1;
reductionRules = {x^y_?EvenQ -> q[x]^(y/2),  x^y_?OddQ -> x q[x]^((y - 1)/2)};
t = Table[FixedPoint[Expand[#1 /. reductionRules] &, Fibonacci[n, x]], {n, 1, 40}];
Table[Coefficient[Part[t, n], x, 0], {n, 1, 40}]
  (* A192232 *)
Table[Coefficient[Part[t, n], x, 1], {n, 1, 40}]
(* A112576 *)
(* Peter J. C. Moses, Jun 25 2011 *)
LinearRecurrence[{1, 3, -1, -1}, {1, 0, 2, 1}, 60] (* Vladimir Joseph Stephan Orlovsky, Feb 08 2012 *)
  • PARI
    Vec((1-x-x^2)/(1-x-3*x^2+x^3+x^4)+O(x^99)) \\ Charles R Greathouse IV, Jan 08 2013

Formula

Empirical G.f.: -x*(x^2+x-1)/(x^4+x^3-3*x^2-x+1). - Colin Barker, Sep 11 2012
The above formula is correct. - Charles R Greathouse IV, Jan 08 2013
a(n) = A265752(A206296(n)). - Antti Karttunen, Dec 15 2015
a(n) = A112576(n) -A112576(n-1) -A112576(n-2). - R. J. Mathar, Dec 16 2015

Extensions

Example corrected by Clark Kimberling, Dec 18 2017

A180662 The Golden Triangle: T(n,k) = A001654(k) for n>=0 and 0<=k<=n.

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 0, 1, 2, 6, 0, 1, 2, 6, 15, 0, 1, 2, 6, 15, 40, 0, 1, 2, 6, 15, 40, 104, 0, 1, 2, 6, 15, 40, 104, 273, 0, 1, 2, 6, 15, 40, 104, 273, 714, 0, 1, 2, 6, 15, 40, 104, 273, 714, 1870, 0, 1, 2, 6, 15, 40, 104, 273, 714, 1870, 4895
Offset: 0

Views

Author

Johannes W. Meijer, Sep 21 2010

Keywords

Comments

The terms in the n-th row of the Golden Triangle are the first (n+1) golden rectangle numbers. The golden rectangle numbers are A001654(n)=F(n)*F(n+1), with F(n) the Fibonacci numbers. The mirror image of the Golden Triangle is A180663.
We define below 24 mostly new triangle sums. The Row1 and Row2 sums are the ordinary and alternating row sums respectively and the Kn11 and Kn12 sums are commonly known as antidiagonal sums. Each of the names of these sums, except for the row sums, comes from a (fairy) chess piece that moves in its own peculiar way over a chessboard, see Hooper and Whyld. All pieces are leapers: knight (sqrt(5) or 1,2), fil (sqrt(8) or 2,2), camel (sqrt(10) or 3,1), giraffe (sqrt(17) or 4,1) and zebra (sqrt(13) or 3,2). Information about the origin of these chess sums can be found in "Famous numbers on a chessboard", see Meijer.
Each triangle or chess sum formula adds up numbers on a chessboard using the moves of its namesake. Converting a number triangle to a square array of numbers shows this most clearly (use the table button!). The formulas given below are for number triangles.
The chess sums of the Golden Triangle lead to six different sequences, see the crossrefs. As could be expected all these sums are related to the golden rectangle numbers.
Some triangles with complete sets of triangle sums are: A002260 (Natural Numbers), A007318 (Pascal), A008288 (Delannoy) A013609 (Pell-Jacobsthal), A036561 (Nicomachus), A104763 (Fibonacci(n)), A158405 (Odd Numbers) and of course A180662 (Golden Triangle).
#..Name....Type..Code....Definition of triangle sums.
1. Row......1....Row1.. a(n) = Sum_{k=0..n} T(n, k).
2. Row Alt..2....Row2.. a(n) = Sum_{k=0..n} (-1)^(n+k)*T(n, k).
3. Knight...1....Kn11.. a(n) = Sum_{k=0..floor(n/2)} T(n-k, k).
4. Knight...1....Kn12.. a(n) = Sum_{k=0..floor(n/2)} T(n-k+1, k+1).
5. Knight...1....Kn13.. a(n) = Sum_{k=0..floor(n/2)} T(n-k+2, k+2).
6. Knight...2....Kn21.. a(n) = Sum_{k=0..floor(n/2)} T(n-k, n-2*k).
7. Knight...2....Kn22.. a(n) = Sum_{k=0..floor(n/2)} T(n-k+1, n-2*k).
8. Knight...2....Kn23.. a(n) = Sum_{k=0..floor(n/2)} T(n-k+2, n-2*k).
9. Knight...3....Kn3... a(n) = Sum_{k=0..n} T(n+k, 2*k).
10. Knight...4....Kn4... a(n) = Sum_{k=0..n} T(n+k, n-k).
11. Fil......1....Fi1... a(n) = Sum_{k=0..floor(n/2)} T(n, 2*k).
12. Fil......2....Fi2... a(n) = Sum_{k=0..floor(n/2)} T(n, n-2*k).
13. Camel....1....Ca1... a(n) = Sum_{k=0..floor(n/3)} T(n-2*k, k).
14. Camel....2....Ca2... a(n) = Sum_{k=0..floor(n/3)} T(n-2*k, n-3*k).
15. Camel....3....Ca3... a(n) = Sum_{k=0..n} T(n+2*k, 3*k).
16. Camel....4....Ca4... a(n) = Sum_{k=0..n} T(n+2*k, n-k).
17. Giraffe..1....Gi1... a(n) = Sum_{k=0..floor(n/4)} T(n-3*k, k).
18. Giraffe..2....Gi2... a(n) = Sum_{k=0..floor(n/4)} T(n-3*k, n-4*k).
19. Giraffe..3....Gi3... a(n) = Sum_{k=0..n} T(n+3*k, 4*k).
20. Giraffe..4....Gi4... a(n) = Sum_{k=0..n} T(n+3*k, n-k).
21. Zebra....1....Ze1... a(n) = Sum_{k=0..floor(n/2)} T(n+k, 3*k).
22. Zebra....2....Ze2... a(n) = Sum_{k=0..floor(n/2)} T(n+k, n-2*k).
23. Zebra....3....Ze3... a(n) = Sum_{k=0..floor(n/3)} T(n-k, 2*k).
24. Zebra....4....Ze4... a(n) = Sum_{k=0..floor(n/3)} T(n-k, n-3*k).

Examples

			The first few rows of the Golden Triangle are:
  0;
  0, 1;
  0, 1, 2;
  0, 1, 2, 6;
  0, 1, 2, 6, 15;
  0, 1, 2, 6, 15, 40;

References

  • David Hooper and Kenneth Whyld, The Oxford Companion to Chess, p. 221, 1992.

Links

Crossrefs

Cf. A180663 (Mirror), A001654 (Golden Rectangle), A000045 (F(n)).
Triangle sums: A064831 (Row1, Kn11, Kn12, Kn4, Ca1, Ca4, Gi1, Gi4), A077916 (Row2), A180664 (Kn13), A180665 (Kn21, Kn22, Kn23, Fi2, Ze2), A180665(2*n) (Kn3, Fi1, Ze3), A115730(n+1) (Ca2, Ze4), A115730(3*n+1) (Ca3, Ze1), A180666 (Gi2), A180666(4*n) (Gi3).

Programs

  • Haskell
    import Data.List (inits)
a180662 n k = a180662_tabl !! n !! k
a180662_row n = a180662_tabl !! n
a180662_tabl = tail $ inits a001654_list
-- Reinhard Zumkeller, Jun 08 2013
  • Magma
    [Fibonacci(k)*Fibonacci(k+1): k in [0..n], n in [0..12]]; // G. C. Greubel, May 25 2021
  • Maple
    F:= combinat[fibonacci]:
T:= (n, k)-> F(k)*F(k+1):
seq(seq(T(n, k), k=0..n), n=0..10); # revised Johannes W. Meijer, Sep 13 2012
  • Mathematica
    Table[Times @@ Fibonacci@ {k, k + 1}, {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Aug 18 2016 *)
Module[{nn=20,f},f=Times@@@Partition[Fibonacci[Range[0,nn]],2,1];Table[Take[f,n],{n,nn}]]//Flatten (* Harvey P. Dale, Nov 26 2022 *)
  • PARI
    T(n,k)=fibonacci(k)*fibonacci(k+1) \\ Charles R Greathouse IV, Nov 07 2016
  • Sage
    flatten([[fibonacci(k)*fibonacci(k+1) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 25 2021

Formula

T(n, k) = F(k)*F(k+1) with F(n) = A000045(n), for n>=0 and 0<=k<=n.
From Johannes W. Meijer, Jun 22 2015: (Start)
Kn1p(n) = Sum_{k=0..floor(n/2)} T(n-k+p-1, k+p-1), p >= 1.
Kn1p(n) = Kn11(n+2*p-2) - Sum_{k=0..p-2} T(n-k+2*p-2, k), p >= 2.
Kn2p(n) = Sum_{k=0..floor(n/2)} T(n-k+p-1, n-2*k), p >= 1.
Kn2p(n) = Kn21(n+2*p-2) - Sum_{k=0..p-2} T(n+k+p, n+2*k+2), p >= 2. (End)
G.f. as triangle: xy/((1-x)(1+xy)(1-3xy+x^2 y^2)). - Robert Israel, Sep 06 2015

A001654 Golden rectangle numbers: F(n) * F(n+1), where F(n) = A000045(n) (Fibonacci numbers).

Original entry on oeis.org

0, 1, 2, 6, 15, 40, 104, 273, 714, 1870, 4895, 12816, 33552, 87841, 229970, 602070, 1576239, 4126648, 10803704, 28284465, 74049690, 193864606, 507544127, 1328767776, 3478759200, 9107509825, 23843770274, 62423800998, 163427632719, 427859097160, 1120149658760
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

a(n)/A007598(n) ~= golden ratio, especially for larger n. - Robert Happelberg (roberthappelberg(AT)yahoo.com), Jul 25 2005
Let phi be the golden ratio (cf. A001622). Then 1/phi = phi - 1 = Sum_{n>=1} (-1)^(n-1)/a(n), an alternating infinite series consisting solely of unit fractions. - Franz Vrabec, Sep 14 2005
a(n+2) is the Hankel transform of A005807 aerated. - Paul Barry, Nov 04 2008
A more exact name would be: Golden convergents to rectangle numbers. These rectangles are not actually golden (ratio of sides is not phi) but are golden convergents (sides are numerator and denominator of convergents in the continued fraction expansion of phi, whence ratio of sides converges to phi). - Daniel Forgues, Nov 29 2009
The Kn4 sums (see A180662 for definition) of the "Races with Ties" triangle A035317 lead to this sequence. - Johannes W. Meijer, Jul 20 2011
Numbers m such that m(5m+2)+1 or m(5m-2)+1 is a square. - Bruno Berselli, Oct 22 2012
In pairs, these numbers are important in finding binomial coefficients that appear in at least six places in Pascal's triangle. For instance, the pair (m,n) = (40, 104) finds the numbers binomial(n-1,m) = binomial(n,m-1). Two additional numbers are found on the other side of the triangle. The final two numbers appear in row binomial(n-1,m). See A003015. - T. D. Noe, Mar 13 2013
For n>1, a(n) is one-half the area of the trapezoid created by the four points (F(n),L(n)), (L(n),F(n)), (F(n+1), L(n+1)), (L(n+1), F(n+1)) where F(n) = A000045(n) and L(n) = A000032(n). - J. M. Bergot, May 14 2014
[Note on how to calculate: take the two points (a,b) and (c,d) with a
a(n) = A067962(n-1) / A067962(n-2), n > 1. - Reinhard Zumkeller, Sep 24 2015
Can be obtained (up to signs) by setting x = F(n)/F(n+1) in g.f. for Fibonacci numbers - see Pongsriiam. - N. J. A. Sloane, Mar 23 2017

Examples

			G.f. = x + 2*x^2 + 6*x^3 + 15*x^4 + 40*x^5 + 104*x^6 + 273*x^7 + 714*x^8 + ...

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 9.
  • 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).

Links

Crossrefs

Cf. A001655, A001656, A001657, A001658, A010048, A067962.
Cf. A005968, A005969, A098531, A098532, A098533, A119283, A128697.
Cf. A000071, A079472, A080145, A239798, A290565.
Bisection of A006498, A070550, A080239.
First differences of A064831.
Partial sums of A007598.

Programs

  • Haskell
    a001654 n = a001654_list !! n
a001654_list = zipWith (*) (tail a000045_list) a000045_list
-- Reinhard Zumkeller, Jun 08 2013
  • Magma
    I:=[0,1,2]; [n le 3 select I[n] else 2*Self(n-1) + 2*Self(n-2) - Self(n-3): n in [1..30]]; // G. C. Greubel, Jan 17 2018
  • Maple
    with(combinat): A001654:=n->fibonacci(n)*fibonacci(n+1):
seq(A001654(n), n=0..28); # Zerinvary Lajos, Oct 07 2007
  • Mathematica
    LinearRecurrence[{2,2,-1}, {0,1,2}, 100] (* Vladimir Joseph Stephan Orlovsky, Jul 03 2011 *)
Times@@@Partition[Fibonacci[Range[0,30]],2,1] (* Harvey P. Dale, Aug 18 2011 *)
Accumulate[Fibonacci[Range[0, 30]]^2] (* Paolo Xausa, May 31 2024 *)
  • PARI
    A001654(n)=fibonacci(n)*fibonacci(n+1);
  • PARI
    b(n, k)=prod(j=1, k, fibonacci(n+j)/fibonacci(j));
vector(30, n, b(n-1, 2))  \\ Joerg Arndt, May 08 2016
  • Python
    from sympy import fibonacci as F
def a(n): return F(n)*F(n + 1)
[a(n) for n in range(101)] # Indranil Ghosh, Aug 03 2017
  • Python
    from math import prod
from gmpy2 import fib2
def A001654(n): return prod(fib2(n+1)) # Chai Wah Wu, May 19 2022

Formula

a(n) = A010048(n+1, 2) = Fibonomial(n+1, 2).
a(n) = A006498(2*n-1).
a(n) = a(n - 1) + A007598(n) = a(n - 1) + A000045(n)^2 = Sum_{j <= n} Fibonacci(j)^2. - Henry Bottomley, Feb 09 2001 [corrected by Ridouane Oudra, Apr 12 2025]
For n > 0, 1 - 1/a(n+1) = Sum_{k=1..n} 1/(F(k)*F(k+2)) where F(k) is the k-th Fibonacci number. - Benoit Cloitre, Aug 31 2002.
G.f.: x/(1-2*x-2*x^2+x^3) = x/((1+x)*(1-3*x+x^2)). (Simon Plouffe in his 1992 dissertation; see Comments to A055870),
a(n) = 3*a(n-1) - a(n-2) - (-1)^n = -a(-1-n).
Let M = the 3 X 3 matrix [1 2 1 / 1 1 0 / 1 0 0]; then a(n) = the center term in M^n *[1 0 0]. E.g., a(5) = 40 since M^5 * [1 0 0] = [64 40 25]. - Gary W. Adamson, Oct 10 2004
a(n) = Sum{k=0..n} Fibonacci(k)^2. The proof is easy. Start from a square (1*1). On the right side, draw another square (1*1). On the above side draw a square ((1+1)*(1+1)). On the left side, draw a square ((1+2)*(1+2)) and so on. You get a rectangle (F(n)*F(1+n)) which contains all the squares of side F(1), F(2), ..., F(n). - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 19 2007
With phi = (1+sqrt(5))/2, a(n) = round((phi^(2*n+1))/5) = floor((1/2) + (phi^(2*n+1))/5), n >= 0. - Daniel Forgues, Nov 29 2009
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3), a(1)=1, a(2)=2, a(3)=6. - Sture Sjöstedt, Feb 06 2010
a(n) = (A002878(n) - (-1)^n)/5. - R. J. Mathar, Jul 22 2010
a(n) = 1/|F(n+1)/F(n) - F(n)/F(n-1)| where F(n) = Fibonacci numbers A000045. b(n) = F(n+1)/F(n) - F(n)/F(n-1): 1/1, -1/2, 1/6, -1/15, 1/40, -1/104, ...; c(n) = 1/b(n) = a(n)*(-1)^(n+1): 1, -2, 6, -15, 40, -104, ... (n=1,2,...). - Thomas Ordowski, Nov 04 2010
a(n) = (Fibonacci(n+2)^2 - Fibonacci(n-1)^2)/4. - Gary Detlefs, Dec 03 2010
Let d(n) = n mod 2, a(0)=0 and a(1)=1. For n > 1, a(n) = d(n) + 2*a(n-1) + Sum_{k=0..n-2} a(k). - L. Edson Jeffery, Mar 20 2011
From Tim Monahan, Jul 11 2011: (Start)
a(n+1) = ((2+sqrt(5))*((3+sqrt(5))/2)^n+(2-sqrt(5))*((3-sqrt(5))/2)^n+(-1)^n)/5.
a(n) = ((1+sqrt(5))*((3+sqrt(5))/2)^n+(1-sqrt(5))*((3-sqrt(5))/2)^n-2*(-1)^n)/10. (End)
From Wolfdieter Lang, Jul 21 2012: (Start)
a(n) = (2*A059840(n+2) - A027941(n))/3, n >= 0, with A059840(n+2) = Sum_{k=0..n} F(k)*F(k+2) and A027941(n) = A001519(n+1) - 1, n >= 0, where A001519(n+1) = F(2*n+1). (End)
a(n) = (-1)^n * Sum_{k=0..n} (-1)^k*F(2*k), n >= 0. - Wolfdieter Lang, Aug 11 2012
a(-1-n) = -a(n) for all n in Z. - Michael Somos, Sep 19 2014
0 = a(n)*(+a(n+1) - a(n+2)) + a(n+1)*(-2*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 19 2014
a(n) = (L(2*n+1) - (-1)^n)/5 with L(k) = A000032(k). - J. M. Bergot, Apr 15 2016
E.g.f.: ((3 + sqrt(5))*exp((5+sqrt(5))*x/2) - 2*exp((2*x)/(3+sqrt(5))+x) - 1 - sqrt(5))*exp(-x)/(5*(1 + sqrt(5))). - Ilya Gutkovskiy, Apr 15 2016
From Klaus Purath, Apr 24 2019: (Start)
a(n) = A061646(n) - Fibonacci(n-1)^2.
a(n) = (A061646(n+1) - A061646(n))/2. (End)
a(n) = A226205(n+1) + (-1)^(n+1). - Flávio V. Fernandes, Apr 23 2020
Sum_{n>=1} 1/a(n) = A290565. - Amiram Eldar, Oct 06 2020
Product_{n>=2} (1 + (-1)^n/a(n)) = phi^2/2 (A239798). - Amiram Eldar, Dec 02 2024
G.f.: x * exp( Sum_{k>=1} F(3*k)/F(k) * x^k/k ), where F(n) = A000045(n). - Seiichi Manyama, May 07 2025

Extensions

Extended by Wolfdieter Lang, Jun 27 2000

A202453 Fibonacci self-fusion matrix, by antidiagonals.

Original entry on oeis.org

1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 5, 5, 6, 5, 5, 8, 8, 9, 9, 8, 8, 13, 13, 15, 15, 15, 13, 13, 21, 21, 24, 24, 24, 24, 21, 21, 34, 34, 39, 39, 40, 39, 39, 34, 34, 55, 55, 63, 63, 64, 64, 63, 63, 55, 55, 89, 89, 102, 102, 104, 104, 104, 102, 102, 89, 89, 144, 144, 165, 165
Offset: 1

Views

Author

Clark Kimberling, Dec 19 2011

Keywords

Comments

The Fibonacci self-fusion matrix, F, is the fusion P**Q, where P and Q are the lower and upper triangular Fibonacci matrices. See A193722 for the definition of fusion of triangular arrays.
Every term F(n,k) of F is a product of two Fibonacci numbers; indeed,
F(n,k)=F(n)*F(k+1) if k is even;
F(n,k)=F(n+1)*F(k) if k is odd.
antidiagonal sums: (1,2,6,12,...), A054454
diagonal (1,2,6,15,...), A001654
diagonal (1,3,9,24,...), A064831
diagonal (2,5,15,39,..), A059840
diagonal (3,8,24,63,..), A080097
diagonal (5,13,39,102,...), A080143
diagonal (8,21,63,165,...), A080144
principal submatrix sums, A202462
All the principal submatrices are invertible, and the terms in the inverses are in {-3,-2,-1,0,1,2,3}.

Examples

			Northwest corner:
1...1....2....3....5....8....13
1...2....3....5....8...13....21
2...3....6....9...15...24....39
3...5....9...15...24...39....63
5...8...15...24...40...64...104

Links

Crossrefs

Cf. A000045, A202451, A202452, A202503 (Fibonacci fission array).

Programs

  • Mathematica
    n = 12;
Q = NestList[Most[Prepend[#, 0]] &, #, Length[#] - 1] &[Table[Fibonacci[k], {k, 1, n}]];
P = Transpose[Q]; F = P.Q;
Flatten[Table[P[[i]][[k + 1 - i]], {k, 1, n}, {i, 1, k}]] (* A202452 as a sequence *)
Flatten[Table[Q[[i]][[k + 1 - i]], {k, 1, n}, {i, 1, k}]] (* A202451 as a sequence *)
Flatten[Table[F[[i]][[k + 1 - i]], {k, 1, n}, {i, 1, k}]] (* A202453 as a sequence *)
TableForm[Q]  (* A202451, upper tri. Fibonacci array *)
TableForm[P]  (* A202452, lower tri. Fibonacci array *)
TableForm[F]  (* A202453, Fibonacci fusion array *)
TableForm[FactorInteger[F]]

Formula

Matrix product P*Q, where P, Q are the lower and upper triangular Fibonacci matrices, A202451 and A202452.

A192872 Constant term in the reduction by (x^2 -> x+1) of the polynomial p(n,x) given in Comments.

Original entry on oeis.org

1, 0, 3, 4, 13, 30, 81, 208, 547, 1428, 3741, 9790, 25633, 67104, 175683, 459940, 1204141, 3152478, 8253297, 21607408, 56568931, 148099380, 387729213, 1015088254, 2657535553, 6957518400
Offset: 0

Views

Author

Clark Kimberling, Jul 11 2011

Keywords

Comments

The polynomial p(n,x) is defined by p(0,x)=1, p(1,x)=x, and p(n,x) = x*p(n-1,x) + (x^2)*p(n-1,x) + 1. The resulting sequence typifies a general class which we shall describe here. Suppose that u,v,a,b,c,d,e,f are numbers used to define these polynomials:
...
q(x) = x^2
s(x) = u*x + v
p(0,x) = a, p(1,x) = b*x + c
p(n,x) = d*x*p(n-1,x) + e*(x^2)*p(n-2,x) + f.
...
We shall assume that u is not 0 and that {d,e} is not {0}. The reduction of p(n,x) by the repeated substitution q(x)->s(x), as defined and described at A192232 and A192744, has the form h(n)+k(n)*x. The numerical sequences h and k are, formally, linear recurrence sequences of order 5. The second Mathematica program below shows initial terms and the recurrence coefficients, which are too long to be included here, which imply these properties:
(1) The numbers a,b,c,f affect initial terms but not the recurrence coefficients, which depend only on u,v,d,e.
(2) If v=0 or e=0, the order of recurrence is <= 3.
(3) If v=0 and e=0, the order of recurrence is 2, and the coefficients are 1+d*u and d*u.
(See A192904 for similar results for other p(n,x).)
...
Examples:
u v a b c d e f seq h.....seq k
1 1 1 2 0 1 1 0 -A121646..A059929
1 1 1 3 0 1 1 0 A128533...A081714
1 1 2 1 0 1 1 0 A081714...A001906
1 1 1 1 1 1 1 0 A000045...A001906
1 1 2 1 1 1 1 0 A129905...A192879
1 1 1 2 1 1 1 0 A061646...A079472
1 1 1 1 0 1 1 1 A192872...A192873
1 1 1 1 1 2 1 1 A192874...A192875
1 1 1 1 1 2 1 1 A192876...A192877
1 1 1 1 1 1 2 1 A192880...A192882
1 1 1 1 1 1 1 1 A166536...A064831
The terms of several of these sequences are products of Fibonacci numbers (A000045), or Fibonacci numbers and Lucas numbers (A000032).

Examples

			The coefficients in all the polynomials p(n,x) are Fibonacci numbers (A000045).  The first six and their reductions:
p(0,x) = 1 -> 1
p(1,x) = x -> x
p(2,x) = 1 + 2*x^2 -> 3 + 2*x
p(3,x) = 1 + x + 3*x^3 -> 4 + 7*x
p(4,x) = 1 + x + 2*x^2 + 5*x^4 -> 13 + 18*x
p(5,x) = 1 + x + 2*x^2 + 3*x^3 + 8*x^5 -> 30 + 49*x

Links

Crossrefs

Cf. A192232, A192744, A192873, A192908 (sums of adjacent terms).

Programs

  • GAP
    a:=[1,0,3,4];; for n in [5..30] do a[n]:=3*a[n-1]-3*a[n-3]+a[n-4]; od; a; # G. C. Greubel, Jan 06 2019
  • Magma
    m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (2*x-1)*(x^2-x+1)/((x-1)*(1+x)*(x^2-3*x +1)) )); // G. C. Greubel, Jan 06 2019
  • Mathematica
    (* First program *)
q = x^2; s = x + 1; z = 26;
p[0, x_] := 1; p[1, x_] := x;
p[n_, x_] := p[n - 1, x]*x + p[n - 2, x]*x^2 + 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}] (* A192872 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192873 *)
(* End of 1st program *)
(* ******************************************** *)
(* Second program: much more general *)
(* u = 1; v = 1; a = 1; b = 1; c = 0; d = 1; e = 1; f = 1; Nine degrees of freedom for user; shown values generate A192872. *)
q = x^2; s = u*x + v; z = 11;
(* will apply reduction (x^2 -> u*x+v) to p(n,x) *)
p[0, x_] := a; p[1, x_] := b*x + c;
(* initial values of polynomial sequence p(n,x) *)
p[n_, x_] := d*x*p[n - 1, x] + e*(x^2)*p[n - 2, x] + f;
(* recurrence for p(n,x) *)
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}];
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}];
Simplify[FindLinearRecurrence[u1]] (* for 0-sequence *)
Simplify[FindLinearRecurrence[u2]] (* for 1-sequence *)
u1 = Table[Coefficient[Part[t, n], x, 0], {n, 1, 4}]
(* initial values for 0-sequence *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, 4}]
(* initial values for 1-sequence *)
LinearRecurrence[{3,0,-3,1},{1,0,3,4},26] (* Ray Chandler, Aug 02 2015 *)
  • PARI
    my(x='x+O('x^30)); Vec((2*x-1)*(x^2-x+1)/((x-1)*(1+x)*(x^2-3*x +1))) \\ G. C. Greubel, Jan 06 2019
  • Sage
    ((2*x-1)*(x^2-x+1)/((x-1)*(1+x)*(x^2-3*x +1))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jan 06 2019

Formula

a(n) = 3*a(n-1) - 3*a(n-3) + a(n-4).
G.f.: (2*x-1)*(x^2-x+1) / ( (x-1)*(1+x)*(x^2-3*x+1) ). - R. J. Mathar, Oct 26 2011

A080097 a(n) = Fibonacci(n+2)^2 - 1.

Original entry on oeis.org

0, 3, 8, 24, 63, 168, 440, 1155, 3024, 7920, 20735, 54288, 142128, 372099, 974168, 2550408, 6677055, 17480760, 45765224, 119814915, 313679520, 821223648, 2149991423, 5628750624, 14736260448, 38580030723, 101003831720
Offset: 0

Views

Author

Mario Catalani (mario.catalani(AT)unito.it), Jan 29 2003

Keywords

Comments

a(n), a(n)+1 and a(n)+2 are consecutive members of A049997.

Links

Crossrefs

Equals A007598(n+2) - 1.
Cf. A000032, A000045, A001622, A049997, A059840, A064831, A080144.

Programs

Formula

If n is odd, then a(n) = F(n+1)*F(n+3) = F(n)*F(n+4) - 2, else a(n) = F(n)*F(n+4) = F(n+1)*F(n+3) - 2, where F(n) = Fibonacci numbers (A000045).
a(n) = (Lucas(2*n+4) - 2*(-1)^n - 5)/5.
O.g.f.: x*(3-x)/((1-x^2)*(1-3*x+x^2)) (see a comment on A080144). - Wolfdieter Lang, Jul 30 2012
a(n) = Sum_{k=1..n} F(k+3)*F(k) = A027941(n) + 2*A001654(n), n>=0. - Wolfdieter Lang, Jul 27 2012
Sum_{n>=1} 1/a(n) = (43 - 15*sqrt(5))/18 = 29/9 - 5*phi/3, where phi is the golden ratio (A001622). - Amiram Eldar, Oct 20 2020
a(n) = 3*a(n-1) - 3*a(n-3) + a(n-4). - Joerg Arndt, Nov 13 2023

Extensions

Edited by Ralf Stephan, May 15 2005

A059840 a(n) = F(n)*F(n-1) if n odd otherwise F(n)*F(n-1)-1, where F = Fibonacci numbers A000045.

Original entry on oeis.org

0, 0, 2, 5, 15, 39, 104, 272, 714, 1869, 4895, 12815, 33552, 87840, 229970, 602069, 1576239, 4126647, 10803704, 28284464, 74049690, 193864605, 507544127, 1328767775, 3478759200, 9107509824, 23843770274, 62423800997, 163427632719, 427859097159, 1120149658760, 2932589879120
Offset: 1

Views

Author

N. J. A. Sloane, Feb 26 2001

Keywords

Links

Crossrefs

Cf. A000045, A001654, A059248, A064831, A119996.

Programs

  • GAP
    List([1..30],n->Sum([1..n-2],k->Fibonacci(k)*Fibonacci(k+2))); # Muniru A Asiru, Aug 09 2018
  • Magma
    F:=Fibonacci; [(n mod 2) eq 0 select F(n)*F(n-1)-1 else F(n)*F(n-1): n in [1..30]]; // G. C. Greubel, Jul 23 2019
  • Maple
    seq(coeff(series(x^3*(2-x)/((1-x^2)*(1-3*x+x^2)), x,n+1),x,n),n=1..30); # Muniru A Asiru, Aug 09 2018
  • Mathematica
    Table[If[OddQ[n],Fibonacci[n]Fibonacci[n-1],Fibonacci[n] Fibonacci[n-1]-1],{n,30}]  (* Harvey P. Dale, Apr 20 2011 *)
  • PARI
    a(n) = { fibonacci(n)*fibonacci(n-1) - (n%2 == 0) } \\ Harry J. Smith, Jun 29 2009
  • Sage
    a=(x^3*(2-x)/((1-x^2)*(1-3*x+x^2))).series(x, 30).coefficients(x, sparse=False); a[1:] # G. C. Greubel, Jul 23 2019

Formula

G.f.: x^3*(2 - x)/((1 - x^2)*(1 - 3*x + x^2)). See a comment on A080144. - Wolfdieter Lang, Jul 30 2012
a(n) = Sum_{k=1..n-2} F(k)*F(k+2). - Alexander Adamchuk, May 17 2007
a(n+2) = (3*A001654(n) + A027941(n))/2, n >= 0. - Wolfdieter Lang, Jul 21 2012
a(n+2) = (3*(-1)^(n+1) - 5 + 2*Lucas(2*n + 3))/10, n >= 0. - Ehren Metcalfe, Aug 21 2017
a(n) = floor(1/(Sum_{k>=n} 1/Fibonacci(k)^2)) [Ohtsuka and Nakamura]. - Michel Marcus, Aug 09 2018
For n > 2, 2 * A000217(a(n)) = A228873(n-2). - Diego Rattaggi, Jan 27 2021

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

Views

Author

N. J. A. Sloane, Jun 09 2002

Keywords

Comments

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

Examples

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

References

  • 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.

Links

Crossrefs

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

Programs

  • Magma
    [Fibonacci(2*n)*Fibonacci(2*n+2): n in [0..30]]; // Vincenzo Librandi, Apr 18 2011
  • Maple
    fs4:=n->sum(fibonacci(4*k),k=0..n):seq(fs4(n),n=0..21); # Gary Detlefs, Dec 07 2010
  • Mathematica
    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 *)
  • PARI
    a(n)=fibonacci(2*n)*fibonacci(2*n+2) \\ Charles R Greathouse IV, Jul 02 2013

Formula

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
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