A006327 -id:A006327 - cp's OEIS frontend

A157728 a(n) = Fibonacci(n) - 4.

1, 4, 9, 17, 30, 51, 85, 140, 229, 373, 606, 983, 1593, 2580, 4177, 6761, 10942, 17707, 28653, 46364, 75021, 121389, 196414, 317807, 514225, 832036, 1346265, 2178305, 3524574, 5702883, 9227461, 14930348, 24157813, 39088165, 63245982, 102334151

Offset: 5

N. J. A. Sloane, Jun 26 2010

Partial sums of A071679. - R. J. Mathar, Oct 12 2010

Vincenzo Librandi, Table of n, a(n) for n = 5..287
Paul Barry, The Triple Riordan Group, arXiv:2412.05461 [math.CO], 2024. See pp. 5, 10.
Index entries for linear recurrences with constant coefficients, signature (2,0,-1).

Cf. A000045, A001611, A000071, A157725, A001911, A157726, A006327, A157727, A157728, A157729, A167616.

a157728 = subtract 4 . a000045  -- Reinhard Zumkeller, Oct 31 2013

[ Fibonacci(n) - 4: n in [5..50] ]; // Vincenzo Librandi, Apr 24 2011

Fibonacci[Range[5,40]]-4 (* or *) LinearRecurrence[{2,0,-1},{1,4,9},40] (* Harvey P. Dale, Jan 17 2022 *)

a(n)=fibonacci(n) - 4 \\ Charles R Greathouse IV, Jun 11 2015

From R. J. Mathar, Oct 12 2010: (Start)
a(n) = 2*a(n-1) - a(n-3).
G.f.: x^5*(1+x)^2/((x-1)*(x^2+x-1)). (End)

A157729 a(n) = Fibonacci(n) + 5.

Original entry on oeis.org

5, 6, 6, 7, 8, 10, 13, 18, 26, 39, 60, 94, 149, 238, 382, 615, 992, 1602, 2589, 4186, 6770, 10951, 17716, 28662, 46373, 75030, 121398, 196423, 317816, 514234, 832045, 1346274, 2178314, 3524583, 5702892, 9227470, 14930357, 24157822, 39088174, 63245991, 102334160

Offset: 0

N. J. A. Sloane, Jun 26 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..285
Ivana Jovović and Branko Malešević, Some enumerations of non-trivial composition of the differential operations and the directional derivative, Notes on Number Theory and Discrete Mathematics, Vol. 23, 2017, No. 1, 28-38.
Index entries for linear recurrences with constant coefficients, signature (2,0,-1).

Cf. A000045, A001611, A000071, A157725, A001911, A157726, A006327, A157727, A157728, A157729, A167616.

a157729 = (+ 5) . a000045
a157729_list = 5 : 6 : map (subtract 5)
                       (zipWith (+) a157729_list $ tail a157729_list)
-- Reinhard Zumkeller, Jul 30 2013

[ Fibonacci(n) + 5: n in [0..40] ]; // Vincenzo Librandi, Apr 24 2011

Fibonacci[Range[0,40]]+5 (* or *) LinearRecurrence[{2,0,-1},{5,6,6},50] (* Harvey P. Dale, Aug 17 2012 *)

a(n)=fibonacci(n)+5 \\ Charles R Greathouse IV, Jul 02 2013

G.f.: ( 5-4*x-6*x^2 ) / ( (x-1)*(x^2+x-1) ). - R. J. Mathar, Aug 09 2012
a(0)=5, a(1)=6, a(2)=6, a(n)=2*a(n-1)+0*a(n-2)-a(n-3). - Harvey P. Dale, Aug 17 2012
a(0) = 5, a(1) = 6, a(n) = a(n - 2) + a(n - 1) - 5. - Reinhard Zumkeller, Jul 30 2013

A255870 a(n) is the total number of pentagrams in a pentagram fractal after n iterations.

Original entry on oeis.org

1, 6, 26, 76, 191, 411, 816, 1521, 2726, 4741, 8081, 13566, 22536, 37146, 60896, 99436, 161921, 263151, 427086, 692481, 1122056, 1817281, 2942351, 4762926, 7708866, 12475686, 20188766, 32668996, 52862651, 85536891, 138405156, 223948041, 362359586, 586314421

Offset: 0

Kival Ngaokrajang, Mar 08 2015

Inspired by the "calice" (see detail in the CNRS link).

The pentagrams appearing in the calice are scaled down by a factor of 1/phi = 0.61803398... from the pentagrams whose vertex-to-vertex length = d. See illustration of the nested pentagrams in the links.

Colin Barker, Table of n, a(n) for n = 0..1000
Shalom Eliahou, Pavages, symétrie d'ordre 5 et Fibonacci: un amateur passionné, Images des Mathématiques, CNRS, 2015 (in French).
Frédéric Mansuy, Supersymétrie d'ordre cinq périodique (in French).
Kival Ngaokrajang, Illustration of initial terms, Excel calculation sheet
Index entries for linear recurrences with constant coefficients, signature (3,-1,-4,3,1,-1).

Cf. A000071, A006327, A249852, A253688.

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I:=[1,6,26,76,191,411,816]; [n le 7 select I[n] else 3*Self(n-1)-Self(n-2)-4*Self(n-3)+3*Self(n-4)+Self(n-5)-Self(n-6): n in [1..40]]; // Vincenzo Librandi, Mar 18 2015

CoefficientList[Series[(5 x^6 + x^5 - 10 x^4 - 8 x^3 - 9 x^2 - 3 x - 1)/((1-x)^3 (x+1) (x^2+x-1)), {x, 0, 40}], x] (* Vincenzo Librandi, Mar 18 2015 *)

Vec(-(5*x^6+x^5-10*x^4-8*x^3-9*x^2-3*x-1)/((x-1)^3*(x+1)*(x^2+x-1)) + O(x^100)) \\ Colin Barker, Mar 12 2015

a(n) = 3*a(n-1)-a(n-2)-4*a(n-3)+3*a(n-4)+a(n-5)-a(n-6) for n>6. - Colin Barker, Mar 12 2015
G.f.: -(5*x^6+x^5-10*x^4-8*x^3-9*x^2-3*x-1) / ((x-1)^3*(x+1)*(x^2+x-1)). - Colin Barker, Mar 12 2015
a(n) = (-307 + 5*(-1)^n - 3*2^(2-n)*s*((11-5*s)*(1-s)^n - (1+s)^n*(11+5*s)) - 150*(1+n) - 50*(1+n)*(2+n)) / 8 for n>0 where s=sqrt(5). - Colin Barker, Mar 12 2017
E.g.f.: 3*exp(x/2)*(25*cosh(sqrt(5)*x/2) + 11*sqrt(5)*sinh(sqrt(5)*x/2)) - (12 + 5*x)*(23 + 5*x)*cosh(x)/4 - (281 + 25*x*(7 + x))*sinh(x)/4 - 5. - Stefano Spezia, Dec 07 2022

A079502 Triangle T(n,k) read by rows; related to number of preorders.

Original entry on oeis.org

1, 1, 2, 1, 5, 5, 1, 10, 24, 16, 1, 18, 79, 122, 61, 1, 31, 223, 602, 680, 272, 1, 52, 579, 2439, 4682, 4155, 1385, 1, 86, 1432, 8856, 25740, 38072, 27776, 7936, 1, 141, 3434, 30030, 124146, 272416, 326570, 202084, 50521, 1, 230, 8071, 97332

Offset: 0

N. J. A. Sloane, Jan 21 2003

			Triangle T(n,k) begins:
  1;
  1,  2;
  1,  5,    5;
  1, 10,   24,  16;
  1, 18,   79,  122,    61;
  1, 31,  223,  602,   680,   272;
  1, 52,  579, 2439,  4682,  4155,  1385;
  1, 86, 1432, 8856, 25740, 38072, 27776, 7936;

Sean A. Irvine, Table of n, a(n) for n = 0..999
Germain Kreweras, Les préordres totaux compatibles avec un ordre partiel, Math. Sci. Humaines No. 53 (1976), 5-30. (The numbers u_r^n on page 20.)
Germain Kreweras, Les préordres totaux compatibles avec un ordre partiel, Math. Sci. Humaines No. 53 (1976), 5-30. (Annotated scanned copy)
Igor Pak, Boris Shapiro, Ilya Smirnov, and Ken-ichi Yoshida, Hilbert-Kunz multiplicity of quadrics via the Ehrhart theory, Stockholm Univ. (Sweden, 2025). See pp. 5, 10.

Diagonals give A000111, A006326, A006327, A006328. Cf. A050447.

t[n_, m_] := t[n, m] = If[m == 0, 1, t[n, m - 1] + Sum[t[2 k, m - 1] t[n - 1 - 2 k, m], {k, 0, (n - 1)/2}]]; Map[Function[s, Rest@ Reverse@ Map[Abs@ Fold[#2 - #1 &, Reverse@ Take[s, #]] &, Range@ Length@ s]]@ Reverse@ Map[First, NestList[Differences@ # &, {First@ #}~Join~Differences@ #, Length@ # - 2]] &, Table[t[n, k], {n, 2, 11}, {k, 0, n}]] (* Michael De Vlieger, Mar 13 2017, after Jean-François Alcover at A050447 *)

From Sean A. Irvine, Mar 12 2017: (Start)
A079502 can be constructed one row at a time from the corresponding row of A050447. For row n, construct up to the n-th difference sequence of row n in A050447, retaining the first element of each difference sequence. Row n of A079502 is then constructed backwards (i.e., starting with A079502(n,n) and computing down to A079502(n,2)) from the first element of the n-th difference sequence, then successively subtracting the first element of the previous difference sequences. More precisely, let R_n denote the n-th row of A050447 augmented with R_n(1) = 0, and R_n^(d) the d-th difference of that row, such that R_n^(0)(m) = R_n(m) and R_n^(k)(m) = R_n^(k-1)(m+1) - R_n^(k-1)(m). Row n of A079502 is then T(n,n) = R_n^(n)(0) and for m < n, T(n,m) = R_n^(n)(0) - T(n,m+1).
For example, starting with row 4 of A050447: [0], 1, 8, 31, 85, 190, 371, ..., we construct up to order 4 difference sequences: first-differences 1, 7, 23, 54, 105, 181, ...; second-differences 6, 16, 31, 51, 76, ...; third-differences 10, 15, 20, 25, ...; fourth-differences 5, 5, 5, ... (constant). Only the first elements of these difference sequences are needed. Thus T(4,4) = 5, T(4,3) = 10 - 5 = 5, T(4,2) = 6 - (10 - 5) = 1, T(4,1) = 1 - (6 - (10 - 5)) = 0. (End)

More terms from Sean A. Irvine, Mar 12 2017

A180671 a(n) = Fibonacci(n+6) - Fibonacci(6).

Original entry on oeis.org

0, 5, 13, 26, 47, 81, 136, 225, 369, 602, 979, 1589, 2576, 4173, 6757, 10938, 17703, 28649, 46360, 75017, 121385, 196410, 317803, 514221, 832032, 1346261, 2178301, 3524570, 5702879, 9227457, 14930344, 24157809, 39088161, 63245978, 102334147, 165580133

Offset: 0

Johannes W. Meijer, Sep 21 2010

The a(n+1) (terms doubled) are the Kn15 sums of the Fibonacci(n) triangle A104763. See A180662 for information about these knight and other chess sums.

Vincenzo Librandi, Table of n, a(n) for n = 0..280
Index entries for linear recurrences with constant coefficients, signature (2,0,-1).

Cf. A000045.

Cf. A131524 (Kn11), A001911 (Kn12), A006327 (Kn13), A167616 (Kn14), A180671 (Kn15), A180672 (Kn16), A180673 (Kn17), A180674 (Kn18).

List([0..40], n-> Fibonacci(n+6)-8); # G. C. Greubel, Jul 13 2019

[Fibonacci(n+6)-Fibonacci(6): n in [0..40]]; // Vincenzo Librandi, Apr 24 2011

nmax:=40: with(combinat): for n from 0 to nmax do a(n):=fibonacci(n+6)-fibonacci(6) od: seq(a(n),n=0..nmax);

f[n_]:= Fibonacci[n+6] - Fibonacci[6]; Array[f, 40, 0] (* or *)
LinearRecurrence[{2,0,-1}, {0,5,13}, 41] (* or *)
CoefficientList[Series[x(3x+5)/(x^3-2x+1), {x,0,40}], x] (* Robert G. Wilson v, Apr 11 2017 *)

for(n=1,40,print(fibonacci(n+6)-fibonacci(6))); \\ Anton Mosunov, Mar 02 2017

concat(0, Vec(x*(5+3*x)/((1-x)*(1-x-x^2)) + O(x^40))) \\ Colin Barker, Apr 20 2017

[fibonacci(n+6)-8 for n in (0..40)] # G. C. Greubel, Jul 13 2019

a(n) = F(n+6) - F(6) with F = A000045.
a(n) = a(n-1) + a(n-2) + 8 for n>1, a(0)=0, a(1)=5, and where 8 = F(6).
From Colin Barker, Apr 13 2012: (Start)
G.f.: x*(5 + 3*x)/((1 - x)*(1 - x - x^2)).
a(n) = 2*a(n-1) - a(n-3). (End)
a(n) = (-8 + (2^(-n)*((1-sqrt(5))^n*(-9+4*sqrt(5)) + (1+sqrt(5))^n*(9+4*sqrt(5)))) / sqrt(5)). - Colin Barker, Apr 20 2017

A180672 a(n) = Fibonacci(n+7) - Fibonacci(7).

Original entry on oeis.org

0, 8, 21, 42, 76, 131, 220, 364, 597, 974, 1584, 2571, 4168, 6752, 10933, 17698, 28644, 46355, 75012, 121380, 196405, 317798, 514216, 832027, 1346256, 2178296, 3524565, 5702874, 9227452, 14930339, 24157804, 39088156, 63245973

Offset: 0

Johannes W. Meijer, Sep 21 2010

The a(n+1) (terms doubled) are the Kn16 sums of the Fibonacci(n) triangle A104763. See A180662 for information about these knight and other chess sums.

Vincenzo Librandi, Table of n, a(n) for n = 0..280
Index entries for linear recurrences with constant coefficients, signature (2,0,-1).

Cf. A000045, A000071.

Cf. A131524 (Kn11), A001911 (Kn12), A006327 (Kn13), A167616 (Kn14), A180671 (Kn15), A180672 (Kn16), A180673 (Kn17), A180674 (Kn18).

List([0..40], n-> Fibonacci(n+7)-13 ); # G. C. Greubel, Jul 13 2019

[Fibonacci(n+7) - Fibonacci(7): n in [0..40]]; // Vincenzo Librandi, Apr 24 2011

nmax:=40: with(combinat): for n from 0 to nmax do a(n):=fibonacci(n+7)-fibonacci(7) od: seq(a(n),n=0..nmax);

Fibonacci[7 +Range[0, 40]] -13 (* G. C. Greubel, Jul 13 2019 *)

concat(0, Vec(x*(8+5*x)/((1-x)*(1-x-x^2)) + O(x^40))) \\ Colin Barker, Feb 24 2017

a(n)=fibonacci(n+7)-fibonacci(7) \\ Charles R Greathouse IV, Feb 24 2017

[fibonacci(n+7)-13 for n in (0..40)] # G. C. Greubel, Jul 13 2019

a(n) = F(n+7) - F(7) with F = A000045.
a(n) = a(n-1) + a(n-2) + 13 for n>1, a(0)=0, a(1)=8, and where 13 = F(7).
G.f.: x*(8 + 5*x)/((1 - x)*(1 - x - x^2)). - Ilya Gutkovskiy, Feb 24 2017
From Colin Barker, Feb 24 2017: (Start)
a(n) = (-13 + (2^(-1-n)*((1-sqrt(5))^n*(-29+13*sqrt(5)) + (1+sqrt(5))^n*(29+13*sqrt(5)))) / sqrt(5)).
a(n) = 2*a(n-1) - a(n-3) for n>2. (End)
a(n) = 8*A000071(n+2) + 5*A000071(n+1). - Bruno Berselli, Feb 24 2017

A180673 a(n) = Fibonacci(n+8) - Fibonacci(8).

Original entry on oeis.org

0, 13, 34, 68, 123, 212, 356, 589, 966, 1576, 2563, 4160, 6744, 10925, 17690, 28636, 46347, 75004, 121372, 196397, 317790, 514208, 832019, 1346248, 2178288, 3524557, 5702866, 9227444, 14930331, 24157796, 39088148, 63245965, 102334134

Offset: 0

Johannes W. Meijer, Sep 21 2010

The a(n+1) (terms doubled) are the Kn17 sums of the Fibonacci(n) triangle A104763. See A180662 for information about these knight and other chess sums.

Vincenzo Librandi, Table of n, a(n) for n = 0..275
Index entries for linear recurrences with constant coefficients, signature (2,0,-1).

Cf. A000045, A000071.

Cf. A131524 (Kn11), A001911 (Kn12), A006327 (Kn13), A167616 (Kn14), A180671 (Kn15), A180672 (Kn16), A180673 (Kn17), A180674 (Kn18).

List([0..40], n-> Fibonacci(n+8)-21); # G. C. Greubel, Jul 13 2019

[Fibonacci(n+8) - Fibonacci(8): n in [0..40]]; // Vincenzo Librandi, Apr 24 2011

nmax:=40: with(combinat): for n from 0 to nmax do a(n):=fibonacci(n+8)-fibonacci(8) od: seq(a(n),n=0..nmax);

Fibonacci[8 +Range[0, 40]] -21 (* G. C. Greubel, Jul 13 2019 *)

concat(0, Vec(x*(13+8*x)/((1-x)*(1-x-x^2)) + O(x^40))) \\ Colin Barker, Feb 24 2017

a(n)=fibonacci(n+8)-21 \\ Charles R Greathouse IV, Feb 24 2017

[fibonacci(n+8)-21 for n in (0..40)] # G. C. Greubel, Jul 13 2019

a(n) = F(n+8) - F(8) with F(n) the Fibonacci numbers A000045.
a(n) = a(n-1) + a(n-2) + 21 for n>1, a(0)=0, a(1)=13, and where 21 = F(8).
G.f.: x*(13 + 8*x)/((1 - x)*(1 - x - x^2)). - Ilya Gutkovskiy, Feb 24 2017
a(n) = 13*A000071(n+2) + 8*A000071(n+1). - Bruno Berselli, Feb 24 2017
From Colin Barker, Feb 24 2017: (Start)
a(n) = (-21 + (2^(-1-n)*((1-sqrt(5))^n*(-47+21*sqrt(5)) + (1+sqrt(5))^n*(47+21*sqrt(5)))) / sqrt(5)).
a(n) = 2*a(n-1) - a(n-3) for n>2. (End)

A180674 a(n) = Fibonacci(n+9) - Fibonacci(9).

Original entry on oeis.org

0, 21, 55, 110, 199, 343, 576, 953, 1563, 2550, 4147, 6731, 10912, 17677, 28623, 46334, 74991, 121359, 196384, 317777, 514195, 832006, 1346235, 2178275, 3524544, 5702853, 9227431, 14930318, 24157783, 39088135, 63245952, 102334121

Offset: 0

Johannes W. Meijer, Sep 21 2010

The a(n+1) (terms doubled) are the Kn18 sums of the Fibonacci(n) triangle A104763. See A180662 for information about these knight and other chess sums.

Vincenzo Librandi, Table of n, a(n) for n = 0..275
Index entries for linear recurrences with constant coefficients, signature (2,0,-1).

Cf. A000045, A000071.

Cf. A131524 (Kn11), A001911 (Kn12), A006327 (Kn13), A167616 (Kn14), A180671 (Kn15), A180672 (Kn16), A180673 (Kn17), A180674 (Kn18).

List([0..40], n-> Fibonacci(n+9)-34); # G. C. Greubel, Jul 13 2019

[Fibonacci(n+9) - Fibonacci(9): n in [0..40]]; // Vincenzo Librandi, Apr 24 2011

nmax:=31: with(combinat): for n from 0 to nmax do a(n):=fibonacci(n+9)-fibonacci(9) od: seq(a(n),n=0..nmax);

Fibonacci[9 +Range[0, 40]] -34 (* G. C. Greubel, Jul 13 2019 *)
LinearRecurrence[{2,0,-1},{0,21,55},40] (* Harvey P. Dale, Aug 24 2024 *)

concat(0, Vec(x*(21+13*x)/((1-x)*(1-x-x^2)) + O(x^40))) \\ Colin Barker, Feb 24 2017

a(n) = fibonacci(n+9) - fibonacci(9) \\ Charles R Greathouse IV, Feb 24 2017

[fibonacci(n+9)-34 for n in (0..40)] # G. C. Greubel, Jul 13 2019

a(n) = F(n+9) - F(9) with F = A000045.
a(n) = a(n-1) + a(n-2) + 34 for n>1, a(0)=0, a(1)=21, and where 34 = F(9).
G.f.: x*(21 + 13*x)/((1 - x)*(1 - x - x^2)). - Ilya Gutkovskiy, Feb 24 2017
a(n) = 21*A000071(n+2) + 13*A000071(n+1). - Bruno Berselli, Feb 24 2017
From Colin Barker, Feb 24 2017: (Start)
a(n) = (-34 + (2^(-n)*((1-sqrt(5))^n*(-38+17*sqrt(5)) + (1+sqrt(5))^n*(38+17*sqrt(5)))) / sqrt(5)).
a(n) = 2*a(n-1) - a(n-3) for n>2. (End)

A256107 Irregular triangle read by rows, T(n,k) is the number of pentagrams on the k layers at n iterations of a pentagram fractal (see comment).

Original entry on oeis.org

1, 2, 4, 1, 7, 2, 2, 12, 4, 5, 2, 1, 20, 7, 10, 4, 2, 2, 33, 12, 18, 8, 4, 4, 2, 1, 54, 20, 31, 14, 7, 8, 4, 2, 2, 88, 33, 52, 24, 12, 14, 8, 4, 4, 2, 1, 143, 54, 86, 40, 20, 24, 14, 7, 8, 4, 2, 2, 232, 88, 141, 66, 33, 40, 24, 12, 14, 8, 4, 4, 2, 1, 376, 143, 230, 108, 54, 66

Offset: 0

Kival Ngaokrajang, Mar 14 2015

Refer to A255870, the number of pentagrams on one side of the outer layer (including pentagrams on two vertices) at n iterations would be T(n,0), the next layers k >= 1 T(n,k) are the number of pentagrams toward the center. For k >= 2, the row length is A032766. The first differences of A255870 = 5*(rows sum - 1). T(n,k) = A000071 with a shift for k = 0 or k mod 3 = 1. T(n,2) = A006327 with a shift. For k >= 3, T(n,k) = 2*A000071 with a shift for k mod 3 = 0 or 2. See illustration in the links.

			Irreuglar triangle begins:
n/k  0  1  2  3  4  5  6  7  8 ...
0   1
1   2
2   4  1
3   7  2  2
4  12  4  5  2  1
5  20  7 10  4  2  2
6  33 12 18  8  4  4  2  1
7  54 20 31 14  7  8  4  2  2
8  88 33 52 24 12 14  8  4  4
...

Kival Ngaokrajang, Illustration of initial terms, T(n,k) for n = 0..10, k = 0..13

Cf. A255870, A032766, A000071, A006327.

{for(n=0, 20, if(n<2, lk=0, lk=floor(3*(n-2)/2)+1); for (k=0, lk, if(k<>0, if(k<>2, if(Mod(k,3)==1, t=fibonacci(n+1-2*(k-1)/3)-1, t=2*(fibonacci(n+2-ceil((2*k+1)/3))-1)), t=fibonacci(n+2)-3), t=fibonacci(n+3-2*k/3)-1); print1(t, ", ")))}

cp's OEIS Frontend

A157728 a(n) = Fibonacci(n) - 4.

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A157729 a(n) = Fibonacci(n) + 5.

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A255870 a(n) is the total number of pentagrams in a pentagram fractal after n iterations.

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A079502 Triangle T(n,k) read by rows; related to number of preorders.

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A180671 a(n) = Fibonacci(n+6) - Fibonacci(6).

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A180672 a(n) = Fibonacci(n+7) - Fibonacci(7).

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