A004134 -id:A004134 - cp's OEIS frontend

A101191 G.f.: A(x) = Sum_{n>=0}a(n)/2^A004134(n)*x^n = limit_{n->oo} F(n)^(1/2^(n+1)) where F(n) is the n-th iteration of: F(0) = 1, F(n) = F(n-1)^2 + x^(2^n-1) for n>=1.

1, 1, -3, 23, -525, 2695, -29687, 191991, -10488701, 70977675, -968279181, 6752850945, -191225421641, 1363019302883, -19538003443615, 140961586090743, -16379289413266717, 119621607825995891, -1755802638936696081, 12944528671963135869, -383361262914445548739

Offset: 0

Paul D. Hanna, Dec 03 2004

sign

Although the power series for the g.f. A(x) diverges at x=1, the Euler transform of the power series A(x) at x=1 converges to the constant A076949: Sum_{n>=0}[Sum_{k=0..n}C(n,k)*a(k))/2^A004134(n) ]/2^(n+1) = 1.2259024435...

			The iteration begins:
F(0) = 1,
F(1) = F(0)^2 + x^(2^1-1) = 1 +x,
F(2) = F(1)^2 + x^(2^2-1) = 1 +2*x +x^2 +x^3,
F(3) = F(2)^2 + x^(2^3-1) = 1 +4*x +6*x^2 +6*x^3 +5*x^4 +2*x^5 +x^6 +x^7.
The 2^(n+1)-th roots of F(n) tend to the limit of the g.f.:
F(1)^(1/2^2) = 1 +1/4*x -3/32*x^2 +7/128*x^3 -77/2048*x^4 +231/8192*x^5 +...
F(2)^(1/2^3) = 1 +1/4*x -3/32*x^2 +23/128*x^3 -525/2048*x^4 +2695/8192*x^5 +...
F(3)^(1/2^4) = 1 +1/4*x -3/32*x^2 +23/128*x^3 -525/2048*x^4 +2695/8192*x^5 +...
The limit of this process is the g.f. A(x) of this sequence.
The coefficients of x^k in the 2^n powers of the g.f. A(x) begin:
A^(2^0)=[1,1/4,-3/32,23/128,-525/2048,2695/8192,-29687/65536,...],
A^(2^1)=[1,1/2,-1/8,5/16,-53/128,127/256,-677/1024,2221/2048,...],
A^(2^2)=[1,1,0,1/2,-1/2,1/2,-5/8,9/8,-2,53/16,-89/16,155/16,...],
A^(2^3)=[1,2,1,1,0,0,0,1/2,-1,3/2,-5/2,9/2,-8,14,-197/8,44,...],
A^(2^4)=[1,4,6,6,5,2,1,1,0,0,0,0,0,0,0,1/2,-2,5,...],
A^(2^5)=[1,8,28,60,94,116,114,94,69,44,26,14,5,2,1,1,0,0,...].
Note: the sum of the coefficients of x^k in F(n) equals A003095(n+1):
1, 2=1+1, 5=1+2+1+1, 26=1+4+6+6+5+2+1+1, ...
The last n coefficients in F(n) read backwards are Catalan numbers (A000108).

Cf. A101189, A101190, A005187, A003095, A076949.

{a(n)=local(F=1,A,L);if(n==0,A=1,L=ceil(log(n+1)/log(2)); for(k=1,L,F=F^2+x^(2^k-1)); A=polcoeff(F^(1/2^(L+1))+x*O(x^n),n));numerator(A)}

G.f. A(x) satisfies: A(x)^2 = Sum_{n>=0} A101190(n)/2^A005187(n)*x^n. G.f. A(x) satisfies: A(2*x)^4 = Sum_{n>=0} A101189(n)*(2x)^n.

A005187 a(n) = a(floor(n/2)) + n; also denominators in expansion of 1/sqrt(1-x) are 2^a(n); also 2n - number of 1's in binary expansion of 2n.

Original entry on oeis.org

0, 1, 3, 4, 7, 8, 10, 11, 15, 16, 18, 19, 22, 23, 25, 26, 31, 32, 34, 35, 38, 39, 41, 42, 46, 47, 49, 50, 53, 54, 56, 57, 63, 64, 66, 67, 70, 71, 73, 74, 78, 79, 81, 82, 85, 86, 88, 89, 94, 95, 97, 98, 101, 102, 104, 105, 109, 110, 112, 113, 116, 117, 119, 120, 127, 128

Offset: 0

N. J. A. Sloane, May 20 1991; Allan Wilks, Dec 11 1999

Also exponent of the largest power of 2 dividing (2n)! (A010050) and (2n)!! (A000165).
Write n in binary: 1ab..yz, then a(n) = 1ab..yz + ... + 1ab + 1a + 1. - Ralf Stephan, Aug 27 2003
Also numbers having a partition into distinct Mersenne numbers > 0; A079559(a(n))=1; complement of A055938. - Reinhard Zumkeller, Mar 18 2009
Wikipedia's article on the "Whitney Immersion theorem" mentions that the a(n)-dimensional sphere arises in the Immersion Conjecture proved by Ralph Cohen in 1985. - Jonathan Vos Post, Jan 25 2010
For n > 0, denominators for consecutive pairs of integral numerator polynomials L(n+1,x) for the Legendre polynomials with o.g.f. 1 / sqrt(1-tx+x^2). - Tom Copeland, Feb 04 2016
a(n) is the total number of pointers in the first n elements of a perfect skip list. - Alois P. Heinz, Dec 14 2017
a(n) is the position of the n-th a (indexing from 0) in the fixed point of the morphism a -> aab, b -> b. - Jeffrey Shallit, Dec 24 2020
Numbers that can be expressed as the sum of distinct numbers of the form 2^k - 1 (lenient Mersenne numbers, A000225). This follows from the 2N - Hamming weight definition. A corollary is that these are the numbers with no 2 in their skew-binary representation (cf. A169683). - Allan C. Wechsler, Feb 25 2025

			G.f. = x + 3*x^2 + 4*x^3 + 7*x^4 + 8*x^5 + 10*x^6 + 11*x^7 + 15*x^8 + ...

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

N. J. A. Sloane, Table of n, a(n) for n = 0..20000 (first 1000 terms from T. D. Noe)
Jean-Paul Allouche, Jean Bétréma, and Jeffrey Shallit, Sur des points fixes de morphismes d'un monoïde libre, RAIRO-Theor. Inf. Appl. 23 (1989), 235-249.
Laurent Alonso, Edward M. Reingold, and René Schott, Determining the majority, Inform. Process. Lett. 47 (1993), no. 5, 253-255.
Laurent Alonso, Edward M. Reingold, and René Schott, The average-case complexity of determining the majority, SIAM J. Comput. 26 (1997), no. 1, 1-14.
Barry Brent, On the Constant Terms of Certain Laurent Series, Preprints (2023) 2023061164.
Sung-Hyuk Cha, On Integer Sequences Derived from Balanced k-ary Trees, Applied Mathematics in Electrical and Computer Engineering, 2012.
Sung-Hyuk Cha, On Complete and Size Balanced k-ary Tree Integer Sequences, International Journal of Applied Mathematics and Informatics, Issue 2, Volume 6, 2012, pp. 67-75. - From _N. J. A. Sloane_, Dec 24 2012
Ralph L. Cohen, The Immersion Conjecture for Differentiable Manifolds, The Annals of Mathematics, 1985: 237-328.
Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications, preprint, 2016.
Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, 13:4 (2017), #47; DOI: 10.1145/3127585.
Keith Johnson and Kira Scheibelhut, Rational Polynomials That Take Integer Values at the Fibonacci Numbers, American Mathematical Monthly 123.4 (2016): 338-346. See p. 340.
Tanya Khovanova, There are no coincidences, arXiv preprint 1410.2193 [math.CO], 2014.
Anunay Kulshrestha, On Hamming Distance between base-n representations of whole numbers, arXiv:1203.4547 [cs.DM], 2012.
Michael E. Saks and Michael Werman, On computing majority by comparisons, Combinatorica 11 (1991), no. 4, 383-387.
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions
Wikipedia, Whitney Immersion Theorem.
Allan Wilks, Email to N. J. A. Sloane, Jul 7 1988.

Cf. A001790, A011371, A067080, A098844, A132027, A004128, A054899, A046161.
Cf. A001511 (first differences), A122247 (partial sums), A055938 (complement).
Cf. A004134, A010050, A000165.
Cf. A000120, A079559, A085354, A030308, A000225.

a005187 n = a005187_list !! n
a005187_list = 0 : zipWith (+) [1..] (map (a005187 . (`div` 2)) [1..])
-- Reinhard Zumkeller, Nov 07 2011, Oct 05 2011

[n + Valuation(Factorial(n), 2): n in [0..70]]; // Vincenzo Librandi, Jun 11 2019

A005187 := n -> 2*n - add(i, i=convert(n, base, 2)):
seq(A005187(n), n=0..65); # Peter Luschny, Apr 08 2014

a[0] = 0; a[n_] := a[n] = a[Floor[n/2]] + n; Table[ a[n], {n, 0, 50}] (* or *)
Table[IntegerExponent[(2n)!, 2], {n, 0, 65}] (* Robert G. Wilson v, Apr 19 2006 *)
Table[2n-DigitCount[2n,2,1],{n,0,70}] (* Harvey P. Dale, May 24 2014 *)

{a(n) = if( n<0, 0, valuation((2*n)!, 2))}; /* Michael Somos, Oct 24 2002 */

{a(n) = if( n<0, 0, sum(k=1, n, (2*n)\2^k))}; /* Michael Somos, Oct 24 2002 */

{a(n) = if( n<0, 0, 2*n - subst( Pol( binary( n ) ), x, 1) ) }; /* Michael Somos, Aug 28 2007 */

a(n)=my(s=n);while(n>>=1,s+=n);s \\ Charles R Greathouse IV, Apr 07 2012

a(n)=2*n-hammingweight(n) \\ Charles R Greathouse IV, Jan 07 2013

def A005187(n): return 2*n-bin(n).count('1') # Chai Wah Wu, Jun 03 2021

@CachedFunction
def A005187(n): return A005187(n//2) + n if n > 0 else 0
[A005187(n) for n in range(66)]  # Peter Luschny, Dec 13 2012

a(n) = A011371(2n+1) = A011371(n) + n, n >= 0.
A046161(n) = 2^a(n).
For m>0, let q = floor(log_2(m)); a(2m+1) = 2^q + 3m + Sum_{k>=1} floor((m-2^q)/2^k); a(2m) = a(2m+1) - 1. - Len Smiley
a(n) = Sum_{k >= 0} floor(n/2^k) = n + A011371(n). - Henry Bottomley, Jul 03 2001
G.f.: A(x) = Sum_{k>=0} x^(2^k)/((1-x)*(1-x^(2^k))). - Ralf Stephan, Dec 24 2002
a(n) = Sum_{k=1..n} A001511(k), sum of binary Hamming distances between consecutive integers up to n. - Gary W. Adamson, Jun 15 2003
Conjecture: a(n) = 2n + O(log(n)). - Benoit Cloitre, Oct 07 2003 [true as a(n) = 2*n - hamming_weight(2*n). Joerg Arndt, Jun 10 2019]
Sum_{n=2^k..2^(k+1)-1} a(n) = 3*4^k - (k+4)*2^(k-1) = A085354(k). - Philippe Deléham, Feb 19 2004
From Hieronymus Fischer, Aug 14 2007: (Start)
Recurrence: a(n) = n + a(floor(n/2)); a(2n) = 2n + a(n); a(n*2^m) = 2*n*(2^m-1) + a(n).
  a(2^m) = 2^(m+1) - 1, m >= 0.
Asymptotic behavior: a(n) = 2n + O(log(n)), a(n+1) - a(n) = O(log(n)); this follows from the inequalities below.
a(n) <= 2n-1; equality holds for powers of 2.
a(n) >= 2n-1-floor(log_2(n)); equality holds for n = 2^m-1, m > 0.
lim inf (2n - a(n)) = 1, for n-->oo.
lim sup (2n - log_2(n) - a(n)) = 0, for n-->oo.
lim sup (a(n+1) - a(n) - log_2(n)) = 1, for n-->oo. (End)
a(n) = 2n - A000120(n). - Paul Barry, Oct 26 2007
PURRS demo results: Bounds for a(n) = n + a(n/2) with initial conditions a(1) = 1: a(n) >= -2 + 2*n - log(n)*log(2)^(-1), a(n) <= 1 + 2*n for each n >= 1. - Alexander R. Povolotsky, Apr 06 2008
If n = 2^a_1 + 2^a_2 + ... + 2^a_k, then a(n) = n-k. This can be used to prove that 2^n maximally divides (2n!)/n!. - Jon Perry, Jul 16 2009
a(n) = Sum_{k>=0} A030308(n,k)*A000225(k+1). - Philippe Deléham, Oct 16 2011
a(n) = log_2(denominator(binomial(-1/2,n))). - Peter Luschny, Nov 25 2011
a(2n+1) = a(2n) + 1. - M. F. Hasler, Jan 24 2015
a(n) = A004134(n) - n. - Cyril Damamme, Aug 04 2015
G.f.: (1/(1 - x))*Sum_{k>=0} (2^(k+1) - 1)*x^(2^k)/(1 + x^(2^k)). - Ilya Gutkovskiy, Jul 23 2017

A123854 Denominators in an asymptotic expansion for the cubic recurrence sequence A123851.

Original entry on oeis.org

1, 4, 32, 128, 2048, 8192, 65536, 262144, 8388608, 33554432, 268435456, 1073741824, 17179869184, 68719476736, 549755813888, 2199023255552, 140737488355328, 562949953421312, 4503599627370496, 18014398509481984, 288230376151711744, 1152921504606846976

Offset: 0

Petros Hadjicostas and Jonathan Sondow, Oct 15 2006

A cubic analog of the asymptotic expansion A116603 of Somos's quadratic recurrence sequence A052129. Numerators are A123853.
Equals 2^A004134(n); also the denominators in expansion of (1-x)^(-1/4). - Alexander Adamchuk, Oct 27 2006
All terms are powers of 2 and log_2 a(n) = A004134(n) = 3*n - A000120(n). - Alexander Adamchuk, Oct 27 2006 [Edited by Petros Hadjicostas, May 14 2020]
Is this the same sequence as A088802? - N. J. A. Sloane, Mar 21 2007
Almost certainly this is the same as A088802. - Michael Somos, Aug 23 2007
Denominators of Gegenbauer_C(2n,1/4,2). The denominators of Gegenbauer_C(n,1/4,2) give the doubled sequence. - Paul Barry, Apr 21 2009
If the Greubel formula in A088802 and the Luschny formula here are correct (they are the same), the sequence is a duplicate of A088802. - R. J. Mathar, Aug 02 2023

			A123851(n) ~ c^(3^n)*n^(- 1/2)/(1 + 3/(4*n) - 15/(32*n^2) + 113/(128*n^3) - 5397/(2048*n^4) + ...) where c = 1.1563626843322... is the cubic recurrence constant A123852.

Steven R. Finch, Mathematical Constants, Cambridge University Press, Cambridge, 2003, p. 446.

G. C. Greubel, Table of n, a(n) for n = 0..1000
T. M. Apostol, On the Lerch zeta function, Pacific J. Math. 1 (1951), 161-167. [In Eq. (3.7), p. 166, the index in the summation for the Apostol-Bernoulli numbers should start at s = 0, not at s = 1. - _Petros Hadjicostas_, Aug 09 2019]
Jonathan Sondow and Petros Hadjicostas, The generalized-Euler-constant function gamma(z) and a generalization of Somos's quadratic recurrence constant, arXiv:math/0610499 [math.CA], 2006.
Jonathan Sondow and Petros Hadjicostas, The generalized-Euler-constant function gamma(z) and a generalization of Somos's quadratic recurrence constant, J. Math. Anal. Appl. 332 (2007), 292-314.
Eric Weisstein's World of Mathematics, Somos's Quadratic Recurrence Constant.
Aimin Xu, Asymptotic expansion related to the Generalized Somos Recurrence constant, International Journal of Number Theory 15(10) (2019), 2043-2055. [The author gives recurrences and other formulas for the coefficients of the asymptotic expansion using the Apostol-Bernoulli numbers (see the reference above) and the Bell polynomials. - _Petros Hadjicostas_, Aug 09 2019]

Cf. A052129, A112302, A116603, A123851, A123852, A123853 (numerators).

Cf. A004134, A004130, A000120.

f:=proc(t,x) exp(sum(ln(1+m*x)/t^m,m=1..infinity)); end; for j from 0 to 29 do denom(coeff(series(f(3,x),x=0,30),x,j)); od;
# Alternatively:
A123854 := n -> denom(binomial(1/4,n)):
seq(A123854(n), n=0..25); # Peter Luschny, Apr 07 2016

Denominator[CoefficientList[Series[ 1/Sqrt[Sqrt[1-x]], {x, 0, 25}], x]] (* Robert G. Wilson v, Mar 23 2014 *)

vector(25, n, n--; denominator(binomial(1/4,n)) ) \\ G. C. Greubel, Aug 08 2019

# uses[A000120]
def A123854(n): return 1 << (3*n-A000120(n))
[A123854(n) for n in (0..25)]  # Peter Luschny, Dec 02 2012

From Alexander Adamchuk, Oct 27 2006: (Start)
a(n) = 2^A004134(n).
a(n) = 2^(3n - A000120(n)). (End)
a(n) = denominator(binomial(1/4,n)). - Peter Luschny, Apr 07 2016

A220002 Numerators of the coefficients of an asymptotic expansion in even powers of the Catalan numbers.

Original entry on oeis.org

1, 5, 21, 715, -162877, 19840275, -7176079695, 1829885835675, -5009184735027165, 2216222559226679575, -2463196751104762933637, 1679951011110471133453965, -5519118103058048675551057049, 5373485053345792589762994345215, -12239617587594386225052760043303511

Offset: 0

Peter Luschny, Dec 27 2012

sign

Let N = 4*n+3 and A = sum_{k>=0} a(k)/(A123854(k)*N^(2*k)) then
C(n) ~ 8*4^n*A/(N*sqrt(N*Pi)), C(n) = (4^n/sqrt(Pi))*(Gamma(n+1/2)/ Gamma(n+2)) the Catalan numbers A000108.
The asymptotic expansion of the Catalan numbers considered here is based on the Taylor expansion of square root of the sine cardinal. This asymptotic series involves only even powers of N, making it more efficient than the asymptotic series based on Stirling's approximation to the central binomial which involves all powers (see for example: D. E. Knuth, 7.2.1.6 formula (16)). The series is discussed by Kessler and Schiff but is included as a special case in the asymptotic expansion given by J. L. Fields for quotients Gamma(x+a)/Gamma(x+b) and discussed by Y. L. Luke (p. 34-35), apparently overlooked by Kessler and Schiff.

			With N = 4*n+3 the first few terms of A are A = 1 + 5/(4*N^2) + 21/(32*N^4) + 715/(128*N^6) - 162877/(2048*N^8) + 19840275/(8192*N^10). With this A C(n) = round(8*4^n*A/(N*sqrt(N*Pi))) for n = 0..39 (if computed with sufficient numerical precision).

Donald E. Knuth, The Art of Computer Programming, Volume 4, Fascicle 4: Generating All Trees—History of Combinatorial Generation, 2006.
Y. L. Luke, The Special Functions and their Approximations, Vol. 1. Academic Press, 1969.

Peter Luschny, Table of n, a(n) for n = 0..99
J. L. Fields, A note on the asymptotic expansion of a ratio of gamma functions, Proc. Edinburgh Math. Soc. 15 (1) (1966), 43-45.
D. Kessler and J. Schiff, The asymptotics of factorials, binomial coefficients and Catalan numbers. April 2006.

The logarithmic version is A220422. Appears in A193365 and A220466.

Cf. A220412.

A220002 := proc(n) local s; s := n -> `if`(n > 0, s(iquo(n,2))+n, 0);
(-1)^n*mul(4*i+2, i = 1..2*n)*2^s(iquo(n,2))*coeff(taylor(sqrt(sin(x)/x), x,2*n+2), x, 2*n) end: seq(A220002(n), n = 0..14);
# Second program illustrating J. L. Fields expansion of gamma quotients.
A220002 := proc(n) local recF, binSum, swing;
binSum := n -> add(i,i=convert(n,base,2));
swing := n -> n!/iquo(n, 2)!^2;
recF := proc(n, x) option remember; `if`(n=0, 1, -2*x*add(binomial(n-1,2*k+1)*bernoulli(2*k+2)/(2*k+2)*recF(n-2*k-2,x),k=0..n/2-1)) end: recF(2*n,-1/4)*2^(3*n-binSum(n))*swing(4*n+1) end:

max = 14; CoefficientList[ Series[ Sqrt[ Sinc[x]], {x, 0, 2*max+1}], x^2][[1 ;; max+1]]*Table[ (-1)^n*Product[ (2*k+1), {k, 1, 2*n}], {n, 0, max}] // Numerator (* Jean-François Alcover, Jun 26 2013 *)

length = 15; T = taylor(sqrt(sin(x)/x),x,0,2*length+2)
def A005187(n): return A005187(n//2) + n if n > 0 else 0
def A220002(n):
    P = mul(4*i+2 for i in (1..2*n)) << A005187(n//2)
    return (-1)^n*P*T.coefficient(x, 2*n)
[A220002(n) for n in range(length)]

# Second program illustrating the connection with the Euler numbers.
def A220002_list(n):
    S = lambda n: sum((4-euler_number(2*k))/(4*k*x^(2*k)) for k in (1..n))
    T = taylor(exp(S(2*n+1)),x,infinity,2*n-1).coefficients()
    return [t[0].numerator() for t in T][::-1]
A220002_list(15)

Let [x^n]T(f(x)) denote the coefficient of x^n in the Taylor expansion of f(x) then r(n) = (-1)^n*prod_{i=1..2n}(2i+1)*[x^(2*n)]T(sqrt(sin(x)/x)) is the rational coefficient of the asymptotic expansion (in N=4*n+3) and a(n) = numerator(r(n)) = r(n)*2^(3*n-bs(n)), where bs(n) is the binary sum of n (A000120).
Also a(n) = numerator([x^(2*n)]T(exp(S))) where S = sum_{k>=1}((4-E(2*k))/ (4*k)*x^(2*k)) and E(n) the Euler numbers A122045.
Also a(n) = sf(4*n+1)*2^(3*n-bs(n))*F_{2*n}(-1/4) where sf(n) is the swinging factorial A056040, bs(n) the binary sum of n and F_{n}(x) J. L. Fields' generalized Bernoulli polynomials A220412.
In terms of sequences this means
r(n) = (-1)^n*A103639(n)*A008991(n)/A008992(n),
a(n) = (-1)^n*A220371(n)*A008991(n)/A008992(n).
Note that a(n) = r(n)*A123854(n) and A123854(n) = 2^A004134(n) = 8^n/2^A000120(n).
Formula from Johannes W. Meijer:
a(n) = d(n+1)*A098597(2*n+1)*(A008991(n)/A008992(n)) with d(1) = 1 and
d(n+1) = -4*(2*n+1)*A161151(n)*d(n),
d(n+1) = (-1)^n*2^(-1)*(2*(n+1))!*A060818(n)*A048896(n).

A004130 Numerators in expansion of (1-x)^{-1/4}.

Original entry on oeis.org

1, 1, 5, 15, 195, 663, 4641, 16575, 480675, 1762475, 13042315, 48612265, 729183975, 2748462675, 20809788825, 79077197535, 4823709049635, 18443593425075, 141400882925575, 543277076503525, 8366466978154285, 32270658344309385

Offset: 0

N. J. A. Sloane

Numerators in expansion of sqrt(1/sqrt(1-4x)). - Paul Barry, Jul 12 2005

Denominators are in A088802. - Michael Somos, Aug 23 2007

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Cf. A004134, A004981, A034255, A034385, A048882, A007696, A000265, A049606.

Table[Numerator[Binomial[-1/4, n] (-1)^n], {n, 0, 20}]

{a(n) = if( n<0, 0, numerator( polcoeff( (1 - x +x*O(x^n))^(-1/4), n ) ) ) } /* Michael Somos, Aug 23 2007 */

a(n) = prod(k=1, n, (4k-3)/k * 2^A007814(k)), proved by Mitch Harris, following a conjecture by Ralf Stephan.
a(n) = 2^(e_2((2n)!)-n)/n! Product[4k+1,{k,0,n-1}], where e_2((2n)!) is the highest power of 2 that divides (2n)! (sequence A005187). - Emanuele Munarini, Jan 25 2011
Numerators in (1-4t)^(-1/4) = 1 + t + (5/2)t^2 + (15/2)t^3 + (195/8)t^4 + (663/8)t^5 + (4641/16)t^6 + (16575/16)t^7 + ... = 1 + t + 5*t^2/2! + 45*t^3/3! + 585*t^4/4! + ... = e.g.f. for the quartic factorials A007696 (cf. A094638). - Tom Copeland, Dec 04 2013

A101189 G.f. A(x) is defined as the limit A(x) = lim_{n->oo} F(n)^(1/2^(n-1)) where F(n) is defined by F(n) = F(n-1)^2 + (2*x)^(2^n-1) for n >= 1 with F(0) = 1.

Original entry on oeis.org

1, 2, 0, 4, -8, 16, -40, 144, -512, 1696, -5696, 19840, -70048, 247744, -880128, 3152768, -11386624, 41389568, -151273728, 555794944, -2052141056, 7610274816, -28331018240, 105833345024, -396594444800, 1490425179136, -5615651143680, 21209004267520, -80276663808000

Offset: 0

Paul D. Hanna, Dec 03 2004

sign

Sequences A101190 and A101191 are related to doubly exponential numbers A003095 and to Catalan numbers (A000108).

			G.f.: A(x) = 1 + 2*x + 4*x^3 - 8*x^4 + 16*x^5 - 40*x^6 + 144*x^7 - 512*x^8 + 1696*x^9 - 5696*x^10 + 19840*x^11 - 70048*x^12 + ...
GENERATING METHOD.
We can illustrate the generating method for g.f. A(x) as follows.
Given F(n) = F(n-1)^2 + (2*x)^(2^n-1) for n >= 1 with F(0) = 1,
the first few polynomials generated by F(n) begin
F(0) = 1,
F(1) = F(0)^2 + (2*x)^(2^1-1) = 1 + 2*x,
F(2) = F(1)^2 + (2*x)^(2^2-1) = 1 + 4*x + 4*x^2 + 8*x^3,
F(3) = F(2)^2 + (2*x)^(2^3-1) = 1 + 8*x + 24*x^2 + 48*x^3 + 80*x^4 + 64*x^5 + 64*x^6 + 128*x^7,
F(4) = F(3)^2 + (2*x)^(2^4-1) = = 1 + 16*x + 112*x^2 + 480*x^3 + 1504*x^4 + 3712*x^5 + 7296*x^6 + 12032*x^7 + 17664*x^8 + 22528*x^9 + 26624*x^10 + 28672*x^11 + 20480*x^12 + 16384*x^13 + 16384*x^14 + 32768*x^15,
...
and the 2^(n-1)-th root of F(n) yields the series shown by
F(1)^(1/2^0) = 1 + 2*x,
F(2)^(1/2^1) = 1 + 2*x + 4*x^3 - 8*x^4 + 16*x^5 - 40*x^6 + 112*x^7 - 320*x^8 + 928*x^9 - 2752*x^10 + 8320*x^11 - 25504*x^12 + ...,
F(3)^(1/2^2) = 1 + 2*x + 4*x^3 - 8*x^4 + 16*x^5 - 40*x^6 + 144*x^7 - 512*x^8 + 1696*x^9 - 5696*x^10 + 19840*x^11 - 70048*x^12 + ...,
F(4)^(1/2^3) = 1 + 2*x + 4*x^3 - 8*x^4 + 16*x^5 - 40*x^6 + 144*x^7 - 512*x^8 + 1696*x^9 - 5696*x^10 + 19840*x^11 - 70048*x^12 + ...,
...
The limit of this process tends to the g.f. A(x).

Paul D. Hanna, Table of n, a(n) for n = 0..1025

Cf. A101190, A101191, A005187, A004134, A003095.

{a(n) = my(F=1,A,L); if(n==0,A=1, L = ceil(log(n+1)/log(2)); for(k=1,L, F = F^2 + (2*x)^(2^k-1)  +x*O(x^n)); A = polcoeff(F^(1/(2^(L-1))),n)); A}
for(n=0,32, print1(a(n),", "))

G.f. A(x) = ( Sum_{n>=0} A101190(n)/2^A005187(n) * (2*x)^n )^2.

G.f. A(x) = ( Sum_{n>=0} A101191(n)/2^A004134(n) * (2*x)^n )^4.

Entry revised by Paul D. Hanna, Mar 05 2024

A082691 a(1)=1, a(2)=2, then if the first 32^k-1 terms are a(1), a(2), ..., a(32^k - 1), the first 32^(k+1)-1 terms are a(1), a(2), ..., a(32^k - 1), a(1), a(2), ..., a(32^k - 1), a(32^k-1) + 1.

Original entry on oeis.org

1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 4, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 4, 5, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 4, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 4, 5, 6, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 4, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 4, 5, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 4, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 4, 5, 6, 7, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3

Offset: 1

Benoit Cloitre, Apr 12 2003

nonn

Consider the subsequence b(k) such that a(b(k))=1. Then 3k - b(k) = A063787(k+1) and b(k) = 1 + A004134(k-1).
From Sam Alexander, Nov 27 2010: (Start)
A naive way to try and guess whether a sequence is periodic, based on its first k terms (n1, ..., nk), is to look at all sequences which have period less than k, and guess "periodic" if any of them extend (n1, ..., nk), "nonperiodic" otherwise.
a(1)=1, a(2)=2. Suppose a(1), ..., a(n) have been defined, n > 1.
1. If the above guessing method guesses that (a(1), ..., a(n)) is an initial segment of a periodic sequence, then let a(n+1) be the least nonzero number not appearing in (a(1), ..., a(n)).
2. Otherwise, let (a(n+1), ..., a(2n)) be a copy of (a(1), ..., a(n)).
This sequence thwarts the guessing attempt, tricking the guesser into changing his mind infinitely many times as n->infinity. (End)
As n increases, the average value of the first n terms approaches 7/3 = 2.333... - Maxim Skorohodov, Dec 15 2022

			To construct the sequence: start with (1, 2); concatenating those 2 terms gives (1,2,1,2). Appending 3 gives the first 5 terms: (1,2,1,2,3). Concatenating those 5 terms gives (1,2,1,2,3,1,2,1,2,3). Appending 4 gives the first 11 terms: (1,2,1,2,3,1,2,1,2,3,4), etc.

Maxim Skorohodov, Table of n, a(n) for n = 1..10000
Samuel Alexander, On Guessing Whether A Sequence Has A Certain Property, arxiv:1011.6626 [math.LO], 2010-2012; J. Int. Seq. 14 (2011) # 11.4.4

Cf. A082692 (partial sums), A182659, A182660 (other sequences engineered to spite naive guessers).

A061161 Numerators in expansion of Euler transform of b(n) = 1/4.

Original entry on oeis.org

1, 1, 13, 55, 1235, 4615, 55801, 200343, 8977475, 36804235, 367235363, 1444888289, 32062742231, 120729974115, 1205864254225, 5201022002071, 395884671433315, 1603069490974835, 15989295873680415, 64312573140322525, 1250332447587844829, 5262481040435242585

Offset: 0

Vladeta Jovovic, Apr 17 2001

Denominators of c(n) are 2^A004134(n).

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Geoffrey B. Campbell and A. Zujev, Some almost partition theoretic identities, Preprint, 2016.
N. J. A. Sloane, Transforms

Cf. A000712, A061159, A061160.

b:= proc(n) option remember; `if`(n=0, 1, add(add(
      d/4, d=numtheory[divisors](j))*b(n-j), j=1..n)/n)
    end:
a:= n-> numer(b(n)):
seq(a(n), n=0..30);  # Alois P. Heinz, Jul 28 2017

c[n_] := c[n] = If[n == 0, 1,
     (1/(4n)) Sum[c[n-k] DivisorSigma[1, k], {k, 1, n}]];
a[n_] := Numerator[c[n]];
Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Apr 24 2022 *)

Numerators of c(n), where c(n) = (1/(4*n))*Sum_{k=1..n} c(n-k)*sigma(k), n>0, c(0)=1.

A101190 G.f.: A(x) = Sum_{n>=0} a(n)/2^A005187(n) * x^n = lim_{n->oo} F(n)^(1/2^n) where F(n) is defined by F(n) = F(n-1)^2 + x^(2^n-1) for n >= 1 with F(0) = 1.

Original entry on oeis.org

1, 1, -1, 5, -53, 127, -677, 2221, -61133, 205563, -1394207, 4852339, -68586849, 243751723, -1741612525, 6265913725, -363239625661, 1323861506899, -9699189175227, 35700526467479, -527987675255931, 1960112858076289, -14606721595781139, 54604708004873403

Offset: 0

Paul D. Hanna, Dec 03 2004

sign

			G.f.: A(x) = 1 + 1/2*x - 1/8*x^2 + 5/16*x^3 - 53/128*x^4 + 127/256*x^5 - 677/1024*x^6 + 2221/2048*x^7 + ... + a(n)/2^A005187(n)*x^n + ...
where 2^A005187(n) is also the denominator of [x^n] 1/sqrt(1-x).
GENERATING METHOD.
We can illustrate the generating method for g.f. A(x) as follows.
Given F(n) = F(n-1)^2 + (2*x)^(2^n-1) for n >= 1 with F(0) = 1,
the first few polynomials generated by F(n) begin
F(0) = 1,
F(1) = F(0)^2 + x^(2^1-1) = 1 + x,
F(2) = F(1)^2 + x^(2^2-1) = 1 + 2*x + x^2 + x^3,
F(3) = F(2)^2 + x^(2^3-1) = 1 + 4*x + 6*x^2 + 6*x^3 + 5*x^4 + 2*x^5 + x^6 + x^7.
...
The 2^n-th roots of F(n) tend to the limit of the g.f.:
F(1)^(1/2^1) = 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 - 429/32768*x^8 + ...
F(2)^(1/2^2) = 1 + 1/2*x - 1/8*x^2 + 5/16*x^3 - 53/128*x^4 + 127/256*x^5 - 677/1024*x^6 + 1965/2048*x^7 - 46797/32768*x^8 + ...
F(3)^(1/2^3) = 1 + 1/2*x - 1/8*x^2 + 5/16*x^3 - 53/128*x^4 + 127/256*x^5 - 677/1024*x^6 + 2221/2048*x^7 - 61133/32768*x^8 + ...
...
The limit of this process equals the g.f. A(x) of this sequence.
Note: the sum of the coefficients in F(n) equals A003095(n):
1, 2 = 1 + 1, 5 = 1 + 2 + 1 + 1, 26 = 1 + 4 + 6 + 6 + 5 + 2 + 1 + 1, ...
The last n coefficients in F(n) read backwards are Catalan numbers (A000108).
POWERS OF A(x).
The coefficients of x^k in the 2^n powers of the g.f. A(x) begin:
A^(2^0) = [1, 1/2, -1/8, 5/16, -53/128, 127/256, -677/1024, 2221/2048, ...],
A^(2^1) = [1, 1, 0, 1/2, -1/2, 1/2, -5/8, 9/8, -2, 53/16, -89/16, 155/16, ...],
A^(2^2) = [1, 2, 1, 1, 0, 0, 0, 1/2, -1, 3/2, -5/2, 9/2, -8, 14, -197/8, 44, ...],
A^(2^3) = [1, 4, 6, 6, 5, 2, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1/2, -2, 5, ...],
A^(2^4) = [1, 8, 28, 60, 94, 116, 114, 94, 69, 44, 26, 14, 5, 2, 1, 1, 0, 0, ...].

Cf. A101189, A101191, A005187, A003095, A000108.

{a(n) = my(F=1,A,L); if(n==0,A=1,L=ceil(log(n+1)/log(2)); for(k=1,L, F = F^2 + x^(2^k-1) +x*O(x^n)); A = polcoeff(F^(1/2^L),n)); numerator(A)}
for(n=0,32, print1(a(n),", "))

G.f. A(x) = ( Sum_{n>=0} A101191(n)/2^A004134(n) * x^n )^2.

G.f. A(x) satisfies A(2*x)^2 = Sum_{n>=0} A101189(n)*(2*x)^n.

Entry revised by Paul D. Hanna, Mar 05 2024

cp's OEIS Frontend

A101191 G.f.: A(x) = Sum_{n>=0}a(n)/2^A004134(n)*x^n = limit_{n->oo} F(n)^(1/2^(n+1)) where F(n) is the n-th iteration of: F(0) = 1, F(n) = F(n-1)^2 + x^(2^n-1) for n>=1.

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A005187 a(n) = a(floor(n/2)) + n; also denominators in expansion of 1/sqrt(1-x) are 2^a(n); also 2n - number of 1's in binary expansion of 2n.

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A123854 Denominators in an asymptotic expansion for the cubic recurrence sequence A123851.

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A220002 Numerators of the coefficients of an asymptotic expansion in even powers of the Catalan numbers.

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A004130 Numerators in expansion of (1-x)^{-1/4}.

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A101189 G.f. A(x) is defined as the limit A(x) = lim_{n->oo} F(n)^(1/2^(n-1)) where F(n) is defined by F(n) = F(n-1)^2 + (2*x)^(2^n-1) for n >= 1 with F(0) = 1.

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A082691 a(1)=1, a(2)=2, then if the first 32^k-1 terms are a(1), a(2), ..., a(32^k - 1), the first 32^(k+1)-1 terms are a(1), a(2), ..., a(32^k - 1), a(1), a(2), ..., a(32^k - 1), a(32^k-1) + 1.