

A054413


a(n) = 7*a(n1) + a(n2), with a(0)=1 and a(1)=7.


43



1, 7, 50, 357, 2549, 18200, 129949, 927843, 6624850, 47301793, 337737401, 2411463600, 17217982601, 122937341807, 877779375250, 6267392968557, 44749530155149, 319514104054600, 2281348258537349, 16288951913816043, 116304011655249650, 830417033500563593
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OFFSET

0,2


COMMENTS

In general, sequences with recurrence a(n) = k*a(n1) + a(n2) and a(0)=1 (and a(1)=0) have the generating function 1/(1k*xx^2). If k is odd (k>=3) they satisfy a(3n) = b(5n), a(3n+1) = b(5n+3), a(3n+2) = 2*b(5n+4) where b(n) is the sequence of denominators of continued fraction convergents to sqrt(k^2+4). [If k is even then a(n) is the sequence of denominators of continued fraction convergents to sqrt(k^2/4+1).]
a(p) == 53^((p1)/2)) (mod p), for odd primes p.  Gary W. Adamson, Feb 22 2009
For the sequence given above k=7 which implies that it is associated with A041091.
For a similar statement about sequences with recurrence a(n) = k*a(n1) + a(n2) but with a(0) = 2, and a(1) = 0, see A086902; a sequence that is associated with A041090.
(End)
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 7's along the main diagonal and 1's along the superdiagonal and the subdiagonal.  John M. Campbell, Jul 08 2011
a(n) equals the number of words of length n on alphabet {0,1,...,7} avoiding runs of zeros of odd lengths.  Milan Janjic, Jan 28 2015
Also called the 7metallonacci sequence; the g.f. 1/(1k*xx^2) gives the kmetallonacci sequence.
a(n) is the number of tilings of an nboard (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 7 kinds of squares available. (End)


LINKS



FORMULA

G.f.: 1/(1  7x  x^2).
a(n) = U(n, 7*i/2)*(i)^n with i^2=1 and Chebyshev's U(n, x/2) = S(n, x) polynomials. See A049310.
a(n) = F(n, 7), the nth Fibonacci polynomial evaluated at x=7.  T. D. Noe, Jan 19 2006
a(n) = (sigma^n  (sigma)^(n))/(sqrt(53)) with sigma = (7+sqrt(53))/2;
a(n) = Sum_{i=0..floor((n1)/2)} binomial(n1i,i)*7^(n12i). (End)
a(n) = (7/106)*sqrt(53)*(7/2  (1/2)*sqrt(53))^n + (1/2)*(7/2 + (1/2)*sqrt(53))^n + (1/2)*(7/2  (1/2)*sqrt(53))^n + (7/106)*(7/2 + (1/2)*sqrt(53))^n*sqrt(53), with n >= 0.  Paolo P. Lava, Jun 25 2008
a(n) = ((7 + sqrt(53))^n  (7  sqrt(53))^n)/(2^n*sqrt(53)). Offset 1. a(3)=50.  Al Hakanson (hawkuu(AT)gmail.com), Jan 17 2009
Lim_{k>infinity} a(n+k)/a(k) = (A086902(n) + A054413(n1)*sqrt(53))/2.
(End)
Sum_{m>=0} 1/(a(m)*a(m+2)) = 1/49.
Sum_{m>=0} 1/(a(2*m)*a(2*m+2)) = (sqrt(53)7)/14.
In general, for sequences with recurrence f(n)= k*f(n1)+f(n2) and f(0)=1,
Sum_{m>=0} 1/(f(m)*f(m+2)) = 1/(k^2).
Sum_{m>=0} 1/(f(2*m)*f(2*m+2)) = (sqrt(k^2+4)  k)/(2*k). (End)
E.g.f.: (1/53)*exp(7*x/2)*(53*cosh(sqrt(53)*x/2) + 7*sqrt(53)*sinh(sqrt(53)*x/2)).  Stefano Spezia, Feb 26 2020


MATHEMATICA



PROG

(Sage) [lucas_number1(n, 7, 1) for n in range(1, 19)] # Zerinvary Lajos, Apr 24 2009
(Magma) I:=[1, 7]; [n le 2 select I[n] else 7*Self(n1)+Self(n2): n in [1..25]]; // Vincenzo Librandi, Feb 23 2013


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



EXTENSIONS



STATUS

approved



