Displaying 1-10 of 11 results found.
Triangle T read by rows: diagonal differences of triangle A037027.
+10
31
1, 1, 0, 2, 1, 0, 3, 3, 1, 0, 5, 7, 4, 1, 0, 8, 15, 12, 5, 1, 0, 13, 30, 31, 18, 6, 1, 0, 21, 58, 73, 54, 25, 7, 1, 0, 34, 109, 162, 145, 85, 33, 8, 1, 0, 55, 201, 344, 361, 255, 125, 42, 9, 1, 0, 89, 365, 707, 850, 701, 413, 175, 52, 10, 1, 0, 144, 655, 1416, 1918, 1806, 1239, 630, 236, 63, 11, 1, 0
COMMENTS
Or, coefficients of a generalized Lucas-Pell polynomial read by rows. - Philippe Deléham, Nov 05 2006
FORMULA
G.f.: (1-y*z) / (1-y*(1+y+z)).
T(i, j) = R(i-j, j), where R(0, 0)=1, R(0, j)=0 for j >= 1, R(1, j)=1 for j >= 0, R(i, j) = Sum_{k=0..j} (R(i-2, k) + R(i-1, k)) for i >= 1, j >= 1.
Sum_{k=0..n} x^k*T(n,k) = A039834(n-2), A000012(n), A000045(n+1), A001333(n), A003688(n), A015448(n), A015449(n), A015451(n), A015453(n), A015454(n), A015455(n), A015456(n), A015457(n) for x= -2,-1,0,1,2,3,4,5,6,7,8,9,10. - Philippe Deléham, Oct 22 2006 Triangle T(n,k), 0 <= k <= n, given by [1, 1, -1, 0, 0, 0, 0, 0, ...] DELTA [0, 1, 0, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 05 2006 T(n,0) = Fibonacci(n+1) = A000045(n+1). Sum_{k=0..n} T(n,k) = A001333(n). T(n,k)=0 if k > n or if k < 0, T(0,0)=1, T(1,1)=0, T(n,k) = T(n-1,k-1) + T(n-1,k) + T(n-2,k). - Philippe Deléham, Nov 05 2006
EXAMPLE
Triangle begins:
1
1, 0
2, 1, 0
3, 3, 1, 0
5, 7, 4, 1, 0
8, 15, 12, 5, 1, 0
13, 30, 31, 18, 6, 1, 0
21, 58, 73, 54, 25, 7, 1, 0
34, 109, 162, 145, 85, 33, 8, 1, 0
55, 201, 344, 361, 255, 125, 42, 9, 1, 0
...
MAPLE
with(combinat);
T:= proc(n, k) option remember;
if k<0 or k>n then 0
elif k=0 then fibonacci(n+1)
elif n=1 and k=1 then 0
else T(n-1, k-1) + T(n-1, k) + T(n-2, k)
fi; end:
MATHEMATICA
T[n_, k_]:= T[n, k]= If[k<0 || k>n, 0, If[k==0, Fibonacci[n+1], If[n==1 && k==1, 0, T[n-1, k-1] + T[n-1, k] + T[n-2, k]]]]; Table[T[n, k], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Dec 19 2017 *)
PROG
(PARI) T(n, k) = if(k<0 || k>n, 0, if(k==0, fibonacci(n+1), if(n==1 && k==1, 0, T(n-1, k-1) + T(n-1, k) + T(n-2, k) )));
for(n=0, 12, for(k=0, n, print1(T(n, k), ", "))) \\ G. C. Greubel, Jan 21 2020 (Magma)
function T(n, k)
if k lt 0 or k gt n then return 0;
elif k eq 0 then return Fibonacci(n+1);
elif n eq 1 and k eq 1 then return 0;
else return T(n-1, k-1) + T(n-1, k) + T(n-2, k);
end if; return T; end function;
[T(n, k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 21 2020 (Sage)
@CachedFunction
def T(n, k):
if (k<0 or k>n): return 0
elif (k==0): return fibonacci(n+1)
elif (n==1 and k==1): return 0
else: return T(n-1, k-1) + T(n-1, k) + T(n-2, k)
[[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Jan 21 2020
CROSSREFS
Row sums: A001333 (numerators of continued fraction convergents to sqrt(2)).
Power ceiling-floor sequence of (golden ratio)^4.
+10
20
7, 47, 323, 2213, 15169, 103969, 712615, 4884335, 33477731, 229459781, 1572740737, 10779725377, 73885336903, 506417632943, 3471038093699, 23790849022949, 163064905066945, 1117663486445665, 7660579500052711
COMMENTS
Let f = floor and c = ceiling. For x > 1, define four sequences as functions of x, as follows:
p1(0) = f(x), p1(n) = f(x*p1(n-1));
p2(0) = f(x), p2(n) = c(x*p2(n-1)) if n is odd and p2(n) = f(x*p1(n-1)) if n is even;
p3(0) = c(x), p3(n) = f(x*p3(n-1)) if n is odd and p3(n) = c(x*p3(n-1)) if n is even;
p4(0) = c(x), p4(n) = c(x*p4(n-1)).
The present sequence is given by a(n) = p3(n).
Following the terminology at A214986, call the four sequences power floor, power floor-ceiling, power ceiling-floor, and power ceiling sequences. In the table below, a sequence is identified with an A-numbered sequence if they appear to agree except possibly for initial terms. Notation: S(t)=sqrt(t), r = (1+S(5))/2 = golden ratio, and Limit = limit of p3(n)/p2(n). x ......p1..... p2..... p3..... p4.......Limit
...
Properties of p1, p2, p3, p4:
(1) If x > 2, the terms of p2 and p3 interlace: p2(0) < p3(0) < p2(1) < p3(1) < p2(2) < p3(2)... Also, p1(n) <= p2(n) <= p3(n) <= p4(n) <= p1(n+1) for all x>0 and n>=0.
(2) If x > 2, the limits L(x) = limit(p/x^n) exist for the four functions p(x), and L1(x) <= L2(x) <= L3(x) <= L4 (x). See the Mathematica programs for plots of the four functions; one of them also occurs in the Odlyzko and Wilf article, along with a discussion of the special case x = 3/2.
(3) Suppose that x = u + sqrt(v) where v is a nonsquare positive integer. If u = f(x) or u = c(x), then p1, p2, p3, p4 are linear recurrence sequences. Is this true for sequences p1, p2, p3, p4 obtained from x = (u + sqrt(v))^q for every positive integer q?
(4) Suppose that x is a Pisot-Vijayaraghavan number. Must p1, p2, p3, p4 then be linearly recurrent? If x is also a quadratic irrational b + c*sqrt(d), must the four limits L(x) be in the field Q(sqrt(d))?
(5) The Odlyzko and Wilf article (page 239) raises three interesting questions about the power ceiling function; it appears that they remain open.
FORMULA
a(n) = floor(r*a(n-1)) if n is odd and a(n) = ceiling(r*a(n-1)) if n is even, where a(0) = ceiling(r), r = (golden ratio)^4 = (7 + sqrt(5))/2.
a(n) = 6*a(n-1) + 6*a(n-2) - a(n-3).
G.f.: (7 + 5*x - x^2)/((1 + x)*(1 - 7*x + x^2)).
a(n) = (10*(-2)^n+(10+3*sqrt(5))*(7-3*sqrt(5))^(n+2)+(10-3*sqrt(5))*(7+3*sqrt(5))^(n+2))/(90*2^n). - Bruno Berselli, Nov 14 2012 E.g.f.: exp(-x)*(5 + 2*exp(9*x/2)*(155*cosh(3*sqrt(5)*x/2) + 69*sqrt(5)*sinh(3*sqrt(5)*x/2)))/45. - Stefano Spezia, Oct 28 2024
EXAMPLE
a(0) = ceiling(r) = 7, where r = ((1+sqrt(5))/2)^4 = 6.8...; a(1) = floor(7*r) = 47; a(2) = ceiling(47) = 323.
MATHEMATICA
(* Program 1. A214992 and related sequences *) x = GoldenRatio^4; z = 30; (* z = # terms in sequences *)
z1 = 100; (* z1 = # digits in approximations *)
f[x_] := Floor[x]; c[x_] := Ceiling[x];
p1[0] = f[x]; p2[0] = f[x]; p3[0] = c[x]; p4[0] = c[x];
p1[n_] := f[x*p1[n - 1]]
p2[n_] := If[Mod[n, 2] == 1, c[x*p2[n - 1]], f[x*p2[n - 1]]]
p3[n_] := If[Mod[n, 2] == 1, f[x*p3[n - 1]], c[x*p3[n - 1]]]
p4[n_] := c[x*p4[n - 1]]
Table[p1[n], {n, 0, z}] (* A049685 *) Table[p2[n], {n, 0, z}] (* A157335 *) Table[p3[n], {n, 0, z}] (* A214992 *) Table[p4[n], {n, 0, z}] (* A004187 *) Table[p4[n] - p1[n], {n, 0, z}] (* A004187 *) Table[p3[n] - p2[n], {n, 0, z}] (* A098305 *) (* Program 2. Plot of power floor and power ceiling functions, p1(x) and p4(x) *)
f[x_] := f[x] = Floor[x]; c[x_] := c[x] = Ceiling[x];
p1[x_, 0] := f[x]; p1[x_, n_] := f[x*p1[x, n - 1]];
p4[x_, 0] := c[x]; p4[x_, n_] := c[x*p4[x, n - 1]];
Plot[Evaluate[{p1[x, 10]/x^10, p4[x, 10]/x^10}], {x, 2, 3}, PlotRange -> {0, 4}]
(* Program 3. Plot of power floor-ceiling and power ceiling-floor functions, p2(x) and p3(x) *)
f[x_] := f[x] = Floor[x]; c[x_] := c[x] = Ceiling[x];
p2[x_, 0] := f[x]; p3[x_, 0] := c[x];
p2[x_, n_] := If[Mod[n, 2] == 1, c[x*p2[x, n - 1]], f[x*p2[x, n - 1]]]
p3[x_, n_] := If[Mod[n, 2] == 1, f[x*p3[x, n - 1]], c[x*p3[x, n - 1]]]
Plot[Evaluate[{p2[x, 10]/x^10, p3[x, 10]/x^10}], {x, 2, 3}, PlotRange -> {0, 4}]
Square array read by antidiagonals: row m (m >= 1) satisfies b(0) = b(1) = 1; b(n) = m*b(n-1) + b(n-2):
+10
11
1, 1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 7, 5, 1, 1, 5, 13, 17, 8, 1, 1, 6, 21, 43, 41, 13, 1, 1, 7, 31, 89, 142, 99, 21, 1, 1, 8, 43, 161, 377, 469, 239, 34, 1, 1, 9, 57, 265, 836, 1597, 1549, 577, 55, 1, 1, 10, 73, 407, 1633, 4341, 6765, 5116, 1393, 89, 1, 1, 11, 91, 593, 2906
COMMENTS
For n > 1, the number of independent vertex sets in the graph K_m X P_{n-1}. For example, in K_3 X P_1 there are 4 independent vertex sets. - Andrew Howroyd, May 23 2017
FORMULA
O.g.f. row m: (mx-x-1)/(x^2+mx-1). - R. J. Mathar, Apr 21 2008
EXAMPLE
Array begins:
========================================================
m\n| 0 1 2 3 4 5 6 7 8 9
---|----------------------------------------------------
1 | 1 1 2 3 5 8 13 21 34 55 ...
2 | 1 1 3 7 17 41 99 239 577 1393 ...
3 | 1 1 4 13 43 142 469 1549 5116 16897 ...
4 | 1 1 5 21 89 377 1597 6765 28657 121393 ...
5 | 1 1 6 31 161 836 4341 22541 117046 607771 ...
6 | 1 1 7 43 265 1633 10063 62011 382129 2354785 ...
7 | 1 1 8 57 407 2906 20749 148149 1057792 7552693 ...
8 | 1 1 9 73 593 4817 39129 317849 2581921 20973217 ...
...
MAPLE
A135597 := proc(m, c) coeftayl( (m*x-x-1)/(x^2+m*x-1), x=0, c) ; end: for d from 1 to 15 do for c from 0 to d-1 do printf("%d, ", A135597(d-c, c)) ; od: od: # R. J. Mathar, Apr 21 2008
MATHEMATICA
a[_, 0] = a[_, 1] = 1; a[m_, n_] := m*a[m, n-1] + a[m, n-2]; Table[a[m-n+1, n], {m, 0, 11}, {n, 0, m}] // Flatten (* Jean-François Alcover, Jan 20 2014 *)
Eight white queens and one red queen on a 3 X 3 chessboard. G.f.: (1 + 3*x)/(1 - 6*x - 3*x^2).
+10
10
1, 9, 57, 369, 2385, 15417, 99657, 644193, 4164129, 26917353, 173996505, 1124731089, 7270376049, 46996449561, 303789825513, 1963728301761, 12693739287105, 82053620627913, 530402941628793, 3428578511656497
COMMENTS
The a(n) represent the number of n-move routes of a fairy chess piece starting in the center square (m = 5) on a 3 X 3 chessboard. This fairy chess piece behaves like a white queen on the eight side and corner squares but on the central square the queen explodes with fury and turns into a red queen.
On a 3 X 3 chessboard there are 2^9 = 512 ways to explode with fury on the center square (off the center square the piece behaves like a normal queen). The red queen is represented by the A[5] vector in the fifth row of the adjacency matrix A, see the Maple program and A180140. For the center square the 512 red queens lead to 17 red queen sequences, see the overview of red queen sequences and the crossreferences. The sequence above corresponds to just one red queen vector, i.e., A[5] = [111 111 111] vector. The other squares lead for this vector to A090018. Inverse binomial transform of A107903.
REFERENCES
Gary Chartrand, Introductory Graph Theory, pp. 217-221, 1984.
FORMULA
G.f.: (1+3*x)/(1 - 6*x - 3*x^2).
a(n) = 6*a(n-1) + 3*a(n-2) with a(0) = 1 and a(1) = 9.
a(n) = ((1-A)*A^(-n-1) + (1-B)*B^(-n-1))/4 with A=(-1+2*sqrt(3)/3) and B=(-1-2*sqrt(3)/3).
Lim_{k->infinity} a(n+k)/a(k) = (-1)^(n-1)* A108411(n+1)/( A041017(n-1)*sqrt(12) - A041016(n-1)) for n >= 1.
MAPLE
nmax:=19; m:=5; A[1]:=[0, 1, 1, 1, 1, 0, 1, 0, 1]: A[2]:=[1, 0, 1, 1, 1, 1, 0, 1, 0]: A[3]:=[1, 1, 0, 0, 1, 1, 1, 0, 1]: A[4]:=[1, 1, 0, 0, 1, 1, 1, 1, 0]: A[5]:=[1, 1, 1, 1, 1, 1, 1, 1, 1]: A[6]:=[0, 1, 1, 1, 1, 0, 0, 1, 1]: A[7]:=[1, 0, 1, 1, 1, 0, 0, 1, 1]: A[8]:=[0, 1, 0, 1, 1, 1, 1, 0, 1]: A[9]:=[1, 0, 1, 0, 1, 1, 1, 1, 0]: A:=Matrix([A[1], A[2], A[3], A[4], A[5], A[6], A[7], A[8], A[9]]): for n from 0 to nmax do B(n):=A^n: a(n):= add(B(n)[m, k], k=1..9): od: seq(a(n), n=0..nmax);
PROG
(Magma) I:=[1, 9]; [n le 2 select I[n] else 6*Self(n-1)+3*Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 15 2011
CROSSREFS
Cf. A180032 (Corner and side squares). Cf. Red queen sequences center square [decimal value A[5]]: A180028 [511], A180029 [255], A180031 [495], A015451 [127], A152240 [239], A000400 [63], A057088 [47], A001653 [31], A122690 [15], A180034 [23], A180036 [7], A084120 [19], A180038 [3], A154626 [17], A015449 [1], A000012 [16], A000007 [0].
1, 2, 8, 40, 208, 1088, 5696, 29824, 156160, 817664, 4281344, 22417408, 117379072, 614604800, 3218112512, 16850255872, 88229085184, 461973487616, 2418924584960, 12665653559296, 66318223015936, 347246723858432, 1818207451086848, 9520257811087360
COMMENTS
Hankel transform of 1,1,3,11,45,... (see A026375). Binomial transform of A015448. A production matrix for the sequence is M =
1, 1, 0, 0, 0, ...
1, 0, 5, 0, 0, ...
1, 0, 0, 5, 0, ...
1, 0, 0, 0, 5, ...
...
Take powers of M, extracting the upper left terms; getting
the sequence starting (1, 1, 2, 8, 40, 208, ...). (End)
The sequence is N=5 in an infinite set of INVERT transforms of powers of N prefaced with a "1". (1, 2, 8, 40, ...) is the INVERT transform of (1, 1, 5, 25, 125, ...). The first six of such sequences are shown in A006012 (N=3). - Gary W. Adamson, Jul 24 2016 The sequence is the first in an infinite set in which we perform the operation for matrix M (Cf. Jul 22 2016), but change the left border successively from (1, 1, 1, 1, ...) then to (1, 2, 2, 2, ...), then (1, 3, 3, 3, ...) ...; generally (1, N, N, N, ...). Extracting the upper left terms of each matrix operation, we obtain the infinite set beginning:
N=1 ( A154626): 1, 2, 8, 40, 208, 1088, ... N=2 ( A084120): 1, 3, 15, 81, 441, 1403, ... N=3 ( A180034): 1, 4, 22, 124, 700, 3952, ... N=4 ( A001653): 1, 5, 29, 169, 985, 5741, ... N=5 ( A000400): 1, 6, 36, 216, 1296, 7776, ... N=6 ( A015451): 1, 7, 43, 265, 1633, 10063, ... N=7 ( A180029): 1, 8, 50, 316, 1996, 12608, ... N=8 ( A180028): 1, 9, 57, 369, 1285, 15417, ... N=9 (.......): 1, 10, 64, 424, 2800, 18496, ...
N=10 ( A123361): 1, 11, 71, 481, 3241, 21851, ... N=11 (.......): 1, 12, 78, 540, 3708, 25488, ...
... Each of the sequences begins (1, (N+1), (7*N + 1),
(40*N + (N-1)^2), ... (End)
The set of infinite sequences shown (Cf. comment of Jul 27 2016), can be generated from the matrices P = [(1,N; 1,5]^n, (N=1,2,3,...) by extracting the upper left terms. Example: N=6 sequence ( A015451): (1, 7, 43, 265, ...) can be generated from the matrix P = [(1,6); (1,5)]^n. - Gary W. Adamson, Jul 28 2016
FORMULA
G.f.: (1 - 4*x) / (1 - 6*x + 4*x^2).
a(n) = (((3-sqrt(5))^n*(1+sqrt(5)) + (-1+sqrt(5))*(3+sqrt(5))^n)) / (2*sqrt(5)).
a(n) = 6*a(n-1) - 4*a(n-2) for n>1.
(End)
PROG
(Magma) [n le 2 select (n) else 6*Self(n-1)-4*Self(n-2): n in [1..25]]; // Vincenzo Librandi, May 15 2015 (PARI) Vec((1-4*x) / (1-6*x+4*x^2) + O(x^30)) \\ Colin Barker, Sep 22 2017
Least d for which the number with continued fraction [n,n,n,n...] is in Q(sqrt(d)).
+10
6
5, 2, 13, 5, 29, 10, 53, 17, 85, 26, 5, 37, 173, 2, 229, 65, 293, 82, 365, 101, 445, 122, 533, 145, 629, 170, 733, 197, 5, 226, 965, 257, 1093, 290, 1229, 13, 1373, 362, 61, 401, 1685, 442, 1853, 485, 2029, 530, 2213, 577, 2405, 626, 2605, 677, 2813, 730, 3029, 785, 3253
MATHEMATICA
z = 5000; u = Table[{p, e} = Transpose[FactorInteger[n]];
CROSSREFS
a(n) = 2 is equivalent to "n is in the sequence A077444", a(n) = 5 is equivalent to "n is in the sequence A002878".
a(0) = 1, a(1) = 2; a(n+1) = 6*a(n) + a(n-1) for n>1.
+10
4
1, 2, 13, 80, 493, 3038, 18721, 115364, 710905, 4380794, 26995669, 166354808, 1025124517, 6317101910, 38927735977, 239883517772, 1478228842609, 9109256573426, 56133768283165, 345911866272416, 2131604965917661, 13135541661778382, 80944854936587953
COMMENTS
a(n)/a(n-1) converges to 1/(sqrt(10) - 3) = 6.16227766017... = A176398.
FORMULA
Let M = the 2x2 matrix [2,3; 3,4]. a(n) = term (1,1) in M^n.
a(n) = ((3-sqrt(10))^n*(1+sqrt(10))+(-1+sqrt(10))*(3+sqrt(10))^n)/(2*sqrt(10)). - Colin Barker, Oct 13 2015
EXAMPLE
a(5) = 3038 = 6*a(5) + a(4) = 6*493 + 80.
a(5) = term (1,1) in M^5 where M^5 = [3038, 4215, 4215, 5848].
MATHEMATICA
CoefficientList[Series[(-1 + 4 x)/(-1 + 6 x + x^2), {x, 0, 33}], x] (* Vincenzo Librandi, Oct 13 2015 *)
PROG
(PARI) Vec((-1+4*x)/(-1+6*x+x^2) + O(x^40)) \\ Colin Barker, Oct 13 2015 (Magma) I:=[1, 2]; [n le 2 select I[n] else 6*Self(n-1)+Self(n-2): n in [1..40]]; // Vincenzo Librandi, Oct 13 2015
Fixed points of a sequence h(n) defined by the minimum number of 6's in the relation n*[n,6,6,...,6,n] = [x,...,x] between simple continued fractions.
+10
3
7, 11, 23, 47, 127, 139, 211, 223, 251, 331, 367, 379, 383, 463, 487, 499, 607, 619, 691, 727, 739, 743, 811, 823, 863, 887, 967, 971, 983, 1051, 1063, 1087, 1171, 1291, 1303, 1327, 1367, 1423, 1447, 1451, 1459
COMMENTS
In a variant of A213891, multiply n by a number with simple continued fraction [n,6,6,...,6,n] and increase the number of 6's until the continued fraction of the product has the same first and last entry (called x in the NAME). Examples are 2 * [2, 6, 2] = [4, 3, 4],
3 * [3, 6, 3] = [9, 2, 9],
4 * [4, 6, 6, 6, 4] = [16, 1, 1, 1, 5, 1, 1, 1, 16],
5 * [5, 6, 6, 6, 6, 5] = [25, 1, 4, 3, 3, 4, 1, 25],
6 * [6, 6, 6] = [36, 1, 36],
7 * [7, 6, 6, 6, 6, 6, 6, 6, 7] = [50, 7, 2, 1, 4, 4, 4, 1, 2, 7, 50].
The number of 6's needed defines the sequence h(n) = 1, 1, 3, 4, 1, 7, 7, 5, 9, ... (n>=2).
The current sequence contains the fixed points of h, i.e., those n where h(n)=n.
We conjecture that this sequence contains numbers is analogous to the sequence of prime numbers A000057, in the sense that, instead of referring to the Fibonacci sequences (sequences satisfying f(n) = f(n-1) + f(n-2) with arbitrary positive integer values for f(1) and f(2)) it refers to the generalized Fibonacci sequences satisfying f(n) = 6*f(n-1) + f(n-2), A005668, A015451, A179237, etc. This would mean that a prime is in the sequence if and only if it divides some term in each of the sequences satisfying f(n) = 6*f(n-1) + f(n-2).
MATHEMATICA
f[m_, n_] := Block[{c, k = 1}, c[x_, y_] := ContinuedFraction[x FromContinuedFraction[Join[{x}, Table[m, {y}], {x}]]]; While[First@ c[n, k] != Last@ c[n, k], k++]; k]; Select[Range[2, 1000], f[6, #] == # &] (* Michael De Vlieger, Sep 16 2015 *)
PROG
(PARI)
{a(n) = local(t, m=1); if( n<2, 0, while( 1,
t = contfracpnqn( concat([n, vector(m, i, 6), n]));
t = contfrac(n*t[1, 1]/t[2, 1]);
if(t[1]<n^2 || t[#t]<n^2, m++, break));
m)};
for(k=1, 1500, if(k==a(k), print1(a(k), ", ")));
Power floor sequence of 3+sqrt(10).
+10
3
6, 36, 221, 1361, 8386, 51676, 318441, 1962321, 12092366, 74516516, 459191461, 2829665281, 17437183146, 107452764156, 662153768081, 4080375372641, 25144406003926, 154946811396196, 954825274381101, 5883898457682801
COMMENTS
See A214992 for a discussion of power floor sequence and the power floor function, p1(x) = lim_{n->oo} a(n,x)/x^n. The present sequence is a(n,r), where r = 3+sqrt(10), and the limit p1(r) = 5.815421188487681054332319082... See A218992 for the power floor function, p4. For comparison with p1, we have lim_{r->oo} p4(r)/p1(r) = (3+sqrt(10))/5 = 1.23245553....
FORMULA
a(n) = floor(r*a(n-1)), where r=3+sqrt(10), a(0) = floor(r).
a(n) = 7*a(n-1) - 5*a(n-2) - a(n-3).
G.f.: (6 - 6*x - x^2)/(1 - 7*x + 5*x^2 + x^3).
a(n) = ((5+sqrt(10))*(3-sqrt(10))^(n+2) + (5-sqrt(10))*(3+sqrt(10))^(n+2)+2)/12. - Bruno Berselli, Nov 22 2012
EXAMPLE
a(0) = floor(r) = 6, where r = 3+sqrt(10);
a(1) = floor(6*r) = 36;
a(2) = floor(36*r) = 221.
MATHEMATICA
x = 3 + Sqrt[10]; z = 30; (* z = # terms in sequences *)
f[x_] := Floor[x]; c[x_] := Ceiling[x];
p1[0] = f[x]; p2[0] = f[x]; p3[0] = c[x]; p4[0] = c[x];
p1[n_] := f[x*p1[n - 1]]
p2[n_] := If[Mod[n, 2] == 1, c[x*p2[n - 1]], f[x*p2[n - 1]]]
p3[n_] := If[Mod[n, 2] == 1, f[x*p3[n - 1]], c[x*p3[n - 1]]]
p4[n_] := c[x*p4[n - 1]]
t1 = Table[p1[n], {n, 0, z}] (* A218991 *) t2 = Table[p2[n], {n, 0, z}] (* A005668 *) t3 = Table[p3[n], {n, 0, z}] (* A015451 *) t4 = Table[p4[n], {n, 0, z}] (* A218992 *)
PROG
(Magma) [IsZero(n) select Floor(r) else Floor(r*Self(n)) where r is 3+Sqrt(10): n in [0..20]]; // Bruno Berselli, Nov 22 2012
Power ceiling sequence of 3+sqrt(10).
+10
3
7, 44, 272, 1677, 10335, 63688, 392464, 2418473, 14903303, 91838292, 565933056, 3487436629, 21490552831, 132430753616, 816075074528, 5028881200785, 30989362279239, 190965054876220, 1176779691536560, 7251643204095581
COMMENTS
See A214992 for a discussion of power ceiling sequence and the power ceiling function, p4(x) = limit of a(n,x)/x^n. The present sequence is a(n,r), where r = 3+sqrt(10), and the limit p4(r) = 7.16724801485749657... See A218991 for the power floor function, p1(x); for comparison of p1 and p4, we have limit(p4(r)/p1(r) = (3+sqrt(10))/5 = 1.23245553...
FORMULA
a(n) = ceiling(r*a(n-1)), where r=3+sqrt(10), a(0) = ceiling(r).
a(n) = 7*a(n-1) - 5*a(n-2) - a(n-3).
G.f.: (7 - 5*x - x^2)/(1 - 7*x + 5*x^2 + x^3).
a(n) = ((5+sqrt(10))*(3-sqrt(10))^(n+3)+(5-sqrt(10))*(3+sqrt(10))^(n+3)-10)/60. [ Bruno Berselli, Nov 22 2012]
EXAMPLE
a(0) = ceiling(r) = 7, where r = 3+sqrt(10);
a(1) = ceiling(7*r) = 44;
a(2) = ceiling(44*r) = 272.
MATHEMATICA
LinearRecurrence[{7, -5, -1}, {7, 44, 272}, 20] (* Harvey P. Dale, Sep 22 2016 *)
PROG
(Magma) [IsZero(n) select Ceiling(r) else Ceiling(r*Self(n)) where r is 3+Sqrt(10): n in [0..20]]; // Bruno Berselli, Nov 22 2012
Search completed in 0.010 seconds
|