cp's OEIS Frontend

A002596 Numerators in expansion of sqrt(1+x). Absolute values give numerators in expansion of sqrt(1-x).

%I A002596 M3768 N1538 #97 Feb 16 2025 08:32:26
%S A002596 1,1,-1,1,-5,7,-21,33,-429,715,-2431,4199,-29393,52003,-185725,334305,
%T A002596 -9694845,17678835,-64822395,119409675,-883631595,1641030105,
%U A002596 -6116566755,11435320455,-171529806825,322476036831,-1215486600363,2295919134019
%N A002596 Numerators in expansion of sqrt(1+x). Absolute values give numerators in expansion of sqrt(1-x).
%C A002596 Also, absolute values are numerators of (2n-3)!!/n! or the odd part of the (n-1)-th Catalan number.
%C A002596 From _Dimitri Papadopoulos_, Oct 28 2016: (Start)
%C A002596 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.
%C A002596 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.
%C A002596 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.
%C A002596 (End)
%D A002596 B. D. Hughes, Random Walks and Random Environments, Oxford 1995, vol. 1, p. 513, Eq. (7.281).
%D A002596 M. Kauers and P. Paule, The Concrete Tetrahedron, Springer 2011, p. 88.
%D A002596 Eli Maor, e: The Story of a Number. Princeton, New Jersey: Princeton University Press (1994): 72.
%D A002596 N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D A002596 N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%D A002596 Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 6, equation 6:14:6 at page 51.
%H A002596 T. D. Noe, <a href="/A002596/b002596.txt">Table of n, a(n) for n = 0..200</a>
%H A002596 Tom Copeland, <a href="http://tcjpn.wordpress.com/2015/10/12/the-elliptic-lie-triad-kdv-and-ricattt-equations-infinigens-and-elliptic-genera/">Addendum to Elliptic Lie Triad</a>.
%H A002596 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/LegendrePolynomial.html">Legendre Polynomial</a>.
%F A002596 a(n+2) = C(n+1)/2^k(n+1), n >= 0; where C(n) = A000108(n), k(n) = A048881(n).
%F A002596 From _Johannes W. Meijer_, Jun 08 2009: (Start)
%F A002596 a(n) = (-1)^n*numerator((1/(1-2*n))*binomial(2*n,n)/(4^n)).
%F A002596 (1+x)^(1/2) = Sum_{n>=0} (1/(1-2*n))*binomial(2*n,n)/(4^n)*(-x)^n.
%F A002596 (1-x)^(1/2) = Sum_{n>=0} (1/(1-2*n))*binomial(2*n,n)/(4^n)*(x)^n. (End)
%F A002596 a(n) = numerator(Product_{k=1..n} (3-2*k)/(2*k)). - _Dimitri Papadopoulos_, Oct 22 2016
%e A002596 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 + ...
%e A002596 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, ...
%p A002596 seq(numer(subs(k=1/2,expand(binomial(k,n)))),n=0..50); # _James R. Buddenhagen_, Aug 16 2014
%t A002596 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 *)
%t A002596 Numerator[CoefficientList[Series[Sqrt[1+x],{x,0,30}],x]] (* _Harvey P. Dale_, Oct 22 2011 *)
%t A002596 Table[Numerator[Product[(3 - 2 k)/(2 k) , {k, j}]], {j, 0, 30}] (* _Dimitri Papadopoulos_, Oct 22 2016 *)
%o A002596 (PARI) x = 'x + O('x^40); apply(x->numerator(x), Vec(sqrt(1+x))) \\ _Michel Marcus_, Jan 14 2016
%o A002596 (Magma) [(-1)^n*Numerator((1/(1-2*n))*Binomial(2*n,n)/(4^n)): n in [0..30]]; // _Vincenzo Librandi_, Jan 14 2016
%Y A002596 Denominators are A046161.
%Y A002596 Cf. A001795.
%Y A002596 Equals A000265(A000108(n-1)), n>0.
%Y A002596 Absolute values are essentially A098597.
%Y A002596 From _Johannes W. Meijer_, Jun 08 2009: (Start)
%Y A002596 Cf. A161200 [(1-x)^(3/2)] and A161202 [(1-x)^(5/2)], A001803 [(1-x)^(-3/2)].
%Y A002596 Cf. A161198 = triangle related to the series expansions of (1-x)^((-1-2*n)/2) for all values of n. (End)
%K A002596 easy,nice,frac,sign
%O A002596 0,5
%A A002596 _N. J. A. Sloane_
%E A002596 Minor correction to definition from _Johannes W. Meijer_, Jun 05 2009