A064831 -id:A064831 - cp's OEIS frontend

A180663 Mirror image of the Golden Triangle: T(n,k) = A001654(n-k) for n>=0 and 0<=k<=n.

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

Johannes W. Meijer, Sep 21 2010

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.

			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;

Reinhard Zumkeller, Rows n = 0..120 of triangle, flattened

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).

a180663 n k = a180663_tabl !! n !! k
a180663_row n = a180663_tabl !! n
a180663_tabl = map reverse a180662_tabl
-- Reinhard Zumkeller, Jun 08 2013

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

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 *)

T(n,k) = F(n-k)*F(n-k+1) with F(n) = A000045(n), for n>=0 and 0<=k<=n.

A215580 Partial sums of A215602.

Original entry on oeis.org

2, 5, 17, 45, 122, 320, 842, 2205, 5777, 15125, 39602, 103680, 271442, 710645, 1860497, 4870845, 12752042, 33385280, 87403802, 228826125, 599074577, 1568397605, 4106118242, 10749957120, 28143753122, 73681302245, 192900153617, 505019158605, 1322157322202, 3461452808000, 9062201101802, 23725150497405, 62113250390417, 162614600673845

Offset: 0

J. M. Bergot, Aug 16 2012

Dividing the terms of this sequence by Fibonacci or Lucas numbers yields symmetric sets of remainders of determinable lengths. For F(n) beginning at n=3: (a) F(2n) will have a set of remainders of length 2n in which the sum of the remainders is 3*(F(2n)-n). Example for F(2*6)=144: the set of remainders is {2,5,17,45,122,32,122,45,17,5,2,0} with 2*6=12 terms and a sum of 3*(144-6)=414. (b) For F(2n+1) there will be 2*(2n+1) terms having a sum equal to (2n+1)*(F(2n+1)-3). Example for F(2*4+1)=34: the remainders are {2,5,7,11,20,14,26,29,31,29,26,14,20,11,17,5,2,0} with 2*9 terms and a sum of 9*(34-1)=279.

Using Lucas numbers starting at n=2: (a) L(2n) has 4n remainders with sum (2n+1)*(L(2n)-6*n). Example for n=4 giving L(2*4)=47, has remainders {2,5,17,45,28,38,43,43,43,38,28,45,17,5,2,0} with a sum of (8+1)*(47)-6*4=399. (B) For L(2n+1) the length of the period is 2*(2n+1) and the sum of the remainders is 4*L(2n+1)-3*(2n+1). Example for n=3 for L(2*3+1)=29 has remainders {2,5,17,16,6,1,11,6,16,17,5,2,0} with length 2*7 and sum of terms 4*29-3*7=95.

Paolo Xausa, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,0,-3,1).

Cf. A000032, A075269, A064831, A215602.

Cf. A005248, A000034.

LinearRecurrence[{3, 0, -3, 1}, {2, 5, 17, 45}, 35] (* Paolo Xausa, Feb 22 2024 *)

a(2n) = L(4*n)-2, a(2*n+1) = L(4*n+2)-1, where L() are the Lucas numbers A000032.
G.f. ( -2+x-2*x^2 ) / ( (x-1)*(1+x)*(x^2-3*x+1) ). - R. J. Mathar, Aug 21 2012
a(n) = A005248(n+1)-A000034(n). - R. J. Mathar, Aug 21 2012

Edited by N. J. A. Sloane, Aug 17 2012

A230448 T(n, k) = T(n-1, k-1) + T(n-1, k) with T(n, 0) = 1 and T(n, n) = A226205(n+1), n >= 0 and 0 <= k <= n.

Original entry on oeis.org

1, 1, 0, 1, 1, 3, 1, 2, 4, 5, 1, 3, 6, 9, 16, 1, 4, 9, 15, 25, 39, 1, 5, 13, 24, 40, 64, 105, 1, 6, 18, 37, 64, 104, 169, 272, 1, 7, 24, 55, 101, 168, 273, 441, 715, 1, 8, 31, 79, 156, 269, 441, 714, 1156, 1869, 1, 9, 39, 110, 235, 425, 710, 1155, 1870, 3025, 4896

Offset: 0

Johannes W. Meijer, Oct 19 2013

Triangle T(n, k) is related to the Kn1p sums of the ‘Races with Ties’ triangle A035317. See A230447 for the Kn1p sums and A180662 for the definitions of these sums.
The row sums equal ((-1)^n*3*A083581(n) + A022379(2*n+2))/15.
Note that the partial fraction expansion of the G.f. of the terms in the n-th row of the square array Tsq(n, k) = T(n+k, k) is related to A014334, the exponential convolution of the Fibonacci numbers with themselves, and to A000032, the Lucas numbers.

			The first few rows of triangle T(n, k), n >= 0 and 0 <= k <= n.
n/k 0   1   2    3    4     5     6     7
------------------------------------------------
0|  1
1|  1,  0
2|  1,  1,  3
3|  1,  2,  4,   5
4|  1,  3,  6,   9,  16
5|  1,  4,  9,  15,  25,   39
6|  1,  5, 13,  24,  40,   64,  105
7|  1,  6, 18,  37,  64,  104,  169,   272
The triangle as a square array Tsq(n, k) = T(n+k, k), n >= 0 and k >= 0.
n/k 0   1   2    3    4    5      6     7
------------------------------------------------
0|  1,  0,  3,   5,  16,  39,   105,  272
1|  1,  1,  4,   9,  25,  64,   169,  441
2|  1,  2,  6,  15,  40,  104,  273,  714
3|  1,  3,  9,  24,  64,  168,  441, 1155
4|  1,  4, 13,  37, 101,  269,  710, 1865
5|  1,  5, 18,  55, 156,  425, 1135, 3000
6|  1,  6, 24,  79, 235,  660, 1795, 4795
7|  1,  7, 31, 110, 345, 1005, 2800, 7595

Cf. (Triangle columns) A000012, A001477, A152950, A226205, A007598, A001654, A064831, A080145

T := proc(n, k) option remember: if k=0 then return(1) elif k=n then return(combinat[fibonacci](n+2)*combinat[fibonacci](n-1)) else procname(n-1, k-1) + procname(n-1, k) fi: end: seq(seq(T(n, k), k=0..n), n=0..10); # End first program.
T := proc(n, k): add(A035317(n+k-p-2, p), p=0..k) end: A035317 := proc(n, k): add((-1)^(i+k) * binomial(i+n-k+1, i), i=0..k) end: seq(seq(T(n, k), k=0..n), n=0..10); # End second program.

T(n, k) = T(n-1, k-1) + T(n-1, k) with T(n, 0) = 1 and T(n, n) = F(n+2) * F(n-1) = A226205(n+1) with F(n) = A000045(n), the Fibonacci numbers, n >= 0 and 0 <= k <= n.
T(n, k) = sum(A035317(n+k-p-2, p), p=0..k), n >= 0 and 0 <= k <= n.
T(n+p+2, p-2) = A080239(n+2*p-1) - sum(A035317(n-k+p-1, k+p-1), k=0..floor(n/2)), n >= 0 and p >= 2.
The triangle as a square array Tsq(n, k) = T(n+k, k), n >= 0 and k >= 0.
Tsq(n, k) = sum(Tsq(n-1, i), i=0..k), n >= 1 and k >= 0, with Tsq(0, k) = A226205(k+1).
The two G.f.’s given below generate the terms in the n-th row of the square array Tsq(n, k). The remarkable second G.f. is the partial fraction expansion of the first G.f..
G.f.: 1/((1-x)^(n-2)*(1+x)*(x^2-3*x+1)), n >= 0.
G.f.: sum((-1)^(n+k-1)*A014334(k+2)/(2^(k+2)*(x-1)^(n-k-2)), k=0..n-3) + 1/(5*2^(n-2)*(1+x)) + (A000032(n+1) - A000032(n-1)*x)/(5*(x^2-3*x+1)), n >= 0.

A214729 Member m=6 of the m-family of sums b(m,n) = Sum_{k=0..n} F(k+m)*F(k), m >= 0, n >= 0, with the Fibonacci numbers F.

Original entry on oeis.org

0, 13, 34, 102, 267, 712, 1864, 4893, 12810, 33550, 87835, 229968, 602064, 1576237, 4126642, 10803702, 28284459, 74049688, 193864600, 507544125, 1328767770, 3478759198, 9107509819, 23843770272, 62423800992, 163427632717, 427859097154, 1120149658758

Offset: 0

Wolfdieter Lang, Jul 27 2012

See the comment section on A080144 for the general formula and the o.g.f. for b(m,n).

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,0,-3,1).

Cf. A001654, A027941, A059840(n+2), A064831, A080097, A080143 and A080144 for the m=0,1,...,5 members.

Cf. A027941.

[(9*(-1)^(n+1)-20+Lucas(2*n+7))/5: n in [0..40]]; // Vincenzo Librandi, Aug 26 2017

With[{m = 6}, Table[Sum[Fibonacci[k + m]*Fibonacci[k], {k, 0, n}], {n, 0, 25}]] (* or *)
Table[(9 (-1)^(n + 1) - 20 + LucasL[2 n + 7])/5, {n, 0, 25}] (* Michael De Vlieger, Aug 23 2017 *)
LinearRecurrence[{3,0,-3,1},{0,13,34,102},40] (* Harvey P. Dale, Jun 13 2022 *)

concat(0, Vec(x*(13 - 5*x) / ((1 - x)*(1 + x)*(1 - 3*x + x^2)) + O(x^30))) \\ Colin Barker, Aug 25 2017

[fibonacci(n+3)*fibonacci(n+4) - 2*(2+(-1)^n) for n in range(41)] # G. C. Greubel, Dec 31 2023

a(n) = b(6,n) = 4*A027941(n) + 9*A001654(n), with A027941(n) = Fibonacci(2*n+1) - 1 and A001654(n) = Fibonacci(n+1)*Fibonacci(n), n >= 0. 4 = Fibonacci(6)/2 and 9 = LucasL(6)/2.
O.g.f.: x*(13-5*x)/((1-x^2)*(1-3*x+x^2)) (see a comment above). - Wolfdieter Lang, Jul 30 2012
a(n) = (9*(-1)^(n+1) - 20 + Lucas(2*n + 7))/5. - Ehren Metcalfe, Aug 21 2017
From Colin Barker, Aug 25 2017: (Start)
a(n) = (1/10)*((29 - 13*sqrt(5))*((3 - sqrt(5))/2)^n + (29 + 13*sqrt(5))*((3 + sqrt(5))/2)^n - 2*(20 + 9*(-1)^n) ).
a(n) = 2*a(n-1) + 2*a(n-2) + 2*a(n-3) - a(n-4) for n>3. (End)
a(n) = A001654(n+3) - 2*(2 + (-1)^n). - G. C. Greubel, Dec 31 2023

A216243 Partial sums of the squares of Lucas numbers (A000032).

Original entry on oeis.org

4, 5, 14, 30, 79, 200, 524, 1365, 3574, 9350, 24479, 64080, 167764, 439205, 1149854, 3010350, 7881199, 20633240, 54018524, 141422325, 370248454, 969323030, 2537720639, 6643838880, 17393796004, 45537549125, 119218851374, 312119004990, 817138163599, 2139295485800

Offset: 0

R. J. Mathar, Mar 14 2013

Index entries for linear recurrences with constant coefficients, signature (3,0,-3,1).

Cf. A001654.

A001254 := proc(n)
        A000032(n)^2 ;
end proc;
A := proc(n)
        add( A001254(i),i=0..n) ;
end proc:

Accumulate[LucasL[Range[0,30]]^2] (* or *) LinearRecurrence[{3,0,-3,1},{4,5,14,30},30] (* Harvey P. Dale, Oct 13 2019 *)

a(n) = Sum_{i=0..n} A001254(i) = A002878(n) +A176040(n) = A215602(n)+2.
G.f.: ( -4+7*x+x^2 ) / ( (x-1)*(1+x)*(x^2-3*x+1) ).
a(n) = -7*A064831(n) -A064831(n-1) +4*A064831(n+1).
a(n) = L(2*n+1) + 2 + (-1)^n, for L(n) the Lucas sequence A000032(n). - Greg Dresden, Jan 26 2021

A110032 A characteristic triangle for the Euler totient function (A000010).

Original entry on oeis.org

1, -1, 1, 1, -3, 1, 0, 5, -7, 1, -4, 22, 23, -13, 1, -64, 20, 172, 29, -19, 1, -240, -1120, -496, 354, 111, -29, 1, 2656, 1760, -4952, -1816, 1054, 239, -41, 1, 15360, 24800, -28640, -29040, 2384, 3938, 203, -49, 1, 88064, 587136, 601216, -63776, -191224, -21360, 6658, 511, -65, 1, -1714176, -4244224, 4907392

Offset: 0

Paul Barry, Jul 08 2005

Form the n X n Hankel matrices phi(i-j+1), 1<=i,j<=n for the Euler totient function and take characteristic polynomials. Triangle rows give coefficients of the characteristic polynomials of these matrices. (Construction described by Michael Somos in A064831). First column is a signed version of Hankel transform of Euler totient function, A056229.

			Rows begin
1;
-1,1;
1,-3,1;
0,5,-7,1;
-4,22,23,-13,1;
-64,20,172,29,-19,1;
-240,-1120,-496,354,111,-29,1;

cp's OEIS Frontend

A180663 Mirror image of the Golden Triangle: T(n,k) = A001654(n-k) for n>=0 and 0<=k<=n.

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A215580 Partial sums of A215602.

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A230448 T(n, k) = T(n-1, k-1) + T(n-1, k) with T(n, 0) = 1 and T(n, n) = A226205(n+1), n >= 0 and 0 <= k <= n.

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A214729 Member m=6 of the m-family of sums b(m,n) = Sum_{k=0..n} F(k+m)*F(k), m >= 0, n >= 0, with the Fibonacci numbers F.

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A216243 Partial sums of the squares of Lucas numbers (A000032).

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A110032 A characteristic triangle for the Euler totient function (A000010).

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