Displaying 1-4 of 4 results found.
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1, 6, 6, 20, 15, 42, 28, 72, 45, 110, 66, 156, 91, 210, 120, 272, 153, 342, 190, 420, 231, 506, 276, 600, 325, 702, 378, 812, 435, 930, 496, 1056, 561, 1190, 630, 1332, 703, 1482, 780, 1640, 861, 1806, 946, 1980, 1035, 2162, 1128, 2352, 1225, 2550, 1326, 2756, 1431
COMMENTS
Also period length divided by 2 of pairs (a,b), where a has period 2*n-2 and b has period n.
0, 1/6, 1/6, 3/20, 2/15, 5/42, ...
-1/6, -1/60, 0, 1/420, 1/420, 1/504, ...
3/20, 1/60, 1/420, 0, -1/2520, -1/2520, ...
-2/15, -1/70, -1/420, -1/2520, 0, 1/13860, ...
5/42, 1/84, 1/504, 1/2520, -1/13860, 0, ...
Autosequence of the first kind. The main diagonal is A000004. The first two upper diagonals are equal. Their denominators are A000911. (End)
FORMULA
G.f.: (2*x^3+3*x^2+6*x+1)/(1-x^2)^3.
a(n) = (n+1)*(n+2) if n odd; or (n+1)*(n+2)/2 if n even = (n+1)*(n+2)*(3-(-1)^n)/4. - C. Ronaldo (aga_new_ac(AT)hotmail.com), Dec 16 2004
a(n) = lcm(2*n+2, n+2)/2.
E.g.f.: (2 + 8*x + x^2)*cosh(x)/2 + (2 + 2*x + x^2)*sinh(x). - Stefano Spezia, Apr 24 2024
MAPLE
seq((n+1)*(n+2)*(3-(-1)^n)/4, n=0..20); # C. Ronaldo
with(combinat): seq(lcm(n+1, binomial(n+2, n)), n=0..50); # Zerinvary Lajos, Apr 20 2008
MATHEMATICA
Table[LCM[2*n + 2, n + 2]/2, {n, 0, 40}] (* corrected by Amiram Eldar, Sep 14 2022 *)
Denominator[#[[1]]/(#[[2]]#[[3]])&/@Partition[Range[0, 60], 3, 1]] (* Harvey P. Dale, Aug 15 2013 *)
PROG
(Haskell)
import Data.Ratio ((%), denominator)
a045896 n = denominator $ n % ((n + 1) * (n + 2))
a(n) = (2*n+3)!/(6*n!*(n+1)!).
(Formerly M4724 N2019)
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1, 10, 70, 420, 2310, 12012, 60060, 291720, 1385670, 6466460, 29745716, 135207800, 608435100, 2714556600, 12021607800, 52895074320, 231415950150, 1007340018300, 4365140079300, 18839025605400, 81007810103220, 347176329013800, 1483389769422600
COMMENTS
For n >= 1 a(n) is also the number of rooted bicolored unicellular maps of genus 1 on n+2 edges. - Ahmed Fares (ahmedfares(AT)my-deja.com), Aug 20 2001
a(n) is half the number of (n+2) X 2 Young tableaux with a three horizontal walls between the first and second column. If there is a wall between two cells, the entries may be decreasing; see [Banderier, Wallner 2021], A000984 for one horizontal wall, and A002457 for two. - Michael Wallner, Jan 31 2022
Call B(p,g) the number of genus g partitions of a set with p elements (genus-dependent Bell number). Up to an appropriate shift the given sequence counts the genus 1 partitions of a set: we have a(n) = B(n+4,1), with a(0)= B(4,1)=1.
When shifted with an offset 4 (i.e., defining b(p)=a(p-4), which starts with 0,0,0,1,10,70, etc., and b(4)=1)), the given sequence reads b(p) = (1/( 2^4 3 )) * (1/( (2 p - 1) (2 p - 3))) * (1/(p - 4)!) * (2p)!/p!. In this form it appears as a generalization of Catalan numbers (that indeed count the genus 0 partitions).
Call C[p, [alpha], g] the number of partitions of a set with p elements, of cyclic type [alpha], and of genus g (genus g Faa di Bruno coefficients of type [alpha]). Up to an appropriate shift the given sequence also counts the genus 1 partitions of p=2k into k parts of length 2, which is then called C[2k, [2^k], 1], and we have a(n) = C[2k, [2^k], 1] for k=n+2.
The two previous interpretations of this sequence, leading to a(n) = B(n+4, 1) and to a(n) = C[2(n+2), [2^(n+2)], 1] are not related in any obvious way. (End)
REFERENCES
C. Jordan, Calculus of Finite Differences. Röttig and Romwalter, Budapest, 1939; Chelsea, NY, 1965, p. 449.
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).
LINKS
R. Cori and G. Hetyei, How to count genus one partitions, FPSAC 2014, Chicago, Discrete Mathematics and Theoretical Computer Science (DMTCS), Nancy, France, 2014, 333-344.
FORMULA
G.f.: (1 - 4*x)^(-5/2) = 1F0(5/2;;4x).
Asymptotic expression for a(n) is a(n) ~ (n+2)^(3/2) * 4^(n+2) / (sqrt(Pi) * 48).
a(n) = Sum_{a+b+c+d+e=n} f(a)*f(b)*f(c)*f(d)*f(e) with f(n) = binomial(2n, n) = A000984(n). - Philippe Deléham, Jan 22 2004
a(n-1) = (1/4)*Sum_{k=1..n} k*(k+1)*binomial(2*k, k). - Benoit Cloitre, Mar 20 2004
a(n) = ((2n+3)(2n+1)/(3*1)) * binomial(2n, n).
a(n) = binomial(2n+4, 4) * binomial(2n, n) / binomial(n+2, 2).
a(n) = binomial(n+2, 2) * binomial(2n+4, n+2) / binomial(4, 2).
a(n) = binomial(2n+4, n+2) * (n+2)*(n+1) / 12. (End)
D-finite with recurrence: n*a(n) - 2*(2*n+3)*a(n-1) = 0. - R. J. Mathar, Jan 31 2014
a(n) = 4^n*hypergeom([-n,-3/2], [1], 1). - Peter Luschny, Apr 26 2016
Boas-Buck recurrence: a(n) = (10/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, a(0) = 1. Proof from a(n) = A046521(n+2, 2). See a comment there. - Wolfdieter Lang, Aug 10 2017
Sum_{n>=0} 1/a(n) = 12 - 2*sqrt(3)*Pi. - Amiram Eldar, Oct 13 2020
EXAMPLE
G.f. = 1 + 10*x + 70*x^2 + 420*x^3 + 2310*x^4 + 12012*x^5 + 60060*x^6 + ...
MAPLE
seq(simplify(4^n*hypergeom([-n, -3/2], [1], 1)), n=0..25); # Peter Luschny, Apr 26 2016
PROG
(PARI) {a(n) = if( n<0, 0, (2*n + 3)! / (6 * n! * (n+1)!))}; /* Michael Somos, Sep 16 2013 */
(PARI) {a(n) = 2^(n+3) * polcoeff( pollegendre(n+4), n) / 3}; /* Michael Somos, Sep 16 2013 */
(Magma) F:=Factorial; [F(2*n+3)/(6*F(n)*F(n+1)): n in [0..25]]; // G. C. Greubel, Jul 20 2019
(Sage) f=factorial; [f(2*n+3)/(6*f(n)*f(n+1)) for n in (0..25)] # G. C. Greubel, Jul 20 2019
(GAP) F:=Factorial;; List([0..25], n-> F(2*n+3)/(6*F(n)*F(n+1)) ); # G. C. Greubel, Jul 20 2019
Denominator of c(n) = (n^2+n+2)/((n+1)*(n+2)*(n+3)).
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3, 6, 15, 60, 105, 21, 126, 360, 495, 330, 429, 1092, 1365, 420, 1020, 2448, 2907, 1710, 1995, 4620, 5313, 759, 3450, 7800, 8775, 4914, 5481, 12180, 13485, 3720, 8184, 17952, 19635, 10710, 11655, 25308, 27417, 3705, 15990, 34440, 37023, 19866, 21285, 45540
COMMENTS
All terms are multiples of 3.
Difference table of c(n):
1/3, 1/6, 2/15, 7/60, 2/21,...
-1/6, -1/30, -1/60, -1/84, -1/105,...
2/15, 1/60, 1/210, 1/420, 1/630,...
-7/60, -1/84, -1/420, -1/1260, -1/2520,... .
This is an autosequence of the second kind; the inverse binomial transform is the signed sequence. The main diagonal is the first upper diagonal multiplied by 2.
Denominators of the main diagonal: A051133(n+1).
Denominators of the first upper diagonal; A000911(n).
Based on the Akiyama-Tanigawa transform applied to 1/(n+1) which yields the Bernoulli numbers A164555(n)/ A027642(n).
Are the numerators of the main diagonal (-1)^n? If yes, what is the value of 1/3 - 1/30 + 1/210,... or 1 - 1/10 + 1/70 - 1/420, ... , from A002802(n)?
Is a(n+40) - a(n) divisible by 10?
Are the common divisors to A014206(n) and A007531(n+3) of period 16: repeat 2, 4, 4, 2, 2, 16, 4, 2, 2, 4, 4, 2, 2, 8, 4, 2?
Reduce c(n) = f(n) = b(n)/a(n) = 1/3, 1/6, 2/15, 7/60, 11/105, 2/21, 11/126, 29/360, ... .
Consider the successively interleaved autosequences (also called eigensequences) of the second kind and of the first kind
1, 1/2, 1/3, 1/4, 1/5, 1/6, ...
0, 1/6, 1/6, 3/20, 2/15, 5/42, ...
1/3, 1/6, 2/15, 7/60, 11/105, 2/21, ...
0, 1/10, 1/10, 13/140, 3/35, 5/63, ...
1/5, 1/10, 3/35, 11/140, 23/315, 43/630, ...
0, 1/14, 1/14, 17/252, 4/63, ...
This array is Au1(m,n). Au1(0,0)=1, Au1(0,1)=1/2.
Au1(m+1,n) = 2*Au1(m,n+1) - Au1(m,n).
First row: see A003506, Leibniz's Harmonic Triangle.
a(n) is the denominator of the third row f(n).
The first column is 1, 0, 1/3, 0, 1/5, 0, 1/7, 0, ... . Numerators: A093178(n+1). This incites, considering tan(1), to introduce before the first row
Ta0(n) = 0, 1/2, 1/2, 5/12, 1/3, 4/15, 13/60, 151/840, ... .
FORMULA
The sum of the difference table main diagonal is 1/3 - 1/30 + 1/210 - ... = 10* A086466-4 = 4*(sqrt(5)*log(phi)-1) = 0.3040894... - Jean-François Alcover, Apr 22 2014
a(n) = (n+1)*(n+2)*(n+3)/gcd(4*n - 4, n^2 + n + 2), where gcd(4*n - 4, n^2 + n + 2) is periodic with period 16. - Robert Israel, Jul 17 2023
MAPLE
seq(denom((n^2+n+2)/((n+1)*(n+2)*(n+3))), n=0..1000);
MATHEMATICA
Denominator[Table[(n^2+n+2)/Times@@(n+{1, 2, 3}), {n, 0, 50}]] (* Harvey P. Dale, Mar 27 2015 *)
PROG
(PARI) for(n=0, 100, print1(denominator((n^2+n+2)/((n+1)*(n+2)*(n+3))), ", ")) \\ Colin Barker, Apr 18 2014
a(n) = binomial(2n,n)*n*(2n+1)/2.
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0, 3, 30, 210, 1260, 6930, 36036, 180180, 875160, 4157010, 19399380, 89237148, 405623400, 1825305300, 8143669800, 36064823400, 158685222960, 694247850450, 3022020054900, 13095420237900, 56517076816200, 243023430309660, 1041528987041400
FORMULA
(-n+1)*a(n) + 2*(2*n+1)*a(n-1) = 0. - R. J. Mathar, Feb 05 2013
Sum_{n>=1} 1/a(n) = 4 - 2*Pi/sqrt(3). - Amiram Eldar, Oct 22 2020
EXAMPLE
G.f. = 3*x + 30*x^2 + 210*x^3 + 1260*x^4 + 6930*x^5 + 36036*x^6 + ...
MAPLE
seq(binomial(2*n, n)*binomial(n, (n-2))/2, n=1..23); # Zerinvary Lajos, May 05 2007
MATHEMATICA
a[ n_]:= SeriesCoefficient[ 3x(1-4x)^(-5/2), {x, 0, n}]; (* Michael Somos, Sep 09 2013 *)
Table[Binomial[2*n, n]*n*(2*n + 1)/2, {n, 0, 22}] (* Amiram Eldar, Oct 22 2020 *)
PROG
(PARI) {a(n) = if( n<1, 0, (2*n + 1)! / (2 * n! *(n-1)!))}; /* Michael Somos, Sep 09 2013 */
(PARI) {a(n) = 2^(n+2) * polcoeff( pollegendre( n+3), n-1)}; /* Michael Somos, Sep 09 2013 */
(Magma) [Binomial(2*n, n)*n*(2*n+1)/2: n in [0..25]]; // G. C. Greubel, Feb 10 2019
(Sage) [binomial(2*n, n)*n*(2*n+1)/2 for n in (0..25)] # G. C. Greubel, Feb 10 2019
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