A046126 -id:A046126 - cp's OEIS frontend

A079484 a(n) = (2n-1)!! * (2n+1)!!, where the double factorial is A006882.

1, 3, 45, 1575, 99225, 9823275, 1404728325, 273922023375, 69850115960625, 22561587455281875, 9002073394657468125, 4348001449619557104375, 2500100833531245335015625, 1687568062633590601135546875, 1321365793042101440689133203125

Offset: 0

Benoit Cloitre, Jan 17 2003

nonn

a(n) is the determinant of M(2n+1) where M(k) is the k X k matrix with m(i,j)=j if i+j=k m(i,j)=i otherwise. - Adapted to offset 0, Rainer Rosenthal, Jun 19 2024
In the following two comments on the calculation of the terms using permanents, offset 1 is assumed. In the corresponding PARI code, this is implemented with offset 0. - Hugo Pfoertner, Jun 23 2024
(-1)^n*a(n)/2^(2n-1) is the permanent of the (m X m) matrix {1/(x_i-y_j), 1<=i<=m, 1<=j<=m}, where x_1,x_2,...,x_m are the zeros of x^m-1 and y_1,y_2,...,y_m the zeros of y^m+1 and m=2n-1.
In 1881, R. F. Scott posed a conjecture that the absolute value of permanent of square matrix with elements a(i,j)= (x_i - y_j)^(-1), where x_1,...,x_n are roots of x^n=1, while y_1,...,y_n are roots of y^n=-1, equals a((n-1)/2)/2^n, if n>=1 is odd, and 0, if n>=2 is even. After a century (in 1979), the conjecture was proved by H. Minc. - Vladimir Shevelev, Dec 01 2013
a(n) is the number of permutations in S_{2n+1} in which all cycles have odd length. - José H. Nieto S., Jan 09 2012
Number of 3-bundled increasing bilabeled trees with 2n labels. - Markus Kuba, Nov 18 2014
a(n) is the number of rooted, binary, leaf-labeled topologies with 2n+2 leaves that have n+1 cherry nodes. - Noah A Rosenberg, Feb 12 2019

			G.f. = 1 + 3*x + 45*x^2 + 1575*x^3 + 99225*x^4 + 9823275*x^5 + ...
M(5) =
  [1, 2, 3, 1, 5]
  [1, 2, 2, 4, 5]
  [1, 3, 3, 4, 5]
  [4, 2, 3, 4, 5]
  [1, 2, 3, 4, 5].
Integral_{x=0..oo} x^3*BesselK(1, sqrt(x)) = 1575*Pi. - _Olivier Gérard_, May 20 2009

Miklós Bóna, A walk through combinatorics, World Scientific, 2006.

Alois P. Heinz, Table of n, a(n) for n = 0..224
Cyril Banderier, Markus Kuba, and Michael Wallner, Analytic Combinatorics of Composition schemes and phase transitions with mixed Poisson distributions, arXiv:2103.03751 [math.PR], 2021.
Guo-Niu Han and Christian Krattenthaler, Rectangular Scott-type permanents, arXiv:math/0003072 [math.RA], 2000.
Markus Kuba and Alois Panholzer, Combinatorial families of multilabelled increasing trees and hook-length formulas, arXiv:1411.4587 [math.CO], 17 Nov 2014.
MathOverflow, Geometric / physical / probabilistic interpretations of Riemann zeta(n>1)?, answer by Tom Copeland posted in Aug 2021.
Henryk Minc, On a conjecture of R. F. Scott (1881), Linear Algebra Appl., Vol. 28 (1979), pp. 141-153.
Theodoros Theodoulidis, On the Closed-Form Expression of Carson’s Integral, Period. Polytech. Elec. Eng. Comp. Sci., Vol. 59, No. 1 (2015), pp. 26-29.
Eric Weisstein's World of Mathematics, Struve function.

Cf. A001818, A000165.
Bisection of A000246, A053195, |A013069|, |A046126|. Cf. A000909.
Cf. A001044, A010791, |A129464|, A114779, are also values of similar moments.
Equals the row sums of A162005.
Cf. A316728.
Diagonal elements of A306364 in even-numbered rows.

I:=[1, 3]; [n le 2 select I[n] else (4*n^2-8*n+3)*Self(n-1): n in [1..20]]; // Vincenzo Librandi, Nov 18 2014

a:= n-> (d-> d(2*n-1)*d(2*n+1))(doublefactorial):
seq(a(n), n=0..15);  # Alois P. Heinz, Jan 30 2013
# second Maple program:
A079484 := n-> LinearAlgebra[Determinant](Matrix(2*n+1, (i, j)-> `if`(i+j=2*n+1, j, i))): seq(A079484(n), n=0..14); # Rainer Rosenthal, Jun 18 2024

a[n_] := (2n - 1)!!*(2n + 1)!!; Table[a[n], {n, 0, 13}] (* Jean-François Alcover, Jan 30 2013 *)

/* Formula using the zeta function and a log integral:*/
L(n)= intnum(t=0, 1, log(1-1/t)^n);
Zetai(n)= -I*I^n*(2*Pi)^(n-1)/(n-1)*L(1-n);
a(m)={my(n=m+1);round(real(-I*2^(2*n-1)*Zetai(1/2-n)*L(-1/2+n)/(Zetai(-1/2+n)*L(1/2-n))))};
/* Gerry Martens, Mar 07 2011, adapted to offset 0 by Hugo Pfoertner, Jun 19 2024 */

{a(n) = if( n<0, -1 / self()(-1-n), (2*n + 1)! * (2*n)! / (n! * 2^n)^2 )}; /* Michael Somos, May 04 2017 */

{a(n) = if( n<0, -1 / self()(-1-n), my(m = 2*n + 1); m! * polcoeff( x / sqrt( 1 - x^2 + x * O(x^m) ), m))}; /* Michael Somos, May 04 2017 */

\\ using the Pochhammer symbol
a(n) = {my(P(x,k)=gamma(x+k)/gamma(x)); 4^n*round(P(1/2,n)*P(3/2,n))} \\ Hugo Pfoertner, Jun 20 2024

\\ Scott's (1881) method
a(n) = {my(m=2*n+1, X = polroots(x^m-1), Y = polroots(x^m+1), M = matrix(m, m, i, j, 1/(X[i]-Y[j]))); (-1)^n * round(2^m * real(matpermanent(M)))}; \\ Hugo Pfoertner, Jun 23 2024

D-finite with recurrence a(n) = (4*n^2 - 1) * a(n-1) for all n in Z.
a(n) = A001147(n)*A001147(n+1).
E.g.f.: 1/(1-x^2)^(3/2) (with interpolated zeros). - Paul Barry, May 26 2003
a(n) = (2n+1)! * C(2n, n) / 2^(2n). - Ralf Stephan, Mar 22 2004.
Alternatingly signed values have e.g.f. sqrt(1+x^2).
a(n) is the value of the n-th moment of (1/Pi)*BesselK(1, sqrt(x)) on the positive part of the real line. - Olivier Gérard, May 20 2009
a(n) = -2^(2*n-1)*exp(i*n*Pi)*gamma(1/2+n)/gamma(3/2-n). - Gerry Martens, Mar 07 2011
E.g.f. (odd powers) tan(arcsin(x)) = Sum_{n>=0} (2n-1)!!*(2n+1)!!*x^(2*n+1)/(2*n+1)!. - Vladimir Kruchinin, Apr 22 2011
G.f.: 1 + x*(G(0) - 1)/(x-1) where G(k) = 1 - ((2*k+2)^2-1)/(1-x/(x - 1/G(k+1))); ( continued fraction ). - Sergei N. Gladkovskii, Jan 15 2013
a(n) = (2^(2*n+3)*Gamma(n+3/2)*Gamma(n+5/2))/Pi. - Jean-François Alcover, Jul 20 2015
Limit_{n->oo} 4^n*(n!)^2/a(n) = Pi/2. - Daniel Suteu, Feb 05 2017
From Michael Somos, May 04 2017: (Start)
a(n) = (2*n + 1) * A001818(n).
E.g.f.: Sum_{n>=0} a(n) * x^(2*n+1) / (2*n+1)! = x / sqrt(1 - x^2) = tan(arcsin(x)).
Given e.g.f. A(x) = y, then x * y' = y + y^3.
a(n) = -1 / a(-1-n) for all n in Z.
0 = +a(n)*(+288*a(n+2) -60*a(n+3) +a(n+4)) +a(n+1)*(-36*a(n+2) -4*a(n+3)) +a(n+2)*(+3*a(n+2)) for all n in Z. (End)
a(n) = Sum_{k=0..2n} (k+1) * A316728(n,k). - Alois P. Heinz, Jul 12 2018
From Amiram Eldar, Mar 18 2022: (Start)
Sum_{n>=0} 1/a(n) = 1 + L_1(1)*Pi/2, where L is the modified Struve function.
Sum_{n>=0} (-1)^n/a(n) = 1 - H_1(1)*Pi/2, where H is the Struve function. (End)

Simpler description from Daniel Flath (deflath(AT)yahoo.com), Mar 05 2004

A086639 Write decimal expansion of Pi in triangular form; sequence gives left edge.

Original entry on oeis.org

3, 1, 1, 2, 5, 3, 2, 2, 4, 9, 9, 7, 8, 3, 8, 7, 2, 1, 8, 9, 5, 3, 6, 6, 3, 5, 7, 6, 2, 2, 9, 9, 4, 0, 4, 2, 3, 0, 4, 1, 6, 7, 8, 9, 9, 1, 2, 3, 0, 1, 7, 2, 2, 4, 7, 8, 3, 1, 8, 3, 0, 2, 7, 9, 1, 6, 2, 2, 6, 7, 6, 8, 1, 5, 7, 3, 7, 7, 2, 4, 9, 3, 2, 1, 9, 8, 9, 1, 2, 7, 7, 9, 4, 0, 9, 2, 9, 8, 4, 9, 9, 2, 0, 7, 0

Offset: 1

Cino Hilliard, Jul 24 2003

In the second formula, "if" can most probably be strengthened to "if and only if": Indeed, a(n) = 0 can be equal to A000030(A090897(n)) only if A090897(n) = 0, i.e., there would be a string of n consecutive zeros in the decimals of Pi from position T(n-1)+1 to position T(n). The probability that this happens appears to be zero. (Notice how A096764(n), first occurrence of n consecutive zeros, grows incredibly much faster than T(n).) Maybe this could be proved considering, e.g., a continued fraction expansion of Pi whose coefficients follow some pattern of moderate growth (as e.g. in A046126), while a very long string of zeros in the decimal expansion would mean that it is exceptionally close to the rational number given by the truncation. - M. F. Hasler, Jan 06 2023

			Triangle is
  3
  14
  159
  2653
  58979
  323846
  2643383
  27950288
  419716939
  9375105820
a(34) = 0 because in the decimals of Pi there is a 0 at position 562, following the triangular number A000217(33) = 561, i.e., in the first column of the 34th row in the above triangle. - _Michel Marcus_ and _M. F. Hasler_, Jan 06 2023

Cf. A000796, A083437.

Cf. A000030, A090897.

pi = RealDigits[Pi, 10, 5461][[1]]; Table[ pi[[n(n + 1)/2 + 1]], {n, 0, 104}]
Module[{nn=110,pid},pid=RealDigits[Pi,10,(nn(nn+1))/2][[1]];TakeList[ pid,Range[ nn]]][[;;,1]] (* Harvey P. Dale, Mar 06 2023 *)

a(n) = A000796(1-n(n-1)/2). - M. F. Hasler, Oct 20 2011

a(n) = A000030(A090897(n)) if (and probably only if) a(n) is nonzero. - Michel Marcus and M. F. Hasler, Jan 06 2023

Edited by Robert G. Wilson v, Jul 26 2003

cp's OEIS Frontend

A079484 a(n) = (2n-1)!! * (2n+1)!!, where the double factorial is A006882.

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A086639 Write decimal expansion of Pi in triangular form; sequence gives left edge.

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Extensions

A013069 Expansion of e.g.f.: exp(arcsinh(x)+log(x+1))=1+2x+3/2!x^2+3/3!x^3-3/4!x^4-15/5!*x^5...

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cp's OEIS Frontend

A079484 a(n) = (2n-1)!! * (2n+1)!!, where the double factorial is A006882.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

PARI

PARI

PARI

PARI

Formula

Extensions

A086639 Write decimal expansion of Pi in triangular form; sequence gives left edge.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Formula

Extensions

A013069 Expansion of e.g.f.: exp(arcsinh(x)+log(x+1))=1+2*x+3/2!*x^2+3/3!*x^3-3/4!*x^4-15/5!*x^5...

Views

Author

Keywords

Crossrefs

Programs

Mathematica

Extensions

A013069 Expansion of e.g.f.: exp(arcsinh(x)+log(x+1))=1+2x+3/2!x^2+3/3!x^3-3/4!x^4-15/5!*x^5...