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 1-4 of 4 results.

A099924 Self-convolution of Lucas numbers.

Original entry on oeis.org

4, 4, 13, 22, 45, 82, 152, 274, 491, 870, 1531, 2676, 4652, 8048, 13865, 23798, 40713, 69446, 118144, 200510, 339559, 573894, 968183, 1630632, 2742100, 4604572, 7721797, 12933334, 21637221, 36159610, 60367976, 100687786
Offset: 0

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Author

Ralf Stephan, Nov 01 2004

Keywords

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 57.

Crossrefs

Cf. A001629, A000032. Bisection: A203573 (even), 2*A203574 (odd).

Programs

  • Mathematica
    Table[Sum[LucasL[k]LucasL[n-k],{k,0,n}],{n,0,40}] (* or *) LinearRecurrence[ {2,1,-2,-1},{4,4,13,22},40] (* Harvey P. Dale, Mar 06 2012 *)

Formula

a(n) = (n+1)*L(n) + 2F(n+1) = Sum_{k=0..n} L(k)*L(n-k).
G.f.: (2-x)^2/(1-x-x^2)^2, corrected Aug 23 2022
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) - a(n-4), a(0)=4, a(1)=4, a(2)=13, a(3)=22. - Harvey P. Dale, Mar 06 2012
a(n) = 2*A099920(n+1)-A099920(n). - R. J. Mathar, Aug 23 2022

A203573 Bisection of A099924 (convolution of Lucas numbers); even arguments.

Original entry on oeis.org

4, 13, 45, 152, 491, 1531, 4652, 13865, 40713, 118144, 339559, 968183, 2742100, 7721797, 21637221, 60367976, 167787107, 464776435, 1283571068, 3535240289, 9713031489, 26627195728, 72847698655, 198929987567, 542305383076, 1476061431421
Offset: 0

Views

Author

Wolfdieter Lang, Jan 03 2012

Keywords

Comments

One half of the odd part of the bisection of A099924 is found in A203574.

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(4-11x+11x^2+x^3)/(1-3x+x^2)^2,{x,0,30}],x] (* or *) LinearRecurrence[{6,-11,6,-1},{4,13,45,152},30] (* Harvey P. Dale, Jan 11 2014 *)

Formula

a(n) = A099924(2*n), n>=0.
O.g.f.: (4-11*x+11*x^2+x^3)/(1-3*x+x^2)^2.
a(n) = 4*(n+1)*F(2*n+1)-(2*n+1)*F(2*n), n>=0, with the Fibonacci numbers F(n)=A000045(n). From the partial fraction decomposition of the o.g.f. and the Fibonacci recurrence.
a(0)=4, a(1)=13, a(2)=45, a(3)=152, a(n) = 6*a(n-1)-11*a(n-2)+6*a(n-3)-a(n-4). - Harvey P. Dale, Jan 11 2014

A201207 Half-convolution of sequence A000032 (Lucas) with itself.

Original entry on oeis.org

4, 2, 7, 11, 27, 41, 84, 137, 270, 435, 826, 1338, 2488, 4024, 7353, 11899, 21461, 34723, 61960, 100255, 177344, 286947, 503892, 815316, 1422892, 2302286, 3996619, 6466667, 11173935, 18079805, 31114236
Offset: 0

Views

Author

Wolfdieter Lang, Jan 03 2012

Keywords

Comments

For the definition of the half-convolution of a sequence with itself see a comment on A201204. There the rule for the o.g.f. is given. Here the o.g.f. is (L(x)^2 + L2(x^2))/2, with the o.g.f. L(x)=(2-x)/(1-x-x^2) of A000032, and L2(x) = (4-7*x-x^2)/((1+x)*(1-3*x+x^2)) the o.g.f. of A001254. This leads to the o.g.f given in the formula section.
For the bisection of this sequence see A203570 and A203574.

Crossrefs

Formula

a(n) = Sum_{k=0..floor(n/2)} L(k)*L(n-k), n >= 0, with the Lucas numbers L(n)=A000032(n).
O.g.f.: (4-2*x-7*x^2+6*x^3-x^4+3*x^5)/((1-3*x^2+x^4)*(1+x^2)*(1-x-x^2)). See a comment above.
a(n) = (1/4)*(2*(2*n+5+(-1)^n)*F(n+1)-(2*n+3+(-1)^n)*F(n)) +(i^n+(-i)^n)/2, n >= 0, with the Fibonacci numbers F(n)=A000045(n) and the imaginary unit i=sqrt(-1). From the partial fraction decomposition of the o.g.f. and the Fibonacci recurrence.

A203570 Bisection of A201207 (half-convolution of the Lucas sequence A000032 with itself); even part.

Original entry on oeis.org

4, 7, 27, 84, 270, 826, 2488, 7353, 21461, 61960, 177344, 503892, 1422892, 3996619, 11173935, 31114236, 86328978, 238764238, 658478176, 1811322045, 4970928809, 13613135152, 37208048132, 101518052904, 276527670100, 752102592271
Offset: 0

Views

Author

Wolfdieter Lang, Jan 03 2012

Keywords

Comments

The odd part of the bisection of A201207 is given in A203574.
See a comment on A201204 for the definition of the half-convolution of a sequence with itself, and the rule for the o.g.f.s of the bisection. Here the o.g.f. is (Lconve(x) + L2(x))/2, with the o.g.f. Lconve(x) = (4-11*x+11*x^2+x^3)/
(1-3*x+x^2)^2 of A203573 and the o.g.f. L2(x)= (4-7*x-x^2)/ ((1+x)*(1-3*x+x^2)) of A001254. This leads to the o.g.f. given in the formula section.

Crossrefs

Formula

a(n) = A201207(2*n), n>=0.
a(n) = (2*(4*n+6)*F(2*n+1)-4*(n+1)*F(2*n))/4 + (-1)^n, with the Fibonacci numbers F(n)=A000045(n).
O.g.f.: (4-13*x+4*x^3+12*x^2)/((1-3*x+x^2)^2*(1+x)). See a comment above.
Showing 1-4 of 4 results.