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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A222472 Numerator sequence of the n-th convergent of the continued fraction 1/(1+3/(2+3/(3+3/(4+...

Original entry on oeis.org

1, 2, 9, 42, 237, 1548, 11547, 97020, 907821, 9369270, 105785433, 1297533006, 17185285377, 244486594296, 3718854770571, 60235136112024, 1035153878216121, 18813475216226250, 360561490742947113, 7267670240507621010
Offset: 1

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Author

Gary Detlefs and Wolfdieter Lang, Mar 09 2013

Keywords

Comments

The corresponding denominator sequence is A213190.
a(n) = Phat(n,3) with the numerator polynomials Phat of A221913. All the given formulas follow from there and from the comments given under A084950. The limit of the continued fraction (0 + K_{k=1..oo} (3/k))/3 = 1/(1+3/(2+3/(3+3/(4+... is (1/3)*sqrt(3)*BesselI(1,2*sqrt(3))/BesselI(0,2*sqrt(3)) = 0.484516174987404...
For a combinatorial interpretation in terms of labeled Morse codes see a comment on A221913. Here each dash has label x=3, and the dots have label j if they are at position j. Labels are multiplied and all codes on positions [2,...,n+1] are summed.

Examples

			a(4) = 4*a(3) + 2*a(2) = 4*9 + 3*2 = 42.
Continued fraction convergent: 1/(1+3/(2+3/(3+3/4))) = 14/29 = 42/87 = a(4)/A213190(4).
Morse code: a(5) = 237 from the sum of all 5 labeled codes on [2,3,4,5], one with no dash, three with one dash and one with two dashes: 5!/1 + (4*5 + 2*5 + 2*3)*(3) +3^2 = 237.

Crossrefs

Cf. A084950, A221913, A222467, A001040(n+1) (x=1), A058798 (x=-1).

Programs

  • PARI
    a=vector(50); a[1]=1;a[2]=2; for(n=3, #a, a[n]=n*a[n-1]+3*a[n-2]); a \\ Altug Alkan, Apr 20 2018

Formula

Recurrence: a(n) = n*a(n-1) + 3*a(n-2), with a(-1) = 1/3, a(0) = 0, n >= 1.
As a sum: a(n) = Sum_{m=0..floor(n/2)} b(n-m,m)*3^m, n >= 1, with b(n,m) = binomial(n,m)*(n+1)!/(m+1)! = |A066667(n,m)| (Laguerre coefficients, parameter alpha =1).
Explicit form: a(n) = -2*(sqrt(3))^n*(BesselK(1, -2*sqrt(3))*BesselI(n+1, -2*sqrt(3)) + (-1)^(n+1)*BesselI(1,-2*sqrt(3))*BesselK(n+1,-2*sqrt(3))).
E.g.f.: Pi*(BesselJ(1, 2*I*sqrt(3)*sqrt(1-z))*BesselY(1, 2*I*sqrt(3)) - BesselY(1, (2*I)*sqrt(3)*sqrt(1-z))*BesselJ(1, 2*I*sqrt(3)))/sqrt(1-z). Here Phat(0,x) = 0.
Asymptotics: lim_{n->oo} a(n)/n! = BesselI(1,2*sqrt(3))/(sqrt(3)) = 3.468649618760...

A303224 a(0)=0, a(1)=1; for n>1, a(n) = n*a(n-1) - 3*a(n-2).

Original entry on oeis.org

0, 1, 2, 3, 6, 21, 108, 693, 5220, 44901, 433350, 4632147, 54285714, 691817841, 9522592632, 140763435957, 2223647197416, 37379712048201, 666163875275370, 12544974494087427, 248900998255922430, 5189286039892108749, 113417589882858625188, 2593036709186072053077, 61892628250817153398284
Offset: 0

Views

Author

Bruno Berselli, Apr 20 2018

Keywords

Comments

a(n) is divisible by 3^floor(n/3).

Links

Crossrefs

Cf. A058798: a(n) = n*a(n-1) - a(n-2).
Cf. A222470: a(n) = n*a(n-1) - 2*a(n-2), without 0.
Cf. A222472: a(n) = n*a(n-1) + 3*a(n-2), without 0.
Cf. A221913.

Programs

  • Mathematica
    RecurrenceTable[{a[0] == 0, a[1] == 1, a[n] == n a[n - 1] - 3 a[n - 2]}, a, {n, 0, 30}]
Flatten[{0, Table[n!*HypergeometricPFQ[{1/2 - n/2, 1 - n/2}, {2, 1 - n, -n}, -12], {n, 1, 25}]}] (* Vaclav Kotesovec, Apr 20 2018 *)
Round[Table[-2 I^n 3^(n/2) (BesselI[1 + n, -2 I Sqrt[3]] BesselK[1, -2 I Sqrt[3]] + (-1)^n BesselI[1, 2 I Sqrt[3]] BesselK[1 + n, -2 I Sqrt[3]]), {n, 0, 25}]] (* Vaclav Kotesovec, Apr 20 2018 *)
  • PARI
    a=vector(30); a[1]=0; a[2]=1; for(n=3, #a, a[n]=(n-1)*a[n-1]-3*a[n-2]); a

Formula

From Peter Bala, Apr 20 2018: (Start)
a(n) = Sum_{k = 0..floor((n-1)/2)} (-3)^k*binomial(n-k,k+1)*binomial(n-k-1,k)*(n-2*k-1)!.
a(n)/n! ~ BesselJ(1, 2*sqrt(3)) / sqrt(3). (End)
a(n) = -2 * i^n * 3^(n/2) * (BesselI(1+n, -2*i*sqrt(3)) * BesselK(1,-2*i*sqrt(3)) + (-1)^n * BesselI(1, 2*i*sqrt(3)) * BesselK(1+n, -2*i*sqrt(3))), where i is the imaginary unit. - Vaclav Kotesovec, Apr 20 2018

A330609 T(n, k) = binomial(n-k-1, k-1)*(n-k)!/k! for n >= 0 and 0 <= k <= floor(n/2). Irregular triangle read by rows.

Original entry on oeis.org

1, 0, 0, 1, 0, 2, 0, 6, 1, 0, 24, 6, 0, 120, 36, 1, 0, 720, 240, 12, 0, 5040, 1800, 120, 1, 0, 40320, 15120, 1200, 20, 0, 362880, 141120, 12600, 300, 1, 0, 3628800, 1451520, 141120, 4200, 30, 0, 39916800, 16329600, 1693440, 58800, 630, 1
Offset: 0

Views

Author

Peter Luschny, Dec 27 2019

Keywords

Comments

Also the antidiagonals of the Lah triangle A271703.

Examples

			Triangle begins:
[0] 1
[1] 0
[2] 0, 1
[3] 0, 2
[4] 0, 6,     1
[5] 0, 24,    6
[6] 0, 120,   36,    1
[7] 0, 720,   240,   12
[8] 0, 5040,  1800,  120,  1
[9] 0, 40320, 15120, 1200, 20

Links

Crossrefs

Variants: A180047, A221913. Row sums: A001053.
Cf. A271703.

Programs

  • Maple
    T := (n, k) -> binomial(n-k-1, k-1)*(n-k)!/k!:
seq(seq(T(n, k), k=0..floor(n/2)), n=0..12);
# Alternative:
T := proc(n, k) option remember;
if (n=0 and k=0) or (n=2 and k=1) then 1 elif (k < 1) or (k > ceil(n/2)) then 0
else (n-1)*T(n-1, k) + T(n-2, k-1) fi end: seq(seq(T(n, k), k=0..n/2), n=0..12);
  • Mathematica
    Table[Binomial[n-k-1,k-1] (n-k)!/k!,{n,0,20},{k,0,Floor[n/2]}]//Flatten (* Harvey P. Dale, Oct 19 2021 *)

Formula

T(0,0) = T(2,1) = 1. If k < 1 or k > ceiling(n/2) then T(n,k) = 0. Otherwise:
T(n, k) = (n-1)*T(n-1, k) + T(n-2, k-1)
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