A059480 -id:A059480

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A180049

Coefficient triangle of the numerators of the (n-th convergents to) the continued fraction 1/(w+2/(w+3/(w+4/... .

+10
4

1, 0, 1, 3, 0, 1, 0, 7, 0, 1, 15, 0, 12, 0, 1, 0, 57, 0, 18, 0, 1, 105, 0, 141, 0, 25, 0, 1, 0, 561, 0, 285, 0, 33, 0, 1, 945, 0, 1830, 0, 510, 0, 42, 0, 1, 0, 6555, 0, 4680, 0, 840, 0, 52, 0, 1, 10395, 0, 26685, 0, 10290, 0, 1302, 0, 63, 0, 1, 0, 89055, 0, 82845, 0, 20370, 0, 1926, 0, 75, 0, 1

(list; table; graph; refs; listen; history; text; internal format)

OFFSET

1,4

COMMENTS

Equivalence to the recurrence formula needs formal proof. This continued fraction converges to 0.525135276160981... for w=1. A conjecture by Ramanujan puts this equal to -1 + 1/(sqrt(e*Pi/2) - Sum_{k>=1} 1/(2k-1)!!). Row sums equal A059480.

LINKS

Table of n, a(n) for n=1..78.

Hungarian discussion forum

FORMULA

b(0)=1; b(1)=w; b(n) = w*b(n-1) + (n+1)*b(n-2) (conjecture).

EXAMPLE

The numerator of 1/(w+2/(w+3/(w+4/(w+5/(w+6/w))))) equals 57w + 18w^3 + w^5.

From Philippe Deléham, Nov 06 2013: (Start)

Triangle begins:

0, 1;

3, 0, 1;

0, 7, 0, 1;

15, 0, 12, 0, 1;

0, 57, 0, 18, 0, 1;

105, 0, 141, 0, 25, 0, 1;

0, 561, 0, 285, 0, 33, 0, 1;

945, 0, 1830, 0, 510, 0, 42, 0, 1;

0, 6555, 0, 4680, 0, 840, 0, 52, 0, 1;

10395, 0, 26685, 0, 10290, 0, 1302, 0, 63, 0, 1;

... (End)

[extended by M. F. Hasler, Oct 21 2014]

MATHEMATICA

Table[ CoefficientList[ Numerator[ Together[ Fold[ #2/(w+#1) &, Infinity, Reverse @ Table[ k, {k, 1, n} ] ] ] ], w ], {n, 2, 16} ] or equivalently Clear[ b ]; b[ 0 ]=1; b[ 1 ]=w; b[ n_ ]:=b[ n ] = w b[ n-1 ]+(n+1) b[ n-2 ]; Table[ CoefficientList[ b[ k ]//Expand, w ], {k, 0, 14} ]

PROG

(PARI) t=x-w; for(n=1, 12, t=substpol(t, x, w+n/x); print(Vecrev(numerator(substpol(t, x, w))))) \\ M. F. Hasler, Oct 21 2014

CROSSREFS

Cf. A059480, A084950, A180047, A180048, A230698.

KEYWORD

nonn,tabl

AUTHOR

Wouter Meeussen, Aug 08 2010

STATUS

approved

A070895

Triangle read by rows where T(n+1,k)=T(n,k)+n*T(n-1,k) starting with T(n,n)=1 and T(n,k)=0 if n<k.

+10
1

1, 1, 1, 2, 1, 1, 4, 3, 1, 1, 10, 6, 4, 1, 1, 26, 18, 8, 5, 1, 1, 76, 48, 28, 10, 6, 1, 1, 232, 156, 76, 40, 12, 7, 1, 1, 764, 492, 272, 110, 54, 14, 8, 1, 1, 2620, 1740, 880, 430, 150, 70, 16, 9, 1, 1, 9496, 6168, 3328, 1420, 636, 196, 88, 18, 10, 1, 1, 35696, 23568

(list; table; graph; refs; listen; history; text; internal format)

OFFSET

0,4

COMMENTS

For n>k+1, T(n,k) is a multiple of k+2.

Eigentriangle of inverse of (-1)^(n-k)*A094587. Row sums are A187044. - Paul Barry, Mar 02 2011

LINKS

Table of n, a(n) for n=0..67.

FORMULA

T(n, k+1)=(T(n, k-1)-T(n-1, k))/k for 0<k<n.

EXAMPLE

Rows start: 1; 1,1; 2,1,1; 4,3,1,1; 10,6,4,1,1; etc.

Triangle begins

1, 1,

2, 1, 1,

4, 3, 1, 1,

10, 6, 4, 1, 1,

26, 18, 8, 5, 1, 1,

76, 48, 28, 10, 6, 1, 1,

232, 156, 76, 40, 12, 7, 1, 1

Production matrix begins

1, 1,

1, 0, 1,

1, 1, 0, 1,

2, 1, 1, 0, 1,

4, 3, 1, 1, 0, 1,

10, 6, 4, 1, 1, 0, 1,

26, 18, 8, 5, 1, 1, 0, 1,

76, 48, 28, 10, 6, 1, 1, 0, 1,

232, 156, 76, 40, 12, 7, 1, 1, 0, 1

Inverse begins

-1, 1,

-1, -1, 1,

0, -2, -1, 1,

0, 0, -3, -1, 1,

0, 0, 0, -4, -1, 1,

0, 0, 0, 0, -5, -1, 1,

0, 0, 0, 0, 0, -6, -1, 1

- Paul Barry, Mar 02 2011

CROSSREFS

Columns include A000085, A000932, A059480. Right hand columns effectively include A000012 (twice), A000027, A005843, A028552. Cf. A062323 for a triangle with similar formulas.

KEYWORD

easy,nonn,tabl

AUTHOR

Henry Bottomley, May 23 2002

STATUS

approved

A247376

Triangular array: row n gives the coefficients of the polynomial p(n,x) defined in Comments.

+10
1

1, 2, 2, 3, 5, 5, 15, 8, 8, 35, 33, 13, 80, 131, 48, 21, 171, 409, 279, 34, 355, 1180, 1375, 384, 55, 715, 3128, 5335, 2895, 89, 1410, 7858, 18510, 17029, 3840, 144, 2730, 18851, 58253, 78609, 35685, 233, 5208, 43629, 171059, 317758, 243873, 46080, 377, 9810

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,2

COMMENTS

The polynomial p(n,x) is the numerator of the rational function given by f(n,x) = 1 + (2*x + 1)/f(n-1,x), where f(0,x) = 1.

(Sum of numbers in row n) = A059480(n+1) for n >= 0.

(Column 1) is essentially A000045 (Fibonacci numbers).

LINKS

Clark Kimberling, Rows 0..100, flattened

FORMULA

f(0,x) = 1/1, so that p(0,x) = 1

f(1,x) = (2 + 2 x)/1, so that p(1,x) = 2 + 2 x;

f(2,x) = (3 + 5 x)/(2 + 2 x), so that p(2,x) = 3 + 5 x.

First 6 rows of the triangle of coefficients:

2 2

3 5

5 15 8

8 35 33

13 80 131 48

MATHEMATICA

z = 15; f[x_, n_] := 1 + (2 x + 1)/f[x, n - 1]; f[x_, 1] = 1;

t = Table[Factor[f[x, n]], {n, 1, z}]

u = Numerator[t]

TableForm[Table[CoefficientList[u[[n]], x], {n, 1, z}]] (* A247376 array *)

Flatten[CoefficientList[u, x]] (* A247376 sequence *)

PROG

(PARI) rown(n) = if (n==0, 1, 1 + (2*x+1)/rown(n-1));

tabl(nn) = for (n=0, nn, print(Vecrev(numerator(rown(n))))); \\ Michel Marcus, Oct 28 2014

CROSSREFS

Cf. A059480, A000045.

KEYWORD

nonn,tabf,easy

AUTHOR

Clark Kimberling, Oct 23 2014

STATUS

approved

A371943

Number of permutations that end with a consecutive pattern 123, and avoid consecutive patterns 123 and 213 elsewhere.

+10
0

0, 0, 0, 1, 2, 10, 28, 116, 388, 1588, 5960, 25168, 102856, 453608, 1985008, 9163360, 42486128, 205065136, 1000056928, 5035366208, 25689681760, 134588839648, 715328668736, 3889568161408, 21463055829568, 120839175460160, 690344333849728, 4015753752384256

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,5

COMMENTS

(This can be proved by observing the possible positions of n.)

LINKS

Table of n, a(n) for n=0..27.

Sergi Elizalde and Yixin Lin, Penney's game for permutations, arXiv:2404.06585 [math.CO], 2024.

FORMULA

a(0)=a(1)=a(2)=0, a(3)=1, a(4)=2; a(n) = a(n-1)+(n-1)*a(n-2)+b(n-1), where b(n) = b(n-1)+(n-1)*b(n-2) is the same sequence as A059480, up to the first initial terms. Here, our b(n) has initial terms 0, 0, 0, 1, 4.

From Vaclav Kotesovec, Apr 14 2024: (Start)

a(n) ~ c * n^((n+1)/2) * exp(sqrt(n) - n/2), where c = exp(-1/4) / sqrt(2) - exp(1/4) * sqrt(Pi) * erfc(1/sqrt(2)) / 2 = 0.189615662815288097469466802437...

E.g.f.: -1 + exp(x*(2 + x)/2) * (1 + x) + exp((1 + x)^2/2) * sqrt(Pi/2) * (2 + x) * (erf(1/sqrt(2)) - erf((1 + x)/sqrt(2))). (End)

EXAMPLE

For n=0, 1, 2, there are no permutations ending with 123. Hence, a(0)=a(1)=a(2)=0. For n=3, a(3)=1 since 123 is the only permutation that ends with 123. For n=4, a(4)=2 with qualified permutations 3124, 4123. For n=5, a(5)=10 with qualified permutations 14235, 15234, 24135, 25134, 34125, 35124, 43125, 45123, 53124, 54123.

MAPLE

a:= proc(n) option remember; `if`(n<3, 0, `if`(n=3, 1,

2*a(n-1)+2*(n-2)*(a(n-2)-a(n-3))-(n-2)*(n-3)*a(n-4)))

end:

seq(a(n), n=0..30); # Alois P. Heinz, Apr 13 2024

MATHEMATICA

a[n_]:=a[n]=If[n<3, 0, If[n==3, 1, 2*a[n-1]+2*(n-2)*(a[n-2]-a[n-3])-(n-2)*(n-3)*a[n-4]]]; Table[a[n], {n, 0, 27}] (* Stefano Spezia, Apr 13 2024 *)

RecurrenceTable[{a[n] == 2*a[n-1] + 2*(n-2)*(a[n-2] - a[n-3]) - (n-2)*(n-3)*a[n-4], a[0] == 0, a[1] == 0, a[2] == 0, a[3] == 1}, a, {n, 0, 30}] (* Vaclav Kotesovec, Apr 14 2024 *)

nmax = 30; FullSimplify[CoefficientList[Series[-1 + E^(x*(2 + x)/2) * (1 + x) + E^((1 + x)^2/2) * Sqrt[Pi/2] * (2 + x)*(Erf[1/Sqrt[2]] - Erf[(1 + x)/Sqrt[2]]), {x, 0, nmax}], x] * Range[0, nmax]!] (* Vaclav Kotesovec, Apr 14 2024 *)

PROG

(Python)

def aList(len):

b = [0, 0, 0, 1, 4]

a = [0, 0, 0, 1, 2]

for i in range(4, len):

b.append(b[i] + i * b[i - 1])

a.append(a[i] + i * a[i - 1] + b[i])

return a

print(aList(27))

CROSSREFS

Cf. A059480.

KEYWORD

nonn

AUTHOR

Yixin Lin, Apr 13 2024

STATUS

approved

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