A035513 Wythoff array read by falling antidiagonals.
1, 2, 4, 3, 7, 6, 5, 11, 10, 9, 8, 18, 16, 15, 12, 13, 29, 26, 24, 20, 14, 21, 47, 42, 39, 32, 23, 17, 34, 76, 68, 63, 52, 37, 28, 19, 55, 123, 110, 102, 84, 60, 45, 31, 22, 89, 199, 178, 165, 136, 97, 73, 50, 36, 25, 144, 322, 288, 267, 220, 157, 118, 81, 58, 41, 27, 233, 521
Offset: 1
A022086 Fibonacci sequence beginning 0, 3.
0, 3, 3, 6, 9, 15, 24, 39, 63, 102, 165, 267, 432, 699, 1131, 1830, 2961, 4791, 7752, 12543, 20295, 32838, 53133, 85971, 139104, 225075, 364179, 589254, 953433, 1542687, 2496120, 4038807, 6534927, 10573734, 17108661, 27682395, 44791056, 72473451, 117264507
Offset: 0
Comments
First differences of A111314. - Ross La Haye, May 31 2006
Pisano period lengths: 1, 3, 1, 6, 20, 3, 16, 12, 8, 60, 10, 6, 28, 48, 20, 24, 36, 24, 18, 60, ... . - R. J. Mathar, Aug 10 2012
For n>=6, a(n) is the number of edge covers of the union of two cycles C_r and C_s, r+s=n, with a single common vertex. - Feryal Alayont, Oct 17 2024
References
- A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 7,17.
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- Tanya Khovanova, Recursive Sequences
- Index entries for linear recurrences with constant coefficients, signature (1,1).
Crossrefs
Programs
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Magma
[3*Fibonacci(n): n in [0..40]]; // Vincenzo Librandi, Dec 31 2016
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Maple
BB := n->if n=0 then 3; > elif n=1 then 0; > else BB(n-2)+BB(n-1); > fi: > L:=[]: for k from 1 to 34 do L:=[op(L),BB(k)]: od: L; # Zerinvary Lajos, Mar 19 2007 with (combinat):seq(sum((fibonacci(n,1)),m=1..3),n=0..32); # Zerinvary Lajos, Jun 19 2008
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Mathematica
LinearRecurrence[{1, 1}, {0, 3}, 40] (* Arkadiusz Wesolowski, Aug 17 2012 *) Table[Fibonacci[n + 4] + Fibonacci[n - 4] - 4 Fibonacci[n], {n, 0, 40}] (* Bruno Berselli, Dec 30 2016 *) Table[3 Fibonacci[n], {n, 0, 40}] (* Vincenzo Librandi, Dec 31 2016 *) -
PARI
a(n)=3*fibonacci(n) \\ Charles R Greathouse IV, Nov 06 2014
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SageMath
def A022086(n): return 3*fibonacci(n) print([A022086(n) for n in range(41)]) # G. C. Greubel, Apr 10 2025
Formula
a(n) = 3*Fibonacci(n).
a(n) = F(n-2) + F(n+2) for n>1, with F=A000045.
a(n) = round( ((6*phi-3)/5) * phi^n ) for n>2. - Thomas Baruchel, Sep 08 2004
a(n) = A119457(n+1,n-1) for n>1. - Reinhard Zumkeller, May 20 2006
G.f.: 3*x/(1-x-x^2). - Philippe Deléham, Nov 19 2008
a(n) = A187893(n) - 1. - Filip Zaludek, Oct 29 2016
E.g.f.: 6*sinh(sqrt(5)*x/2)*exp(x/2)/sqrt(5). - Ilya Gutkovskiy, Oct 29 2016
a(n) = F(n+4) + F(n-4) - 4*F(n), F = A000045. - Bruno Berselli, Dec 29 2016
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.
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
Examples
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;
Links
- G. C. Greubel, Rows n = 1..50 of the triangle, flattened
- Eric Weisstein's World of Mathematics, Fibonacci Number
Crossrefs
Main diagonal: A023607(n).
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).
Programs
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Magma
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
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Mathematica
(* 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 *) -
SageMath
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
Formula
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).
A280154 a(n) = 5*Lucas(n).
10, 5, 15, 20, 35, 55, 90, 145, 235, 380, 615, 995, 1610, 2605, 4215, 6820, 11035, 17855, 28890, 46745, 75635, 122380, 198015, 320395, 518410, 838805, 1357215, 2196020, 3553235, 5749255, 9302490, 15051745, 24354235, 39405980, 63760215, 103166195, 166926410, 270092605, 437019015
Offset: 0
Comments
Fibonacci sequence beginning 10, 5.
After 5, the sequence provides the 3rd column of the rectangular array in A213590.
After 5, all terms belong to A191921 because a(n) = Lucas(n+4) - 3*Lucas(n-1).
From G. C. Greubel, Dec 27 2016: (Start)
{a(n) mod 3} yields (1,2,0,2,2,1,0,1), repeated, and is given as A082115.
{a(n) mod 6} yields (4,5,3,2,5,1,0,1,1,2,3,5,2,1,3,4,1,5,0,5,5,4,3,1) and is given as A082117. (End)
Links
- Bruno Berselli, Table of n, a(n) for n = 0..1000
- Tanya Khovanova, Recursive Sequences
- Index entries for linear recurrences with constant coefficients, signature (1,1).
Crossrefs
Programs
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Magma
[5*Lucas(n): n in [0..40]];
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Maple
F := n -> combinat:-fibonacci(n): seq(F(n+5) + F(n-5), n=0..38); # Peter Luschny, Dec 29 2016
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Mathematica
Table[5 LucasL[n], {n, 0, 40}] -
PARI
vector(40, n, n--; fibonacci(n+5)+fibonacci(n-5))
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Sage
def A280154(): x, y = 10, 5 while True: yield x x, y = y, x + y a = A280154(); print([next(a) for in range(39)]) # _Peter Luschny, Dec 29 2016
Formula
G.f.: 5*(2 - x)/(1 - x - x^2).
a(n) = a(n-1) + a(n-2) for n>1.
a(n) = Fibonacci(n+5) + Fibonacci(n-5), with Fibonacci(-k) = -(-1)^k*Fibonacci(k) for the negative indices.
A294157 Fibonacci sequence beginning 2, 8.
2, 8, 10, 18, 28, 46, 74, 120, 194, 314, 508, 822, 1330, 2152, 3482, 5634, 9116, 14750, 23866, 38616, 62482, 101098, 163580, 264678, 428258, 692936, 1121194, 1814130, 2935324, 4749454, 7684778, 12434232, 20119010, 32553242, 52672252, 85225494, 137897746, 223123240
Offset: 0
References
- Steven Vajda, Fibonacci and Lucas Numbers, and the Golden Section: Theory and Applications, Dover Publications (2008), page 24 (formula 8).
Links
- Colin Barker, Table of n, a(n) for n = 0..1000
- Tanya Khovanova, Recursive Sequences.
- Index entries for linear recurrences with constant coefficients, signature (1,1).
Programs
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Magma
a0:=2; a1:=8; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]];
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Maple
f:= gfun:-rectoproc({a(n)=a(n-1)+a(n-2),a(0)=2,a(1)=8},a(n),remember): map(f, [$0..100]); # Robert Israel, Oct 24 2017 -
Mathematica
LinearRecurrence[{1, 1}, {2, 8}, 40] -
PARI
Vec(2*(1 + 3*x)/(1 - x - x^2) + O(x^40)) \\ Colin Barker, Oct 25 2017
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Sage
a = BinaryRecurrenceSequence(1, 1, 2, 8) print([a(n) for n in range(38)]) # Peter Luschny, Oct 25 2017
Formula
G.f.: 2*(1 + 3*x)/(1 - x - x^2).
a(n) = a(n-1) + a(n-2).
a(n) = 2*A000285(n).
Let g(r,s;n) be the n-th generalized Fibonacci number with initial values r, s. We have:
a(n) = Lucas(n) + g(0,7;n), see A022090;
a(n) = Fibonacci(n) + g(2,7;n), see A022113;
a(n) = 2*g(1,8;n) - g(0,8;n);
a(n) = g(1,k;n) + g(1,8-k;n) for all k in Z.
a(h+k) = a(h)*Fibonacci(k-1) + a(h+1)*Fibonacci(k) for all h, k in Z (see S. Vajda in References section). For h=0 and k=n:
a(n) = 2*Fibonacci(n-1) + 8*Fibonacci(n).
Sum_{j=0..n} a(j) = a(n+2) - 8.
a(n) = (2^(-n)*((1-sqrt(5))^n*(-7+sqrt(5)) + (1+sqrt(5))^n*(7+sqrt(5)))) / sqrt(5). - Colin Barker, Oct 25 2017
A168622 Triangle read by rows: T(n, k) = [x^k]( 7*(1+x)^n - 6*(1+x^n) ) with T(0, 0) = 1.
1, 1, 1, 1, 14, 1, 1, 21, 21, 1, 1, 28, 42, 28, 1, 1, 35, 70, 70, 35, 1, 1, 42, 105, 140, 105, 42, 1, 1, 49, 147, 245, 245, 147, 49, 1, 1, 56, 196, 392, 490, 392, 196, 56, 1, 1, 63, 252, 588, 882, 882, 588, 252, 63, 1, 1, 70, 315, 840, 1470, 1764, 1470, 840, 315, 70, 1
Offset: 0
Examples
Triangle begins as: 1; 1, 1; 1, 14, 1; 1, 21, 21, 1; 1, 28, 42, 28, 1; 1, 35, 70, 70, 35, 1; 1, 42, 105, 140, 105, 42, 1; 1, 49, 147, 245, 245, 147, 49, 1; 1, 56, 196, 392, 490, 392, 196, 56, 1; 1, 63, 252, 588, 882, 882, 588, 252, 63, 1; 1, 70, 315, 840, 1470, 1764, 1470, 840, 315, 70, 1;
Links
- G. C. Greubel, Rows n = 0..50 of the triangle, flattened
Crossrefs
Programs
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Magma
A168622:= func< n,k | k eq 0 or k eq n select 1 else 7*Binomial(n,k) >; [A168622(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Apr 10 2025
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Mathematica
(* First program *) p[x_, n_]:= With[{m=3}, If[n==0, 1, (2*m+1)(1+x)^n - 2*m*(1+x^n)]]; Table[CoefficientList[p[x,n], x], {n,0,12}]//Flatten (* Second program *) A168622[n_, k_]:= If[k==0 || k==n, 1, 7*Binomial[n,k]]; Table[A168622[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Apr 10 2025 *) -
SageMath
def A168622(n,k): if k==0 or k==n: return 1 else: return 7*binomial(n,k) print(flatten([[A168622(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Apr 10 2025
Formula
From G. C. Greubel, Apr 10 2025: (Start)
T(n, k) = 7*binomial(n, k), with T(n, 0) = T(n, n) = 1.
T(n, n-k) = T(n, k).
Sum_{k=0..n} T(n, k) = 2*A048489(n-1) + 6*[n=0].
Sum_{k=0..n} (-1)^k*T(n, k) = -6*(1 + (-1)^n) + 13*[n=0].
Sum_{k=0..floor(n/2)} T(n-k, k) = A022090(n+1) - 3*(3 + (-1)^n) + 6*[n=0].
Sum_{k=0..floor(n/2)} (-1)^k*T(n-k, k) = (14/sqrt(3))*(-1)^n*cos((4*n+1)*Pi/6) - 6*(1 + (-1)^n*cos(n*Pi/2)) + 6*[n=0]. (End)
Comments
Examples
The Wythoff array begins: 1 2 3 5 8 13 21 34 55 89 144 ... 4 7 11 18 29 47 76 123 199 322 521 ... 6 10 16 26 42 68 110 178 288 466 754 ... 9 15 24 39 63 102 165 267 432 699 1131 ... 12 20 32 52 84 136 220 356 576 932 1508 ... 14 23 37 60 97 157 254 411 665 1076 1741 ... 17 28 45 73 118 191 309 500 809 1309 2118 ... 19 31 50 81 131 212 343 555 898 1453 2351 ... 22 36 58 94 152 246 398 644 1042 1686 2728 ... 25 41 66 107 173 280 453 733 1186 1919 3105 ... 27 44 71 115 186 301 487 788 1275 2063 3338 ... ... The extended Wythoff array has two extra columns, giving the row number n and A000201(n), separated from the main array by a vertical bar: 0 1 | 1 2 3 5 8 13 21 34 55 89 144 ... 1 3 | 4 7 11 18 29 47 76 123 199 322 521 ... 2 4 | 6 10 16 26 42 68 110 178 288 466 754 ... 3 6 | 9 15 24 39 63 102 165 267 432 699 1131 ... 4 8 | 12 20 32 52 84 136 220 356 576 932 1508 ... 5 9 | 14 23 37 60 97 157 254 411 665 1076 1741 ... 6 11 | 17 28 45 73 118 191 309 500 809 1309 2118 ... 7 12 | 19 31 50 81 131 212 343 555 898 1453 2351 ... 8 14 | 22 36 58 94 152 246 398 644 1042 1686 2728 ... 9 16 | 25 41 66 107 173 280 453 733 1186 1919 3105 ... 10 17 | 27 44 71 115 186 301 487 788 1275 2063 3338 ... 11 19 | 30 49 79 ... 12 21 | 33 54 87 ... 13 22 | 35 57 92 ... 14 24 | 38 62 ... 15 25 | 40 65 ... 16 27 | 43 70 ... 17 29 | 46 75 ... 18 30 | 48 78 ... 19 32 | 51 83 ... 20 33 | 53 86 ... 21 35 | 56 91 ... 22 37 | 59 96 ... 23 38 | 61 99 ... 24 40 | 64 ... 25 42 | 67 ... 26 43 | 69 ... 27 45 | 72 ... 28 46 | 74 ... 29 48 | 77 ... 30 50 | 80 ... 31 51 | 82 ... 32 53 | 85 ... 33 55 | 88 ... 34 56 | 90 ... 35 58 | 93 ... 36 59 | 95 ... 37 61 | 98 ... 38 63 | ... ... Each row of the extended Wythoff array also satisfies the Fibonacci recurrence, and may be extended to the left using this recurrence backwards. From _Peter Munn_, Jun 11 2021: (Start) The Wythoff array appears to have the following relationship to the traditional Fibonacci rabbit breeding story, modified for simplicity to be a story of asexual reproduction. Give each rabbit a number, 0 for the initial rabbit. When a new round of rabbits is born, allocate consecutive numbers according to 2 rules (the opposite of many cultural rules for inheritance precedence): (1) newly born child of Rabbit 0 gets the next available number; (2) the descendants of a younger child of any given rabbit precede the descendants of an older child of the same rabbit. Row n of the Wythoff array lists the children of Rabbit n (so Rabbit 0's children have the Fibonacci numbers: 1, 2, 3, 5, ...). The generation tree below shows rabbits 0 to 20. It is modified so that each round of births appears on a row. 0 : ,-------------------------: : : ,---------------: 1 : : : ,--------: 2 ,---------: : : : : : ,-----: 3 ,-----: ,-----: 4 : : : : : : : : ,--: 5 ,--: ,---: 6 ,---: 7 ,---: : : : : : : : : : : : : : ,--: 8 ,--: ,--: 9 ,--: 10 ,--: ,--: 11 ,--: ,--: 12 : : : : : : : : : : : : : : : : : : : : : : 13 : : 14 : 15 : : 16 : : 17 : 18 : : 19 : 20 : The extended array's nontrivial extra column (A000201) gives the number that would have been allocated to the first child of Rabbit n, if Rabbit n (and only Rabbit n) had started breeding one round early. (End)References
Links
Crossrefs
Programs
Maple
Mathematica
W[n_, k_] := Fibonacci[k + 1] Floor[n*GoldenRatio] + (n - 1) Fibonacci[k]; Table[ W[n - k + 1, k], {n, 12}, {k, n, 1, -1}] // FlattenPARI
Python
Python
Formula
Extensions