A006490 -id:A006490 - cp's OEIS frontend

A050946 "Stirling-Bernoulli transform" of Fibonacci numbers.

0, 1, 1, 7, 13, 151, 421, 6847, 25453, 532231, 2473141, 63206287, 352444093, 10645162711, 69251478661, 2413453999327, 17943523153933, 708721089607591, 5927841361456981, 261679010699505967, 2431910546406522973, 118654880542567722871, 1212989379862721528101

Offset: 0

N. J. A. Sloane, Jan 02 2000

nonn

From Paul Curtz, Oct 11 2013: (Start)
Differences table:
     0,    1,    1,    7,   13,  151,  421, 6847, ...
     1,    0,    6,    6,  138,  270, 6426, ...
    -1,    6,    0,  132,  132, 6156, ...
     7,   -6,  132,    0, 6024, ...
   -13,  138, -132, 6024, ...
   151, -270, 6156, ...
  -421, 6426, ...
  6847, ... .
a(n) is an autosequence of first kind: the inverse binomial transform is the sequence signed, the main diagonal is A000004=0's.
The "Stirling-Bernoulli transform" applied to an autosequence of first kind is an autosequence of first kind.
Now consider the Akiyama-Tanigawa transform or algorithm applied to A000045(n):
     0,   1,   1,   2,   3,   5, 8, ...
    -1,   0,  -3,  -4, -10, -18, ...    = -A006490
    -1,   6,   3,  24,  40, ...
    -7,   6, -63, -64, ...
   -13, 138,   3, ...
  -151, 270, ...
  -421, ... .
Hence -a(n). The Akiyama-Tanigawa algorithm applied to an autosequence of first kind is an autosequence of first kind.
a(n+5) - a(n+1) = 150, 420, 6840, ... is divisible by 30.
For an autosequence of the second kind, the inverse binomial transform is the sequence signed with the main diagonal double of the first upper diagonal.
The Akiyama-Tanigawa algorithm applied to an autosequence leads to an autosequence of the same kind. Example: the A-T algorithm applied to the autosequence of second kind 1/n leads to the autosequence of the second kind A164555(n)/A027642(n).
Note that a2(n) = 2*a1(n+1) - a1(n) applied to the autosequence of the first kind a1(n) is a corresponding autosequence of the second kind. (End)

Alois P. Heinz, Table of n, a(n) for n = 0..447
C. J. Pita Ruiz V., Some Number Arrays Related to Pascal and Lucas Triangles, J. Int. Seq. 16 (2013) #13.5.7

Cf. A000045, A051782, A105796, A163626.

with(combinat):
a:= n-> add((-1)^(k+1) *k! *stirling2(n+1, k+1)*fibonacci(k), k=0..n):
seq(a(n), n=0..30);  # Alois P. Heinz, May 17 2013

CoefficientList[Series[E^x*(1-E^x)/(1-3*E^x+E^(2*x)), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Aug 13 2013 *)
t[0, k_] := Fibonacci[k]; t[n_, k_] := t[n, k] = (k+1)*(t[n-1, k] - t[n-1, k+1]); a[n_] := t[n, 0] // Abs; Table[a[n], {n, 0, 22}] (* Jean-François Alcover, Oct 22 2013, after Paul Curtz *)

{a(n)=polcoeff(sum(m=0, n, fibonacci(m)*m!*x^m/prod(k=1, m, 1+k*x+x*O(x^n))), n)} /* Paul D. Hanna, Jul 20 2011 */

O.g.f.: Sum_{n>=1} Fibonacci(n) * n! * x^n / Product_{k=1..n} (1+k*x). - Paul D. Hanna, Jul 20 2011
A100872(n)=a(2*n) and A100868(n)=a(2*n-1).
From Paul Barry, Apr 20 2005: (Start)
E.g.f.: exp(x)*(1-exp(x))/(1-3*exp(x)+exp((2*x))).
a(n) = Sum_{k=0..n} (-1)^(n-k)*S2(n, k)*k!*Fibonacci(k). [corrected by Ilya Gutkovskiy, Apr 04 2019] (End)
a(n) ~ c * n! / (log((3+sqrt(5))/2))^(n+1), where c = 1/sqrt(5) if n is even and c = 1 if n is odd. - Vaclav Kotesovec, Aug 13 2013
a(n) = -1 * Sum_{k = 0..n} A163626(n,k)*A000045(k). - Philippe Deléham, May 29 2015

A119458 Triangle read by rows: T(n,k) is the number of (marked) circular binary words of length n having k occurrences of 00 (0 <= k <= n).

Original entry on oeis.org

1, 1, 1, 3, 0, 1, 4, 3, 0, 1, 7, 4, 4, 0, 1, 11, 10, 5, 5, 0, 1, 18, 18, 15, 6, 6, 0, 1, 29, 35, 28, 21, 7, 7, 0, 1, 47, 64, 60, 40, 28, 8, 8, 0, 1, 76, 117, 117, 93, 54, 36, 9, 9, 0, 1, 123, 210, 230, 190, 135, 70, 45, 10, 10, 0, 1, 199, 374, 440, 396, 286, 187, 88, 55, 11, 11, 0, 1

Offset: 0

Emeric Deutsch, May 20 2006

Sum of entries in row n is 2^n (A000079).
In Carlitz and Scoville (p. 252) the first term is 2.
From Petros Hadjicostas, Jan 05 2019: (Start)
Note that T(n,k) is the number of marked circular binary words of length n having k occurrences of 00 (0 <= k <= n). Let W(n,k) be the number of binary necklaces (= unmarked circular binary words) of length n with exactly k occurrences of the pattern 00 provided wrapping around the circle is allowed when n=1. More precisely, when n=1, we allow the string to wrap around itself on the circle to form a circular string of length 2. We have W(n, k) = A320341(n, k) for 0 <= k <= n.
Fortunately, since T(n=1, k=0) = 1 = T(n=1, k=1), the author of the sequence (indirectly) assumes that the string 0 has one occurrence of the pattern 00 (if allowed to wrap around itself on the circle only once), while the string 1 has zero occurrences of the pattern 00.
It makes sense to define T(n,k) = 0 when k > n. Note that G(z,t) = Sum_{n,k >=0} T(n,k)*z^n*t^k = 1 - z*(d(1-A(z,t))/dz)/(1-A(z,t)), where A(z,t) = z*(t+1) + z^2*(1-t) is the "Flajolet-Soria polynomial (g.f.)" in z of some (difficult to find) linear combinatorial structure that is defined very abstractly at the beginning of their paper and in the book by Flajolet and Sedgewick (2009). (Since the series here converge for z and t close to 0, we have that 1-t is positive for all t in a "small" interval around 0.)
Using the theory in Flajolet and Soria (1991) and the one in Hadjicostas and Zhang (2018), we can prove that the g.f. of the numbers W(n,k) is F(z,t) = Sum_{n >= 1, k >= 0} W(n,k)*z^n*t^k = -Sum_{d>=1} (phi(d)/d)*log(1-A(z^d,t^d)). We can also prove that W(n,k) = (1/n)*Sum_{d|gcd(n,k)} phi(d)*T(n/d, k/d) for n>=1 and 0 <= k <= n. (We omit the details.)
For  k=0, we get that T(n, k=0) = A000204(n) and W(n, k=0) = A000358(n) for n >= 1, and the Flajolet-Soria polynomial is A(z,t=0) = z + z^2 (which was discovered by Joerg Arndt).
To get univariate g.f.'s of the sequences (T(n,k): n >= 1) and (W(n, k): n >= 1) when k >= 1, we have to differentiate the previous two g.f.'s k times with respect to t, set t=0, and divide by k!. (Obviously, the log now disappears.)
For k=1, we get T(n, k=1) = A006490(n) and W(n, k=1) = A212804(n-1) = A006490(n)/n for n >= 1.
For k=2, we get T(n, k=2) = A006491(n-1) for n >= 1 (with A006491(0) := 0) and W(n, k=2) = (1/n)*(T(n,2) + I(2|n)*T(n/2,1)) for n >= 1, where I(2|n) = 1 if 2|n, and 0 otherwise.
For k=3, we get T(n, k=3) = A006492(n-2) for n >=1 (with A006492(m) = 0 for m = 1,2) and W(n, k=3) = (1/n)*(T(n,3) + 2*I(3|n)*T(n/3, 1)) for n >= 1.
This theory can be generalized for any pattern P of 0's and 1's provided one discovers the correct "Flajolet-Soria" polynomial A_P(z,t). In other words, if P is a finite pattern of zeros and ones of length L, and we let T_P(n,k) be the number of (marked) circular binary words of length n having k occurrences of P (0 <= k <= n) and we allow strings of length n, with 1 <= n < L, to wrap around themselves on the circle to form a string of length n*ceiling(L/n), then (most probably) we can express the g.f. of T_p(n,k), say G_p(z,t), in the form 1-z*(d(1-A_P(z,t))/dz)/(1-A_P(z,t)). In such a case, if W_P(n,k) is the number of binary necklaces (= unmarked circular binary words) of length n with exactly k occurrences of the pattern P, we have that its generating function is Sum_{n >= 1, k >= 0} W_P(n,k)*z^n*t^k = -Sum_{d>=1} (phi(d)/d)*log(1-A_P(z^d,t^d)). We can also prove that W_P(n,k) = (1/n)*Sum_{d|gcd(n,k)} phi(d)*T_P(n/d, k/d) for n>=1 and 0 <= k <= n.
One final note: it seems that the theory of Flajolet and Soria (1991) applies only to the case k=0 and to the case we consider all k simultaneously. (For fixed k >= 2, W_P(n,k) depends not only on T_P(n,k), but also on all T(n/d, k/d) where d ranges over the common divisors of n and k. Thus, for fixed k >= 2, it seems there is no linear combinatorial structure whose list of cycles of elements corresponds to the collection of necklaces we want.) See also Flajolet and Sedgewick (2009), pp. 84-85 and 729-730.
(End)

			T(5,3) = 5 because we have 10000, 01000, 00100, 00010 and 00001.
Triangle for T(n,k) begins:
n=0:    1;
n=1:    1,   1;
n=2:    3,   0,   1;
n=3:    4,   3,   0,   1;
n=4:    7,   4,   4,   0,   1;
n=5:   11,  10,   5,   5,   0,  1
n=6:   18,  18,  15,   6,   6,  0,  1;
n=7:   29,  35,  28,  21,   7,  7,  0,  1;
n=8:   47,  64,  60,  40,  28,  8,  8,  0,  1;
n=9:   76, 117, 117,  93,  54, 36,  9,  9,  0, 1;
n=10: 123, 210, 230, 190, 135, 70, 45, 10, 10, 0, 1;
  ...
From _Petros Hadjicostas_, Jan 06 2019: (Start)
If we take the Taylor expansion of g.f. G(z,t) of T(n,k) around z=0, we get G(z,t) = 1 + (1+t)*z + (3+0*t+t^2)*z^2 + (4+3*t+0*t^2+t^3)*z^3 + (7+4*t+4*t^2+0*t^3+t^4)*z^4 +(11+10*t+5*t^2+5*t^3+0*t^4+t^5)*z^5 + ...
One the other hand, if we take the Taylor expansion of the g.f. F(z,t) of W(n,k) = A320341(n, k) around z=0, we get F(z,t) = (1+t)*z + (2+0*t+t^2)*z^2 + (2+t+0*t^2+t^3)*z^3 + (3+t+t^2+0*t^3+t^4)*z^4 + (3+2*t+t^2+t^3+0*t^4+t^5)*z^5 + ...
Triangle for W(n,k) begins:
n=1:    1,  1;
n=2:    2,  0,  1;
n=3:    2,  1,  0,  1;
n=4:    3,  1,  1,  0,  1;
n=5:    3,  2,  1,  1,  0,  1;
n=6:    5,  3,  3,  1,  1,  0,  1;
n=7:    5,  5,  4,  3,  1,  1,  0,  1;
n=8:    8,  8,  8,  5,  4,  1,  1,  0,  1;
n=9:   10, 13, 13, 11,  6,  4,  1,  1,  0, 1;
n=10:  15, 21, 24, 19, 14,  7,  5,  1,  1, 0, 1;
...
For example, for n=4, we have the following marked and unmarked circular binary words (the square brackets denote equivalence classes):
k=0: [1111], [1110,1101,1011,0111], [1010,0101], T(4,0) = 7 and W(4,0) = 3;
k=1: [1100,1001,0011,0110], T(4,1) = 4 and W(4,1) = 1;
k=2: [0001,0010,0100,1000], T(4,2) = 4 and W(4,2) = 1;
k=3: none, T(4,3) = 0 = W(4,3);
k=4: [0000], T(4,4) = 1 = W(4,4).
(End)

L. Carlitz and R. Scoville, Zero-one sequences and Fibonacci numbers, Fibonacci Quarterly, 15 (1977), 246-254.
P. Flajolet and R. Sedgewick, Analytic Combinatorics, 2009.
P. Flajolet and M. Soria, The Cycle Construction, SIAM J. Discr. Math., vol. 4 (1), 1991, pp. 58-60.
P. Flajolet and M. Soria, The Cycle Construction, SIAM J. Discr. Math., vol. 4 (1), 1991, pp. 58-60.
P. Hadjicostas and L. Zhang, On cyclic strings avoiding a pattern, Discrete Mathematics, 341 (2018), 1662-1674.
L. Zhang and P. Hadjicostas, On sequences of independent Bernoulli trials avoiding the pattern '11..1', Math. Scientist, 40 (2015), 89-96.

Cf. A000079, A000204, A000358, A006490, A006491, A006492, A057711, A212804, A320341.

T:=proc(n,k): if k>n or k<0 then 0 elif n=0 and k=0 then 1 elif n=1 then 1 elif n=2 and k=0 then 3 elif n=2 and k=1 then 0 else T(n-1,k)+T(n-2,k)+T(n-1,k-1)-T(n-2,k-1) fi end: for n from 0 to 12 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form

T[n_, k_] := T[n, k] = Which[k > n || k < 0, 0, n == 0 && k == 0, 1, n == 1, 1, n == 2 && k == 0, 3, n == 2 && k == 1, 0, True, T[n-1, k] + T[n-2, k] + T[n-1, k-1] - T[n-2, k-1]];
Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 11 2019 *)

T(n,k) = T(n-1,k) + T(n-2,k) + T(n-1,k-1) - T(n-2,k-1) for n >= 3 and k >= 1.
G.f.: G(z,t) = Sum_{n,k >=0} T(n,k)*z^n*t^k = (1 + z^2 - t*z^2)/(1 - z - z^2 - t*z + t*z^2). [edited by Petros Hadjicostas, Jan 05 2019]
T(n,0) = A000204(n) for n >= 1 (Lucas numbers).
T(n,1) = A006490(n) for n >= 1.
T(n,2) = A006491(n-1) for n >= 1 (with A006491(0):=0).
Sum_{k=0..n} k*T(n,k) = A057711(n).

Name was edited by Petros Hadjicostas, Jan 06 2019

Values of T(9,5) and T(9,6) were corrected in the example by Petros Hadjicostas, Jan 06 2019

A320341 Triangle read by rows: T(n,k) is the number of unmarked circular binary words (necklaces) of length n having k occurrences of the pattern 00 (n >= 0 and 0 <= k <= n).

Original entry on oeis.org

1, 1, 1, 2, 0, 1, 2, 1, 0, 1, 3, 1, 1, 0, 1, 3, 2, 1, 1, 0, 1, 5, 3, 3, 1, 1, 0, 1, 5, 5, 4, 3, 1, 1, 0, 1, 8, 8, 8, 5, 4, 1, 1, 0, 1, 10, 13, 13, 11, 6, 4, 1, 1, 0, 1, 15, 21, 24, 19, 14, 7, 5, 1, 1, 0, 1, 19, 34, 40, 36, 26, 17, 8, 5, 1, 1, 0, 1, 31, 55, 71, 67, 54, 34, 22, 9, 6, 1, 1, 0, 1

Offset: 0

Petros Hadjicostas, Jan 07 2019

The array T(n,k) is the number of unmarked circular binary words (necklaces) of length n having exactly k occurrences of 00 (n >= 0 and 0 <= k <= n). Let V(n,k) be the number of marked circular binary words of length n with exactly k occurrences of the pattern 00 provided wrapping around the circle is allowed when n=1. More precisely, when n=1, we allow the string to wrap around itself on the circle to form a circular string of length 2. It turns out that V(n,k) = A119458(n,k). The properties of the array V(n,k) were studied by Carlitz and Scoville (1977).
Both for this array and for array V(n,k) = A119458(n,k) we have T(n=1, k=0) = T(n=1, k=1) = V(n=1, k=0) = V(n=1, k=1) = 1, which means in both arrays we (indirectly) assume that the string 0 has one occurrence of the pattern 00 (if allowed to wrap around itself on the circle only once), while the string 1 has zero occurrences of the pattern 00.
It makes sense to define T(n,k) = 0 = V(n,k) when k > n. Using the theory in Flajolet and Soria (1991), Flajolet and Sedgewick (2009), and Hadjicostas and Zhang (2018), we can prove that the g.f. of the numbers T(n,k) is F(z,t) = Sum_{n >= 0, k >= 0} T(n,k)*z^n*t^k = 1 - Sum_{d>=1} (phi(d)/d)*log(1-A(z^d,t^d)) while the g.f. of the numbers V(n,k) is K(z,t) = 1 - z*(d(1-A(z,t))/dz)/(1-A(z,t)), where A(z,t) = z*(t+1) + z^2*(1-t). (The latter g.f. was proved in Carlitz and Scoville (1977).)
We can also prove that T(n,k) = (1/n)*Sum_{d|gcd(n,k)} phi(d)*V(n/d, k/d) for n>=1 and 0 <= k <= n.
For k=0, we get that T(n, k=0) = A000358(n) and V(n, k=0) = A000204(n) and for n >= 1. To get univariate g.f.'s of the sequences (T(n,k): n >= 1) and (V(n, k): n >= 1) when k >= 1, we have to differentiate the previous two g.f.'s k times with respect to t, set t=0, and divide by k!. (Obviously, the log now disappears from the g.f. of T(n,k).)
For k=1, we get T(n, k=1) = A212804(n-1) = A006490(n)/n and V(n, k=1) = A006490(n) for n >= 1.
For k=2, we get T(n, k=2) = (1/n)*(V(n,2) + I(2|n)*V(n/2,1)) for n >= 1, where I(2|n) = 1 if 2|n, and 0 otherwise. Also, V(n, k=2) = A006491(n-1) for n >= 1 (with A006491(0) := 0).
For k=3, we get T(n, k=3) = (1/n)*(V(n,3) + 2*I(3|n)*V(n/3, 1)) for n >= 1. Also,
V(n, k=3) = A006492(n-2) for n >=1 (with A006492(m) = 0 for m = 1, 2). To get the g.f. for (T(n,k=3): n >= 0), we differentiate F(z,t) three times w.r.t. t, set t=0, and divide by 3! = 6. We get: 2*z^3*(1-z^3)/(3*(1-z^3-z^6)) + z^3*(1-z)^3/(3*(1-z-z^2)^3) = z^3 + 0*z^4 + z^5 + z^6 + 3*z^7 + 5*z^8 + 11*z^9 + 19*z^10 + 36*z^11 + 67*z^12 + 122*z^13 + 222*z^14 + ...
For 0 <= k <= n, T(n,k) - I(k=0) is the number of cyclic compositions of n with exactly k ones, where I(k=0) is 1 if k=0, and 0 otherwise. This can be proved using MacMahon's bijection between binary necklaces of length n and (unmarked) cyclic compositions of n. (We exclude the binary necklace consisting only of 1's, and that is why we need the term I(k=0).)
Given a binary necklace of 0's and 1's with length n (with at least one 0), starting with a 0 and going clockwise, let b_1 be the number of 1's until before the next zero plus one (for the initial 0); starting with the next zero, let b_2 be the number of 1's plus one (for the 2nd 0); continue this process until you reach the last 0 (say the m-th 0), and denote by b_m the number of 1's plus one before the first 0. Then b_1 + b_2 + ... + b_m is a cyclic composition of n. The process can be reversed (since b_1, b_2, ..., b_m >= 1). The only necklace that cannot be obtained from a cyclic composition is the one having all 1's, which we exclude. In the above process, b_i = 1 if and only if the i-th 0 in the above process is followed by 0 (to the right of it on the circle). Hence, for 0 <= k <= n, T(n,k) - I(k=0) is the number of cyclic compositions of n with exactly k ones.
Note that array A105422(n,k) is the linear version of this array: A105422(n,k) is the number of (linear) compositions of n having exactly k parts equal to 1. The denominator of the bivariate g.f. of array A105422(n,k) is indeed 1 - A(z,t), where A(z,t) = z*(t+1) + z^2*(1-t) (see above), and this is no coincidence.

			Triangle for T(n,k) begins:
n=0:    1;
n=1:    1,  1;
n=2:    2,  0,  1;
n=3:    2,  1,  0,  1;
n=4:    3,  1,  1,  0,  1;
n=5:    3,  2,  1,  1,  0,  1;
n=6:    5,  3,  3,  1,  1,  0,  1;
n=7:    5,  5,  4,  3,  1,  1,  0,  1;
n=8:    8,  8,  8,  5,  4,  1,  1,  0,  1;
n=9:   10, 13, 13, 11,  6,  4,  1,  1,  0, 1;
n=10:  15, 21, 24, 19, 14,  7,  5,  1,  1, 0, 1;
...
If we take the Taylor expansion of g.f. F(z,t) of T(n,k) around z=0, we get F(z,t) = 1 + (1+t)*z + (2+0*t+t^2)*z^2 + (2+t+0*t^2+t^3)*z^3 + (3+t+t^2+0*t^3+t^4)*z^4 + (3+2*t+t^2+t^3+0*t^4+t^5)*z^5 + ...
For example, for n=4, we have the following marked and unmarked circular binary words (the square brackets denote equivalence classes):
k=0: [1111], [1110,1101,1011,0111], [1010,0101], V(4,0) = 7 and T(4,0) = 3;
k=1: [1100,1001,0011,0110], V(4,1) = 4 and T(4,1) = 1;
k=2: [0001,0010,0100,1000], V(4,2) = 4 and T(4,2) = 1;
k=3: none, V(4,3) = 0 = T(4,3);
k=4: [0000], V(4,4) = 1 = T(4,4).
The corresponding cyclic compositions of n=4 under MacMahon's bijection are the following:
k=0 (no 1's): [none], [4], [2+2], T(4,1) - 1 = 3 - 1 = 2;
k=1 (one 1): [1+3], T(4,1) = 1;
k=2 (two 1's): [1+1+2], T(4,2) = 1;
k=3 (three 1's): none, T(4,3) = 0;
k=4 (four 1's): [1+1+1+1], T(4,4) = 1.

L. Carlitz and R. Scoville, Zero-one sequences and Fibonacci numbers, Fibonacci Quarterly, 15 (1977), 246-254.
P. Flajolet and R. Sedgewick, Analytic Combinatorics, 2009.
P. Flajolet and M. Soria, The Cycle Construction, SIAM J. Discr. Math., vol. 4 (1), 1991, pp. 58-60.
P. Flajolet and M. Soria, The Cycle Construction, SIAM J. Discr. Math., vol. 4 (1), 1991, pp. 58-60.
P. Hadjicostas and L. Zhang, On cyclic strings avoiding a pattern, Discrete Mathematics, 341 (2018), 1662-1674.
L. Zhang and P. Hadjicostas, On sequences of independent Bernoulli trials avoiding the pattern '11..1', Math. Scientist, 40 (2015), 89-96.

Cf. A000079, A000204, A000358, A006490, A006491, A006492, A057711, A105422, A119458, A212804.

T(n,k) = (1/n)*Sum_{d|gcd(n,k)} phi(d)*A119458(n/d, k/d) for n>=1 and 0 <= k <= n.

G.f.: F(z,t) = Sum_{n >= 0, k >= 0} T(n,k)*z^n*t^k = 1 - Sum_{d>=1} (phi(d)/d)*log(1-A(z^d,t^d)), where A(z,t) = z*(t+1) + z^2*(1-t).

A006491 Generalized Lucas numbers.

Original entry on oeis.org

1, 0, 4, 5, 15, 28, 60, 117, 230, 440, 834, 1560, 2891, 5310, 9680, 17527, 31545, 56468, 100590, 178395, 315106, 554530, 972564, 1700400, 2964325, 5153868, 8938300, 15465497, 26700915, 46004620, 79112304, 135801105, 232715006, 398151740

Offset: 1

N. J. A. Sloane

For n>2 note that (n+1)|a(n) unless n is prime, in which case (n+1)|2*a(n). This sequence is not the better-known generalized Lucas numbers V(n,a,b) defined for fixed integers a and b such that D = a^2 + 4*b is nonnegative, V(0) = 2, V(1) = a and for n>1 the recurrence V(n) = V(n-1) + V(n-2). The a = b = 1 case gives the Lucas Numbers. - Jonathan Vos Post, Mar 16 2005
Number of circular binary words of length n+1 having exactly two occurrences of 00. Example: a(4) = 5 because we have 00011, 10001, 11000, 00110 and 01100. Column 2 of A119458. - Emeric Deutsch, May 20 2006
From Petros Hadjicostas, Jan 10 2019: (Start)
In view of the comment by Emeric Deutsch above, we clarify the previous comment by Jonathan Vos Post. We have that 25 + 1 = 26 does not divide a(25) = 2964325 even though n = 25 is not a prime. What he probably meant is that, for n >= 1, we have (n+1) | a(n) unless n is odd, in which case (n+1)|2*a(n). (Of course, for some odd numbers n, we do have (n+1)|a(n), but not for all of them.)
From Emeric Deutsch's comment above, we have a(n) = A119458(n+1, k=2), while from the theory of marked and unmarked circular words, we have A320341(n+1, k=2) = (1/(n+1))*Sum_{d|gcd(n+1, 2)} phi(d)*A119458((n+1)/d, 2/d).
If n is even, then n+1 is odd, and hence  A320341(n+1, k=2) = (1/(n+1)) * A119458(n+1, 2) = a(n)/(n+1), i.e., (n+1)|a(n).
If n=1, then (1+1)|2*a(1). Let n be odd >= 3, in which case n+1 is even and 2*A320341(n+1, k=2) = (1/(n+1))*(2*A119458(n+1, k=2) + 2*A119458((n+1)/2, k=1)). Thus,  2*A320341(n+1, k=2) = (1/(n+1))*(2*a(n) + 2*A006490((n+1)/2)) = (1/(n+1))*(2*a(n) + 2*((n+1)/2)*Fibonacci((n-3)/2)). It follows that  2*A320341(n+1, k=2) = 2*a(n)/(n+1) + Fibonacci((n-3)/2). Thus, 2*a(n)/(n+1) = 2*A320341(n+1, k=2) - Fibonacci((n-3)/2), and thus, (n+1)|2*a(n). (End)

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

G. C. Greubel, Table of n, a(n) for n = 1..1000
L. Carlitz and R. Scoville, Zero-one sequences and Fibonacci numbers, Fibonacci Quarterly, 15 (1977), 246-254.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Index entries for linear recurrences with constant coefficients, signature (3,0,-5,0,3,1).

Cf. A006490, A119458, A320341.

I:=[1,0,4,5,15,28]; [n le 6 select I[n] else 3*Self(n-1) -5*Self(n-3) +3*Self(n-5)+Self(n-6): n in [1..30]]; // G. C. Greubel, Jan 01 2018

G:=x*(1-x)*(1-2*x+2*x^2)/(1-x-x^2)^3: Gser:=series(G,x=0,45): seq(coeff(Gser,x^n),n=1..40); # Emeric Deutsch, Feb 07 2006
with(combinat): a[1]:=1: a[2]:=0: for n from 3 to 40 do a[n]:=a[n-1]+a[n-2]+n*fibonacci(n-2)-(n-1)*fibonacci(n-3) od: seq(a[n],n=1..40); # Emeric Deutsch, May 20 2006
A006491:=(z-1)*(1-2*z+2*z**2)/(z**2+z-1)**3; # conjectured by Simon Plouffe in his 1992 dissertation

LinearRecurrence[{3, 0, -5, 0, 3, 1}, {1, 0, 4, 5, 15, 28}, 50] (* G. C. Greubel, Jan 01 2018 *)

x='x+O('x^30); Vec(x*(1-x)*(1-2*x+2*x^2)/(1-x-x^2)^3) \\ G. C. Greubel, Jan 01 2018

G.f.: x*(1-x)*(1-2*x+2*x^2)/(1-x-x^2)^3. - Ralf Stephan, Apr 23 2004, corrected Feb 08 2006
a(n) = a(n-1) + a(n-2) + n*Fibonacci(n-2) - (n-1)*Fibonacci(n-3) for n >= 3; a(1)=1, a(2)=0. - Emeric Deutsch, May 20 2006
a(n) = 3*a(n-1) - 5*a(n-3) + 3*a(n-5) + a(n-6). - G. C. Greubel, Jan 01 2018

More terms from Emeric Deutsch, Feb 07 2006

A100224 Triangle, read by rows, of the coefficients of [x^k] in G100224(x)^n such that the row sums are 2^n-1 for n>0, where G100224(x) is the g.f. of A100224.

Original entry on oeis.org

1, 1, 0, 1, 0, 2, 1, 0, 3, 3, 1, 0, 4, 4, 6, 1, 0, 5, 5, 10, 10, 1, 0, 6, 6, 15, 18, 17, 1, 0, 7, 7, 21, 28, 35, 28, 1, 0, 8, 8, 28, 40, 60, 64, 46, 1, 0, 9, 9, 36, 54, 93, 117, 117, 75, 1, 0, 10, 10, 45, 70, 135, 190, 230, 210, 122

Offset: 0

Paul D. Hanna, Nov 28 2004

Diagonals are: T(n,n)=A001610(n-1) for n>0, with T(0,0)=1, T(n+1,n)=A006490(n), T(n+2,n)=A006491(n), T(n+3,n)=A006492(n), T(n+4,n)=A006493(n). The ratio of the generating functions of any two adjacent diagonals gives: (1-x)/(1-x-x^2) = 1+ x^2+ x^3+ 2*x^4+ 3*x^5+ 5*x^6+ 8*x^7+ 13*x^8+...

			Rows begin:
  [1],
  [1,0],
  [1,0,2],
  [1,0,3,3],
  [1,0,4,4,6],
  [1,0,5,5,10,10],
  [1,0,6,6,15,18,17],
  [1,0,7,7,21,28,35,28],
  [1,0,8,8,28,40,60,64,46],...
where row sums form 2^n-1 for n>0:
2^1-1 = 1+0 = 1
2^2-1 = 1+0+2 = 3
2^3-1 = 1+0+3+3 = 7
2^4-1 = 1+0+4+4+6 = 15
2^5-1 = 1+0+5+5+10+10 = 31.
The main diagonal forms A001610 = [0,2,3,6,10,17,...], where Sum_{n>=1} (A001610(n-1)/n)*x^n = log((1-x)/(1-x-x^2)).

Cf. A100223, A001610, A006490, A006491, A006492, A006493.

PARI
```
T(n,k)=if(n
				
```

G.f.: A(x, y)=(1-2*x*y+2*x^2*y^2)/((1-x*y)*(1-x*y-x^2*y^2-x*(1-x*y))).

A345378 Number of terms m <= n, where m is a term in A006497.

Original entry on oeis.org

0, 0, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4

Offset: 0

Ovidiu Bagdasar, Jun 16 2021

Table 1 of Andrica 2021 paper (p. 24), refers to A006497 as "bronze Lucas" numbers.

			a(0)=a(1)=0, since the least term in A006497 is 2.
a(2)=1 since A006497(0) = 2 is followed in that sequence by 3.
a(k)=3 for 3 <= k <= 11 since the first terms of A006490 are {0, 2, 3, 11}.

Michael De Vlieger, Table of n, a(n) for n = 0..10000
Dorin Andrica, Ovidiu Bagdasar, and George Cătălin Tųrcąs, On some new results for the generalised Lucas sequences, An. Şt. Univ. Ovidius Constanţa (Romania, 2021) Vol. 29, No. 1, 17-36. See Section 5.3, pp. 33, Table 4.

Cf. A006497, A108852 (Fibonacci), A130245 (Lucas), A345377.

Block[{a = 3, b = -1, nn = 105, u, v = {}}, u = {0, 1}; Do[AppendTo[u, Total[{-b, a} u[[-2 ;; -1]]]]; AppendTo[v, Count[u, _?(# <= i &)]], {i, nn}]; {Boole[First[u] <= 0]}~Join~v] (* or *)
{0}~Join~Accumulate@ ReplacePart[ConstantArray[0, Last[#]], Map[# -> 1 &, #]] &@ LucasL[Range[0, 4], 3] (* Michael De Vlieger, Jun 16 2021 *)

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A050946 "Stirling-Bernoulli transform" of Fibonacci numbers.

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A119458 Triangle read by rows: T(n,k) is the number of (marked) circular binary words of length n having k occurrences of 00 (0 <= k <= n).

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A320341 Triangle read by rows: T(n,k) is the number of unmarked circular binary words (necklaces) of length n having k occurrences of the pattern 00 (n >= 0 and 0 <= k <= n).

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A006491 Generalized Lucas numbers.

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A100224 Triangle, read by rows, of the coefficients of [x^k] in G100224(x)^n such that the row sums are 2^n-1 for n>0, where G100224(x) is the g.f. of A100224.

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A345378 Number of terms m <= n, where m is a term in A006497.

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