a(n+1) = (2*k+1)*a(n) - a(n-1) where k = floor(a(n-1)/a(n)). -
David S. Newman, Mar 04 2001
Let e(n) =
A007814(n) = exponent of highest power of 2 dividing n. Then a(n+1) = (2k+1)*a(n)-a(n-1), n > 0, where k = e(n). Moreover, floor(a(n-1)/a(n)) = e(n), in agreement with D. Newman's formula. - Dragutin Svrtan (dsvrtan(AT)math.hr) and Igor Urbiha (urbiha(AT)math.hr), Jan 10 2002
Calkin and Wilf showed 0.9588 <= limsup a(n)/n^(log(phi)/log(2)) <= 1.1709 where phi is the golden mean. Does this supremum limit = 1? -
Benoit Cloitre, Jan 18 2004. Coons and Tyler show the limit is
A246765 = 0.9588... -
Kevin Ryde, Jan 09 2021
a(n) = Sum_{k=0..floor((n-1)/2)} (binomial(n-k-1, k) mod 2). -
Paul Barry, Sep 13 2004
a(n) = Sum_{k=0..n-1} (binomial(k, n-k-1) mod 2). -
Paul Barry, Mar 26 2005
G.f. A(x) satisfies 0 = f(A(x), A(x^2), A(x^4)) where f(u, v, w) = v^3 + 2*u*v*w - u^2*w. -
Michael Somos, May 02 2005
G.f. A(x) satisfies 0 = f(A(x), A(x^2), A(x^3), A(x^6)) where f(u1, u2, u3, u6) = u1^3*u6 - 3*u1^2*u2*u6 + 3*u2^3*u6 - u2^3*u3. -
Michael Somos, May 02 2005
G.f.: x * Product_{k>=0} (1 + x^(2^k) + x^(2^(k+1))) [Carlitz].
a(n) = a(n-2) + a(n-1) - 2*(a(n-2) mod a(n-1)). -
Mike Stay, Nov 06 2006
a(n) = Sum_{k=1..n} K(k, n-k)*a(n - k), where K(n,k) = 1 if 0 <= k AND k <= n AND n-k <= 2 and K(n,k) = 0 else. (When using such a K-coefficient, several different arguments to K or several different definitions of K may lead to the same integer sequence. For example, if we drop the condition k <= n in the above definition, then we arrive at
A002083 = Narayana-Zidek-Capell numbers.) -
Thomas Wieder, Jan 13 2008
a(k+1)*a(2^n - k) - a(k)*a(2^n - (k+1)) = 1; a(2^n - k) + a(k) = a(2^(n+1) + k). Both formulas hold for 0 <= k <= 2^n - 1. G.f.: G(z) = a(1) + a(2)*z + a(3)*z^2 + ... + a(k+1)*z^k + ... Define f(z) = (1 + z + z^2), then G(z) = lim f(z)*f(z^2)*f(z^4)* ... *f(z^(2^n))*... = (1 + z + z^2)*G(z^2). - Arie Werksma (werksma(AT)tiscali.nl), Apr 11 2008
a(k+1)*a(2^n - k) - a(k)*a(2^n - (k+1)) = 1 (0 <= k <= 2^n - 1). - Arie Werksma (werksma(AT)tiscali.nl), Apr 18 2008
a(2^n + k) = a(2^n - k) + a(k) (0 <= k <= 2^n). - Arie Werksma (werksma(AT)tiscali.nl), Apr 18 2008
Let g(z) = a(1) + a(2)*z + a(3)*z^2 + ... + a(k+1)*z^k + ..., f(z) = 1 + z + z^2. Then g(z) = lim_{n->infinity} f(z)*f(z^2)*f(z^4)*...*f(z^(2^n)), g(z) = f(z)*g(z^2). - Arie Werksma (werksma(AT)tiscali.nl), Apr 18 2008
For 0 <= k <= 2^n - 1, write k = b(0) + 2*b(1) + 4*b(2) + ... + 2^(n-1)*b(n-1) where b(0), b(1), etc. are 0 or 1. Define a 2 X 2 matrix X(m) with entries x(1,1) = x(2,2) = 1, x(1,2) = 1 - b(m), x(2,1) = b(m). Let P(n)= X(0)*X(1)* ... *X(n-1). The entries of the matrix P are members of the sequence: p(1,1) = a(k+1), p(1,2) = a(2^n - (k+1)), p(2,1) = a(k), p(2,2) = a(2^n - k). - Arie Werksma (werksma(AT)tiscali.nl), Apr 20 2008
Let f(x) =
A030101(x); if 2^n + 1 <= x <= 2^(n + 1) and y = 2^(n + 1) - f(x - 1) then a(x) = a(y). - Arie Werksma (Werksma(AT)Tiscali.nl), Jul 11 2008
Equals infinite convolution product of [1,1,1,0,0,0,0,0,0] aerated
A000079 - 1 times, i.e., [1,1,1,0,0,0,0,0,0] * [1,0,1,0,1,0,0,0,0] * [1,0,0,0,1,0,0,0,1]. -
Mats Granvik and
Gary W. Adamson, Oct 02 2009; corrected by
Mats Granvik, Oct 10 2009
a(k+1)*a(2^m+k) - a(k)*a(2^m+(k+1)) = 1 for m >= 0, 0 <= k < 2^m. -
Yosu Yurramendi, Nov 07 2014
a(2^(m+1)+(k+1))*a(2^m+k) - a(2^(m+1)+k)*a(2^m+(k+1)) = 1 for m >= 0, 0 <= k < 2^m. -
Yosu Yurramendi, Nov 07 2014
a(5*2^k) = 3. a(7*2^k) = 3. a(9*2^k) = 4. a(11*2^k) = 5. a(13*2^k) = 5. a(15*2^k) = 4. In general: a((2j-1)*2^k) =
A007306(j), j > 0, k >= 0 (see Adamchuk's comment). -
Yosu Yurramendi, Mar 05 2016
a(2^m+2^m'+k') = a(2^m'+k')*(m-m'+1) - a(k'), m >= 0, m' <= m-1, 0 <= k' < 2^m'. -
Yosu Yurramendi, Jul 13 2016
Let n be a natural number and [b_m b_(m-1) ... b_1 b_0] its binary expansion with b_m=1.
Let L = Sum_{i=0..m} b_i be the number of binary digits equal to 1 (L >= 1).
Let {m_j: j=1..L} be the set of subindices such that b_m_j = 1, j=1..L, and 0 <= m_1 <= m_2 <= ... <= m_L = m.
If L = 1 then c_1 = 1, otherwise let {c_j: j=1..(L-1)} be the set of coefficients such that c_(j) = m_(j+1) - m_j + 1, 1 <= j <= L-1.
Let f be a function defined on {1..L+1} such that f(1) = 0, f(2) = 1, f(j) = c_(j-2)*f(j-1) - f(j-2), 3 <= j <= L+1.
Then a(n) = f(L+1) (see example). (End)
a(2^(m+2) + 2^(m+1) + k) - a(2^(m+1) + 2^m + k) = 2*a(k), m >= 0, 0 <= k < 2^m.
a(2^(m+2) + 2^(m+1) + k) - a(2^(m+1) + k) = a(2^m + k), m >= 0, 0 <= k < 2^m.
a(2^m + k) = a(k)*(m - floor(log_2(k)) - 1) + a(2^(floor(log_2(k))+1) + k), m >= 0, 0 < k < 2^m, a(2^m) = 1, a(0) = 0. (End)
a(2^m) = 1, m >= 0.
a(2^r*(2*k+1)) = a(2^r*(2*k)) + a(2^r*(2*k+2)), r < - m - floor(log_2(k)) - 1, m > 0, 1 <= k < 2^m. (End)
Trow(n) = [card({k XOR (j-k): k=0..j}) for j = 2^(n-1)-1..2^n-2] when regarded as an irregular table (n >= 1). -
Peter Luschny, Sep 29 2024
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