A059840 -id:A059840 - cp's OEIS frontend

A194000 Triangular array: the self-fission of (p(n,x)), where sum{F(k+1)*x^(n-k) : 0<=k<=n}, where F=A000045 (Fibonacci numbers).

1, 2, 3, 3, 5, 9, 5, 8, 15, 24, 8, 13, 24, 39, 64, 13, 21, 39, 63, 104, 168, 21, 34, 63, 102, 168, 272, 441, 34, 55, 102, 165, 272, 440, 714, 1155, 55, 89, 165, 267, 440, 712, 1155, 1869, 3025, 89, 144, 267, 432, 712, 1152, 1869, 3024, 4895, 7920, 144, 233

Offset: 0

Clark Kimberling, Aug 11 2011

See A193917 for the self-fusion of the same sequence of polynomials.  (Fusion is defined at A193822; fission, at A193842; see A202503 and A202453 for infinite-matrix representations of fusion and fission.)
...
First five rows of P (triangle of coefficients of polynomials p(n,x)):
1
1...1
1...1...2
1...1...2...3
1...1...2...3...5
First eight rows of A194000:
1
2....3
3....5....9
5....8....15...24
8....13...24...39...64
13...21...29...63...104...168
21...34...63...102..168...272...441
34...55...102..165..272...440...714..1155
...
col 1:  A000045
col 2:  A000045
col 3:  A022086
col 4:  A022086
col 5:  A022091
col 6:  A022091
right edge, d(n,n):  A064831
d(n,n-1):  A059840
d(n,n-2):  A080097
d(n,n-3):  A080143
d(n,n-4):  A080144
...
Suppose n is an odd positive integer and d(n+1,x) is the polynomial matched to row n+1 of A194000 as in the Mathematica program (and definition of fission at A193842), where the first row is counted as row 0.

			First six rows:
1
2....3
3....5....9
5....8....15...24
8....13...24...39...64
13...21...29...63...104...168
...
Referring to the matrix product for fission at A193842,
the row (5,8,15,24) is the product of P(4) and QQ, where
P(4)=(p(4,4), p(4,3), p(4,2), p(4,1))=(5,3,2,1); and
QQ is the 4x4 matrix
(1..1..2..3)
(0..1..1..2)
(0..0..1..1)
(0..0..0..1).

Cf. A193842, A194001, A193917, A193918.

z = 11;
p[n_, x_] := Sum[Fibonacci[k + 1]*x^(n - k), {k, 0, n}];
q[n_, x_] := p[n, x];
p1[n_, k_] := Coefficient[p[n, x], x^k];
p1[n_, 0] := p[n, x] /. x -> 0;
d[n_, x_] := Sum[p1[n, k]*q[n - 1 - k, x], {k, 0, n - 1}]
h[n_] := CoefficientList[d[n, x], {x}]
TableForm[Table[Reverse[h[n]], {n, 0, z}]]
Flatten[Table[Reverse[h[n]], {n, -1, z}]]  (* A194000 *)
TableForm[Table[h[n], {n, 0, z}]]
Flatten[Table[h[n], {n, -1, z}]]  (* A194001 *)

A228873 a(n) = F(n) * F(n+1) * F(n+2) * F(n+3), the product of four consecutive Fibonacci numbers, A000045.

Original entry on oeis.org

6, 30, 240, 1560, 10920, 74256, 510510, 3495030, 23965920, 164237040, 1125770256, 7715953440, 52886430870, 362487682830, 2484530961360, 17029219589256, 116720030923320, 800010932051760, 5483356663145790, 37583485265670630, 257601041359736256

Offset: 1

T. D. Noe, Sep 24 2013

nonn

Mohanty and Mohanty prove in Corollary 2.5 that these numbers are Pythagorean. The number a(n) is primitive Pythagorean if F(n) and F(n+1) have opposite parity. Every third number, starting at a(1) = 6, is not primitive Pythagorean.

Since a(n) = F(n+1)*F(n+2)*(F(n+2)^2 - F(n+1)^2), a(n) is in A073120. - Robert Israel, Apr 06 2015

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Supriya Mohanty and S. P. Mohanty, Pythagorean Numbers, The Fibonacci Quarterly, Vol. 28, No. 1 (1990), pp. 31-42.
H.-J. Seiffert, Problem B-1020, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 44, No. 4 (2006), p. 278; Two Fibonacci Identities, Solution to Problem B-1020 by Harris Kwong, ibid., Vol. 45, No. 2 (2007), pp. 182-184.
Index entries for linear recurrences with constant coefficients, signature (5,15,-15,-5,1).

Cf. A000045 (Fibonacci numbers), A228874 (similar sequence for Lucas numbers).

Cf. A009112 (Pythagorean numbers), A024365, A073120.

[Fibonacci(n)*Fibonacci(n+1)*Fibonacci(n+2)*Fibonacci(n+3): n in [1..30]]; // Vincenzo Librandi, Oct 04 2013

seq(mul(combinat:-fibonacci(i),i=n..n+3),n=1..30); # Robert Israel, Apr 06 2015

Table[Fibonacci[n] Fibonacci[n+1] Fibonacci[n+2] Fibonacci[n+3], {n, 25}]
CoefficientList[Series[-6/((x - 1) (x^2 + 3 x + 1) (x^2 - 7 x + 1)), {x, 0, 30}], x] (* Vincenzo Librandi, Oct 04 2013 *)
Times@@@Partition[Fibonacci[Range[30]],4,1] (* Harvey P. Dale, Dec 23 2013 *)
LinearRecurrence[{5,15,-15,-5,1},{6,30,240,1560,10920},30] (* Harvey P. Dale, Jul 24 2021 *)

G.f.: -6*x/((x-1)*(x^2+3*x+1)*(x^2-7*x+1)). - Alois P. Heinz, Oct 02 2013
a(n+5) = 5*a(n+4)+15*a(n+3)-15*a(n+2)-5*a(n+1)+a(n). - Robert Israel, Apr 06 2015
a(n) = 2 * A000217(A059840(n+2)). - Diego Rattaggi, Jan 27 2021
Sum_{n>=1} 1/a(n) = (12-5*sqrt(5))/4. - Diego Rattaggi, Aug 16 2021
a(n) = 3 * Sum_{k=1..n} 2^(n-k)*F(k)^2*F(k+1)*F(k+2) (Seiffert, 2006). - Amiram Eldar, Jan 11 2022

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

cp's OEIS Frontend

A194000 Triangular array: the self-fission of (p(n,x)), where sum{F(k+1)*x^(n-k) : 0<=k<=n}, where F=A000045 (Fibonacci numbers).

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A228873 a(n) = F(n) * F(n+1) * F(n+2) * F(n+3), the product of four consecutive Fibonacci numbers, A000045.

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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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A027418 Duplicate of A035508.

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