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.

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A264960 Half-convolution of the central binomial coefficients A000984 with itself.

Original entry on oeis.org

1, 2, 10, 32, 146, 512, 2248, 8192, 35218, 131072, 556040, 2097152, 8815496, 33554432, 140107040, 536870912, 2230302098, 8589934592, 35541690568, 137438953472, 566823203656, 2199023255552, 9044910175520, 35184372088832, 144393718191496
Offset: 0

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Author

Peter Bala, Nov 29 2015

Keywords

Comments

The half-convolution of a sequence {s(n)}n>=0 with itself is defined by r(n) := Sum_{k = 0..floor(n/2)} s(k)*s(n-k). See A201204.

Crossrefs

Programs

  • GAP
    List([0..24],n->Sum([0..Int(n/2)],k->Binomial(2*k,k)*Binomial(2*n-2*k,n-k))); # Muniru A Asiru, Nov 25 2018
    
  • Magma
    [(&+[Binomial(2*k,k)*Binomial(2*n-2*k, n-k): k in [0..Floor(n/2)]]): n in [0..30]]; // G. C. Greubel, Nov 26 2018
    
  • Maple
    A264960:= n-> add(binomial(2*k,k)*binomial(2*n - 2*k, n - k),k = 0..floor(n/2)):
    seq(A264960(n),n = 0..24);
  • Mathematica
    a[n_] := Sum[Binomial[2k, k]*Binomial[2n - 2k, n - k], {k, 0, Floor[n/2]}]; Array[a, 30, 0] (* Amiram Eldar, Nov 25 2018 *)
  • PARI
    a(n) = sum(k = 0, n\2, binomial(2*k,k)*binomial(2*n - 2*k, n - k)); \\ Michel Marcus, Nov 30 2015
    
  • Sage
    [sum(binomial(2*k,k)*binomial(2*n-2*k, n-k) for k in (0..floor(n/2))) for n in range(30)] # G. C. Greubel, Nov 26 2018

Formula

a(n) = Sum_{k = 0..floor(n/2)} binomial(2*k,k)*binomial(2*n - 2*k, n - k).
a(2*n + 1) = 2^(4*n + 1) = A013776(n).
a(2*n) = (1/2)*(binomial(2*n,n)^2 + 16^n) = A112830(2*n,n).
O.g.f.: (1/2)*( 2/Pi*EllipticK(4*x) + 1/(1 - 4*x) ).
E.g.f.: (1/2)*( cosh(4*x) + sinh(4*x) + (BesselI(0,2*x))^2 ).
D-finite with recurrence: - (2*n-3)*n^2*a(n) + 4*(2*n-1)*(n-1)^2*a(n-1) + 16*(2*n-3)*(n-1)^2*a(n-2) - 64*(2*n-1)*(n-2)^2*a(n-3) = 0. - Georg Fischer, Nov 25 2022

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

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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.
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