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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A360313 a(n) = Sum_{k=0..floor(n/2)} (-1)^k * binomial(n-1-k,n-2*k) * binomial(2*k,k).

Original entry on oeis.org

1, 0, -2, -2, 4, 10, -4, -38, -22, 114, 188, -234, -914, -18, 3376, 3338, -9416, -21718, 14416, 96338, 39274, -328558, -471344, 795398, 2586064, -517690, -10453424, -8272658, 32186818, 63596494, -61876584, -307070174, -62655330, 1129250706, 1356328788
Offset: 0

Views

Author

Seiichi Manyama, Feb 03 2023

Keywords

Links

Crossrefs

Cf. A360314, A360315.
Cf. A026585, A360293.

Programs

  • Mathematica
    RecurrenceTable[{a[0]==1,a[1]==0,a[2]==-2,a[n]==1/n (2(n-1)a[n-1]-(5n-6)a[n-2]+2(2n-5)a[n-3])},a,{n,40}] (* Harvey P. Dale, Sep 20 2024 *)
  • PARI
    a(n) = sum(k=0, n\2, (-1)^k*binomial(n-1-k, n-2*k)*binomial(2*k, k));
  • PARI
    my(N=40, x='x+O('x^N)); Vec(1/sqrt(1+4*x^2/(1-x)))

Formula

G.f.: 1 / sqrt(1+4*x^2/(1-x)).
n*a(n) = 2*(n-1)*a(n-1) - (5*n-6)*a(n-2) + 2*(2*n-5)*a(n-3).

A377215 Expansion of 1/(1 - 4*x^2/(1-x))^(5/2).

Original entry on oeis.org

1, 0, 10, 10, 80, 150, 640, 1550, 5190, 13870, 41912, 115650, 333490, 925970, 2607540, 7220062, 20053700, 55230870, 152005380, 416295350, 1137980678, 3100453710, 8429823180, 22862244210, 61882724100, 167159512794, 450739897980, 1213298505770, 3260824389510
Offset: 0

Views

Author

Seiichi Manyama, Oct 20 2024

Keywords

Links

Crossrefs

Cf. A377199, A377216.
Cf. A026585, A377204.

Programs

  • Magma
    R:=PowerSeriesRing(Rationals(), 35); Coefficients(R!( 1/(1 - 4*x^2/(1-x))^(5/2))); // Vincenzo Librandi, May 08 2025
  • Mathematica
    Table[Sum[(-4)^k*Binomial[-5/2,k]*Binomial[n-k-1,n-2*k],{k,0,Floor[n/2]}],{n,0,35}] (* Vincenzo Librandi, May 08 2025 *)
  • PARI
    a(n) = sum(k=0, n\2, (-4)^k*binomial(-5/2, k)*binomial(n-k-1, n-2*k));

Formula

a(n) = (2*(n-1)*a(n-1) + (3*n+14)*a(n-2) - 2*(2*n-1)*a(n-3))/n for n > 2.
a(n) = Sum_{k=0..floor(n/2)} (-4)^k * binomial(-5/2,k) * binomial(n-k-1,n-2*k).
a(n) ~ n^(3/2) * 2^(3*n - 1/2) / (3 * 17^(5/4) * sqrt(Pi) * (sqrt(17) - 1)^(n - 5/2)). - Vaclav Kotesovec, May 03 2025

A026570 a(n) = A026568(n,n-1), also a(n) = number of integer strings s(0),...,s(n) counted by A026568 such that s(n)=1.

Original entry on oeis.org

1, 1, 4, 7, 20, 43, 111, 259, 648, 1565, 3885, 9533, 23662, 58547, 145630, 362151, 903110, 2253615, 5633359, 14094035, 35304658, 88511733, 222115782, 557819793, 1401987930, 3526066273, 8874034647, 22346581133, 56304982154
Offset: 1

Views

Author

Clark Kimberling

Keywords

Comments

Also a(n) = T'(n,n-1), T' given by A026584. Also a(n) = number of integer strings s(0),...,s(n) counted by T' such that s(n)=1.

Crossrefs

Cf. A026568, A026569, A026584, A026585.

Formula

Conjecture: (n+1)*a(n) -2*n*a(n-1) +(-3*n-1)*a(n-2) +2*(2*n-3)*a(n-3)=0. - R. J. Mathar, Jun 23 2013
If recurrence is correct then a(n) = (A026569(n+1)-A026569(n))/2 = A026585(n+1)/2. - Mark van Hoeij, Nov 29 2024
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