A146523 -id:A146523 - cp's OEIS frontend

A003947 Expansion of (1+x)/(1-4*x).

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

Offset: 0

N. J. A. Sloane

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

T. D. Noe, Table of n, a(n) for n = 0..200
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 306
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
Index to divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (4).
Index entries for sequences related to trees

Cf. A003948, A003949. Column 5 in A265583.

Concatenation([1], List([1..30], n-> 5*4^(n-1) )); # G. C. Greubel, Aug 10 2019

[1] cat [5*4^(n-1): n in [1..30]]; // G. C. Greubel, Aug 10 2019

k := 5; if n = 0 then 1 else k*(k-1)^(n-1); fi;

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 *)
LinearRecurrence[{4},{1,5},30] (* Harvey P. Dale, Apr 19 2015 *)

a(n)=5*4^n\4 \\ Charles R Greathouse IV, Sep 08 2011

[1]+[5*4^(n-1) for n in (1..30)] # G. C. Greubel, Aug 10 2019

Binomial transform of A060925. Its binomial transform is A003463 (without leading zero). - Paul Barry, May 19 2003
From Paul Barry, May 19 2003: (Start)
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)
a(n) = Sum_{k=0..n} A029653(n, k)*x^k for x = 3. - Philippe Deléham, Jul 10 2005
a(n) = A146523(n)*A011782(n). - R. J. Mathar, Jul 08 2009
a(n) = 5*A000302(n-1), n>0.
a(n) = 4*a(n-1), n>1. - Vincenzo Librandi, Dec 31 2010
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

Edited by N. J. A. Sloane, Dec 04 2009

A008851 Congruent to 0 or 1 mod 5.

Original entry on oeis.org

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

Offset: 1

N. J. A. Sloane

Numbers k that have the same last digit as k^2.

L. E. Dickson, History of the Theory of Numbers, I, p. 459.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A001622, A003226, A045953, A046831, A046851, A086457.

a008851 n = a008851_list !! (n-1)
a008851_list = [10*n + m | n <- [0..], m <- [0,1,5,6]]
-- Reinhard Zumkeller, Jul 27 2011

[n: n in [0..200] | n mod 5 in {0, 1}]; // Vincenzo Librandi, Nov 17 2014

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

Select[Range[0, 151], MemberQ[{0, 1}, Mod[#, 5]] &] (* T. D. Noe, Mar 31 2013 *)

a(n) = 5*(n\2)+bitand(n,1); /* Joerg Arndt, Mar 31 2013 */

a(n) = floor((5/3)*floor(3*(n-1)/2)); /* Joerg Arndt, Mar 31 2013 */

a(n) = 5*n - a(n-1) - 9, n >= 2. - Vincenzo Librandi, Nov 18 2010 [Corrected for offset by David Lovler, Oct 10 2022]
G.f.: x^2*(1+4*x) / ( (1+x)*(x-1)^2 ). - R. J. Mathar, Oct 07 2011
a(n+1) = Sum_{k>=0} A030308(n,k)*A146523(k). - Philippe Deléham, Oct 17 2011
a(n) = floor((5/3)*floor(3*(n-1)/2)). - Clark Kimberling, Jul 04 2012
a(n) = (10*n - 13 - 3*(-1)^n)/4. - Robert Israel, Nov 17 2014 [Corrected by David Lovler, Sep 21 2022]
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

Offset corrected by Reinhard Zumkeller, Jul 27 2011

A134931 a(n) = (5*3^n-3)/2.

Original entry on oeis.org

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

Offset: 0

Rolf Pleisch, Jan 29 2008

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
A046901(a(n)) = 1. - Reinhard Zumkeller, Jan 31 2013
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

Vincenzo Librandi, Table of n, a(n) for n = 0..500
Eric Weisstein's World of Mathematics, Hanoi Graph
Eric Weisstein's World of Mathematics, Maximal Clique
Index entries for linear recurrences with constant coefficients, signature (4,-3).

Cf. A003462, A007051, A034472, A024023, A067771, A029858.

Cf. A005030, A010685, A046901, A060816, A116952, A146523, A225918.

[(5*3^n-3)/2: n in [0..30]]; // Vincenzo Librandi, Jun 05 2011

seq((5*3^n-3)/2, n= 0..25); # Gary Detlefs, Jun 22 2010

a=1; lst={a}; Do[a=a*3+3; AppendTo[lst,a], {n,0,100}]; lst (* Vladimir Joseph Stephan Orlovsky, Dec 25 2008 *)
Table[(5 3^n - 9)/6, {n, 20}] (* Eric W. Weisstein, Dec 01 2017 *)
(5 3^Range[20] - 9)/6 (* Eric W. Weisstein, Dec 01 2017 *)
LinearRecurrence[{4, -3}, {1, 6}, 20] (* Eric W. Weisstein, Dec 01 2017 *)
CoefficientList[Series[(1 + 2 x)/(1 - 4 x + 3 x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Dec 01 2017 *)

a(n) = (5*3^n-3)/2; /* Joerg Arndt, Apr 14 2013 */

a(n) = 3*(a(n-1) + 1), with a(0)=1.
From R. J. Mathar, Jan 31 2008: (Start)
O.g.f.: (5/2)/(1-3*x) - (3/2)/(1-x).
a(n) = (A005030(n) - 3)/2. (End)
a(n) = A060816(n+1) - 1. - Philippe Deléham, Apr 14 2013
E.g.f.: exp(x)*(5*exp(2*x) - 3)/2. - Stefano Spezia, Aug 28 2023

More terms from Vladimir Joseph Stephan Orlovsky, Dec 25 2008

A257113 a(1) = 2, a(2) = 3; thereafter a(n) is the sum of all the previous terms.

Original entry on oeis.org

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

Offset: 1

Giovanni Teofilatto, Apr 24 2015

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

Colin Barker, Table of n, a(n) for n = 1..1000
E. R. Berlekamp, A contribution to mathematical psychometrics, Unpublished Bell Labs Memorandum, Feb 08 1968 [Annotated scanned copy]
David Eppstein, Making Change in 2048, arXiv:1804.07396 [cs.DM], 2018.
Index entries for linear recurrences with constant coefficients, signature (2).

Cf. A000079, A020714. Essenitally the same as A084215.

[2,3] cat [5*2^n: n in [0..35]]; // Vincenzo Librandi, May 15 2015

t = {2, 3}; For[k = 3, k <= 27, k++, AppendTo[t, Total@ t]]; t (* Michael De Vlieger, May 14 2015 *)
Join[{2, 3}, Table[5 2^n, {n, 0, 40}]] (* Vincenzo Librandi, May 15 2015 *)
Join[{2,3},NestList[2#&,5,40]] (* Harvey P. Dale, Apr 06 2018 *)

a(n) = if(n<3, n+1, 5*2^(n-3)); \\ Altug Alkan, Jul 18 2018

Vec(x*(1 - x)*(2 + x) / (1 - 2*x) + O(x^40)) \\ Colin Barker, Nov 17 2018

a(n) = ceil(5*2^(n-3)) \\ Alan Michael Gómez Calderón, Mar 30 2022

a(n) = A020714(n-3) for n>2.
a(n) = A146523(n-2) for n>2. - R. J. Mathar, May 14 2015
G.f.: x*(1 - x)*(2 + x) / (1 - 2*x). - Colin Barker, Nov 17 2018

A146541 Binomial transform of A010688.

Original entry on oeis.org

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

Offset: 0

Philippe Deléham, Oct 31 2008

Hankel transform is := 1,-48,0,0,0,0,0,0,0,...

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2).

Cf. A000007, A011782, A000079, A003945, A046055, A146523, A082505, A135092.

Essentially the same as A132479, A077552, A020707.

Join[{1},2^Range[3,40]] (* Harvey P. Dale, Feb 28 2016 *)

Vec((1+6*x)/(1-2*x) + O(x^50)) \\ Colin Barker, Mar 17 2016

a(0)=1, a(n) = 2^(n+2) for n>0.
a(n) = Sum_{k, 0..n} A109466(n,k)*A146534(k).
a(n) = A132479(n), n>1. - R. J. Mathar, Nov 02 2008
G.f.: (1+6*x) / (1-2*x). - Colin Barker, Mar 17 2016

Corrected and extended by Harvey P. Dale, Feb 28 2016

A239303 Triangle of compressed square roots of Gray code * bit-reversal permutation.

Original entry on oeis.org

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

Offset: 1

Tilman Piesk, Mar 14 2014

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)

			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).

Tilman Piesk, First 140 rows of the triangle, flattened
Tilman Piesk, Sequency ordered Walsh matrix (Wikiversity)
Tilman Piesk, Calculation in MATLAB

Cf. A239304, A003188, A030109.

A287798 Least k such that A006667(k)/A006577(k) = 1/n.

Original entry on oeis.org

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

Offset: 3

Michel Lagneau, Jun 01 2017

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.
a(n) = {159, 6} union {A020714}.

			a(3) = 159 because A006667(159)/A006577(159) = 18/54 = 1/3.

Colin Barker, Table of n, a(n) for n = 3..1000
Index entries for linear recurrences with constant coefficients, signature (2).

Cf. A006577, A006666, A006667. Essentially the same as A020714, A084215, A146523 and A257113.

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:

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}]

a(n) = if(n < 5, [0,0,159,6][n], 5<<(n-5)) \\ David A. Corneth, Jun 01 2017

Vec(x^3*(159 - 312*x - 7*x^2) / (1 - 2*x) + O(x^50)) \\ Colin Barker, Jun 01 2017

For n >= 5, a(n) = 5*2^n/32. - David A. Corneth, Jun 01 2017
From Colin Barker, Jun 01 2017: (Start)
G.f.: x^3*(159 - 312*x - 7*x^2) / (1 - 2*x).
a(n) = 2*a(n-1) for n>5.
(End)

A140950 a(n) = A140944(n+1) - 3*A140944(n).

Original entry on oeis.org

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

Offset: 0

Paul Curtz, Jul 25 2008

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).

Cf. A000096, A000217, A005015, A001045, A007283, A020988, A077925, A078008, A140503, A146523, A151575, A175805, A084247.

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 *)

More terms and a(19)=-48 instead of 42 corrected by Jean-François Alcover, Dec 22 2014

cp's OEIS Frontend

A003947 Expansion of (1+x)/(1-4*x).

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A008851 Congruent to 0 or 1 mod 5.

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A134931 a(n) = (5*3^n-3)/2.

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A257113 a(1) = 2, a(2) = 3; thereafter a(n) is the sum of all the previous terms.

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A146541 Binomial transform of A010688.

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A239303 Triangle of compressed square roots of Gray code * bit-reversal permutation.

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