A014286 -id:A014286 - cp's OEIS frontend

A263578 Positive values of k such that A014286(k) is divisible by k.

1, 3, 18, 24, 42, 48, 72, 96, 120, 138, 144, 192, 216, 240, 258, 264, 282, 288, 336, 360, 384, 402, 432, 480, 498, 576, 600, 618, 642, 648, 672, 714, 720, 744, 762, 768, 864, 912, 960, 978, 1002, 1008, 1080, 1104, 1152, 1200, 1224, 1296, 1320, 1338, 1344, 1362

Offset: 1

Altug Alkan, Oct 22 2015

Sequence is interesting because of the values of a(n) - a(n-1). For a(n) < 10000, the most common repeated values of a(n) - a(n-1) are 24 and 6. Will this situation continue?

			a(2) = 3 because Fibonacci(1) + 2 * Fibonacci(2) + 3 * Fibonacci(3) = 9, which is divisible by 3.

Cf. A000045, A014286.

Select[Range@ 1584, Divisible[Sum[i Fibonacci@ i, {i, 0, #}], #] &] (* Michael De Vlieger, Oct 22 2015 *)

for(n=1, 2000, if(sum(k=1, n, k*fibonacci(k)) % n == 0, print1(n, ", ")))

A204922 Ordered differences of Fibonacci numbers.

Original entry on oeis.org

1, 2, 1, 4, 3, 2, 7, 6, 5, 3, 12, 11, 10, 8, 5, 20, 19, 18, 16, 13, 8, 33, 32, 31, 29, 26, 21, 13, 54, 53, 52, 50, 47, 42, 34, 21, 88, 87, 86, 84, 81, 76, 68, 55, 34, 143, 142, 141, 139, 136, 131, 123, 110, 89, 55, 232, 231, 230, 228, 225, 220, 212, 199, 178

Offset: 1

Clark Kimberling, Jan 21 2012

For a guide to related sequences, see A204892. For numbers not in A204922, see A050939.
From Emanuele Munarini, Mar 29 2012: (Start)
Diagonal elements = Fibonacci numbers F(n+1) (A000045)
First column = Fibonacci numbers - 1 (A000071);
Second column = Fibonacci numbers - 2 (A001911);
Row sums = n*F(n+3) - F(n+2) + 2 (A014286);
Central coefficients = F(2*n+1) - F(n+1) (A096140).
(End)

			a(1) = s(2) - s(1) = F(3) - F(2) = 2-1 = 1, where F=A000045;
a(2) = s(3) - s(1) = F(4) - F(2) = 3-1 = 2;
a(3) = s(3) - s(2) = F(4) - F(3) = 3-2 = 1;
a(4) = s(4) - s(1) = F(5) - F(2) = 5-1 = 4.
From _Emanuele Munarini_, Mar 29 2012: (Start)
Triangle begins:
   1;
   2,  1;
   4,  3,  2;
   7,  6,  5,  3;
  12, 11, 10,  8,  5;
  20, 19, 18, 16, 13,  8;
  33, 32, 31, 29, 26, 21, 13;
  54, 53, 52, 50, 47, 42, 34, 21;
  88, 87, 86, 84, 81, 76, 68, 55, 34;
  ... (End)

G. C. Greubel, Rows n=1..100 of triangle, flattened

Cf. A204924, A204892.

/* As triangle */ [[Fibonacci(n+2)-Fibonacci(k+1) : k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Aug 04 2015

Mathematica
```
(See the program at A204924.)
```

create_list(fib(n+3)-fib(k+2),n,0,20,k,0,n); /* Emanuele Munarini, Mar 29 2012 */

{T(n,k) = fibonacci(n+2) - fibonacci(k+1)};
for(n=1,15, for(k=1,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Feb 03 2019

[[fibonacci(n+2) - fibonacci(k+1) for k in (1..n)] for n in (1..15)] # G. C. Greubel, Feb 03 2019

From Emanuele Munarini, Mar 29 2012: (Start)
T(n,k) = Fibonacci(n+2) - Fibonacci(k+1).
T(n,k) = Sum_{i=k..n} Fibonacci(i+1). (End)

A045925 a(n) = n*Fibonacci(n).

Original entry on oeis.org

0, 1, 2, 6, 12, 25, 48, 91, 168, 306, 550, 979, 1728, 3029, 5278, 9150, 15792, 27149, 46512, 79439, 135300, 229866, 389642, 659111, 1112832, 1875625, 3156218, 5303286, 8898708, 14912641, 24961200, 41734339, 69705888, 116311074, 193898158, 322961275, 537492672

Offset: 0

Jeff Burch

Number of levels in all compositions of n+1 with only 1's and 2's.
Apart from first term: row sums of the triangle in A131410. - Reinhard Zumkeller, Oct 07 2012
Number of permutations of length n>0 avoiding the partially ordered pattern (POP) {1>3} of length 4. That is, number of length n permutations having no subsequences of length 4 in which the first element is larger than the third one. - Sergey Kitaev, Dec 08 2020

Jean Paul Van Bendegem, The Heterogeneity of Mathematical Research, a chapter in Perspectives on Interrogative Models of Inquiry, Volume 8 of the series Logic, Argumentation & Reasoning pp 73-94, Springer 2015. See Section 2.1.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Russell Euler, Problem B-670, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 28, No. 3 (1990), p. 277; Application of Generating Functions, Solution to Problem B-670 by Russell Jay Hendel, ibid., Vol. 29, No. 3 (1991), p. 278.
Rigoberto Flórez, Robinson Higuita and Alexander Ramírez, The resultant, the discriminant, and the derivative of generalized Fibonacci polynomials, arXiv:1808.01264 [math.NT], 2018.
Alice L. L. Gao and Sergey Kitaev, On partially ordered patterns of length 4 and 5 in permutations, arXiv:1903.08946 [math.CO], 2019.
Alice L. L. Gao and Sergey Kitaev, On partially ordered patterns of length 4 and 5 in permutations, The Electronic Journal of Combinatorics, Vol. 26, No. 3 (2019), Article P3.26.
Silvia Heubach and Toufik Mansour, Counting rises, levels and drops in compositions, arXiv:math/0310197 [math.CO], 2003.
A. G. Shannon, B. Kuloğlu, and E. Özkan, Rhaly terraced sequences their generalizations, properties and applications, Comp. Appl. Math. 44, 226 (2025). See p. 2.
Kai Ting Keshia Yap, David Wehlau and Imed Zaguia, Permutations Avoiding Certain Partially-ordered Patterns, arXiv:2101.12061 [math.CO], 2021.
Index to divisibility sequences.
Index entries for linear recurrences with constant coefficients, signature (2,1,-2,-1).

Partial sums: A014286. Cf. A000045.

Cf. A099920, A023607.

a045925 n = a045925_list !! (n-1)
a045925_list = zipWith (*) [0..] a000045_list
-- Reinhard Zumkeller, Oct 01 2012

[n*Fibonacci(n): n in [0..60]]; // Vincenzo Librandi, Apr 23 2011

a:= n-> n*(<<0|1>, <1|1>>^n)[1,2]:
seq(a(n), n=0..37);  # Alois P. Heinz, May 07 2021

Table[Fibonacci[n]*n, {n, 0, 33}] (* Zerinvary Lajos, Jul 09 2009 *)
LinearRecurrence[{2, 1, -2, -1}, {0, 1, 2, 6}, 34] (* or *)
CoefficientList[ Series[(x + x^3)/(-1 + x + x^2)^2, {x, 0, 35}], x] (* Robert G. Wilson v, Nov 14 2015 *)

Lucas(n)=fibonacci(n-1)+fibonacci(n+1)
a(n)=polcoeff(sum(m=1,n,eulerphi(m)*fibonacci(m)*x^m/(1-Lucas(m)*x^m+(-1)^m*x^(2*m)+x*O(x^n))),n) \\ Paul D. Hanna, Jan 12 2012

a(n)=n*fibonacci(n) \\ Charles R Greathouse IV, Jan 12 2012

concat(0, Vec(x*(1+x^2)/(1-x-x^2)^2 + O(x^100))) \\ Altug Alkan, Oct 28 2015

G.f.: x*(1+x^2)/(1-x-x^2)^2.
G.f.: Sum_{n>=1} phi(n)*Fibonacci(n)*x^n/(1 - Lucas(n)*x^n + (-1)^n*x^(2*n)) = Sum_{n>=1} n*Fibonacci(n)*x^n, where phi(n) = A000010(n) and Lucas(n) = A000204(n). - Paul D. Hanna, Jan 12 2012
a(n) = a(n-1) + a(n-2) + L(n-1). - Gary Detlefs, Dec 29 2012
a(n) = F(n+1) + Sum_{k=1..n-2} F(k)*L(n-k), F = A000045 and L = A000032. - Gary Detlefs, Dec 29 2012
a(n) = F(2*n)/Sum_{k=0..floor(n/2)} binomial(n-k,k)/(n-k). - Gary Detlefs, Jan 19 2013
a(n) = A014965(n) * A104714(n). - Michel Marcus, Oct 24 2013
a(n) = 3*A001629(n+1) - A001629(n+2) + A000045(n-1). - Ralf Stephan, Apr 26 2014
a(n) = 2*n*(F(n-2) + floor(F(n-3)/2)) + (n^3 mod 3*n), F = A000045. - Gary Detlefs, Jun 06 2014
E.g.f.: x*(exp(-x/phi)/phi + exp(x*phi)*phi)/sqrt(5), where phi = (1+sqrt(5))/2. - Vladimir Reshetnikov, Oct 28 2015
This is a divisibility sequence and is generated by  x^4 - 2*x^3 - x^2 + 2*x + 1. - R. K. Guy, Nov 13 2015
a(n) = L'(n, 1), the first derivative of the n-th Lucas polynomial evaluated at 1. - Andrés Ventas, Nov 12 2021
Sum_{n>=0} a(n)/2^n = 10 (Euler, 1990). - Amiram Eldar, Jan 22 2022

Incorrect formula removed by Gary Detlefs, Oct 27 2011

A143061 Triangle read by rows, A000012 * A127647 * A000012.

Original entry on oeis.org

1, 1, 2, 2, 3, 4, 3, 5, 6, 7, 5, 8, 10, 11, 12, 8, 13, 16, 18, 19, 20, 13, 21, 26, 29, 31, 32, 33, 21, 34, 42, 47, 50, 52, 53, 54, 34, 55, 68, 76, 81, 84, 86, 87, 88, 55, 89, 110, 123, 131, 136, 139, 141, 142, 143, 89, 144, 178, 199, 212, 220, 225, 228, 230, 231

Offset: 1

Gary W. Adamson, Jul 20 2008

Row sums = A014286 (1, 3, 9, 21, 46, 94, ...); left border = Fibonacci numbers.

			First few rows of the triangle are:
  1;
  1,  2;
  2,  3,  4;
  3,  5,  6,  7;
  5,  8, 10, 11, 12;
  8, 13, 16, 18, 19, 20;
  ...

Robert Israel, Table of n, a(n) for n = 1..10011 (rows 1 to 141, flattened)

Cf. A000012, A000045, A014286, A127647.

seq(seq(combinat:-fibonacci(i+2)-combinat:-fibonacci(i+2-j),j=1..i),i=1..20); # Robert Israel, Nov 06 2016

From Robert Israel, Nov 06 2016: (Start)
T(n,k) = A000045(n+2) - A000045(n+2-k) for 1 <= k <= n.
G.f. as triangle: x*y*(1+x^2*y)/((1-x*y)*(1-x-x^2)*(1-x*y-x^2*y^2)). (End)

Corrected by Dintle N Kagiso, Nov 06 2016

A282464 a(n) = Sum_{i=0..n} i*Fibonacci(i)^2.

Original entry on oeis.org

0, 1, 3, 15, 51, 176, 560, 1743, 5271, 15675, 45925, 133056, 381888, 1087645, 3077451, 8658951, 24245655, 67602608, 187789616, 519924075, 1435228575, 3951341811, 10852291273, 29740435200, 81340229376, 222058995001, 605201766675, 1646862596223, 4474969884411

Offset: 0

Bruno Berselli, Feb 16 2017

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-4,-10,10,4,-5,1).

Cf. A000045.
Partial sums of A169630.
Cf. A014286: partial sums of i*Fibonacci(i).
Cf. A064831: partial sums of (n+1-i)*Fibonacci(i)^2.

[&+[i*Fibonacci(i)^2: i in [0..n]]: n in [0..30]];

with(combinat): P:=proc(q) local a,n; a:=0; print(a); for n from 1 to q do
a:=a+n*fibonacci(n)^2; print(a); od; end: P(100); # Paolo P. Lava, Feb 17 2017

a[n_] := Sum[i*Fibonacci[i]^2, {i, 0, n}]; Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Feb 16 2017 *)
LinearRecurrence[{5,-4,-10,10,4,-5,1},{0,1,3,15,51,176,560},30] (* Harvey P. Dale, May 15 2021 *)

makelist(sum(i*fib(i)^2, i, 0, n), n, 0, 30);

a(n) = sum(i=0, n, i*fibonacci(i)^2) \\ Colin Barker, Feb 16 2017

[sum(i*fibonacci(i)^2 for i in [0..n]) for n in range(30)]

O.g.f.: x*(1 - 2*x + 4*x^2 - 2*x^3 + x^4)/((1 - x)*(1 + x)^2*(1 - 3*x + x^2)^2).
a(n) = 5*a(n-1) - 4*a(n-2) - 10*a(n-3) + 10*a(n-4) + 4*a(n-5) - 5*a(n-6) + a(n-7).
a(n) = ((n-1)*Fibonacci(n) + n*Fibonacci(n-1))*Fibonacci(n) + (1 - (-1)^n)/2.

A136391 a(n) = nF(n) - (n-1)F(n-1), where the F(j)'s are the Fibonacci numbers (F(0)=0, F(1)=1).

Original entry on oeis.org

1, 1, 4, 6, 13, 23, 43, 77, 138, 244, 429, 749, 1301, 2249, 3872, 6642, 11357, 19363, 32927, 55861, 94566, 159776, 269469, 453721, 762793, 1280593, 2147068, 3595422, 6013933, 10048559, 16773139, 27971549, 46605186, 77587084, 129063117, 214531397, 356346557

Offset: 1

Gary W. Adamson, Dec 28 2007

By definition, the arithmetic mean of a(1) ... a(n) is equal to A000045(n).

Proof of the three-term recurrence formula: a(n+1) - a(n) - a(n-1) = ((n+1)*F(n+1) - n*F(n)) - (n*F(n) - (n-1)*F(n-1)) - ((n-1)*F(n-1) - (n-2)*F(n-2)) = (n+1)*F(n+1) - 2*n*F(n) + (n-2)*F(n-2) = (n+1)*(2*F(n) - F(n-2)) - 2*n*F(n) + (n-2)*F(n-2) = 2*F(n) - 3*F(n-2) = F(n-1) + F(n-3) = L(n-2). - Giuseppe Coppoletta, Sep 01 2014

			a(6) = 23 = 6*F(6) - 5*F(5) = 6*8 - 5*5 = 48 - 25.

Index entries for linear recurrences with constant coefficients, signature (2,1,-2,-1).

Cf. A000045, A246715, A354044.

# The function 'fibrec' is defined in A354044.
function A136391(n)
    a, b = fibrec(n - 1)
    n*b - (n - 1)*a
end
println([A136391(n) for n in 1:35]) # Peter Luschny, May 18 2022

with(combinat): seq(n*fibonacci(n)-(n-1)*fibonacci(n-1),n=1..30); # Emeric Deutsch, Jan 01 2008

Table[n Fibonacci[n] - (n-1) Fibonacci[n-1], {n, 1, 20}] (* Vladimir Reshetnikov, Oct 28 2015 *)

Vec(x*(1-x)*(1+x^2)/(1-x-x^2)^2 + O(x^100)) \\ Altug Alkan, Oct 28 2015

Equals A128064 * A000045.
From R. J. Mathar, Nov 25 2008: (Start)
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) - a(n-4) = A045925(n) - A045925(n-1).
G.f.: x*(1 - x)*(1 + x^2)/(1 - x - x^2)^2.
a(n) = A014286(n-1) - A014286(n-2), n>3. (End)
Recurrence: a(n+1) = a(n) + a(n-1) + L(n-2) for n>1, where L = A000032 (see proof in Comments section). - Giuseppe Coppoletta, Sep 01 2014
E.g.f.: (exp(x*phi)/phi+exp(-x/phi)*phi)*(x+1)/sqrt(5)-1, where phi=(1+sqrt(5))/2. - Vladimir Reshetnikov, Oct 28 2015
a(n) = F(n-1) + n*F(n-2). - Bruno Berselli, Jul 26 2017

More terms from Emeric Deutsch, Jan 01 2008

A263878 a(n) = Sum_{k=0..n} (-1)^kkFibonacci(k), where Fibonacci(k) = A000045(k).

Original entry on oeis.org

0, -1, 1, -5, 7, -18, 30, -61, 107, -199, 351, -628, 1100, -1929, 3349, -5801, 9991, -17158, 29354, -50085, 85215, -144651, 244991, -414120, 698712, -1176913, 1979305, -3323981, 5574727, -9337914, 15623286, -26111053, 43594835, -72716239, 121181919, -201779356

Offset: 0

Vladimir Reshetnikov, Oct 28 2015

			G.f. = - x + x^2 - 5*x^3 + 7*x^4 - 18*x^5 + 30*x^6 - 61*x^7 + 107*x^8 - 199*x^9 + ...

Colin Barker, Table of n, a(n) for n = 0..1000
Eric Weisstein's MathWorld, Fibonacci Number.
Index entries for linear recurrences with constant coefficients, signature (-1,3,1,-3,1).

Cf. A000045, A014286.

[(-1)^n*(Fibonacci(n-3) + n*Fibonacci(n-1)) - 2: n in [0..30]]; // G. C. Greubel, Jul 30 2018

Table[Sum[(-1)^k k Fibonacci[k], {k, 0, n}], {n, 0, 20}]
Table[(-1)^n (Fibonacci[n-3] + n Fibonacci[n-1]) - 2, {n, 0, 20}]

concat(0, Vec(x*(x^2+1)/((x-1)*(x^2-x-1)^2) + O(x^40))) \\ Colin Barker, Oct 31 2015

a(n) = (-1)^n*(fibonacci(n-3) + n*fibonacci(n-1)) - 2; \\ Michel Marcus, Nov 02 2015

a(n) = (-1)^n*(F(n-3) + n*F(n-1)) - 2, where F(n) = A000045(n).
G.f.: x*(x^2+1)/((x-1)*(x^2-x-1)^2).
E.g.f.: (exp(x/phi)*(phi^3+x)+exp(-phi*x)*(1/phi^3-x))/sqrt(5)-2*exp(x), where phi=(1+sqrt(5))/2.
Recurrences:
6-term, homogeneous, constant coefficients: a(0) = 0, a(1) = -1, a(2) = 1, a(3) = -5, a(4) = 7, a(n) = -a(n-1) + 3*a(n-2) + a(n-3) - 3*a(n-4) + a(n-5).
5-term, non-homogeneous, constant coefficients: a(0) = 0, a(1) = -1, a(2) = 1, a(3) = -5, a(n) = -2*a(n-1) + a(n-2) + 2*a(n-3) - a(n-4) - 2.
4-term, homogeneous: a(0) = 0, a(1) = -1, a(2) = 1, (n-1)*(n-2)*a(n) = (2-n)*a(n-1) + n*(2*n-3)*a(n-2) + n*(1-n)*a(n-3).
3-term, non-homogeneous: a(0) = 0, a(1) = -1, (n^2-1)*a(n) = -(n^2+n+1)*a(n-1) + n*(n+2)*a(n-2) - 2*n*(n-1).
0 = a(n)*(-2*a(n) + 15*a(n+1) - 9*a(n+2) + a(n+3) - 3*a(n+4)) + a(n+1)*(-25*a(n+1) + 15*a(n+2) + 15*a(n+3) + 5*a(n+4)) + a(n+2)*(18*a(n+2) - 29*a(n+3) - 13*a(n+4)) + a(n+3)*(+3*a(n+3) + 7*a(n+4)) + a(n+4)*(2*a(n+4)) for all n in Z. - Michael Somos, Nov 02 2015

A104731 Triangle T(n,k) = sum_{j=k..n} (j+1)*binomial(k,j-k), read by rows, 0<=k<=n.

Original entry on oeis.org

1, 1, 2, 1, 5, 3, 1, 5, 11, 4, 1, 5, 16, 19, 5, 1, 5, 16, 37, 29, 6, 1, 5, 16, 44, 71, 41, 7, 1, 5, 16, 44, 103, 121, 55, 8, 1, 5, 16, 44, 112, 211, 190, 71, 9, 1, 5, 16, 44, 112, 261, 390, 281, 89, 10, 1, 5, 16, 44, 112, 272, 555, 666, 397, 109, 11

Offset: 0

Gary W. Adamson, Mar 20 2005

			The first few rows of the triangle are:
1;
1, 2;
1, 5, 3;
1, 5, 11, 4
1, 5, 16, 19, 5;
1, 5, 16, 37, 29, 6;
...

Cf. A014286 (row sums), A045925, A026729.

Product of the triangles A(n,k) = k+1 and B = binomial(k,n-k) = [1; 0, 1; 0, 1, 1; 0, 0, 2, 1; 0, 0, 1, 3, 1;...], the triangular view of A026729.

A117152 Sum of product of Fibonacci and triangular numbers.

Original entry on oeis.org

0, 0, 1, 7, 25, 75, 195, 468, 1056, 2280, 4755, 9650, 19154, 37328, 71635, 135685, 254125, 471317, 866669, 1581620, 2866970, 5165630, 9256871, 16507092, 29304660, 51812160, 91264885, 160207603, 280340161, 489117135, 851054535

Offset: 0

Mitch Harris, Feb 28 2006

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

Cf. A086926, A014286.

Binomial[n, 2]Fibonacci[n + 2] - n Fibonacci[n + 3] + Fibonacci[n + 5] - 5

a(n) = sum(k=2, n, k*(k-1)*fibonacci(k)/2); \\ Michel Marcus, Feb 28 2019

a(n) = Sum_{k=2..n} C(k,2)*F(k), where F(n) = A000045(n), the Fibonacci numbers and C(n, 2) = A000217(n-1), the triangular numbers, n(n-1)/2.
a(n) = C(n,2) F(n+2) - n F(n+3) + F(n+5) - 5.
G.f.: x^2(1 + 3x + x^3)/((1 - x)(1 - x - x^2)^3).
a(n)-a(n-1) = A086926(n). - R. J. Mathar, May 16 2025

cp's OEIS Frontend

A263578 Positive values of k such that A014286(k) is divisible by k.

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A204922 Ordered differences of Fibonacci numbers.

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A045925 a(n) = n*Fibonacci(n).

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A143061 Triangle read by rows, A000012 * A127647 * A000012.

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A282464 a(n) = Sum_{i=0..n} i*Fibonacci(i)^2.

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A136391 a(n) = n*F(n) - (n-1)*F(n-1), where the F(j)'s are the Fibonacci numbers (F(0)=0, F(1)=1).

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Julia

A136391 a(n) = nF(n) - (n-1)F(n-1), where the F(j)'s are the Fibonacci numbers (F(0)=0, F(1)=1).

A263878 a(n) = Sum_{k=0..n} (-1)^kkFibonacci(k), where Fibonacci(k) = A000045(k).