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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A051622 a(n) = (4*n+10)(!^4)/10(!^4), related to A000407 ((4*n+2)(!^4) quartic, or 4-factorials).

Original entry on oeis.org

1, 14, 252, 5544, 144144, 4324320, 147026880, 5587021440, 234654900480, 10794125422080, 539706271104000, 29144138639616000, 1690360041097728000, 104802322548059136000, 6916953288171902976000, 484186730172033208320000
Offset: 0

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Author

Keywords

Comments

Row m=10 of the array A(5; m,n) := ((4*n+m)(!^4))/m(!^4), m >= 0, n >= 0.
From Zerinvary Lajos, Feb 15 2008: (Start)
a(n) = A001813(n+3)/120.
a(n) = A051618(n+1)/10. (End)

Crossrefs

Cf. A047053, A007696(n+1), A000407, A034176(n+1), A034177(n+1), A051617-A051621 (rows m=0..9).

Programs

  • Magma
    m:=30; R:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!(1/(1-4*x)^(14/4))); [Factorial(n-1)*b[n]: n in [1..m]]; // G. C. Greubel, Aug 15 2018
  • Maple
    seq(mul((n+3+k), k=1..n+3)/120, n=0..18); # Zerinvary Lajos, Feb 15 2008
  • Mathematica
    s=1;lst={s};Do[s+=n*s;AppendTo[lst, s], {n, 13, 5!, 4}];lst (* Vladimir Joseph Stephan Orlovsky, Nov 08 2008 *)
    With[{nn = 30}, CoefficientList[Series[1/(1 - 4*x)^(7/2), {x, 0, nn}], x]*Range[0, nn]!] (* G. C. Greubel, Aug 15 2018 *)
  • PARI
    x='x+O('x^30); Vec(serlaplace(1/(1-4*x)^(14/4))) \\ G. C. Greubel, Aug 15 2018
    

Formula

a(n) = ((4*n+10)(!^4))/10(!^4) = A000407(n+2)/(6*10).
E.g.f.: 1/(1-4*x)^(7/2).
G.f.: G(0)/2, where G(k)= 1 + 1/(1 - 2*x/(2*x + 1/(2*k+7)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 02 2013

A265891 a(n) = A099563(A000407(n)); the most significant digit in factorial base representation of (2n+1)! / n!.

Original entry on oeis.org

1, 1, 2, 1, 3, 8, 2, 6, 1, 3, 10, 1, 5, 14, 1, 5, 16, 1, 5, 15, 1, 4, 12, 1, 3, 9, 28, 2, 6, 19, 1, 3, 11, 35, 2, 6, 19, 1, 3, 10, 30, 1, 4, 14, 44, 2, 6, 20, 61, 2, 8, 25, 1, 3, 10, 31, 1, 3, 11, 35, 1, 4, 12, 38, 1, 4, 12, 39, 1, 4, 12, 39, 1, 3, 11, 36, 1, 3, 10, 33, 102, 3, 9, 28, 89, 2, 7, 23, 74, 1, 6, 19, 59
Offset: 0

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Author

Antti Karttunen, Dec 20 2015

Keywords

Examples

			The terms A000407(0) .. A000407(8) in factorial base representation (A007623) look as:
  1, 100, 2200, 110000, 3000000, 82000000, 2374000000, 65500000000, 1550000000000, ...
Taking the first digit (actually: a place holder value) of each gives the terms a(0) .. a(8) of this sequence: 1, 1, 2, 1, 3, 8, 2, 6, 1, ...
		

Crossrefs

Main diagonal of A265890 (apart from the corner term).
Cf. A265897 (positions of ones).
Cf. also A265894.

Programs

  • Mathematica
    a[n_] := Module[{k = (2*n+1)!/n!, m = 2, r, d=0}, While[{k, r} = QuotientRemainder[k, m]; k != 0 || r != 0, If[r > 0, d = r]; m++]; d]; Array[a, 100, 0] (* Amiram Eldar, Feb 14 2024 *)
  • PARI
    allocatemem((2^31)); \\ Enough?
    A099563(n) = { my(i=2,dig=0); until(0==n, dig = n % i; n = (n - dig)/i; i++); return(dig); };
    A265891 = n -> A099563(((2*n)+1)! / n!);
    
  • Scheme
    (define (A265891 n) (A099563 (A000407 n)))
    
  • Scheme
    (define (A265891 n) (A265890bi (+ 1 n) (+ 1 n))) ;; Code for A265890bi given in A265890.

Formula

a(n) = A099563(A000407(n)).
a(n) = A265890(n+1, n+1).

A265897 Numbers n for which (2n+1)! / n! [= A000407(n)] is >= k! but < 2*k! for some k; positions of ones in A265891.

Original entry on oeis.org

0, 1, 3, 8, 11, 14, 17, 20, 23, 30, 37, 41, 52, 56, 60, 64, 68, 72, 76, 89, 93, 97, 106, 110, 119, 128, 132, 137, 146, 155, 164, 169, 178, 183, 197, 202, 216, 221, 226, 231, 260, 265, 270, 275, 280, 285, 290, 295, 300, 331, 336, 341, 346, 351, 367, 372, 377, 393, 398, 403, 414, 419, 435, 440, 451, 456, 467, 472, 483, 494, 499, 510
Offset: 0

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Author

Antti Karttunen, Dec 24 2015

Keywords

Comments

a(0) = 0 is a special case in this sequence, thus the indexing starts from zero.
Numbers n such that A000407(n) is in A265334.

Crossrefs

Cf. also A265898.

A086985 a(n) = A000407(n) - A086984(n).

Original entry on oeis.org

0, 0, 0, 144, 6000, 198000, 6420960, 218453760, 7988440320, 316674489600, 13630070361600, 635880851020800, 32050874586931200, 1738651147833600000, 101100246720625152000, 6277170099060596736000, 414620927349422137344000
Offset: 1

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Author

Jon Perry, Jul 27 2003

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_]:=(2*n-1)!/(n-1)!-n!-(2^(n-1)-1)*n*n!; Array[a,17] (* Stefano Spezia, Mar 07 2025 *)
  • PARI
    a(n)=(2*n-1)!/(n-1)!-n!-sum(i=2,n,binomial(n-1,n-i)*n*n!)

Formula

a(n) = (2*n-1)!/(n-1)! - n! - Sum_{i=2..n} binomial(n-1, n-i)*n*n!.
a(n) = (2*n-1)!/(n-1)! - n! - (2^(n-1) - 1)*n*n!. - Stefano Spezia, Mar 07 2025

A001813 Quadruple factorial numbers: a(n) = (2n)!/n!.

Original entry on oeis.org

1, 2, 12, 120, 1680, 30240, 665280, 17297280, 518918400, 17643225600, 670442572800, 28158588057600, 1295295050649600, 64764752532480000, 3497296636753920000, 202843204931727360000, 12576278705767096320000, 830034394580628357120000, 58102407620643984998400000
Offset: 0

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Author

Keywords

Comments

Counts binary rooted trees (with out-degree <= 2), embedded in plane, with n labeled end nodes of degree 1. Unlabeled version gives Catalan numbers A000108.
Define a "downgrade" to be the permutation which places the items of a permutation in descending order. We are concerned with permutations that are identical to their downgrades. Only permutations of order 4n and 4n+1 can have this property; the number of permutations of length 4n having this property are equinumerous with those of length 4n+1. If a permutation p has this property then the reversal of this permutation also has it. a(n) = number of permutations of length 4n and 4n+1 that are identical to their downgrades. - Eugene McDonnell (eemcd(AT)mac.com), Oct 26 2003
Number of broadcast schemes in the complete graph on n+1 vertices, K_{n+1}. - Calin D. Morosan (cd_moros(AT)alumni.concordia.ca), Nov 28 2008
Hankel transform is A137565. - Paul Barry, Nov 25 2009
The e.g.f. of 1/a(n) = n!/(2*n)! is (exp(sqrt(x)) + exp(-sqrt(x)) )/2. - Wolfdieter Lang, Jan 09 2012
From Tom Copeland, Nov 15 2014: (Start)
Aerated with intervening zeros (1,0,2,0,12,0,120,...) = a(n) (cf. A123023 and A001147), the e.g.f. is e^(t^2), so this is the base for the Appell sequence with e.g.f. e^(t^2) e^(x*t) = exp(P(.,x),t) (reverse A059344, cf. A099174, A066325 also). P(n,x) = (a. + x)^n with (a.)^n = a_n and comprise the umbral compositional inverses for e^(-t^2)e^(x*t) = exp(UP(.,x),t), i.e., UP(n,P(.,t)) = x^n = P(n,UP(.,t)), e.g., (P(.,t))^n = P(n,t).
Equals A000407*2 with leading 1 added. (End)
a(n) is also the number of square roots of any permutation in S_{4*n} whose disjoint cycle decomposition consists of 2*n transpositions. - Luis Manuel Rivera Martínez, Mar 04 2015
Self-convolution gives A076729. - Vladimir Reshetnikov, Oct 11 2016
For n > 1, it follows from the formula dated Aug 07 2013 that a(n) is a Zumkeller number (A083207). - Ivan N. Ianakiev, Feb 28 2017
For n divisible by 4, a(n/4) is the number of ways to place n points on an n X n grid with pairwise distinct abscissae, pairwise distinct ordinates, and 90-degree rotational symmetry. For n == 1 (mod 4), the number of ways is a((n-1)/4) because the center point can be considered "fixed". For 180-degree rotational symmetry see A006882, for mirror symmetry see A000085, A135401, and A297708. - Manfred Scheucher, Dec 29 2017

Examples

			The following permutations of order 8 and their reversals have this property:
  1 7 3 5 2 4 0 6
  1 7 4 2 5 3 0 6
  2 3 7 6 1 0 4 5
  2 4 7 1 6 0 3 5
  3 2 6 7 0 1 5 4
  3 5 1 7 0 6 2 4
		

References

  • D. E. Knuth, The Art of Computer Programming, Vol. 4, Section 7.2.1.6, Eq. 32.
  • L. C. Larson, The number of essentially different nonattacking rook arrangements, J. Recreat. Math., 7 (No. 3, 1974), circa pages 180-181.
  • Eugene McDonnell, "Magic Squares and Permutations" APL Quote-Quad 7.3 (Fall, 1976)
  • R. W. Robinson, Counting arrangements of bishops, pp. 198-214 of Combinatorial Mathematics IV (Adelaide 1975), Lect. Notes Math., 560 (1976).
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • GAP
    List([0..20],n->Factorial(2*n)/Factorial(n)); # Muniru A Asiru, Nov 01 2018
    
  • Magma
    [Factorial(2*n)/Factorial(n): n in [0..20]]; // Vincenzo Librandi, Oct 09 2018
    
  • Maple
    A001813 := n->(2*n)!/n!;
    A001813 := n -> mul(k, k = select(k-> k mod 4 = 2,[$1 .. 4*n])):
    seq(A001813(n), n=0..16);  # Peter Luschny, Jun 23 2011
  • Mathematica
    Table[(2n)!/n!, {n,0,20}] (* Harvey P. Dale, May 02 2011 *)
  • Maxima
    makelist(binomial(n+n, n)*n!,n,0,30); /* Martin Ettl, Nov 05 2012 */
    
  • PARI
    a(n)=binomial(n+n,n)*n! \\ Charles R Greathouse IV, Jun 15 2011
    
  • PARI
    first(n) = x='x+O('x^n); Vec(serlaplace((1 - 4*x)^(-1/2))) \\ Iain Fox, Jan 01 2018 (corrected by Iain Fox, Jan 11 2018)
    
  • Python
    from math import factorial
    def A001813(n): return factorial(n<<1)//factorial(n) # Chai Wah Wu, Feb 14 2023
  • Sage
    [binomial(2*n,n)*factorial(n) for n in range(0, 17)] # Zerinvary Lajos, Dec 03 2009
    

Formula

E.g.f.: (1-4*x)^(-1/2).
a(n) = (2*n)!/n! = Product_{k=0..n-1} (4*k + 2) = A081125(2*n).
Integral representation as n-th moment of a positive function on a positive half-axis: a(n) = Integral_{x=0..oo} x^n*exp(-x/4)/(sqrt(x)*2*sqrt(Pi)) dx, n >= 0. This representation is unique. - Karol A. Penson, Sep 18 2001
Define a'(1)=1, a'(n) = Sum_{k=1..n-1} a'(n-k)*a'(k)*C(n, k); then a(n)=a'(n+1). - Benoit Cloitre, Apr 27 2003
With interpolated zeros (1, 0, 2, 0, 12, ...) this has e.g.f. exp(x^2). - Paul Barry, May 09 2003
a(n) = A000680(n)/A000142(n)*A000079(n) = Product_{i=0..n-1} (4*i + 2) = 4^n*Pochhammer(1/2, n) = 4^n*GAMMA(n+1/2)/sqrt(Pi). - Daniel Dockery (peritus(AT)gmail.com), Jun 13 2003
For asymptotics, see the Robinson paper.
a(k) = (2*k)!/k! = Sum_{i=1..k+1} |A008275(i,k+1)| * k^(i-1). - André F. Labossière, Jun 21 2007
a(n) = 12*A051618(a) n >= 2. - Zerinvary Lajos, Feb 15 2008
a(n) = A000984(n)*A000142(n). - Zerinvary Lajos, Mar 25 2008
a(n) = A016825(n-1)*a(n-1). - Roger L. Bagula, Sep 17 2008
a(n) = (-1)^n*A097388(n). - D. Morosan (cd_moros(AT)alumni.concordia.ca), Nov 28 2008
From Paul Barry, Jan 15 2009: (Start)
G.f.: 1/(1-2x/(1-4x/(1-6x/(1-8x/(1-10x/(1-... (continued fraction);
a(n) = (n+1)!*A000108(n). (End)
a(n) = Sum_{k=0..n} A132393(n,k)*2^(2n-k). - Philippe Deléham, Feb 10 2009
G.f.: 1/(1-2x-8x^2/(1-10x-48x^2/(1-18x-120x^2/(1-26x-224x^2/(1-34x-360x^2/(1-42x-528x^2/(1-... (continued fraction). - Paul Barry, Nov 25 2009
a(n) = A173333(2*n,n) for n>0; cf. A006963, A001761. - Reinhard Zumkeller, Feb 19 2010
From Gary W. Adamson, Jul 19 2011: (Start)
a(n) = upper left term of M^n, M = an infinite square production matrix as follows:
2, 2, 0, 0, 0, 0, ...
4, 4, 4, 0, 0, 0, ...
6, 6, 6, 6, 0, 0, ...
8, 8, 8, 8, 8, 0, ...
...
(End)
a(n) = (-2)^n*Sum_{k=0..n} 2^k*s(n+1,n+1-k), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012
G.f.: 1/Q(0), where Q(k) = 1 + x*(4*k+2) - x*(4*k+4)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 18 2013
G.f.: 2/G(0), where G(k) = 1 + 1/(1 - x*(8*k+4)/(x*(8*k+4) - 1 + 8*x*(k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 30 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - 2*x/(2*x + 1/(2*k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013
D-finite with recurrence: a(n) = (4*n-6)*a(n-2) + (4*n-3)*a(n-1), n>=2. - Ivan N. Ianakiev, Aug 07 2013
Sum_{n>=0} 1/a(n) = (exp(1/4)*sqrt(Pi)*erf(1/2) + 2)/2 = 1 + A214869, where erf(x) is the error function. - Ilya Gutkovskiy, Nov 10 2016
Sum_{n>=0} (-1)^n/a(n) = 1 - sqrt(Pi)*erfi(1/2)/(2*exp(1/4)), where erfi(x) is the imaginary error function. - Amiram Eldar, Feb 20 2021
a(n) = 1/([x^n] hypergeom([1], [1/2], x/4)). - Peter Luschny, Sep 13 2024
a(n) = 2^n*n!*JacobiP(n, -1/2, -n, 3). - Peter Luschny, Jan 22 2025
G.f.: 2F0(1,1/2;;4x). - R. J. Mathar, Jun 07 2025

Extensions

More terms from James Sellers, May 01 2000

A374848 Obverse convolution A000045**A000045; see Comments.

Original entry on oeis.org

0, 1, 2, 16, 162, 3600, 147456, 12320100, 2058386904, 701841817600, 488286500625000, 696425232679321600, 2038348954317776486400, 12259459134020160144810000, 151596002479762016373851690400, 3855806813438155578522841251840000
Offset: 0

Views

Author

Clark Kimberling, Jul 31 2024

Keywords

Comments

The obverse convolution of sequences
s = (s(0), s(1), ...) and t = (t(0), t(1), ...)
is introduced here as the sequence s**t given by
s**t(n) = (s(0)+t(n)) * (s(1)+t(n-1)) * ... * (s(n)+t(0)).
Swapping * and + in the representation s(0)*t(n) + s(1)*t(n-1) + ... + s(n)*t(0)
of ordinary convolution yields s**t.
If x is an indeterminate or real (or complex) variable, then for every sequence t of real (or complex) numbers, s**t is a sequence of polynomials p(n) in x, and the zeros of p(n) are the numbers -t(0), -t(1), ..., -t(n).
Following are abbreviations in the guide below for triples (s, t, s**t):
F = (0,1,1,2,3,5,...) = A000045, Fibonacci numbers
L = (2,1,3,4,7,11,...) = A000032, Lucas numbers
P = (2,3,5,7,11,...) = A000040, primes
T = (1,3,6,10,15,...) = A000217, triangular numbers
C = (1,2,6,20,70, ...) = A000984, central binomial coefficients
LW = (1,3,4,6,8,9,...) = A000201, lower Wythoff sequence
UW = (2,5,7,10,13,...) = A001950, upper Wythoff sequence
[ ] = floor
In the guide below, sequences s**t are identified with index numbers Axxxxxx; in some cases, s**t and Axxxxxx differ in one or two initial terms.
Table 1. s = A000012 = (1,1,1,1...) = (1);
t = A000012; 1 s**t = A000079; 2^(n+1)
t = A000027; n s**t = A000142; (n+1)!
t = A000040, P s**t = A054640
t = A000040, P (1/3) s**t = A374852
t = A000079, 2^n s**t = A028361
t = A000079, 2^n (1/3) s**t = A028362
t = A000045, F s**t = A082480
t = A000032, L s**t = A374890
t = A000201, LW s**t = A374860
t = A001950, UW s**t = A374864
t = A005408, 2*n+1 s**t = A000165, 2^n*n!
t = A016777, 3*n+1 s**t = A008544
t = A016789, 3*n+2 s**t = A032031
t = A000142, n! s**t = A217757
t = A000051, 2^n+1 s**t = A139486
t = A000225, 2^n-1 s**t = A006125
t = A032766, [3*n/2] s**t = A111394
t = A034472, 3^n+1 s**t = A153280
t = A024023, 3^n-1 s**t = A047656
t = A000217, T s**t = A128814
t = A000984, C s**t = A374891
t = A279019, n^2-n s**t = A130032
t = A004526, 1+[n/2] s**t = A010551
t = A002264, 1+[n/3] s**t = A264557
t = A002265, 1+[n/4] s**t = A264635
Sequences (c)**L, for c=2..4: A374656 to A374661
Sequences (c)**F, for c=2..6: A374662, A374662, A374982 to A374855
The obverse convolutions listed in Table 1 are, trivially, divisibility sequences. Likewise, if s = (-1,-1,-1,...) instead of s = (1,1,1,...), then s**t is a divisibility sequence for every choice of t; e.g. if s = (-1,-1,-1,...) and t = A279019, then s**t = A130031.
Table 2. s = A000027 = (0,1,2,3,4,5,...) = (n);
t = A000027, n s**t = A007778, n^(n+1)
t = A000290, n^2 s**t = A374881
t = A000040, P s**t = A374853
t = A000045, F s**t = A374857
t = A000032, L s**t = A374858
t = A000079, 2^n s**t = A374859
t = A000201, LW s**t = A374861
t = A005408, 2*n+1 s**t = A000407, (2*n+1)! / n!
t = A016777, 3*n+1 s**t = A113551
t = A016789, 3*n+2 s**t = A374866
t = A000142, n! s**t = A374871
t = A032766, [3*n/2] s**t = A374879
t = A000217, T s**t = A374892
t = A000984, C s**t = A374893
t = A038608, n*(-1)^n s**t = A374894
Table 3. s = A000290 = (0,1,4,9,16,...) = (n^2);
t = A000290, n^2 s**t = A323540
t = A002522, n^2+1 s**t = A374884
t = A000217, T s**t = A374885
t = A000578, n^3 s**t = A374886
t = A000079, 2^n s**t = A374887
t = A000225, 2^n-1 s**t = A374888
t = A005408, 2*n+1 s**t = A374889
t = A000045, F s**t = A374890
Table 4. s = t;
s = t = A000012, 1 s**s = A000079; 2^(n+1)
s = t = A000027, n s**s = A007778, n^(n+1)
s = t = A000290, n^2 s**s = A323540
s = t = A000045, F s**s = this sequence
s = t = A000032, L s**s = A374850
s = t = A000079, 2^n s**s = A369673
s = t = A000244, 3^n s**s = A369674
s = t = A000040, P s**s = A374851
s = t = A000201, LW s**s = A374862
s = t = A005408, 2*n+1 s**s = A062971
s = t = A016777, 3*n+1 s**s = A374877
s = t = A016789, 3*n+2 s**s = A374878
s = t = A032766, [3*n/2] s**s = A374880
s = t = A000217, T s**s = A375050
s = t = A005563, n^2-1 s**s = A375051
s = t = A279019, n^2-n s**s = A375056
s = t = A002398, n^2+n s**s = A375058
s = t = A002061, n^2+n+1 s**s = A375059
If n = 2*k+1, then s**s(n) is a square; specifically,
s**s(n) = ((s(0)+s(n))*(s(1)+s(n-1))*...*(s(k)+s(k+1)))^2.
If n = 2*k, then s**s(n) has the form 2*s(k)*m^2, where m is an integer.
Table 5. Others
s = A000201, LW t = A001950, UW s**t = A374863
s = A000045, F t = A000032, L s**t = A374865
s = A005843, 2*n t = A005408, 2*n+1 s**t = A085528, (2*n+1)^(n+1)
s = A016777, 3*n+1 t = A016789, 3*n+2 s**t = A091482
s = A005408, 2*n+1 t = A000045, F s**t = A374867
s = A005408, 2*n+1 t = A000032, L s**t = A374868
s = A005408, 2*n+1 t = A000079, 2^n s**t = A374869
s = A000027, n t = A000142, n! s**t = A374871
s = A005408, 2*n+1 t = A000142, n! s**t = A374872
s = A000079, 2^n t = A000142, n! s**t = A374874
s = A000142, n! t = A000045, F s**t = A374875
s = A000142, n! t = A000032, L s**t = A374876
s = A005408, 2*n+1 t = A016777, 3*n+1 s**t = A352601
s = A005408, 2*n+1 t = A016789, 3*n+2 s**t = A064352
Table 6. Arrays of coefficients of s(x)**t(x), where s(x) and t(x) are polynomials
s(x) t(x) s(x)**t(x)
n x A132393
n^2 x A269944
x+1 x+1 A038220
x+2 x+2 A038244
x x+3 A038220
nx x+1 A094638
1 x^2+x+1 A336996
n^2 x x+1 A375041
n^2 x 2x+1 A375042
n^2 x x+2 A375043
2^n x x+1 A375044
2^n 2x+1 A375045
2^n x+2 A375046
x+1 F(n) A375047
x+1 x+F(n) A375048
x+F(n) x+F(n) A375049

Examples

			a(0) = 0 + 0 = 0
a(1) = (0+1) * (1+0) = 1
a(2) = (0+1) * (1+1) * (1+0) = 2
a(3) = (0+2) * (1+1) * (1+1) * (2+0) = 16
As noted above, a(2*k+1) is a square for k>=0. The first 5 squares are 1, 16, 3600, 12320100, 701841817600, with corresponding square roots 1, 4, 60, 3510, 837760.
If n = 2*k, then s**s(n) has the form 2*F(k)*m^2, where m is an integer and F(k) is the k-th Fibonacci number; e.g., a(6) = 2*F(3)*(192)^2.
		

Crossrefs

Programs

  • Maple
    a:= n-> (F-> mul(F(n-j)+F(j), j=0..n))(combinat[fibonacci]):
    seq(a(n), n=0..15);  # Alois P. Heinz, Aug 02 2024
  • Mathematica
    s[n_] := Fibonacci[n]; t[n_] := Fibonacci[n];
    u[n_] := Product[s[k] + t[n - k], {k, 0, n}];
    Table[u[n], {n, 0, 20}]
  • PARI
    a(n)=prod(k=0, n, fibonacci(k) + fibonacci(n-k)) \\ Andrew Howroyd, Jul 31 2024

Formula

a(n) ~ c * phi^(3*n^2/4 + n) / 5^((n+1)/2), where c = QPochhammer(-1, 1/phi^2)^2/2 if n is even and c = phi^(1/4) * QPochhammer(-phi, 1/phi^2)^2 / (phi + 1)^2 if n is odd, and phi = A001622 is the golden ratio. - Vaclav Kotesovec, Aug 01 2024

A006963 Number of planar embedded labeled trees with n nodes: (2*n-3)!/(n-1)! for n >= 2, a(1) = 1.

Original entry on oeis.org

1, 1, 3, 20, 210, 3024, 55440, 1235520, 32432400, 980179200, 33522128640, 1279935820800, 53970627110400, 2490952020480000, 124903451312640000, 6761440164390912000, 393008709555221760000, 24412776311194951680000, 1613955767240110694400000
Offset: 1

Views

Author

Keywords

Comments

For n>1: central terms of the triangle in A173333; cf. A001761, A001813. - Reinhard Zumkeller, Feb 19 2010
Can be obtained from the Vandermonde permanent of the first n positive integers; see A093883. - Clark Kimberling, Jan 02 2012
All trees can be embedded in the plane, but "planar embedded" means that orientation matters but rotation doesn't. For example, the n-star with n-1 edges has n! ways to label it, but rotation removes a factor of n-1. Another example, the n-path has n! ways to label it, but rotation removes a factor of 2. - Michael Somos, Aug 19 2014

Examples

			G.f. = x + x^2 + 3*x^3 + 20*x^4 + 210*x^5 + 3024*x^6 + 55440*x^7 + 1235520*x^8 + ...
a(5) = 210 = 30 + 60 + 120 where 30 is for the star, 60 for the path, and 120 for the tree with one trivalent vertex. - _Michael Somos_, Aug 19 2014
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Magma
    [1] cat [Factorial(2*n-3)/Factorial(n-1): n in [2..20]]; // Vincenzo Librandi, Nov 12 2011
    
  • Maple
    1, seq((2*n-3)!/(n-1)!,n=2..30); # Robert Israel, Aug 20 2014
  • Mathematica
    Join[{1},Table[(2n-3)!/(n-1)!,{n,2,20}]] (* Harvey P. Dale, Nov 03 2011 *)
    a[ n_] := With[{m = n - 1}, If[m < 1, Boole[m == 0], m! SeriesCoefficient[ -Log[(1 + Sqrt[1 - 4 x]) / 2], {x, 0, m}]]] (* Michael Somos, Jul 01 2013 *)
    a[ n_] := If[n < 2, Boole[n == 1], (2 n - 3)! / (n - 1)!]; (* Michael Somos, Aug 19 2014 *)
    a[1] := 1; a[n_] := (-1)^(n - 1)*Sum[(-1)^k*Binomial[2*n - 3, n + k - 2]*StirlingS1[n + k - 1, k + 1], {k, 1, n - 1}]; Flatten[Table[a[n], {n, 1, 19}]] (* Detlef Meya, Jan 18 2024 *)
  • PARI
    {a(n) = n--; if( n<1, n==0, n! * polcoeff( -log( (1 + sqrt(1 - 4*x + x * O(x^n))) / 2), n))}; /* Michael Somos, Jul 01 2013 */
    
  • SageMath
    def A006963(n): return 1 if n==1 else factorial(2*n-3)/factorial(n-1)
    [A006963(n) for n in range(1,31)] # G. C. Greubel, May 23 2023

Formula

E.g.f. for a(n+1), n >= 1, log(c(x)); c(x) = g.f. for Catalan numbers A000108. - Wolfdieter Lang
Integral representation as n-th moment of a positive function on a positive half-axis, in Maple notation: a(n) = int(x^n * erfc(sqrt(x)/2)/2, x=0..infinity), n=0, 1..., where erfc(x) is the complementary error function. - Karol A. Penson, Sep 27 2001
a(n) ~ 2^(-5/2)*n^-2*2^(2*n)*e^-n*n^n. - Joe Keane (jgk(AT)jgk.org), Jun 06 2002
a(n+1) = (n+1)*(n+2)*...*(2n-1) for n>=2. - Jaroslav Krizek, Nov 09 2010
E.g.f. (A(x)-1) is reversion of exp(-x)-exp(-2*x). - Vladimir Kruchinin, Jan 30 2012
G.f.: 1 + x*G(0) where G(k) = 1 + x*(2*k+1)*(4*k+3)/(k + 1 - 4*x*(k+1)^2*(4*k+5)/(4*x*(k+1)*(4*k+5) + (2*k+3)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Feb 02 2013
E.g.f.: 1 + x*E(0) where E(k) = 1 + x*(2*k+1)*(4*k+3)/(2*(k + 1)^2 - 8*x*(k+1)^3*(4*k+5)/(4*x*(k+1)*(4*k+5) + (2*k+3)^2/E(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Feb 02 2013
E.g.f: sqrt(1-4*x)/4 - 1/4 + 3*x/2 - x*log((1+sqrt(1-4*x))/2). - Robert Israel, Aug 20 2014
D-finite with recurrence (-n+1)*a(n) +2*(2*n-3)*(n-2)*a(n-1)=0. - R. J. Mathar, Jan 03 2018
From Amiram Eldar, Apr 03 2022: (Start)
Sum_{n>=1} 1/a(n) = 3/2 + 3*exp(1/4)*sqrt(Pi)*erf(1/2)/4, where erf is the error function.
Sum_{n>=1} (-1)^(n+1)/a(n) = 1/2 - sqrt(Pi)*erfi(1/2)/(4*exp(1/4)), where erfi is the imaginary error function. (End)
a(n) = A000407(n-2)/(n-1). - R. J. Mathar, Mar 30 2023
a(1) = 1; a(n) = (-1)^(n - 1)*Sum_{k=1..n - 1} (-1)^k*binomial(2*n - 3, n + k - 2)*Stirling1(n + k - 1, k + 1). - Detlef Meya, Jan 18 2024

A248897 Decimal expansion of Sum_{i >= 0} (i!)^2/(2*i+1)!.

Original entry on oeis.org

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

Views

Author

Bruno Berselli, Mar 06 2015

Keywords

Comments

Value of the Borwein-Borwein function I_3(a,b) for a = b = 1. - Stanislav Sykora, Apr 16 2015
The area of a circle circumscribing a unit-area regular hexagon. - Amiram Eldar, Nov 05 2020

Examples

			1.2091995761561452337293855050947704881893774987284937170465899569254...
		

References

  • George Boros and Victor H. Moll, Irresistible integrals, Cambridge University Press (2006), pp. 120-121.
  • L. B. W. Jolley, Summation of Series, Dover (1961), No. 261, pp. 48, 49, (and No. 275).

Crossrefs

Cf. A091682 (Sum_{i >= 0} (i!)^2/(2*i)!).

Programs

  • Mathematica
    RealDigits[2 Sqrt[3] Pi/9, 10, 100][[1]]
  • PARI
    a = 2*Pi/(3*sqrt(3)) \\ Stanislav Sykora, Apr 16 2015

Formula

Equals 2*sqrt(3)*Pi/9 = 1 + 1/6 + 1/30 + 1/140 + 1/630 + 1/2772 + 1/12012 + ...
Equals m*I_3(m,m) = m*Integral_{x>=0} (x/(m^3+x^3)), for any m>0. - Stanislav Sykora, Apr 16 2015
Equals Integral_{x>=0} (1/(1+x^3)) dx. - Robert FERREOL, Dec 23 2016
From Peter Bala, Oct 27 2019: (Start)
Equals 3/4*Sum_{n >= 0} (n+1)!*(n+2)!/(2*n+3)!.
Equals Sum_{n >= 1} 3^(n-1)/(n*binomial(2*n,n)).
Equals 2*Sum_{n >= 1} 1/(n*binomial(2*n,n)). See Boros and Moll, pp. 120-121.
Equals Integral_{x = 0..1} 1/(1 - x^3)^(1/3) dx = Sum_{n >= 0} (-1)^n*binomial(-1/3,n) /(3*n + 1).
Equals 2*Sum_{n >= 1} 1/((3*n-1)*(3*n-2)) = 2*(1 - 1/2 + 1/4 - 1/5 + 1/7 - 1/8 + ...) (added Oct 30 2019). (End)
Equals Product_{k>=1} 9*k^2/(9*k^2 - 1). - Amiram Eldar, Aug 04 2020
From Peter Bala, Dec 13 2021: (Start)
Equals (2/3)*A093602.
Conjecture: for k >= 0, 2*sqrt(3)*Pi/9 = (3/2)^k * k!*Sum_{n = -oo..oo} (-1)^n/ Product_{j = 0..k} (3*n + 3*j + 1). (End)
Equals (3/4)*S - 1, where S = A248682. - Peter Luschny, Jul 22 2022
Equals Integral_{x=0..Pi/2} tan(x)^(1/3)/(sin(2*x) + 1) dx. See MIT Link. - Joost de Winter, Aug 26 2023
Continued fraction: 1/(1 - 1/(7 - 12/(12 - 30/(17 - ... - 2*n*(2*n - 1)/((5*n + 2) - ... ))))). See A000407. - Peter Bala, Feb 20 2024
Equals Sum_{n>=2} 1/binomial(n, floor(n/2)); and trivially if "floor" is replaced by "ceiling". - Richard R. Forberg, Aug 30 2024
Equals Product_{k>=2} (1 + (-1)^k/A001651(k)). - Amiram Eldar, Nov 22 2024
Equals 2*A073010 = 1/A086089 = sqrt(A214549) = exp(A256923) = A275486/2. - Hugo Pfoertner, Nov 22 2024
Equals 1 - (1/6) * Sum_{n>=1} A010815(n)/n. - Friedjof Tellkamp, Apr 05 2025
Equals A248181 - 2. - Pontus von Brömssen, Apr 05 2025

A256890 Triangle T(n,k) = t(n-k, k); t(n,m) = f(m)*t(n-1,m) + f(n)*t(n,m-1), where f(x) = x + 2.

Original entry on oeis.org

1, 2, 2, 4, 12, 4, 8, 52, 52, 8, 16, 196, 416, 196, 16, 32, 684, 2644, 2644, 684, 32, 64, 2276, 14680, 26440, 14680, 2276, 64, 128, 7340, 74652, 220280, 220280, 74652, 7340, 128, 256, 23172, 357328, 1623964, 2643360, 1623964, 357328, 23172, 256, 512, 72076, 1637860, 10978444, 27227908, 27227908, 10978444, 1637860, 72076, 512
Offset: 0

Views

Author

Dale Gerdemann, Apr 12 2015

Keywords

Comments

Related triangles may be found by varying the function f(x). If f(x) is a linear function, it can be parameterized as f(x) = a*x + b. With different values for a and b, the following triangles are obtained:
a\b 1.......2.......3.......4.......5.......6
The row sums of these, and similarly constructed number triangles, are shown in the following table:
a\b 1.......2.......3.......4.......5.......6.......7.......8.......9
The formula can be further generalized to: t(n,m) = f(m+s)*t(n-1,m) + f(n-s)*t(n,m-1), where f(x) = a*x + b. The following table specifies triangles with nonzero values for s (given after the slash).
a\b 0 1 2 3
-2 A130595/1
-1
0
With the absolute value, f(x) = |x|, one obtains A038221/3, A038234/4, A038247/5, A038260/6, A038273/7, A038286/8, A038299/9 (with value for s after the slash).
If f(x) = A000045(x) (Fibonacci) and s = 1, the result is A010048 (Fibonomial).
In the notation of Carlitz and Scoville, this is the triangle of generalized Eulerian numbers A(r, s | alpha, beta) with alpha = beta = 2. Also the array A(2,1,4) in the notation of Hwang et al. (see page 31). - Peter Bala, Dec 27 2019

Examples

			Array, t(n, k), begins as:
   1,    2,      4,        8,        16,         32,          64, ...;
   2,   12,     52,      196,       684,       2276,        7340, ...;
   4,   52,    416,     2644,     14680,      74652,      357328, ...;
   8,  196,   2644,    26440,    220280,    1623964,    10978444, ...;
  16,  684,  14680,   220280,   2643360,   27227908,   251195000, ...;
  32, 2276,  74652,  1623964,  27227908,  381190712,  4677894984, ...;
  64, 7340, 357328, 10978444, 251195000, 4677894984, 74846319744, ...;
Triangle, T(n, k), begins as:
    1;
    2,     2;
    4,    12,      4;
    8,    52,     52,       8;
   16,   196,    416,     196,      16;
   32,   684,   2644,    2644,     684,      32;
   64,  2276,  14680,   26440,   14680,    2276,     64;
  128,  7340,  74652,  220280,  220280,   74652,   7340,   128;
  256, 23172, 357328, 1623964, 2643360, 1623964, 357328, 23172,   256;
		

Crossrefs

Programs

  • Magma
    A256890:= func< n,k | (&+[(-1)^(k-j)*Binomial(j+3,j)*Binomial(n+4,k-j)*(j+2)^n: j in [0..k]]) >;
    [A256890(n,k): k in [0..n], n in [0..10]]; // G. C. Greubel, Oct 18 2022
    
  • Mathematica
    Table[Sum[(-1)^(k-j)*Binomial[j+3, j] Binomial[n+4, k-j] (j+2)^n, {j,0,k}], {n,0, 9}, {k,0,n}]//Flatten (* Michael De Vlieger, Dec 27 2019 *)
  • PARI
    t(n,m) = if ((n<0) || (m<0), 0, if ((n==0) && (m==0), 1, (m+2)*t(n-1, m) + (n+2)*t(n, m-1)));
    tabl(nn) = {for (n=0, nn, for (k=0, n, print1(t(n-k, k), ", ");); print(););} \\ Michel Marcus, Apr 14 2015
    
  • SageMath
    def A256890(n,k): return sum((-1)^(k-j)*Binomial(j+3,j)*Binomial(n+4,k-j)*(j+2)^n for j in range(k+1))
    flatten([[A256890(n,k) for k in range(n+1)] for n in range(11)]) # G. C. Greubel, Oct 18 2022

Formula

T(n,k) = t(n-k, k); t(0,0) = 1, t(n,m) = 0 if n < 0 or m < 0 else t(n,m) = f(m)*t(n-1,m) + f(n)*t(n,m-1), where f(x) = x + 2.
Sum_{k=0..n} T(n, k) = A001715(n).
T(n,k) = Sum_{j = 0..k} (-1)^(k-j)*binomial(j+3,j)*binomial(n+4,k-j)*(j+2)^n. - Peter Bala, Dec 27 2019
Modified rule of Pascal: T(0,0) = 1, T(n,k) = 0 if k < 0 or k > n else T(n,k) = f(n-k) * T(n-1,k-1) + f(k) * T(n-1,k), where f(x) = x + 2. - Georg Fischer, Nov 11 2021
From G. C. Greubel, Oct 18 2022: (Start)
T(n, n-k) = T(n, k).
T(n, 0) = A000079(n). (End)

A024199 a(n) = (2n-1)!! * Sum_{k=0..n-1}(-1)^k/(2k+1).

Original entry on oeis.org

0, 1, 2, 13, 76, 789, 7734, 110937, 1528920, 28018665, 497895210, 11110528485, 241792844580, 6361055257725, 163842638377950, 4964894559637425, 147721447995130800, 5066706567801827025, 171002070002301095250, 6548719685561840296125, 247199273204273879989500
Offset: 0

Views

Author

Keywords

Comments

(2*n + 1)!!/a(n+1), n>=0, is the n-th approximant for William Brouncker's continued fraction for 4/Pi = 1 + 1^2/(2 + 3^2/(2 + 5^2/(2 + ... ))) See the C. Brezinski and J.-P. Delahaye references given under A142969 and A142970, respectively. The double factorials (2*n + 1)!! = A001147(n+1) enter. - Wolfdieter Lang, Oct 06 2008

Examples

			a(3) = (2*3 - 1)!! * Sum_{k=0..2} (-1)^k/(2k + 1) = 5!! * (1/(2*0 + 1) - 1/(2*1 + 1) + 1/(2*2 + 1)) = 5*3*1*(1/1 - 1/3 + 1/5) = 15 - 5 + 3 = 13. Notice that the first factor always cancels the common denominator of the sum. - _Michael B. Porter_, Jul 22 2016
		

References

  • A. E. Jolliffe, Continued Fractions, in Encyclopaedia Britannica, 11th ed., pp. 30-33; see p. 31.

Crossrefs

From Johannes W. Meijer, Nov 12 2009: (Start)
Cf. A007509 and A025547.
Equals first column of A167584.
Equals row sums of A167591.
Equals first right hand column of A167594.
(End)
Cf. A167576 and A135457.

Programs

  • Magma
    [0] cat [ n le 2 select (n) else 2*Self(n-1)+(2*n-3)^2*Self(n-2): n in [1..25] ]; // Vincenzo Librandi, Feb 17 2015
  • Maple
    a := proc(n) option remember; if n=0 then 0 elif n=1 then 1 else 2*a(n-1)+(2*n-3)^2* a(n-2) fi end: seq(a(n), n=0..20); # Peter Luschny, Nov 16 2016 after N. J. A. Sloane
  • Mathematica
    f[k_] := (2 k - 1) (-1)^(k + 1)
    t[n_] := Table[f[k], {k, 1, n}]
    a[n_] := SymmetricPolynomial[n - 1, t[n]]
    Table[a[n], {n, 1, 22}]    (* A024199 signed *)
    (* Clark Kimberling, Dec 30 2011 *)
    RecurrenceTable[{a[n+1] == 2*a[n] + (2*n-1)^2*a[n-1],a[0] == 0, a[1] == 1},a,{n,0,20}] (* Vaclav Kotesovec, Mar 18 2014 *)
    CoefficientList[Series[Pi/4/Sqrt[1-2*x] - 1/2*Log[2*x+Sqrt[4*x^2-1]]/Sqrt[2*x-1], {x, 0, 20}], x] * Range[0, 20]! (* Vaclav Kotesovec, Mar 18 2014 *)

Formula

a(n) = s(1)s(2)...s(n)(1/s(1) - 1/s(2) + ... + c/s(n)) where c=(-1)^(n+1) and s(k) = 2k-1 for k = 1, 2, 3, ... (was previous definition). - Clark Kimberling
D-finite with recurrence a(0) = 0, a(1) = 1, a(n+1) = 2*a(n) + (2*n-1)^2*a(n-1). - N. J. A. Sloane, Jul 19 2002
a(n) + A024200(n) = A001147(n) = (2n-1)!!. - Max Alekseyev, Sep 23 2007
a(n)/A024200(n) -> Pi/(4-Pi) as n -> oo. - Max Alekseyev, Sep 23 2007
From Wolfdieter Lang, Oct 06 2008: (Start)
E.g.f. for a(n+1), n>=0: (sqrt(1-2*x)+arcsin(2*x)*sqrt(1+2*x)/2)/((1-4*x^2)^(1/2)*(1-2*x)). From the recurrence, solving (1-4*x^2)y''(x)-2*(8*x+1)*y'(x)-9*y=0 with inputs y(0)=1, y'(0)=2.
a(n+1) = A003148(n) + A143165(n), n>=0 (from the two terms of the e.g.f.). (End)
From Johannes W. Meijer, Nov 12 2009: (Start)
a(n) = (-1)^(n-1)*(2*n-3)!! + (2*n-1)*a(n-1) with a(0) = 0.
a(n) = (2*n-1)!!*sum((-1)^(k)/(2*k+1), k=0..n-1)
(End)
E.g.f.: Pi/4/sqrt(1-2*x) - 1/2*log(2*x+sqrt(4*x^2-1))/sqrt(2*x-1). - Vaclav Kotesovec, Mar 18 2014
a(n) ~ Pi * 2^(n-3/2) * n^n / exp(n). - Vaclav Kotesovec, Mar 18 2014
a(n) = (2*H(n+1/2)-Gamma(n+1/2))*2^(n-2)*sqrt(Pi) with H(x) the Hadamard factorial (see the link section). - Cyril Damamme, Jul 19 2015
a(n) = A135457(n) - (-1)^n A001147(n-1). - Cyril Damamme, Jul 19 2015
a(n) = (Pi + (-1)^n*(Psi(n/2 + 1/4) - Psi(n/2 + 3/4)))*Gamma(n+1/2)*2^(n-2)/sqrt(Pi). - Robert Israel, Jul 20 2015
a(n) = A167576(n) - A135457(n). - Cyril Damamme, Jul 22 2015
a(n)/A001147(n) -> Pi/4 as n -> oo. - Daniel Suteu, Jul 21 2016
From Peter Bala, Nov 15 2016: (Start)
Conjecture: a(n) = 1/2*Sum_{k = 0..2*n-1} (-1)^(n-k+1)*k!*(2*n - 2*k - 3)!!, where the double factorial of an odd integer (positive or negative) may be defined in terms of the gamma function as (2*N - 1)!! = 2^((N+1)/2)*Gamma(N/2 + 1)/sqrt(Pi).
E.g.f. 1/2*arcsin(2*x)/sqrt(1 - 2*x) = x + 2*x^2/2! + 13*x^3/3! + 76*x^4/4! + .... (End)

Extensions

Edited by N. J. A. Sloane, Jul 19 2002
New name from Cyril Damamme, Jul 19 2015
Showing 1-10 of 57 results. Next