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A060165 Number of orbits of length n under the map whose periodic points are counted by A000984. +20
18
2, 2, 6, 16, 50, 150, 490, 1600, 5400, 18450, 64130, 225264, 800046, 2865226, 10341150, 37566720, 137270954, 504171432, 1860277042, 6892317200, 25631327190, 95640829922, 357975249026, 1343650040256, 5056424257500, 19073789328750, 72108867614796 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
The sequence A000984 seems to record the number of points of period n under a map. The number of orbits of length n for this map gives the sequence above.
The number of n-cycles in the graph of overlapping m-permutations where n <= m. - Richard Ehrenborg, Dec 10 2013
a(n) is divisible by n (cf. A268619), 6*a(n) is divisible by n^2 (cf. A268592). - Max Alekseyev, Feb 09 2016
Apparently the number of Lyndon words of length n with a 4-letter alphabet (see A027377) where the first letter of the alphabet appears with the same frequency as the second of the alphabet. E.g a(1)=2 counts the words (2), (3), a(2)= 2 counts (01) (23), a(3)=6 counts (021) (031) (012) (013) (223) (233). R. J. Mathar, Nov 04 2021
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..1669
R. Ehrenborg, S. Kitaev and E. Steingrimsson, Number of cycles in the graph of 312-avoiding permutations, arXiv:1310.1520 [math.CO], 2013.
Y. Puri and T. Ward, Arithmetic and growth of periodic orbits, J. Integer Seqs., Vol. 4 (2001), #01.2.1.
Yash Puri and Thomas Ward, A dynamical property unique to the Lucas sequence, Fibonacci Quarterly, Volume 39, Number 5 (November 2001), pp. 398-402.
FORMULA
a(n) = (1/n) * Sum_{d|n} mu(d) A000984(n/d) with mu = A008683.
a(n) = 2*A022553(n).
a(n) = A007727(n)/n. - R. J. Mathar, Jul 24 2017
G.f.: 2 * Sum_{k>=1} mu(k)*log((1 - sqrt(1 - 4*x^k))/(2*x^k))/k. - Ilya Gutkovskiy, May 18 2019
a(n) ~ 4^n / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Aug 04 2022
EXAMPLE
a(5) = 50 because if a map has A000984 as its periodic points, then it would have 2 fixed points and 252 points of period 5, hence 50 orbits of length 5.
MAPLE
with(numtheory):
a:= n-> add(mobius(n/d)*binomial(2*d, d), d=divisors(n))/n:
seq(a(n), n=1..30); # Alois P. Heinz, Dec 10 2013
MATHEMATICA
a[n_] := (1/n)*Sum[MoebiusMu[d]*Binomial[2*n/d, n/d], {d, Divisors[n]}]; Table[a[n], {n, 1, 30}] (* Jean-François Alcover, Jul 16 2015 *)
PROG
(PARI) a(n)=sumdiv(n, d, moebius(n/d)*binomial(2*d, d))/n \\ Charles R Greathouse IV, Dec 10 2013
(Python)
from sympy import mobius, binomial, divisors
def a(n): return sum(mobius(n//d) * binomial(2*d, d) for d in divisors(n))//n
print([a(n) for n in range(1, 31)]) # Indranil Ghosh, Jul 24 2017
CROSSREFS
Cf. A000984, A007727, A060164, A060166, A060167, A060168, A060169, A060170, A060171, A060172, A060173.
KEYWORD
easy,nonn
AUTHOR
Thomas Ward, Mar 13 2001
STATUS
approved
A054335 A convolution triangle of numbers based on A000984 (central binomial coefficients of even order). +20
12
1, 2, 1, 6, 4, 1, 20, 16, 6, 1, 70, 64, 30, 8, 1, 252, 256, 140, 48, 10, 1, 924, 1024, 630, 256, 70, 12, 1, 3432, 4096, 2772, 1280, 420, 96, 14, 1, 12870, 16384, 12012, 6144, 2310, 640, 126, 16, 1, 48620, 65536, 51480, 28672, 12012, 3840, 924, 160, 18, 1 (list; table; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
In the language of the Shapiro et al. reference (given in A053121) such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to the Bell-subgroup of the Riordan-group. The g.f. for the row polynomials p(n,x) (increasing powers of x) is 1/(sqrt(1-4*z)-x*z).
The column sequences are for m=0..20: A000984, A000302 (powers of 4), A002457, A002697, A002802, A038845, A020918, A038846, A020920, A040075, A020922, A045543, A020924, A054337, A020926, A054338, A020928, A054339, A020930, A054340, A020932.
Riordan array (1/sqrt(1-4x),x/sqrt(1-4x)). - Paul Barry, May 06 2009
The matrix inverse is apparently given by deleting the leftmost column from A206022. - R. J. Mathar, Mar 12 2013
LINKS
G. C. Greubel, Rows n = 0..100 of triangle, flattened
Paul Barry, Embedding structures associated with Riordan arrays and moment matrices, arXiv preprint arXiv:1312.0583 [math.CO], 2013.
Milan Janjić, Pascal Matrices and Restricted Words, J. Int. Seq., Vol. 21 (2018), Article 18.5.2.
FORMULA
a(n, 2*k+1) = binomial(n-k-1, k)*4^(n-2*k-1), a(n, 2*k) = binomial(2*(n-k), n-k)*binomial(n-k, k)/binomial(2*k, k), k >= 0, n >= m >= 0; a(n, m) := 0 if n<m.
Column recursion: a(n, m)=2*(2*n-m-1)*a(n-1, m)/(n-m), n>m >= 0, a(m, m) := 1.
G.f. for column m: cbie(x)*((x*cbie(x))^m, with cbie(x) := 1/sqrt(1-4*x).
G.f.: 1/(1-xy-2x/(1-x/(1-x/(1-x/(1-x/(1-... (continued fraction). - Paul Barry, May 06 2009
Sum_{k>=0} T(n,2k)*(-1)^k*A000108(k) = A000108(n+1). - Philippe Deléham, Jan 30 2012
Sum_{k=0..floor(n/2)} T(n-k,n-2k) = A098615(n). - Philippe Deléham, Feb 01 2012
T(n,k) = 4*T(n-1,k) + T(n-2,k-2) for k>=1. - Philippe Deléham, Feb 02 2012
Vertical recurrence: T(n,k) = 1*T(n-1,k-1) + 2*T(n-2,k-1) + 6*T(n-3,k-1) + 20*T(n-4,k-1) + ... for k >= 1 (the coefficients 1, 2, 6, 20, ... are the central binomial coefficients A000984). - Peter Bala, Oct 17 2015
EXAMPLE
Triangle begins:
1;
2, 1;
6, 4, 1;
20, 16, 6, 1;
70, 64, 30, 8, 1;
252, 256, 140, 48, 10, 1;
924, 1024, 630, 256, 70, 12, 1; ...
Fourth row polynomial (n=3): p(3,x) = 20 + 16*x + 6*x^2 + x^3.
From Paul Barry, May 06 2009: (Start)
Production matrix begins
2, 1;
2, 2, 1;
0, 2, 2, 1;
-2, 0, 2, 2, 1;
0, -2, 0, 2, 2, 1;
4, 0, -2, 0, 2, 2, 1;
0, 4, 0, -2, 0, 2, 2, 1;
-10, 0, 4, 0, -2, 0, 2, 2, 1;
0, -10, 0, 4, 0, -2, 0, 2, 2, 1; (End)
MAPLE
A054335 := proc(n, k)
if k <0 or k > n then
0 ;
elif type(k, odd) then
kprime := floor(k/2) ;
binomial(n-kprime-1, kprime)*4^(n-k) ;
else
kprime := k/2 ;
binomial(2*n-k, n-kprime)*binomial(n-kprime, kprime)/binomial(k, kprime) ;
end if;
end proc: # R. J. Mathar, Mar 12 2013
# Uses function PMatrix from A357368. Adds column 1, 0, 0, 0, ... to the left.
PMatrix(10, n -> binomial(2*(n-1), n-1)); # Peter Luschny, Oct 19 2022
MATHEMATICA
Flatten[ CoefficientList[#1, x] & /@ CoefficientList[ Series[1/(Sqrt[1 - 4*z] - x*z), {z, 0, 9}], z]] (* or *)
a[n_, k_?OddQ] := 4^(n-k)*Binomial[(2*n-k-1)/2, (k-1)/2]; a[n_, k_?EvenQ] := (Binomial[n-k/2, k/2]*Binomial[2*n-k, n-k/2])/Binomial[k, k/2]; Table[a[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Sep 08 2011, updated Jan 16 2014 *)
PROG
(PARI) T(n, k) = if(k%2==0, binomial(2*n-k, n-k/2)*binomial(n-k/2, k/2)/binomial(k, k/2), 4^(n-k)*binomial(n-(k-1)/2-1, (k-1)/2));
for(n=0, 10, for(k=0, n, print1(T(n, k), ", "))) \\ G. C. Greubel, Jul 20 2019
(Magma)
T:= func< n, k | (k mod 2) eq 0 select Binomial(2*n-k, n-Floor(k/2))* Binomial(n-Floor(k/2), Floor(k/2))/Binomial(k, Floor(k/2)) else 4^(n-k)*Binomial(n-Floor((k-1)/2)-1, Floor((k-1)/2)) >;
[[T(n, k): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Jul 20 2019
(Sage)
def T(n, k):
if (mod(k, 2)==0): return binomial(2*n-k, n-k/2)*binomial(n-k/2, k/2)/binomial(k, k/2)
else: return 4^(n-k)*binomial(n-(k-1)/2-1, (k-1)/2)
[[T(n, k) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Jul 20 2019
(GAP)
T:= function(n, k)
if k mod 2=0 then return Binomial(2*n-k, n-Int(k/2))*Binomial(n-Int(k/2), Int(k/2))/Binomial(k, Int(k/2));
else return 4^(n-k)*Binomial(n-Int((k-1)/2)-1, Int((k-1)/2));
fi;
end;
Flat(List([0..10], n-> List([0..n], k-> T(n, k) ))); # G. C. Greubel, Jul 20 2019
CROSSREFS
Cf. A000984, A035324, A054336.
Row sums: A026671.
KEYWORD
easy,nice,nonn,tabl
AUTHOR
Wolfdieter Lang, Mar 13 2000
STATUS
approved
A232606 G.f. A(x) satisfies: the sum of the coefficients of x^k, k=0..n, in A(x)^n equals (2*n)!^2/n!^4, the square of the central binomial coefficients (A000984), for n>=0. +20
11
1, 3, 10, 42, 221, 1379, 9678, 73666, 594326, 5007958, 43641702, 390632678, 3573598539, 33289289533, 314871186248, 3017358158132, 29242725947318, 286209134234602, 2825613061237808, 28111283170770480, 281598654896870051, 2838309465080014489, 28767973963085929656, 293059625830028920012 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
Compare to: Sum_{k=0..n} [x^k] 1/(1-x)^n = (2*n)!/n!^2 = A000984(n).
a(n+1)/a(n) tends to 11.3035... - Vaclav Kotesovec, Jan 23 2014
LINKS
Vaclav Kotesovec, Table of n, a(n) for n = 0..350
FORMULA
Given g.f. A(x), Sum_{k=0..n} [x^k] A(x)^n = (2*n)!^2/n!^4 = A000984(n)^2.
Given g.f. A(x), let G(x) = A(x*G(x)) then (G(x) + x*G'(x)) / (G(x) - x*G(x)^2) = Sum_{n>=0} (2*n)!^2/n!^4 * x^n.
EXAMPLE
G.f.: A(x) = 1 + 3*x + 10*x^2 + 42*x^3 + 221*x^4 + 1379*x^5 + 9678*x^6 +...
ILLUSTRATION OF INITIAL TERMS.
If we form an array of coefficients of x^k in A(x)^n, n>=0, like so:
A^0: [1], 0, 0, 0, 0, 0, 0, 0, 0, ...;
A^1: [1, 3], 10, 42, 221, 1379, 9678, 73666, 594326, ...;
A^2: [1, 6, 29], 144, 794, 4924, 33814, 251544, 1988885, ...;
A^3: [1, 9, 57, 333], 1989, 12669, 86935, 639123, 4979499, ...;
A^4: [1, 12, 94, 636, 4157], 27728, 193504, 1423120, 11006058, ...;
A^5: [1, 15, 140, 1080, 7730, 54538], 391970, 2915490, 22558825, ...;
A^6: [1, 18, 195, 1692, 13221, 99102, 739547], 5612016, 43767477, ...;
A^7: [1, 21, 259, 2499, 21224, 169232, 1317722, 10267666], 81223912, ...;
A^8: [1, 24, 332, 3528, 32414, 274792, 2238492, 17990904, 145096413], ...; ...
then the sum of the coefficients of x^k, k=0..n, in A(x)^n (shown above in brackets) equals the square of the central binomial coefficients:
1^1 = 1;
2^2 = 1 + 3;
6^2 = 1 + 6 + 29;
20^2 = 1 + 9 + 57 + 333;
70^2 = 1 + 12 + 94 + 636 + 4157;
252^2 = 1 + 15 + 140 + 1080 + 7730 + 54538;
924^2 = 1 + 18 + 195 + 1692 + 13221 + 99102 + 739547;
3432^2 = 1 + 21 + 259 + 2499 + 21224 + 169232 + 1317722 + 10267666; ...
RELATED SERIES.
From a main diagonal in the above array we can derive sequence A232607:
[1/1, 6/2, 57/3, 636/4, 7730/5, 99102/6, 1317722/7, 17990904/8, ...] =
[1, 3, 19, 159, 1546, 16517, 188246, 2248863, 27844369, 354576634, ...];
from which we can form the series G(x) = A(x*G(x)):
G(x) = 1 + 3*x + 19*x^2 + 159*x^3 + 1546*x^4 + 16517*x^5 + 188246*x^6 +...
such that
(G(x) + x*G'(x)) / (G(x) - x*G(x)^2) = 1 + 2^2*x + 6^2*x^2 + 20^2*x^3 + 70^2*x^4 + 252^2*x^5 +...+ A000984(n)^2*x^n +...
MATHEMATICA
terms = 24; a[0] = 1; A[x_] = Sum[a[n]*x^n, {n, 0, terms - 1}];
c[n_] := Sum[Coefficient[B[x], x, k], {k, 0, n}] == (2*n)!^2/n!^4 // Solve // First;
Do[B[x_] = A[x]^n + O[x]^(n+1) // Normal; A[x_] = (A[x] /. c[n]) + O[x]^terms, {n, 0, terms-1}];
CoefficientList[A[x], x] (* Jean-François Alcover, Jan 14 2018 *)
PROG
(PARI) /* By Definition: */
{a(n)=if(n==0, 1, ((2*n)!^2/n!^4 - sum(k=0, n, polcoeff(sum(j=0, min(k, n-1), a(j)*x^j)^n + x*O(x^k), k)))/n)}
for(n=0, 20, print1(a(n)*1!, ", "))
(PARI) /* Faster, using series reversion: */
{a(n)=local(CB2=sum(k=0, n, binomial(2*k, k)^2*x^k)+x*O(x^n), G=1+x*O(x^n));
for(i=1, n, G = 1 + intformal( (CB2-1)*G/x - CB2*G^2)); polcoeff(x/serreverse(x*G), n)}
for(n=0, 30, print1(a(n), ", "))
CROSSREFS
Cf. A232683, A232607, A232687, A000984, A002894.
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Nov 26 2013
STATUS
approved
A187364 Trisection of A000984 (central binomial coefficients): binomial(2(3n+1),3n+1)/2, n>=0. +20
10
1, 35, 1716, 92378, 5200300, 300540195, 17672631900, 1052049481860, 63205303218876, 3824345300380220, 232714176627630544, 14226520737620288370, 873065282167813104916, 53753604366668088230810, 3318776542511877736535400, 205397724721029574666088520 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
See a comment under A187363 concerning trisection.
This appears also in the trisection of A001700 (central binomials in the odd numbered Pascal rows): binomial(2*(3*n)+1,3*n+1).
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..554
FORMULA
a(n)=binomial(2*(3*n+1),3*n+1)/2, n>=0.
a(n)=binomial(2*(3*n)+1,3*n+1), n>=0.
O.g.f.: (cb(x^(1/3)) - sqrt(2)*P(x^(1/3))*sqrt(1/P(x^(1/3))-(1+8*x^(1/3))/2))/(6*x^(1/3)),
with cb(x):=1/sqrt(1-4*x) (o.g.f. of A000984) and P(x):=P(-1/2,4*x)=1/sqrt(1+4*x+16*x^2) (o.g.f. of A116091, with P(x,z) the o.g.f. of the Legendre polynomials).
From Peter Bala, Mar 19 2023: (Start)
a(n) = (1/2)*Sum_{k = 0..3*n+1} binomial(3*n+1,k)^2.
a(n) = (1/2)*hypergeom([-1 - 3*n, -1 - 3*n], [1], 1).
a(n) = 8*(2*n - 1)*(6*n + 1)*(6*n - 1)/(n*(3*n + 1)*(3*n - 1)) * a(n-1). (End)
MATHEMATICA
Table[c=3n+1; Binomial[2c, c]/2, {n, 0, 20}] (* Harvey P. Dale, May 10 2012 *)
CROSSREFS
A066802 binomial(6n,3n), A187365 binomial(2(3n+2),3n+2)/3!.
Cf. A002458, A100033.
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Mar 10 2011
STATUS
approved
A081387 Number of non-unitary prime divisors of central binomial coefficient, C(2n,n) = A000984(n), i.e., number of prime factors in C(2n,n) whose exponent is greater than one. +20
9
0, 0, 1, 0, 2, 1, 1, 1, 1, 1, 1, 1, 2, 3, 2, 1, 2, 2, 2, 2, 1, 1, 3, 3, 3, 3, 2, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 2, 3, 2, 3, 2, 3, 3, 2, 3, 3, 4, 3, 2, 2, 2, 2, 1, 1, 2, 1, 1, 2, 2, 3, 3, 3, 2, 4, 2, 2, 3, 3, 3, 3, 4, 4, 5, 4, 4, 2, 3, 3, 2, 2, 2, 4, 3, 3, 4, 3, 4, 5, 4, 2, 2, 2, 3, 5, 5, 5, 5, 3, 2, 3, 2, 3, 3, 3 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,5
LINKS
T. D. Noe, Table of n, a(n) for n = 1..10000
FORMULA
a(n) = A056170(A000984(n)) = A001221(A000984(n)) - A081386(n)= A067434(n) - A081386(n).
EXAMPLE
For n=14: binomial(28,14) = 40116600 = 2*2*2*3*3*3*5*5*17*19*23;
unitary prime divisors: {17,19,23};
non-unitary prime divisors: {2,3,5}, so a(14)=3.
MATHEMATICA
Table[Count[Transpose[FactorInteger[Binomial[2n, n]]][[2]], _?(#>1&)], {n, 110}] (* Harvey P. Dale, Oct 08 2012 *)
CROSSREFS
Cf. A000984, A034444, A048105, A056169, A056170, A067434, A081387-A081389.
KEYWORD
nonn
AUTHOR
Labos Elemer, Mar 27 2003
STATUS
approved
A201555 a(n) = C(2*n^2,n^2) = A000984(n^2), where A000984 is the central binomial coefficients. +20
9
1, 2, 70, 48620, 601080390, 126410606437752, 442512540276836779204, 25477612258980856902730428600, 23951146041928082866135587776380551750, 365907784099042279561985786395502921046971688680, 90548514656103281165404177077484163874504589675413336841320 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
Central coefficients of triangle A228832.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..40
R. Mestrovic, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.
R. Oblath, Congruences with binomial coefficients, Proceedings of the Indian Academy of Science, Section A, Vol. 1 No. 6, 383-386
FORMULA
L.g.f.: ignoring initial term, equals the logarithmic derivative of A201556.
a(n) = (2*n^2)! / (n^2)!^2.
a(n) = Sum_{k=0..n^2} binomial(n^2,k)^2.
For primes p >= 5: a(p) == 2 (mod p^3), Oblath, Corollary II; a(p) == binomial(2*p,p) (mod p^6) - see Mestrovic, Section 5, equation 31. - Peter Bala, Dec 28 2014
A007814(a(n)) = A159918(n). - Antti Karttunen, Apr 27 2017, based on Vladimir Shevelev's Jul 20 2009 formula in A000984.
EXAMPLE
L.g.f.: L(x) = 2*x + 70*x^2/2 + 48620*x^3/3 + 601080390*x^4/4 + ...
where exponentiation equals the g.f. of A201556:
exp(L(x)) = 1 + 2*x + 37*x^2 + 16278*x^3 + 150303194*x^4 + ... + A201556(n)*x^n + ...
MATHEMATICA
Table[Binomial[2n^2, n^2], {n, 0, 10}] (* Harvey P. Dale, Dec 10 2011 *)
PROG
(PARI) a(n) = binomial(2*n^2, n^2)
(Python)
from math import comb
def A201555(n): return comb((m:=n**2)<<1, m) # Chai Wah Wu, Jul 08 2022
CROSSREFS
Cf. A000984, A007814, A159918, A201556, A214441, A228832, A285388, A285717.
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Dec 02 2011
STATUS
approved
A226078 Table read by rows: prime power factors of central binomial coefficients, cf. A000984. +20
9
1, 2, 2, 3, 4, 5, 2, 5, 7, 4, 9, 7, 4, 3, 7, 11, 8, 3, 11, 13, 2, 9, 5, 11, 13, 4, 5, 11, 13, 17, 4, 11, 13, 17, 19, 8, 3, 7, 13, 17, 19, 4, 7, 13, 17, 19, 23, 8, 25, 7, 17, 19, 23, 8, 27, 25, 17, 19, 23, 16, 9, 5, 17, 19, 23, 29, 2, 9, 5, 17, 19, 23, 29, 31 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
T(n,k) = A141809(A000984(n)),k) for k = 0..A067434(n)-1.
LINKS
Reinhard Zumkeller, Rows n = 0..250 of triangle, flattened
EXAMPLE
. n initial rows A000984(n) A226047(n)
. ---+------------------------------+-------------+------------
. 0 [1] 1
. 1 [2] 2 2
. 2 [2,3] 6 3
. 3 [4,5] 20 5
. 4 [2,5,7] 70 7
. 5 [4,9,7] 252 9
. 6 [4,3,7,11] 924 11
. 7 [8,3,11,13] 3432 13
. 8 [2,9,5,11,13] 12870 13
. 9 [4,5,11,13,17] 48620 17
. 10 [4,11,13,17,19] 184756 19
. 11 [8,3,7,13,17,19] 705432 19
. 12 [4,7,13,17,19,23] 2704156 23
. 13 [8,25,7,17,19,23] 10400600 25
. 14 [8,27,25,17,19,23] 40116600 27
. 15 [16,9,5,17,19,23,29] 155117520 29
. 16 [2,9,5,17,19,23,29,31] 601080390 31
. 17 [4,27,5,11,19,23,29,31] 2333606220 31
. 18 [4,3,25,7,11,19,23,29,31] 9075135300 31
. 19 [8,3,25,7,11,23,29,31,37] 35345263800 37
. 20 [4,9,5,7,11,13,23,29,31,37] 137846528820 37 .
MAPLE
f:= n-> add(i[2]*x^i[1], i=ifactors(n)[2]):
b:= proc(n) local p;
p:= add(f(n+i) -f(i), i=1..n);
seq(`if`(coeff(p, x, i)>0,
i^coeff(p, x, i), NULL), i=1..degree(p))
end:
T:= n-> `if`(n=0, 1, b(n)):
seq(T(n), n=0..30); # Alois P. Heinz, May 25 2013
MATHEMATICA
Table[Power @@@ FactorInteger[(2n)!/n!^2] , {n, 0, 30}] // Flatten (* Jean-François Alcover, Jul 29 2015 *)
PROG
(Haskell)
a226078 n k = a226078_tabf !! n !! k
a226078_row n = a226078_tabf !! n
a226078_tabf = map a141809_row a000984_list
CROSSREFS
Cf. A067434 (row lengths), A001316 (left edge), A060308 (right edge), A226047 (row maxima), A226083 (row minima), A000984 (row products).
Cf. A267823.
KEYWORD
nonn,tabf,look
AUTHOR
Reinhard Zumkeller, May 25 2013
STATUS
approved
A081386 Number of unitary prime divisors of central binomial coefficient, C(2n,n) = A000984(n), i.e., number of those prime factors in C(2n,n), whose exponent equals one. +20
8
1, 2, 1, 3, 1, 3, 3, 4, 4, 4, 5, 5, 4, 3, 5, 7, 6, 7, 7, 8, 9, 9, 6, 7, 7, 7, 8, 11, 12, 11, 11, 11, 12, 12, 12, 13, 13, 13, 11, 13, 12, 14, 13, 13, 15, 14, 14, 14, 15, 16, 16, 16, 17, 19, 18, 17, 18, 19, 18, 19, 18, 18, 18, 20, 18, 21, 22, 20, 20, 20, 20, 20, 20, 19, 21, 21, 24, 23 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
LINKS
T. D. Noe, Table of n, a(n) for n=1..2000
FORMULA
a(n) = A056169(A000984(n)).
EXAMPLE
n=10: C(20,10) = 184756 = 2*2*11*13*17*19; unitary-p-divisors = {11,13,17,19}, so a(10)=4.
MATHEMATICA
Table[Function[m, Count[Divisors@ m, k_ /; And[PrimeQ@ k, GCD[k, m/k] == 1]]]@ Binomial[2 n, n], {n, 50}] (* Michael De Vlieger, Dec 17 2016 *)
PROG
(PARI) a(n) = my(f=factor(binomial(2*n, n))); sum(k=1, #f~, f[k, 2] == 1); \\ Michel Marcus, Dec 18 2016
CROSSREFS
Cf. A000984, A067434, A081387-A081389, A034444, A048105, A056170, A056169.
KEYWORD
nonn
AUTHOR
Labos Elemer, Mar 27 2003
STATUS
approved
A134760 a(n) = 2*A000984(n) - 1. +20
8
1, 3, 11, 39, 139, 503, 1847, 6863, 25739, 97239, 369511, 1410863, 5408311, 20801199, 80233199, 310235039, 1202160779, 4667212439, 18150270599, 70690527599, 275693057639, 1076515748879, 4208197927439, 16466861455199, 64495207366199, 252821212875503 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
Inverse binomial transform of this is A134761: (the sequence interpolated with ones): (1, 1, 3, 1, 11, 1, 39, 1, 139, ...).
LINKS
Alois P. Heinz, Table of n, a(n) for n = 0..500
C. J. Fewster and D. Siemssen, Enumerating Permutations by their Run Structure, arXiv preprint arXiv:1403.1723 [math.CO], 2014.
FORMULA
From R. J. Mathar, Mar 23 2015: (Start)
n*a(n) = 2*(3*n-2)*a(n-1) - (9*n-14)*a(n-2) + 2*(2*n-5)*a(n-3).
n*(3*n-5)*a(n) = (15*n^2-31*n+12)*a(n-1) - 2*(3*n-2)*(2*n-3)*a(n-2). (End)
From G. C. Greubel, Apr 06 2024: (Start)
a(n) = 2*(n+1)*A000108(n) - 1.
G.f.: 2/sqrt(1 - 4*x) - 1/(1 - x).
E.g.f.: 2*exp(2*x)*BesselI(0, 2*x) - exp(x). (End)
MAPLE
a:= proc(n) option remember; `if`(n<2, 2*n+1,
((12-31*n+15*n^2) *a(n-1)
-2*(3*n-2)*(2*n-3)*a(n-2)) / (n*(3*n-5)))
end:
seq(a(n), n=0..30); # Alois P. Heinz, Jan 16 2013
MATHEMATICA
a[n_] := 2 Binomial[2n, n] - 1; Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Jul 21 2017 *)
PROG
(Magma) [2*(n+1)*Catalan(n)-1: n in [0..40]]; // G. C. Greubel, Apr 06 2024
(SageMath) [2*binomial(2*n, n)-1 for n in range(41)] # G. C. Greubel, Apr 06 2024
CROSSREFS
Cf. A000984, A000108, A134761.
KEYWORD
nonn,easy
AUTHOR
Gary W. Adamson, Nov 09 2007
STATUS
approved
A187365 Trisection of A000984 (central binomial coefficients): binomial(2(3n+2),3n+2)/3!, n>=0. +20
8
1, 42, 2145, 117572, 6686100, 388934370, 22974421470, 1372238454600, 82653088824684, 5011211083256840, 305437356823765089, 18697712969443807572, 1148770108115543559100, 70797430141465286938140, 4374750896947475198160300, 270950190057528375091435920 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
See a comment under A187357 concerning trisection.
This appears also in the trisection of A001700: binomial(2*(3*n+1)+1,(3*n+1)+1)/3.
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..554
FORMULA
a(n)=binomial(2*(3*n+2),3*n+2)/3!, n>=0.
a(n)=binomial(3*(2*n+1),3*n+2)/3, n>=0.
O.g.f.:(cb(x^(1/3)) - sqrt(2)*P(x^(1/3))*sqrt(1/P(x^(1/3))-(1-4*x^(1/3))/2))/(18*x^(2/3)),
with cb(x):=1/sqrt(1-4*x) (o.g.f. of A000984) and P(x):=P(-1/2,4*x)=1/sqrt(1+4*x+16*x^2) (o.g.f. of A116091, with P(x,z)the o.g.f. of the Legendre polynomials).
From Peter Bala, Mar 19 2023: (Start)
a(n) = (1/6)*Sum_{k = 0..3*n+2} binomial(3*n+2,k)^2.
a(n) = (1/6)*hypergeom([-2 - 3*n, -2 - 3*n], [1], 1).
a(n) = 8*(2*n + 1)*(6*n + 1)*(6*n - 1)/(n*(3*n + 1)*(3*n + 2)) * a(n-1). (End)
CROSSREFS
Cf. A066802 binomial(6n,3n), A187364 binomial(2*(3n+1),3n+1)/2, A002458, A100033.
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Mar 10 2011
STATUS
approved
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Last modified September 13 10:14 EDT 2024. Contains 375904 sequences. (Running on oeis4.)