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The Golden Triangle: T(n,k) = A001654(k) for n>=0 and 0<=k<=n.
+20
144
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
0,6
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).
REFERENCES
David Hooper and Kenneth Whyld, The Oxford Companion to Chess, p. 221, 1992.
LINKS
Verner E. Hoggatt, Jr., A new angle on Pascal’s triangle, The Fibonacci Quarterly, Vol. 6, Number 4, pp. 228-230, Oct. 1968.
Edouard Lucas, Recherches sur plusieurs ouvrages de Léonard de Pise, Ch. 1, pp. 12-14, 1877.
Johannes W. Meijer, Famous numbers on a chessboard, Acta Nova, Volume 4, No.4, December 2010; pp. 589-598.
Johannes W. Meijer, Illustrations of the triangle sums, Mar 07 2013.
S. Northshield, Sums across Pascal's triangle modulo 2, Congressus Numerantium, 200, pp. 35-52, 2010.
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
EXAMPLE
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;
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 *)
PROG
(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
(PARI) T(n, k)=fibonacci(k)*fibonacci(k+1) \\ Charles R Greathouse IV, Nov 07 2016
(Magma) [Fibonacci(k)*Fibonacci(k+1): k in [0..n], n in [0..12]]; // G. C. Greubel, May 25 2021
(Sage) flatten([[fibonacci(k)*fibonacci(k+1) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 25 2021
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).
KEYWORD
easy,nonn,tabl
AUTHOR
Johannes W. Meijer, Sep 21 2010
STATUS
approved
Partial sums of A001654, or sum of the areas of the first n Fibonacci rectangles.
+20
37
0, 1, 3, 9, 24, 64, 168, 441, 1155, 3025, 7920, 20736, 54288, 142129, 372099, 974169, 2550408, 6677056, 17480760, 45765225, 119814915, 313679521, 821223648, 2149991424, 5628750624, 14736260449, 38580030723, 101003831721
OFFSET
0,3
COMMENTS
The n-th rectangle is F(n)*F(n+1), where F(n) = n-th Fibonacci number (F(1)=1, F(2)=1, F(3)=2, etc.), A000045.
If 2*T(a_n) = the oblong number formed by substituting a(n) in the product formula x(x+1), then 2*T(a_n) = F(n-1)*F(n) * F(n)*F(n+1). Thus a(n) equals the integer part of the square root of the right hand side of the given equation. - Kenneth J Ramsey, Dec 19 2006
Contribution from Johannes W. Meijer, Sep 22 2010: (Start)
The a(n) represent several triangle sums of the Golden Triangle A180662: Kn11 (terms doubled), Kn12(n+1) (terms doubled), Kn4, Ca1 (terms tripled), Ca4, Gi1 (terms quadrupled) and Gi4. See A180662 for the definitions of these sums.
(End)
Define a 2 X (n+1) matrix with elements T(r,0)=A000032(r) and T(r,1) = Fibonacci(r), r=0,1,..,n. The matrix times its transposed is a 2 X 2 matrix with one diagonal element A001654(n+1), the other A216243(n), and A027941(n+1) on both outer diagonals. The determinant of this 2 X 2 matrix is 4*a(n). Example: For n=3 the matrix is 2 X 4 with rows 2 1 3 4; 0 1 1 2 to give as a product the 2 X 2 matrix with rows 30 12; 12 6 and determinant 180-144 = 36 =4*a(3). - J. M. Bergot, Feb 13 2013
a(n+1) is equal to the number of ternary strings of length n without any substring of the form 0x1, where x is in {0,1,2}. - John M. Campbell, Apr 03 2016
LINKS
E. Altinisik, A. Keskin, M. Yildiz, M. Demirbuken, On a conjecture of Ilmonen, Haukkanen and Merikoski concerning the smallest eigenvalues of certain GCD related matrices, Linear Algebra and its Applications, Volume 493, 15 March 2016, Pages 1-13
S. Falcon, On the Sequences of Products of Two k-Fibonacci Numbers, American Review of Mathematics and Statistics, March 2014, Vol. 2, No. 1, pp. 111-120.
FORMULA
a(n) = F(n+1)^2 - 1 if n is even, or F(n+1)^2 if n is odd.
a(n) = A005313(n+1) - n.
G.f.: x/((1-x^2)*(1-3*x+x^2)). - N. J. A. Sloane Jul 15 2002
a(n) = Sum_{k=0..floor(n/2)} U(n-2k-1, 3/2). - Paul Barry, Nov 15 2003
Let M_n denote the n X n Hankel matrix M_n(i, j)=F(i+j-1) where F = A000045 is Fibonacci numbers, then the characteristic polynomial of M_n is x^n - F(2n)x^(n-1) + a(n-1)x^(n-2) . - Michael Somos, Nov 14 2002
a(n) = a(n-1) + A001654(n) with a(0)=0. (Partial sums of A001654). - Johannes W. Meijer, Sep 22 2010
a(n) = floor(phi^(2*n+2)/5), where phi =(1+sqrt(5))/2. - Gary Detlefs Mar 12 2011
a(n) = (A027941(n) + A001654(n))/2, n>=0. - Wolfdieter Lang, Jul 23 2012
a(n) = A005248(n+1)/5 -1/2 -(-1)^n/10. - R. J. Mathar, Feb 21 2013
Recurrence: a(0) = 0, a(1) = 1, a(2) = 3, a(3) = 9, a(n) = 3*a(n-1) - 3*a(n-3) + a(n-4). - Vladimir Reshetnikov, Oct 28 2015
a(n) = Sum_{i=0..n} (n+1-i)*Fibonacci(i)^2. - Bruno Berselli, Feb 20 2017
MATHEMATICA
Table[ Sum[ Fibonacci[k]*Fibonacci[k + 1], {k, n} ], {n, 0, 30}]
f[n_] := Floor[GoldenRatio^(2 n + 2)/5]; Array[f, 28, 0] (* Robert G. Wilson v, Oct 25 2001 *)
a[0]= 0; a[1]= 1; a[2]= 3; a[3]= 9; a[n_]:= a[n]= 3a[n-1] - 3a[n-3] + a[n-4]; Table[a[n], {n, 0, 20}] (* Vladimir Reshetnikov, Oct 28 2015 *)
PROG
(PARI) a(n)=if(n<0, 0, fibonacci(n+1)^2-1+n%2)
(PARI) { for (n=0, 200, a=fibonacci(n+1)^2 - 1 + n%2; write("b064831.txt", n, " ", a) ) } \\ Harry J. Smith, Sep 27 2009
(PARI) my(x='x+O('x^30)); concat([0], Vec(x/((1-x^2)*(1-3*x+x^2)))) \\ G. C. Greubel, Jan 09 2019
(Magma) m:=30; R<x>:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!( x/((1-x^2)*(1-3*x+x^2)) )); // G. C. Greubel, Jan 09 2019
(Sage) (x/((1-x^2)*(1-3*x+x^2))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jan 09 2019
(GAP) a:=[0, 1, 3, 9];; 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 09 2019
CROSSREFS
Odd terms of A097083.
Partial sums of A001654.
KEYWORD
nonn,easy
AUTHOR
Howard Stern (hsstern(AT)mindspring.com), Oct 23 2001
EXTENSIONS
More terms from Robert G. Wilson v, Oct 25 2001
STATUS
approved
Golden Triangle sums: a(n) = a(n-1) + A001654(n+1) with a(0)=0.
+20
8
0, 2, 8, 23, 63, 167, 440, 1154, 3024, 7919, 20735, 54287, 142128, 372098, 974168, 2550407, 6677055, 17480759, 45765224, 119814914, 313679520, 821223647, 2149991423, 5628750623, 14736260448, 38580030722, 101003831720
OFFSET
0,2
COMMENTS
The a(n+1) (terms doubled) are the Kn13 sums of the Golden Triangle A180662. See A180662 for information about these knight and other chess sums.
FORMULA
a(n+1) = Sum_{k=0..n} A180662(2*n-k+2, k+2).
a(n) = (-15 + (-1)^n + (6-2*A)*A^(-n-1) + (6-2*B)*B^(-n-1))/10 with A=(3+sqrt(5))/2 and B=(3-sqrt(5))/2.
G.f.: (2*x+2*x^2-x^3)/(1-3*x-x^4+3*x^3).
a(n) = Sum_{i=0..n-1} F(i+2)*F(i+3), where F(i) = A000045(i). - Rigoberto Florez, Jul 07 2020
a(n) = (1/10)*((-1)^n - 15 + 2*Lucas(2*n+4)). - G. C. Greubel, Jan 21 2022
MAPLE
nmax:=26: with(combinat): for n from 0 to nmax+1 do A001654(n):=fibonacci(n) * fibonacci(n+1) od: a(0):=0: for n from 1 to nmax do a(n) := a(n-1)+A001654(n+1) od: seq(a(n), n=0..nmax);
MATHEMATICA
Table[Sum[Fibonacci[i+2]*Fibonacci[i+3], {i, 0, n-1}], {n, 0, 40}] (* Rigoberto Florez, Jul 07 2020 *)
LinearRecurrence[{3, 0, -3, 1}, {0, 2, 8, 23}, 30] (* Harvey P. Dale, Mar 30 2023 *)
PROG
(Magma) [(1/10)*((-1)^n - 15 + 2*Lucas(2*n+4)): n in [0..40]]; // G. C. Greubel, Jan 21 2022
(Sage) [(1/10)*((-1)^n - 15 + 2*lucas_number2(2*n+4, 1, -1)) for n in (0..40)] # G. C. Greubel, Jan 21 2022
KEYWORD
easy,nonn
AUTHOR
Johannes W. Meijer, Sep 21 2010
STATUS
approved
a(n) = a(n-3) + A001654(n-1) with a(0)=0, a(1)=0 and a(2)=1.
+20
6
0, 0, 1, 2, 6, 16, 42, 110, 289, 756, 1980, 5184, 13572, 35532, 93025, 243542, 637602, 1669264, 4370190, 11441306, 29953729, 78419880, 205305912, 537497856, 1407187656, 3684065112, 9645007681, 25250957930, 66107866110
OFFSET
0,4
COMMENTS
The a(n+1) represent the Ca2 and Ze4 sums of the Golden Triangle A180662. Furthermore the a(3*n) represent the Ze1 (terms doubled) and Ca3 sums of the Golden triangle. See A180662 for more information about these and other triangle sums.
FORMULA
a(n) = -floor(g(Fibonacci(n+1)) where g(x) = (1-x^2)^2/(-4*x^2).
G.f.: x^2/( (1-x)*(1+x)*(1+x+x^2)*(1-3*x+x^2) ). - R. J. Mathar, Jun 20 2015
a(n) - a(n-2) = A182890(n-1). - R. J. Mathar, Jun 20 2015
a(n) = (1/60)*((-1)^n*(6 - 5*ChebyshevU(n, 1/2) + 10*ChebyshevU(n-1, 1/2)) - (10 - 9*ChebyshevU(n, 3/2) + 6*ChebyshevU(n-1, 3/2))). - G. C. Greubel, Jan 20 2022
a(n) = floor((2*Fibonacci(2*n+1) + Fibonacci(2*n+2) + 2)/20). - Michael Somos, Sep 05 2023
EXAMPLE
G.f. = x^2 + 2*x^3 + 6*x^4 + 16*x^5 + 42*x^6 + 110*x^7 + 289*x^8 + ... - Michael Somos, Sep 05 2023
MAPLE
nmax:=31: with(combinat): for n from 0 to nmax do A001654(n):=fibonacci(n) * fibonacci(n+1) od: a(0):=0: a(1):=0: a(2):=1: for n from 3 to nmax do a(n):=a(n-3) + A001654(n-1) od: seq(a(n), n=0..nmax);
MATHEMATICA
LinearRecurrence[{2, 2, 0, -2, -2, 1}, {0, 0, 1, 2, 6, 16}, 40] (* modified by G. C. Greubel, Jan 20 2022 *)
a[ n_] := Floor[(2*Fibonacci[2*n+1] + Fibonacci[2*n+2] + 2)/20]; (* Michael Somos, Sep 05 2023 *)
PROG
(Magma)
function A115730(n)
if n lt 3 then return Floor(n/2);
else return A115730(n-3) + Fibonacci(n-1)*Fibonacci(n);
end if; return A115730;
end function;
[A115730(n): n in [0..40]]; // G. C. Greubel, Jan 20 2022
(Sage)
U=chebyshev_U
def A115730(n): return (1/60)*((-1)^n*(6 - 5*U(n, 1/2) + 10*U(n-1, 1/2)) - (10 - 9*U(n, 3/2) + 6*U(n-1, 3/2)))
[A115730(n) for n in (0..40)] # G. C. Greubel, Jan 20 2022
(PARI) {a(n) = (2*fibonacci(2*n+1) + fibonacci(2*n+2) + 2)\20}; /* Michael Somos, Sep 05 2023 */
KEYWORD
nonn,easy
AUTHOR
Roger L. Bagula, Mar 13 2006
EXTENSIONS
Corrected and information added by Johannes W. Meijer, Sep 22 2010
Edited by Editors-in-Chief. - N. J. A. Sloane, Jun 20 2015
STATUS
approved
Golden Triangle sums: a(n)=a(n-2)+A001654(n) with a(0)=0 and a(1)=1.
+20
6
0, 1, 2, 7, 17, 47, 121, 320, 835, 2190, 5730, 15006, 39282, 102847, 269252, 704917, 1845491, 4831565, 12649195, 33116030, 86698885, 226980636, 594243012, 1555748412, 4073002212, 10663258237, 27916772486, 73087059235
OFFSET
0,3
COMMENTS
The a(n) are the Kn21, Kn22, Kn23, Fi2, and Ze2 sums of the Golden Triangle A180662. Furthermore the a(2*n) are the Kn3, Fi1 (terms doubled) and Ze3 (terms tripled) sums. See A180662 for information about these and other chess sums.
FORMULA
a(n) = a(n-2)+A001654(n) with a(0)=0 and a(1)=1.
GF(x) = (-x)/((x-1)*(x+1)^2*(x^2-3*x+1)).
a(n) = ((-1)^(-n-1)*(15+10*n)-25+(16-4*A)*A^(-n-1)+(16-4*B)*B^(-n-1))/100 with A=(3+sqrt(5))/2 and B=(3-sqrt(5))/2.
MAPLE
nmax:=27: with(combinat): for n from 0 to nmax do A001654(n):=fibonacci(n)*fibonacci(n+1) od: a(0):=0: a(1):=1: for n from 2 to nmax do a(n) := a(n-2) + A001654(n) od: seq(a(n), n=0..nmax);
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Johannes W. Meijer, Sep 21 2010
STATUS
approved
Golden Triangle sums: a(n)=a(n-4)+A001654(n) with a(0)=0, a(1)=1, a(2)=2 and a(3)=6.
+20
6
0, 1, 2, 6, 15, 41, 106, 279, 729, 1911, 5001, 13095, 34281, 89752, 234971, 615165, 1610520, 4216400, 11038675, 28899630, 75660210, 198081006, 518582802, 1357667406, 3554419410, 9305590831, 24362353076, 63781468404
OFFSET
0,3
COMMENTS
The a(n) are the Gi2 sums of the Golden Triangle A180662. See A180662 for information about these giraffe and other chess sums.
FORMULA
a(n) = a(n-4)+A001654(n) with a(0)=0, a(1)=1, a(2)=2 and a(3)=6.
G.f.: (-x)/((x^2-3*x+1)*(x-1)*(x+1)^2*(x^2+1)).
a(n) = Sum_{k=0..floor(n/4)} A180662(n-3*k,n-4*k).
120*a(n) = 8*A001519(n) -10*A087960(n) -9*(-1)^n -15 -6*(n+1)*(-1)^n. - R. J. Mathar, Aug 18 2016
MAPLE
nmax:=27: with(combinat): for n from 0 to nmax do A001654(n):=fibonacci(n)*fibonacci(n+1) od: a(0):=0: a(1):=1: a(2):=2: a(3):=6: for n from 4 to nmax do a(n):=a(n-4)+A001654(n) od: seq(a(n), n=0..nmax);
A180666 := proc(n)
option remember;
if n <=3 then
op(n+1, [0, 1, 2, 6]) ;
else
procname(n-4)+A001654(n) ;
end if;
end proc:
seq(A180666(n), n=0..100 ) ; # R. J. Mathar, Aug 18 2016
MATHEMATICA
Take[Total@{#, PadLeft[Drop[#, -4], Length@ #]}, Length@ # - 4] &@ Table[Times @@ Fibonacci@ {n, n + 1}, {n, 0, 31}] (* or *)
CoefficientList[Series[(-x)/((x^2 - 3 x + 1) (x - 1) (x + 1)^2 (x^2 + 1)), {x, 0, 27}], x] (* Michael De Vlieger, Aug 18 2016 *)
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Johannes W. Meijer, Sep 21 2010
STATUS
approved
Primes of the form A001654(k) - 2.
+20
2
13, 271, 28284463, 12610431355437389859745050577282818321295774208230808159, 421002806942919116087256468214897259565893934905313644799059599
OFFSET
1,1
COMMENTS
Generated at the indices k= 4, 7, 19, 133, 151, 205, 385, 475, 991, 1153 etc. of the golden rectangle numbers.
MATHEMATICA
q=0; lst={}; Do[f=Fibonacci[n]; If[PrimeQ[f*q-2], AppendTo[lst, f*q-2]]; q=f, {n, 6!}]; lst
Select[#-2&/@(Times@@@Partition[Fibonacci[Range[3, 300]], 2, 1]), PrimeQ] (* Harvey P. Dale, Aug 18 2011 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved
Mirror image of the Golden Triangle: T(n,k) = A001654(n-k) for n>=0 and 0<=k<=n.
+20
2
0, 1, 0, 2, 1, 0, 6, 2, 1, 0, 15, 6, 2, 1, 0, 40, 15, 6, 2, 1, 0, 104, 40, 15, 6, 2, 1, 0, 273, 104, 40, 15, 6, 2, 1, 0, 714, 273, 104, 40, 15, 6, 2, 1, 0, 1870, 714, 273, 104, 40, 15, 6, 2, 1, 0, 4895, 1870, 714, 273, 104, 40, 15, 6, 2, 1, 0
OFFSET
0,4
COMMENTS
This triangle is the mirror image of the Golden Triangle A180662. The terms in the n-th row of the triangle are the first (n+1) golden rectangle numbers in reversed order. The golden rectangle numbers are A001654(n)=F(n)*F(n+1), with F(n) the Fibonacci numbers.
The chess sums, see A180662 for their definitions, mirror those of the Golden Triangle: Row1 & Row1; Row 2 & Row2; Kn1 and Kn2; Kn3 and Kn4; Fi1 and Fi2; Ca1 and Ca2; Ca3 and Ca4; Gi1 and Gi2; Gi3 and Gi4; Ze1 and Ze2; Ze3 and Ze4.
LINKS
FORMULA
T(n,k) = F(n-k)*F(n-k+1) with F(n) = A000045(n), for n>=0 and 0<=k<=n.
EXAMPLE
The first few rows of this triangle are:
0;
1, 0;
2, 1, 0;
6, 2, 1, 0;
15, 6, 2, 1, 0;
40, 15, 6, 2, 1, 0;
MAPLE
F:= combinat[fibonacci]:
T:= (n, k)-> F(n-k)*F(n-k+1):
seq(seq(T(n, k), k=0..n), n=0..10); # revised Johannes W. Meijer, Sep 13 2012
MATHEMATICA
Module[{nn=20, fb}, fb=Times@@@Partition[Fibonacci[Range[0, (nn(nn+1))/2]], 2, 1]; Table[ Reverse[Take[fb, n]], {n, nn}]]//Flatten (* Harvey P. Dale, Jan 30 2023 *)
PROG
(Haskell)
a180663 n k = a180663_tabl !! n !! k
a180663_row n = a180663_tabl !! n
a180663_tabl = map reverse a180662_tabl
-- Reinhard Zumkeller, Jun 08 2013
CROSSREFS
Cf. A180662 (Golden Triangle), A001654 (Golden Rectangle numbers), A000045 (F(n)).
The triangle sums lead to: A064831 (Row1, Kn21, Kn22, Kn3, Ca2, Ca3, Gi2, Gi3), A077916 (Row2), A180664 (Kn23), A180665 (Kn11, Kn12, Kn13, Fi1, Ze1), A180665(2*n) (Kn4, Fi2, Ze4), A115730(n+1) (Ca1, Ze3), A115730(3*n+1) (Ca4, Ze2), A180666 (Gi1), A180666(4*n) (Gi4).
KEYWORD
easy,nonn,tabl
AUTHOR
Johannes W. Meijer, Sep 21 2010
STATUS
approved
Integers n such that A001654(n) is divisible by n*(n+1)/2.
+20
0
1, 3, 10, 23, 24, 47, 60, 107, 108, 167, 180, 240, 250, 323, 383, 503, 540, 575, 600, 647, 660, 683, 768, 863, 1008, 1103, 1200, 1223, 1320, 1367, 1620, 1728, 1800, 1860, 2160, 2207, 2447, 2520, 2687, 2688, 2736, 3000, 3023, 3060, 3300, 3360, 3527, 3528, 3744, 3863, 3888, 4200, 4703, 4800
OFFSET
1,2
COMMENTS
Odd terms of this sequence are prime most of the time. Odd composite terms of this sequence are 1, 323, 575, 6479, 7055, ...
EXAMPLE
3 is a term because (1^2 + 1^2 + 2^2) / (1 + 2 + 3) = 1.
10 is a term because (1^2 + 1^2 + 2^2 + 3^2 + 5^2 + 8^2 + 13^2 + 21^2 + 34^2 + 55^2) / (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) = 89.
MATHEMATICA
nn = 4800; Function[k, Select[Range@ nn, Divisible[k[[# + 1]], # (# + 1)/2] &]]@ LinearRecurrence[{2, 2, -1}, {0, 1, 2}, nn + 1] (* Michael De Vlieger, Mar 19 2016, after Vladimir Joseph Stephan Orlovsky at A001654 *)
PROG
(PARI) a(n) = fibonacci(n)*fibonacci(n+1);
for(n=1, 1e4, if(a(n) % (n*(n+1)/2) ==0, print1(n, ", ")));
CROSSREFS
KEYWORD
nonn
AUTHOR
Altug Alkan, Mar 17 2016
STATUS
approved
Convolution of A001654 and A007598.
+20
0
0, 0, 1, 3, 12, 38, 122, 372, 1119, 3301, 9624, 27756, 79380, 225384, 636061, 1785639, 4990116, 13889618, 38524238, 106514652, 293668923, 807608137, 2215854384, 6066935640, 16579195560, 45226399440, 123173004985, 334955873739, 909611388732, 2466965351678, 6682629071522
OFFSET
0,4
COMMENTS
Note that A001654(n) = F(n)*F(n+1) and A007598(n) = F(n)^2, for F(n) = A000045(n), the n-th Fibonacci number.
FORMULA
a(n) = Sum_{i=0..n} F(i)*F(i+1)*F(n-i)^2.
a(n) = ((n + 2)/5)*F(n)*F(n+1) - (3/25)*(F(2*n+2) + (n + 1)*(-1)^(n + 1)).
G.f.: x^2*(1-x)/((x+1)*(x^2-3*x+1))^2.
a(n) = 4*a(n-1) - 10*a(n-3) + 4*a(n-5) - a(n-6) for n > 5. - Amiram Eldar, Jan 17 2022
EXAMPLE
For n=2, a(2) = F(0)*F(1)*F(2)^2 + F(1)*F(2)*F(1)^2 + F(2)*F(3)*F(0)^2 = 1.
MATHEMATICA
Table[Sum[Fibonacci[i]*Fibonacci[i + 1]*Fibonacci[n - i]^2, {i, 0, n}], {n, 0, 30}]
PROG
(PARI) a(n) = sum(i=0, n, fibonacci(i)*fibonacci(i+1)*fibonacci(n-i)^2); \\ Michel Marcus, Jan 17 2022
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Greg Dresden, Jan 16 2022
STATUS
approved

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