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.

A130406 Column 1 of triangle A130405.

Original entry on oeis.org

1, 3, 13, 83, 814, 12502, 303102, 11681388, 718217460, 70660085940, 11145552305760, 2823029266531680, 1149529177121700960, 753213189796615454400, 794745942920930023732800
Offset: 0

Views

Author

Paul D. Hanna, May 24 2007, May 25 2007

Keywords

Examples

			a(n) = A003266(n+1)*[F(n+1) + F(n+2)*[1+ 1/2+ 2/3+ 3/5+...+ F(n)/F(n+1)]]:
a(3) = 1*1*2*3*( 3 + 5*(1/1 + 1/2 + 2/3) ) = 83;
a(4) = 1*1*2*3*5*( 5 + 8*(1/1 + 1/2 + 2/3 + 3/5) ) = 814;
a(5) = 1*1*2*3*5*8*( 8 + 13*(1/1 + 1/2 + 2/3 + 3/5 + 5/8) ) = 12502.

Crossrefs

Cf. A130405, A130407, A003266, A000045.

Programs

  • PARI
    a(n)=polcoeff(prod(i=0,n+1,fibonacci(i+1)+x*fibonacci(i)),1)
  • PARI
    /* Recurrence a(n) = F(n+2)*a(n-1) + F(n+1)*A003266(n+1): */ a(n)=if(n==0,1,fibonacci(n+2)*a(n-1)+fibonacci(n+1)*prod(i=1,n+1,fibonacci(i)))
  • PARI
    a(n)=prod(i=1,n+1,fibonacci(i))*(fibonacci(n+1) + fibonacci(n+2)*sum(k=0,n,fibonacci(k)/fibonacci(k+1)))

Formula

a(n) = F(n+2)*a(n-1) + F(n+1)*A003266(n+1), where A003266(n) is the product of the first n nonzero Fibonacci numbers (A000045) and F(n) = A000045(n).
a(n) = A003266(n+1)*[ F(n+1) + F(n+2)*Sum_{k=0..n} F(k)/F(k+1) ] where F(n)=A000045(n) is the n-th Fibonacci number.

A130407 A diagonal of triangle A130405.

Original entry on oeis.org

1, 3, 9, 37, 233, 2254, 34342, 827262, 31730508, 1943441460, 190609515540, 29988517246560, 7579307667005280, 3080578207713982560, 2015291663362285214400, 2123462159890867147060800
Offset: 0

Views

Author

Paul D. Hanna, May 24 2007, May 25 2007

Keywords

Examples

			a(n) = A003266(n)*[F(n+2) + F(n+1)*[1+ 2/1+ 3/2+ 5/3+...+ F(n+1)/F(n)]]:
a(3) = 1*1*2*( 5 + 3*(1/1 + 2/1 + 3/2) ) = 37;
a(4) = 1*1*2*3*( 8 + 5*(1/1 + 2/1 + 3/2 + 5/3) ) = 233;
a(5) = 1*1*2*3*5*( 13 + 8*(1/1 + 2/1 + 3/2 + 5/3 + 8/5) ) = 2254.

Crossrefs

Cf. A130405, A130406, A003266, A000045.

Programs

  • PARI
    a(n)=polcoeff(prod(i=0,n+1,fibonacci(i+1)+x*fibonacci(i)),n)
  • PARI
    /* Recurrence a(n) = F(n+1)*a(n-1) + F(n+2)*A003266(n): */ {a(n)=if(n==0,1,fibonacci(n+1)*a(n-1)+fibonacci(n+2)*prod(i=1,n,fibonacci(i)))}
  • PARI
    a(n)=prod(i=1,n,fibonacci(i))*(fibonacci(n+2) + fibonacci(n+1)*sum(k=1,n,fibonacci(k+1)/fibonacci(k)) )

Formula

a(n) = F(n+1)*a(n-1) + F(n+2)*A003266(n), where A003266(n) is the product of the first n nonzero Fibonacci numbers (A000045) and F(n) = A000045(n).
a(n) = A003266(n)*[ F(n+2) + F(n+1)*Sum_{k=0..n} F(k+1)/F(k) ] where F(n)=A000045(n) is the n-th Fibonacci number.

A153861 Triangle read by rows, binomial transform of triangle A153860.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 3, 6, 4, 1, 4, 10, 10, 5, 1, 5, 15, 20, 15, 6, 1, 6, 21, 35, 35, 21, 7, 1, 7, 28, 56, 70, 56, 28, 8, 1, 8, 36, 84, 126, 126, 84, 36, 9, 1, 9, 45, 120, 210, 252, 210, 120, 45, 10, 1, 10, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1
Offset: 0

Views

Author

Gary W. Adamson, Jan 03 2009

Keywords

Comments

Row sums = A095121: (1, 2, 6, 14, 30, 62, 126,...).
Triangle T(n,k), 0<=k<=n, read by rows, given by [1,1,-1,1,0,0,0,0,0,0,0,...] DELTA [1,0,-1,1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Jan 03 2009
A123110*A007318 as infinite lower triangular matrices. - Philippe Deléham, Jan 06 2009
A153861 is the fusion of polynomial sequences p(n,x)=x^n+x^(n-1)+...+x+1 and q(n,x)=(x+1)^n; see A193722 for the definition of fusion. - Clark Kimberling, Aug 06 2011

Examples

			First few rows of the triangle are:
1;
1, 1;
2, 3, 1;
3, 6, 4, 1;
4, 10, 10, 5, 1;
5, 15, 20, 15, 6, 1;
6, 21, 35, 35, 21, 7, 1;
7, 28, 56, 70, 56, 28, 8, 1;
8, 36, 84, 126, 126, 84, 36, 9, 1;
9, 45, 120, 210, 252, 210, 120, 45, 10, 1;
...

Links

Crossrefs

Cf. A153860, A095121, A130405.
This is A137396 without the initial column and without signs.

Programs

  • Mathematica
    z = 10; c = 1; d = 1;
p[0, x_] := 1
p[n_, x_] := x*p[n - 1, x] + 1; p[n_, 0] := p[n, x] /. x -> 0;
q[n_, x_] := (c*x + d)^n
t[n_, k_] := Coefficient[p[n, x], x^k]; t[n_, 0] := p[n, x] /. x -> 0;
w[n_, x_] := Sum[t[n, k]*q[n + 1 - k, x], {k, 0, n}]; w[-1, x_] := 1
g[n_] := CoefficientList[w[n, x], {x}]
TableForm[Table[Reverse[g[n]], {n, -1, z}]]
Flatten[Table[Reverse[g[n]], {n, -1, z}]]  (* A193815 *)
TableForm[Table[g[n], {n, -1, z}]]
Flatten[Table[g[n], {n, -1, z}]]   (* A153861 *)
(* Clark Kimberling, Aug 06 2011 *)

Formula

Triangle read by rows, A007318 * A153860. Remove left two columns of Pascal's triangle and append (1, 1, 2, 3, 4, 5,...).
As a recursive operation by way of example, row 3 = (3, 6, 4, 1) =
[1, 1, 1, 0] * (flipped Pascal's triangle matrix) = [1, 3, 3, 1]
[1, 2, 1, 0]
[1, 1, 0, 0]
[1, 0, 0, 0].
(Cf. analogous operation in A130405, but in A153861 the linear multiplier = [1,1,1,...,0].)
T(n,k) = 2*T(n-1,k)+T(n-1,k-1)-T(n-2,k)-T(n-2,k-1), T(0,0) = T(1,0) = T(1,1) = T(2,2) = 1, T(2,0)=2, T(2,1)=3, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Dec 15 2013
G.f.: (1-x+x^2+x^2*y)/((x-1)*(-1+x+x*y)). - R. J. Mathar, Aug 11 2015
Showing 1-3 of 3 results.

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