A110034 -id:A110034 - cp's OEIS frontend

A110035 Row sums of an unsigned characteristic triangle for the Fibonacci numbers.

1, 2, 5, 12, 31, 80, 209, 546, 1429, 3740, 9791, 25632, 67105, 175682, 459941, 1204140, 3152479, 8253296, 21607409, 56568930, 148099381, 387729212, 1015088255, 2657535552, 6957518401, 18215019650, 47687540549, 124847601996

Offset: 0

Paul Barry, Jul 08 2005

Rows sums of abs(A110033).

			G.f. = 1 + 2*x + 5*x^2 + 12*x^3 + 31*x^4 + 80*x^5 + 209*x^6 + ... - _Michael Somos_, Mar 03 2023

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

Cf. A110034, A236438.

LinearRecurrence[{3,0,-3,1},{1,2,5,12},50] (* Harvey P. Dale, May 01 2022 *)
a[ n_] := With[{F = Fibonacci}, (1 + F[n+1]*F[n+2] + F[n+n])/2]; (* Michael Somos, Mar 03 2023 *)

{a(n) = my(F = fibonacci); (1 + F(n+1)*F(n+2) + F(n+n))/2}; /* Michael Somos, Mar 03 2023 */

G.f.: (1-x-x^2)/((1-x^2)(1-3x+x^2));
a(n) = 3*a(n-1) - 3*a(n-3) + a(n-4);
a(n) = F(2n) + 1 + Sum_{k=0..n-1} F(k)*F(k+1).
From R. J. Mathar, Jul 22 2010: (Start)
a(n) = Sum_{i=0..n} A061646(i).
a(n) = (5 + (-1)^n + 4*A002878(n))/10. (End)
a(n) = A110034(-n) = 1 - A110034(1+n) = A236438(n) + (n mod 2) = (1 + F(n+1)*F(n+2) + F(2*n))/2 for all n in Z. - Michael Somos, Mar 03 2023

A110033 A characteristic triangle for the Fibonacci numbers.

Original entry on oeis.org

1, -1, 1, 1, -3, 1, 0, 3, -8, 1, 0, 0, 9, -21, 1, 0, 0, 0, 24, -55, 1, 0, 0, 0, 0, 64, -144, 1, 0, 0, 0, 0, 0, 168, -377, 1, 0, 0, 0, 0, 0, 0, 441, -987, 1, 0, 0, 0, 0, 0, 0, 0, 1155, -2584, 1, 0, 0, 0, 0, 0, 0, 0, 0, 3025, -6765, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 7920, -17711, 1

Offset: 0

Paul Barry, Jul 08 2005

			Rows begin
1;
-1,1;
1,-3,1;
0,3,-8,1;
0,0,9,-21,1;
0,0,0,24,-55,1;
0,0,0,0,64,-144,1;
0,0,0,0,0,168,-377,1;

Form the n X n Hankel matrices F(i+j-1), 1<=i, j<=n for the Fibonacci numbers and take the characteristic polynomials of these matrices. Triangle rows give coefficients of these characteristic polynomials. (Construction described by Michael Somos in A064831). Diagonal is (-1)^n*F(2n+2). Subdiagonal is A064831. Row sums are A110034. The unsigned matrix has row sums A110035.

A206282 a(n) = ( a(n-1) * a(n-3) + a(n-2) ) / a(n-4), a(1) = a(2) = 1, a(3) = -1, a(4) = -4.

Original entry on oeis.org

1, 1, -1, -4, -5, 1, 9, 11, -4, -25, -31, 9, 64, 79, -25, -169, -209, 64, 441, 545, -169, -1156, -1429, 441, 3025, 3739, -1156, -7921, -9791, 3025, 20736, 25631, -7921, -54289, -67105, 20736, 142129, 175681, -54289, -372100, -459941, 142129, 974169, 1204139

Offset: 1

Michael Somos, Feb 05 2012

This satisfies the same recurrence as Dana Scott's sequence A048736.

			G.f. = x + x^2 - x^3 - 4*x^4 - 5*x^5 + x^6 + 9*x^7 + 11*x^8 - 4*x^9 - 25*x^10 + ...

Reinhard Zumkeller, Table of n, a(n) for n = 1..5000
Index entries for linear recurrences with constant coefficients, signature (0,0,-2,0,0,2,0,0,1).

Cf. A000045, A048736, A110034, A110035, A126116.

a206282 n = a206282_list !! (n-1)
a206282_list = 1 : 1 : -1 : -4 :
   zipWith div
     (zipWith (+)
       (zipWith (*) (drop 3 a206282_list)
                    (drop 1 a206282_list))
       (drop 2 a206282_list))
     a206282_list
-- Same program as in A048736, see comment.
-- Reinhard Zumkeller, Feb 08 2012

I:=[1,1,-1,-4]; [n le 4 select I[n] else (Self(n-1)*Self(n-3) + Self(n-2))/Self(n-4): n in [1..30]]; // G. C. Greubel, Aug 12 2018

CoefficientList[Series[x*(1+x)*(1-x^2)*(1+x^3)/(1-2*x^2-2*x^4-2*x^6+x^8 ), {x,0,50}], x] (* or *) RecurrenceTable[{a[n] == ( a[n-1]*a[n-3] + a[n-2] )/a[n-4], a[1] == a[2] == 1, a[3] == -1, a[4] == -4}, a, {n,1,50}] (* G. C. Greubel, Aug 12 2018 *)

{a(n) = my(k = n\3); (-1)^k * if( n%3 == 0, fibonacci( k )^2, n%3 == 1, fibonacci( k+2 )^2, fibonacci( k ) * fibonacci( k+3 ) + fibonacci( k+1 ) * fibonacci( k+2 ))};

x='x+O('x^30); Vec(x*(1+x)*(1-x^2)*(1+x^3)/(1-2*x^2-2*x^4 -2*x^6 +x^8 )) \\ G. C. Greubel, Aug 12 2018

G.f.: x * (1 + x - x^2 - 2*x^3 - 3*x^4 - x^5 - x^6 - x^7) / (1 + 2*x^3 - 2*x^6 - x^9).
a(n) = a(-5 - n) = a(n+2) * a(n-2) - a(n+1) * a(n-1) for all n in Z.
a(3*n) = (-1)^n * F(n)^2, a(3*n + 1) = (-1)^n * F(n + 2)^2 where F = Fibonacci A000045.
a(6*n - 4) = - A110034(2*n), a(6*n - 1) = - A110035(2*n), a(3*n + 2) = (-1)^n * A126116(2*n + 3).

A378277 Denominators in a harmonic triangle, based on products of Fibonacci numbers.

Original entry on oeis.org

1, 2, 2, 2, 3, 6, 2, 3, 10, 15, 2, 3, 10, 24, 40, 2, 3, 10, 24, 65, 104, 2, 3, 10, 24, 65, 168, 273, 2, 3, 10, 24, 65, 168, 442, 714, 2, 3, 10, 24, 65, 168, 442, 1155, 1870, 2, 3, 10, 24, 65, 168, 442, 1155, 3026, 4895, 2, 3, 10, 24, 65, 168, 442, 1155, 3026, 7920, 12816

Offset: 1

Werner Schulte, Nov 21 2024

The harmonic triangle uses the terms of this sequence as denominators, numerators = 1.

The inverse of the harmonic triangle has entries -(Fibonacci(k+1))^2 for 1<=k

Row sums of the harmonic triangle are 1.

Conjecture: Alt. row sums of the harmonic triangle are Fibonacci(n-2) / Fibonacci(n+1), where Fibonacci(-1) = 1.

			Triangle T(n, k) for 1 <= k <= n starts:
n\ k :  1  2   3   4   5    6    7     8     9    10     11
===========================================================
   1 :  1
   2 :  2  2
   3 :  2  3   6
   4 :  2  3  10  15
   5 :  2  3  10  24  40
   6 :  2  3  10  24  65  104
   7 :  2  3  10  24  65  168  273
   8 :  2  3  10  24  65  168  442   714
   9 :  2  3  10  24  65  168  442  1155  1870
  10 :  2  3  10  24  65  168  442  1155  3026  4895
  11 :  2  3  10  24  65  168  442  1155  3026  7920  12816
  etc.

Cf. A000045, A110034, A110035, A001654 (main diagonal), A059929 (subdiagonals).

T(n,k)=if(k==n,Fibonacci(n)*Fibonacci(n+1),Fibonacci(k)*Fibonacci(k+2))

T(n, k) = Fibonacci(n) * Fibonacci(n+1) if k = n, and Fibonacci(k) * Fibonacci(k+2) if 1 <= k < n.
Row sums are A110035(n) - 1 = -A110034(n+1).
G.f.: A(t, x) = x*t*(1 + t - x*t^2) / ((1 - t) * (1 + x*t) * (1 - 3*x*t + x^2*t^2)).

Showing 1-4 of 4 results.

cp's OEIS Frontend

A110035 Row sums of an unsigned characteristic triangle for the Fibonacci numbers.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A110033 A characteristic triangle for the Fibonacci numbers.

Views

Author

Keywords

Examples

Formula

A206282 a(n) = ( a(n-1) * a(n-3) + a(n-2) ) / a(n-4), a(1) = a(2) = 1, a(3) = -1, a(4) = -4.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Magma

Mathematica

PARI

PARI

Formula

A378277 Denominators in a harmonic triangle, based on products of Fibonacci numbers.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

Formula