cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A094569 Associated with alternating row sums of array in A094568.

Original entry on oeis.org

2, 11, 78, 532, 3649, 25008, 171410, 1174859, 8052606, 55193380, 378301057, 2592914016, 17772097058, 121811765387, 834910260654, 5722560059188, 39223010153665, 268838511016464, 1842646566961586, 12629687457714635, 86565165637040862, 593326472001571396
Offset: 0

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Author

Clark Kimberling, May 12 2004

Keywords

Examples

			Obtain 11,78,532 from a(0)=2 and Fibonacci numbers 13,89,610: 11=13-2, 78=89-11, 532=610-78.
		

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{6,6,-1},{2,11,78},30] (* Harvey P. Dale, Feb 15 2025 *)
  • PARI
    Vec(-(x-2)/((x+1)*(x^2-7*x+1)) + O(x^100)) \\ Colin Barker, Nov 19 2014
    
  • PARI
    vector(30, n, n--;(fibonacci(4*n+5) + (-1)^n)/3) \\ Michel Marcus, Nov 19 2014

Formula

a(n) = F(4n+3) - a(n-1) for n >= 1, where a(0) = 2.
a(n) = (Fib(4n+5) + (-1)^n )/3. - Ralf Stephan, Dec 04 2004
a(n) = (-1)^n * sum((-1)^k*Fibonacci(4*k+3), k=0..n). - Gary Detlefs, Jan 22 2013
a(n) = 6*a(n-1) + 6*a(n-2) - a(n-3). - Colin Barker, Nov 19 2014
G.f.: -(x-2) / ((x+1)*(x^2-7*x+1)). - Colin Barker, Nov 19 2014

Extensions

More terms from Colin Barker, Nov 19 2014

A094565 Triangle read by rows: binary products of Fibonacci numbers.

Original entry on oeis.org

1, 2, 3, 5, 6, 8, 13, 15, 16, 21, 34, 39, 40, 42, 55, 89, 102, 104, 105, 110, 144, 233, 267, 272, 273, 275, 288, 377, 610, 699, 712, 714, 715, 720, 754, 987, 1597, 1830, 1864, 1869, 1870, 1872, 1885, 1974, 2584, 4181, 4791, 4880, 4893, 4895, 4896, 4901, 4935, 5168, 6765
Offset: 1

Views

Author

Clark Kimberling, May 12 2004

Keywords

Comments

Row n consists of n numbers, first F(2n-1) and last F(2n).
Central numbers: (1,6,40,273,...) = A081016.
Row sums: A001870.
Alternating row sums: 1,1,7,7,48,48,329,329; the sequence b=(1,7,48,329,...) is A004187, given by b(n)=F(4n+2)-b(n-1) for n>=2, with b(1)=1.
In each row, the difference between neighboring terms is a Fibonacci number.

Examples

			Triangle begins:
   1;
   2,   3;
   5,   6    8;
  13,  15,  16,  21;
  34,  39,  40,  42,  55;
  89, 102, 104, 105, 110, 144; ...
		

Crossrefs

Programs

  • GAP
    Flat(List([1..12], n-> List([1..n], k-> Fibonacci(2*k)*Fibonacci(2*n-2*k+1) ))); # G. C. Greubel, Jul 15 2019
  • Magma
    [Fibonacci(2*k)*Fibonacci(2*n-2*k+1): k in [1..n], n in [1..12]]; // G. C. Greubel, Jul 15 2019
    
  • Mathematica
    Table[Fibonacci[2*k]*Fibonacci[2*n-2*k+1], {n,12}, {k,n}]//Flatten (* G. C. Greubel, Jul 15 2019 *)
  • PARI
    row(n) = vector(n, k, fibonacci(2*k)*fibonacci(2*n-2*k+1));
    tabl(nn) = for(n=1, nn, print(row(n))); \\ Michel Marcus, May 03 2016
    
  • Sage
    [[fibonacci(2*k)*fibonacci(2*n-2*k+1) for k in (1..n)] for n in (1..12)] # G. C. Greubel, Jul 15 2019
    

Formula

Row n: F(2)F(2n-1), F(4)F(2n-3), ..., F(2n)F(1).

A094566 Triangle of binary products of Fibonacci numbers.

Original entry on oeis.org

1, 1, 2, 3, 4, 5, 8, 9, 10, 13, 21, 24, 25, 26, 34, 55, 63, 64, 65, 68, 89, 144, 165, 168, 169, 170, 178, 233, 377, 432, 440, 441, 442, 445, 466, 610, 987, 1131, 1152, 1155, 1156, 1157, 1165, 1220, 1597, 2584, 2961, 3016, 3024, 3025, 3026, 3029, 3050, 3194, 4181
Offset: 1

Views

Author

Clark Kimberling, May 12 2004

Keywords

Comments

For n>1, row n consists of n numbers, first F(2n-2) and last F(2n-1).
Central numbers: (1,4,9,25,64,...), essentially A081016.
Row sums: A027991. Alternating row sums: 1,1,4,4,30,30,203,203; the sequence b=(1,4,30,203,1394,...) is A094567.
In each row, the difference between neighboring terms is a Fibonacci number.

Examples

			Rows 1 to 4:
1
1 2
3 4 5
8 9 10 13
		

Crossrefs

Programs

  • PARI
    pef(k, n) = fibonacci(2*k)*fibonacci(2*n-2*k);
    pof(k, n) = fibonacci(2*n-2*k+1)*fibonacci(2*k-1);
    tabl(nn) = {for (n=1, nn, if (n==1, print1(1, ", "), if (n % 2 == 0, for (k=1, n/2, print1(pef(k,n), ", ");); forstep (k=n/2, 1, -1, print1(pof(k,n), ", "););, for (k=1, n\2, print1(pef(k,n), ", ");); forstep (k=n\2+1, 1, -1, print1(pof(k,n), ", ");););); print(););} \\ Michel Marcus, May 04 2016

Formula

Row 1 is the single number 1. For m>=1, Row 2m: F(2)F(4m-2), F(4)F(4m-4), ..., F(2m)F(2m), F(2m+1)F(2m-1), F(2m+3)F(2m-3), ..., F(4m-1)F(1) Row 2m+1: F(2)F(4m), F(4)F(4m-2), ..., F(2m+1)F(2m+1), F(2m+3)F(2m-1), F(2m+5)F(2m-3), ..., F(4m+1)F(1)
Showing 1-3 of 3 results.