Search: a000911 -id:a000911
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A002802
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a(n) = (2*n+3)!/(6*n!*(n+1)!).
(Formerly M4724 N2019)
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+10
43
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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
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OFFSET
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0,2
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COMMENTS
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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)
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REFERENCES
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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).
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LINKS
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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.
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FORMULA
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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
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EXAMPLE
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G.f. = 1 + 10*x + 70*x^2 + 420*x^3 + 2310*x^4 + 12012*x^5 + 60060*x^6 + ...
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MAPLE
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seq(simplify(4^n*hypergeom([-n, -3/2], [1], 1)), n=0..25); # Peter Luschny, Apr 26 2016
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MATHEMATICA
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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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
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OFFSET
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0,2
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COMMENTS
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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)
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LINKS
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FORMULA
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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
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MAPLE
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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
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MATHEMATICA
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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 *)
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PROG
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(Haskell)
import Data.Ratio ((%), denominator)
a045896 n = denominator $ n % ((n + 1) * (n + 2))
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CROSSREFS
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KEYWORD
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nonn,easy,frac,nice
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AUTHOR
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STATUS
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approved
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A241269
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Denominator of c(n) = (n^2+n+2)/((n+1)*(n+2)*(n+3)).
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+10
8
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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
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OFFSET
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0,1
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COMMENTS
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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, ... .
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LINKS
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FORMULA
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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
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MAPLE
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seq(denom((n^2+n+2)/((n+1)*(n+2)*(n+3))), n=0..1000);
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MATHEMATICA
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Denominator[Table[(n^2+n+2)/Times@@(n+{1, 2, 3}), {n, 0, 50}]] (* Harvey P. Dale, Mar 27 2015 *)
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PROG
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(PARI) for(n=0, 100, print1(denominator((n^2+n+2)/((n+1)*(n+2)*(n+3))), ", ")) \\ Colin Barker, Apr 18 2014
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A051133
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a(n) = binomial(2n,n)*n*(2n+1)/2.
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+10
7
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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
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OFFSET
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0,2
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LINKS
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FORMULA
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(-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
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EXAMPLE
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G.f. = 3*x + 30*x^2 + 210*x^3 + 1260*x^4 + 6930*x^5 + 36036*x^6 + ...
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MAPLE
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seq(binomial(2*n, n)*binomial(n, (n-2))/2, n=1..23); # Zerinvary Lajos, May 05 2007
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MATHEMATICA
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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 *)
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PROG
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(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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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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