login
A002596
Numerators in expansion of sqrt(1+x). Absolute values give numerators in expansion of sqrt(1-x).
(Formerly M3768 N1538)
21
1, 1, -1, 1, -5, 7, -21, 33, -429, 715, -2431, 4199, -29393, 52003, -185725, 334305, -9694845, 17678835, -64822395, 119409675, -883631595, 1641030105, -6116566755, 11435320455, -171529806825, 322476036831, -1215486600363, 2295919134019
OFFSET
0,5
COMMENTS
Also, absolute values are numerators of (2n-3)!!/n! or the odd part of the (n-1)-th Catalan number.
From Dimitri Papadopoulos, Oct 28 2016: (Start)
The sum of the coefficients of the expansion of sqrt(1+x) is sqrt(2) (easy). Observation: The sum of the squares of the coefficients is 4/Pi.
Observation/conjecture: If a term of this sequence is divisible by a prime p, then that term is in a block of exactly (p^k-3)/2 consecutive terms all of which are divisible by p. Furthermore, if a(n) is the term preceding such a block then a(p*n-(p-1)/2) also precedes a block of (p^(k+1)-3)/2 terms all divisible by p.
E.g., a(4)=-5 is divisible by 5 and is in a block of (5^1 - 3)/2 = 1 consecutive terms that are all divisible by 5. Then a(5*3 - (5-1)/2) = a(13) = 52003 precedes a block of exactly (5^2 - 3)/2 = 11 terms all divisible by 5.
(End)
REFERENCES
B. D. Hughes, Random Walks and Random Environments, Oxford 1995, vol. 1, p. 513, Eq. (7.281).
M. Kauers and P. Paule, The Concrete Tetrahedron, Springer 2011, p. 88.
Eli Maor, e: The Story of a Number. Princeton, New Jersey: Princeton University Press (1994): 72.
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).
Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 6, equation 6:14:6 at page 51.
LINKS
Eric Weisstein's World of Mathematics, Legendre Polynomial.
FORMULA
a(n+2) = C(n+1)/2^k(n+1), n >= 0; where C(n) = A000108(n), k(n) = A048881(n).
From Johannes W. Meijer, Jun 08 2009: (Start)
a(n) = (-1)^n*numerator((1/(1-2*n))*binomial(2*n,n)/(4^n)).
(1+x)^(1/2) = Sum_{n>=0} (1/(1-2*n))*binomial(2*n,n)/(4^n)*(-x)^n.
(1-x)^(1/2) = Sum_{n>=0} (1/(1-2*n))*binomial(2*n,n)/(4^n)*(x)^n. (End)
a(n) = numerator(Product_{k=1..n} (3-2*k)/(2*k)). - Dimitri Papadopoulos, Oct 22 2016
EXAMPLE
sqrt(1+x) = 1 + (1/2)*x - (1/8)*x^2 + (1/16)*x^3 - (5/128)*x^4 + (7/256)*x^5 - (21/1024)*x^6 + (33/2048)*x^7 + ...
Coefficients are 1, 1/2, -1/8, 1/16, -5/128, 7/256, -21/1024, 33/2048, -429/32768, 715/65536, -2431/262144, 4199/524288, -29393/4194304, 52003/8388608, ...
MAPLE
seq(numer(subs(k=1/2, expand(binomial(k, n)))), n=0..50); # James R. Buddenhagen, Aug 16 2014
MATHEMATICA
1+InverseSeries[Series[2^p*y+y^2/2^q, {y, 0, 24}], x] (* p, q positive integers, then a(n)=numerator(y(n)). - Len Smiley, Apr 13 2000 *)
Numerator[CoefficientList[Series[Sqrt[1+x], {x, 0, 30}], x]] (* Harvey P. Dale, Oct 22 2011 *)
Table[Numerator[Product[(3 - 2 k)/(2 k) , {k, j}]], {j, 0, 30}] (* Dimitri Papadopoulos, Oct 22 2016 *)
PROG
(PARI) x = 'x + O('x^40); apply(x->numerator(x), Vec(sqrt(1+x))) \\ Michel Marcus, Jan 14 2016
(Magma) [(-1)^n*Numerator((1/(1-2*n))*Binomial(2*n, n)/(4^n)): n in [0..30]]; // Vincenzo Librandi, Jan 14 2016
CROSSREFS
Denominators are A046161.
Cf. A001795.
Equals A000265(A000108(n-1)), n>0.
Absolute values are essentially A098597.
From Johannes W. Meijer, Jun 08 2009: (Start)
Cf. A161200 [(1-x)^(3/2)] and A161202 [(1-x)^(5/2)], A001803 [(1-x)^(-3/2)].
Cf. A161198 = triangle related to the series expansions of (1-x)^((-1-2*n)/2) for all values of n. (End)
Sequence in context: A057424 A027152 A076197 * A098597 A097038 A049114
KEYWORD
easy,nice,frac,sign,changed
EXTENSIONS
Minor correction to definition from Johannes W. Meijer, Jun 05 2009
STATUS
approved