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Expansion of (1+x)/(1-4*x).
+10
98
1, 5, 20, 80, 320, 1280, 5120, 20480, 81920, 327680, 1310720, 5242880, 20971520, 83886080, 335544320, 1342177280, 5368709120, 21474836480, 85899345920, 343597383680, 1374389534720, 5497558138880, 21990232555520, 87960930222080, 351843720888320
COMMENTS
Coordination sequence for infinite tree with valency 5.
For n>=1, a(n+1) is equal to the number of functions f:{1,2,...,n+1}->{1,2,3,4,5} such that for fixed, different x_1, x_2,...,x_n in {1,2,...,n+1} and fixed y_1, y_2,...,y_n in {1,2,3,4,5} we have f(x_i)<>y_i, (i=1,2,...,n). - Milan Janjic, May 10 2007 Number of length-n strings of 5 letters with no two adjacent letters identical. The general case (strings of r letters) is the sequence with g.f. (1+x)/(1-(r-1)*x). - Joerg Arndt, Oct 11 2012 Create a rectangular prism with edges of lengths 2^(n-2), 2^(n-1), and 2^(n) starting at n=2; then the surface area = a(n). - J. M. Bergot, Aug 08 2013
FORMULA
a(n) = (5*4^n - 0^n)/4.
G.f.: (1+x)/(1-4*x).
E.g.f.: (5*exp(4*x) - exp(0))/4. (End)
G.f.: 2+x- 2/G(0), where G(k)= 1 + 1/(1 - x*(5*k-4)/(x*(5*k+1) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 04 2013
MAPLE
k := 5; if n = 0 then 1 else k*(k-1)^(n-1); fi;
MATHEMATICA
q = 5; Join[{a = 1}, Table[If[n != 0, a = q*a - a, a = q*a], {n, 0, 25}]] (* and *) Join[{1}, 5*4^Range[0, 25]] (* Vladimir Joseph Stephan Orlovsky, Jul 11 2011 *)
PROG
(Magma) [1] cat [5*4^(n-1): n in [1..30]]; // G. C. Greubel, Aug 10 2019 (Sage) [1]+[5*4^(n-1) for n in (1..30)] # G. C. Greubel, Aug 10 2019 (GAP) Concatenation([1], List([1..30], n-> 5*4^(n-1) )); # G. C. Greubel, Aug 10 2019
Congruent to 0 or 1 mod 5.
+10
38
0, 1, 5, 6, 10, 11, 15, 16, 20, 21, 25, 26, 30, 31, 35, 36, 40, 41, 45, 46, 50, 51, 55, 56, 60, 61, 65, 66, 70, 71, 75, 76, 80, 81, 85, 86, 90, 91, 95, 96, 100, 101, 105, 106, 110, 111, 115, 116, 120, 121, 125, 126, 130, 131, 135, 136, 140, 141, 145, 146, 150, 151
COMMENTS
Numbers k that have the same last digit as k^2.
REFERENCES
L. E. Dickson, History of the Theory of Numbers, I, p. 459.
FORMULA
G.f.: x^2*(1+4*x) / ( (1+x)*(x-1)^2 ). - R. J. Mathar, Oct 07 2011 E.g.f.: 4 + ((10*x - 13)*exp(x) - 3*exp(-x))/4. - David Lovler, Sep 11 2022 Sum_{n>=2} (-1)^n/a(n) = sqrt(1+2/sqrt(5))*Pi/10 + log(phi)/(2*sqrt(5)) + log(5)/4, where phi is the golden ratio ( A001622). - Amiram Eldar, Oct 12 2022
MAPLE
a[0]:=0:a[1]:=1:for n from 2 to 100 do a[n]:=a[n-2]+5 od: seq(a[n], n=0..61); # Zerinvary Lajos, Mar 16 2008
MATHEMATICA
Select[Range[0, 151], MemberQ[{0, 1}, Mod[#, 5]] &] (* T. D. Noe, Mar 31 2013 *)
PROG
(Haskell)
a008851 n = a008851_list !! (n-1)
a008851_list = [10*n + m | n <- [0..], m <- [0, 1, 5, 6]]
(PARI) a(n) = 5*(n\2)+bitand(n, 1); /* Joerg Arndt, Mar 31 2013 */ (PARI) a(n) = floor((5/3)*floor(3*(n-1)/2)); /* Joerg Arndt, Mar 31 2013 */
1, 6, 21, 66, 201, 606, 1821, 5466, 16401, 49206, 147621, 442866, 1328601, 3985806, 11957421, 35872266, 107616801, 322850406, 968551221, 2905653666, 8716961001, 26150883006, 78452649021, 235357947066, 706073841201, 2118221523606
COMMENTS
Numbers n where the recurrence s(0)=1, if s(n-1) >= n then s(n) = s(n-1) - n else s(n) = s(n-1) + n produces s(n)=0. - Hugo Pfoertner, Jan 05 2012 Binomial transform of A146523: (1, 5, 10, 20, 40, ...) and double binomial transform of A010685: (1, 4, 1, 4, 1, 4, ...). - Gary W. Adamson, Aug 25 2016 Also the number of maximal cliques in the (n+1)-Hanoi graph. - Eric W. Weisstein, Dec 01 2017 a(n) is the least k such that f(a(n-1)+1) + ... + f(k) > f(a(n-2)+1) + ... + f(a(n-1)) for n > 1, where f(n) = 1/(n+1). Because Sum_{k=1..5*3^(n-1)} 1/(a(n)+3*k-1) + 1/(a(n)+3*k) + 1/(a(n)+3*k+1) - 1/((a(n)+1+5*3^n)*5*3^(n-1)) < Sum_{k=1..5*3^(n-1)} 1/(a(n-1)+k+1) < Sum_{k=1..5*3^(n-1)} 1/(a(n)+3*k-1) + 1/(a(n)+3*k) + 1/(a(n)+3*k+1), we have 1 < 1/3 + 1/4 + ... + 1/7 < 1/8 + 1/9 + ... + 1/22 < ... . - Jinyuan Wang, Jun 15 2020
FORMULA
a(n) = 3*(a(n-1) + 1), with a(0)=1.
O.g.f.: (5/2)/(1-3*x) - (3/2)/(1-x).
MATHEMATICA
CoefficientList[Series[(1 + 2 x)/(1 - 4 x + 3 x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Dec 01 2017 *)
PROG
(PARI) a(n) = (5*3^n-3)/2; /* Joerg Arndt, Apr 14 2013 */
a(1) = 2, a(2) = 3; thereafter a(n) is the sum of all the previous terms.
+10
12
2, 3, 5, 10, 20, 40, 80, 160, 320, 640, 1280, 2560, 5120, 10240, 20480, 40960, 81920, 163840, 327680, 655360, 1310720, 2621440, 5242880, 10485760, 20971520, 41943040, 83886080, 167772160, 335544320, 671088640, 1342177280, 2684354560, 5368709120, 10737418240
COMMENTS
Except for first three terms, a(n) is 10 times 2^(n-4).
These values comprise the tile values used in the "fives" variant of the game 2048, including 1 as the zeroth term. - Michael De Vlieger, Jul 18 2018
FORMULA
G.f.: x*(1 - x)*(2 + x) / (1 - 2*x). - Colin Barker, Nov 17 2018
MATHEMATICA
t = {2, 3}; For[k = 3, k <= 27, k++, AppendTo[t, Total@ t]]; t (* Michael De Vlieger, May 14 2015 *)
PROG
(PARI) a(n) = if(n<3, n+1, 5*2^(n-3)); \\ Altug Alkan, Jul 18 2018 (PARI) Vec(x*(1 - x)*(2 + x) / (1 - 2*x) + O(x^40)) \\ Colin Barker, Nov 17 2018
1, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144, 524288, 1048576, 2097152, 4194304, 8388608, 16777216, 33554432, 67108864, 134217728, 268435456, 536870912, 1073741824, 2147483648, 4294967296, 8589934592
COMMENTS
Hankel transform is := 1,-48,0,0,0,0,0,0,0,...
FORMULA
a(0)=1, a(n) = 2^(n+2) for n>0.
PROG
(PARI) Vec((1+6*x)/(1-2*x) + O(x^50)) \\ Colin Barker, Mar 17 2016
Triangle of compressed square roots of Gray code * bit-reversal permutation.
+10
2
1, 3, 1, 6, 1, 5, 6, 9, 1, 10, 12, 18, 1, 17, 10, 12, 18, 33, 1, 34, 20, 24, 36, 66, 1, 65, 34, 20, 24, 36, 66, 129, 1, 130, 68, 40, 48, 72, 132, 258, 1, 257, 130, 68, 40, 48, 72, 132, 258, 513, 1, 514, 260, 136, 80
COMMENTS
The permutation that turns a natural ordered into a sequency ordered Walsh matrix of size 2^n is the product of the Gray code permutation A003188(0..2^n-1) and the bit-reversal permutation A030109(n,0..2^n-1). (This permutation of 2^n elements can be represented by the compression vector [2^(n-1), 3*[2^(n-2)..4,2,1]] with n elements.)
This triangle shows the compression vectors of the unique square roots of these permutations, which correspond to symmetric binary matrices with 2n-1 ones.
(These n X n matrices correspond to graphs that can be described by permutations of n elements, which are shown in A239304.) Rows of the square array:
T(1,n) = 1,3,6,6,12,12,24,24,48,48,96,96,192,192,384,384,... (compare A003945) T(2,n) = 1,1,9,18,18,36,36,72,72,144,144,288,288,576,576,... (compare A005010) Columns of the square array:
T(m,1) = 1,1,5,10,10,20,20,40,40,80,80,160,160,320,320,... (compare A146523) T(m,2) = 3,1,1,17,34,34,68,68,136,136,272,272,544,544,... (compare A110287)
EXAMPLE
Triangular array begins:
1
3 1
6 1 5
6 9 1 10
12 18 1 17 10
12 18 33 1 34 20
Square array begins:
1 3 6 6 12 12
1 1 9 18 18 36
5 1 1 33 66 66
10 17 1 1 129 258
10 34 65 1 1 513
20 34 130 257 1 1
The Walsh permutation wp(8,12,6,3) = (0,8,12,4, 6,14,10,2, 3,11,15,7, 5,13,9,1) permutes the natural ordered into the sequency ordered Walsh matrix of size 2^4.
Its square root is wp(6,9,1,10) = (0,6,9,15, 1,7,8,14, 10,12,3,5, 11,13,2,4).
So row 4 of the triangular array is (6,9,1,10).
159, 6, 5, 10, 20, 40, 80, 160, 320, 640, 1280, 2560, 5120, 10240, 20480, 40960, 81920, 163840, 327680, 655360, 1310720, 2621440, 5242880, 10485760, 20971520, 41943040, 83886080, 167772160, 335544320, 671088640, 1342177280, 2684354560, 5368709120, 10737418240
COMMENTS
A006667: number of tripling steps to reach 1 in '3x+1' problem.
A006577: number of halving and tripling steps to reach 1 in '3x+1' problem.
FORMULA
G.f.: x^3*(159 - 312*x - 7*x^2) / (1 - 2*x).
a(n) = 2*a(n-1) for n>5.
(End)
MAPLE
nn:=10^12:
for n from 3 to 35 do:
ii:=0:
for k from 2 to 10^6 while(ii=0) do:
m:=k:s1:=0:s2:=0:
for i from 1 to nn while(m<>1) do:
if irem(m, 2)=0
then
s2:=s2+1:m:=m/2:
else
s1:=s1+1:m:=3*m+1:
fi:
od:
if n*s1=s1+s2
then
ii:=1: printf(`%d, `, k):
else
fi:
od:od:
MATHEMATICA
f[u_]:=Module[{a=u, k=0}, While[a!=1, k++; If[EvenQ[a], a=a/2, a=a*3+1]]; k]; Table[f[u], {u, 10^7}]; g[v_]:=Count[Differences[NestWhileList[If[EvenQ[#], #/2, 3#+1]&, v, #>1&]], _?Positive]; Table[g[v], {v, 10^7}]; Do[k=3; While[g[k]/f[k]!=1/n, k++]; Print[n, " ", k], {n, 3, 35}]
PROG
(PARI) a(n) = if(n < 5, [0, 0, 159, 6][n], 5<<(n-5)) \\ David A. Corneth, Jun 01 2017 (PARI) Vec(x^3*(159 - 312*x - 7*x^2) / (1 - 2*x) + O(x^50)) \\ Colin Barker, Jun 01 2017
1, -3, -1, 5, -6, 3, -11, 10, -12, -5, 21, -22, 20, -24, 11, -43, 42, -44, 40, -48, -21, 85, -86, 84, -88, 80, -96, 43, -171, 170, -172, 168, -176, 160, -192, -85, 341, -342, 340, -344, 336, -352, 320, -384, 171, -683, 682, -684, 680, -688
COMMENTS
Jacobsthal numbers appear twice: 1) A001045(n+2) signed, terms 0, 1, 3, 6, 10 ( A000217); 2) A001045(n+1) signed, terms 0, 2, 5, 9 (n*(n+3)/2= A000096); between them are -3; 5, -6; -11, 10, -12; which appear (opposite sign) by rows in A140503 (1, -1, 2, 3, -2, 4) square. Consider the permutation of the nonnegative numbers
0, 2, 5, 9, 14, 20, 27,
1, 3, 6, 10, 15, 21, 28,
4, 7, 11, 16, 22, 29,
8, 12, 17, 23, 30,
13, 18, 24, 31,
19, 25, 32,
26, 33,
34, etc.
The corresponding distribution of a(n) is
1, -1, 3, -5, 11, -21, 43,
-3, 5, -11, 21, -43, 85, -171,
-6, 10, -22, 42, -86, 170,
-12, 20, -44, 84, -172,
-24, 40, -88, 168,
-48, 80, -176,
-96, 160,
-192, etc.
Column sums: -2, -2, -10, -10, -42, -42, -170, ... duplicate of a bisection of - A078008(n+2). b(n)= 1, -1, 3, -5, 11, 21, ... = (-1)^n* A001045(n+1) = A077925(n). Every row is b(n) or b(n+2) multiplied by 1, -1, -2, -4, -8, -16, ..., essentially - A011782(n).
MATHEMATICA
T[0, 0] = 0; T[1, 0] = T[0, 1] = 1; T[0, n_] := T[0, n] = T[0, n - 1] + 2*T[0, n - 2]; T[d_, d_] = 0; T[d_, n_] := T[d, n] = T[d - 1, n + 1] - T[d - 1, n]; A140944 = Table[T[d, n], {d, 0, 10}, {n, 0, d}] // Flatten; a[n_] := A140944[[n + 2]] - 3* A140944[[n + 1]]; Table[a[n], {n, 0, 60}] (* Jean-François Alcover, Dec 18 2014 *)
CROSSREFS
Cf. A000096, A000217, A005015, A001045, A007283, A020988, A077925, A078008, A140503, A146523, A151575, A175805, A084247.
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