A056451 -id:A056451 - cp's OEIS frontend

A056487 a(n) = 5^(n/2) for n even, a(n) = 3*5^((n-1)/2) for n odd.

1, 3, 5, 15, 25, 75, 125, 375, 625, 1875, 3125, 9375, 15625, 46875, 78125, 234375, 390625, 1171875, 1953125, 5859375, 9765625, 29296875, 48828125, 146484375, 244140625, 732421875, 1220703125, 3662109375, 6103515625, 18310546875, 30517578125, 91552734375

Offset: 0

Marks R. Nester

Apparently identical to A111386! Is this a theorem? - Klaus Brockhaus, Jul 21 2009

For n > 1, number of necklaces with n-1 beads and 5 colors that are the same when turned over and hence have reflection symmetry. - Herbert Kociemba, Nov 24 2016

M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2]

Index entries for linear recurrences with constant coefficients, signature (0,5).

Column 5 of A284855.

Cf. A029744, A038754, A056391, A056451, A087205, A087206, A111386.

[n le 2 select 2*n-1 else 5*Self(n-2): n in [1..28]]; // Bruno Berselli, Mar 24 2011

A056487:=n->3^((1-(-1)^n)/2)*5^((2*n+(-1)^n-1)/4): seq(A056487(n), n=0..40); # Wesley Ivan Hurt, Nov 24 2016

Table[3^((1 - (-1)^n)/2)*5^((2*n + (-1)^n - 1)/4), {n, 0, 30}] (* Wesley Ivan Hurt, Nov 24 2016 *)
CoefficientList[Series[(1 + 3 x)/(1 - 5 x^2), {x, 0, 31}], x] (* Michael De Vlieger, Nov 24 2016 *)
LinearRecurrence[{0, 5}, {1, 3}, 35] (* Vincenzo Librandi, Nov 25 2016 *)
k=5; Table[(k^Floor[(n+1)/2] + k^Ceiling[(n+1)/2]) / 2, {n, -1, 30}] (* Robert A. Russell, Sep 21 2018 *)

a(n)=if(n%2,3,1)*5^(n\2) \\ Charles R Greathouse IV, Oct 07 2015

def A056487(n): return 5**(n>>1)*(3 if n&1 else 1) # Chai Wah Wu, Oct 27 2022

a(n+2) = 5*a(n), a(0)=1, a(2)=3.
Binomial transform of A087205. Binomial transform is A087206. - Paul Barry, Aug 25 2003
G.f.: (1+3*x)/(1-5*x^2); a(n) = 5^(n/2)(1/2 + 3*sqrt(5)/10 + (1/2 - 3*sqrt(5)/10)(-1)^n). - Paul Barry, Mar 19 2004
2nd inverse binomial transform of Fibonacci(3n+2). - Paul Barry, Apr 16 2004
a(n+3) = a(n+2)*a(n+1)/a(n). - Reinhard Zumkeller, Mar 04 2011
a(n) = 3^((1 - (-1)^n)/2) * 5^((2*n + (-1)^n-1)/4). - Bruno Berselli, Mar 24 2011
a(n+1) = (k^floor((n+1)/2) + k^ceiling((n+1)/2)) / 2, where k=5 is the number of possible colors. - Robert A. Russell, Sep 22 2018
E.g.f.: cosh(sqrt(5)*x) + 3*sinh(sqrt(5)*x)/sqrt(5). - Stefano Spezia, Jun 06 2023

Changed one 'even' to 'odd' in the definition. - R. J. Mathar, Oct 06 2010

A074872 Inverse BinomialMean transform of the Fibonacci sequence A000045 (with the initial 0 omitted).

Original entry on oeis.org

1, 1, 5, 5, 25, 25, 125, 125, 625, 625, 3125, 3125, 15625, 15625, 78125, 78125, 390625, 390625, 1953125, 1953125, 9765625, 9765625, 48828125, 48828125, 244140625, 244140625, 1220703125, 1220703125, 6103515625, 6103515625, 30517578125, 30517578125, 152587890625

Offset: 1

John W. Layman, Sep 12 2002

See A075271 for the definition of the BinomialMean transform.

The inverse binomial transform of 2^n*c(n+1), where c(n) is the solution to c(n) = c(n-1) + k*c(n-2), a(0)=0, a(1)=1 is 1, 1, 4k+1, 4k+1, (4k+1)^2, ... - Paul Barry, Feb 12 2004

Vincenzo Librandi, Table of n, a(n) for n = 1..2000
Index entries for linear recurrences with constant coefficients, signature (0,5).

Cf. A000045, A075271, A056451, A063727, A016116, A108411.

[5^Floor((n-1)/2): n in [1..40]]; // Vincenzo Librandi, Aug 16 2011

a[1] := 1; a[2] := 1; a[n_] := 5a[n - 2]; Table[a[n], {n, 30}] (* Alonso del Arte, Mar 04 2011 *)

a(n)=5^((n-1)\2) \\ Charles R Greathouse IV, Oct 03 2016

a(n) = 5^floor((n-1)/2).
a(1)=1, a(2)=1 and, for n > 2, a(n) = 5*a(n-2).
From Paul Barry, Feb 12 2004: (Start)
G.f.: x*(1+x)/(1-5*x^2);
a(n) = (1/(2*sqrt(5))*((1+sqrt(5))*(sqrt(5))^n - (1-sqrt(5))*(-sqrt(5))^n)).
Inverse binomial transform of A063727 (2^n*Fibonacci(n+1)). (End)
a(n+3) = a(n+2)*a(n+1)/a(n). - Reinhard Zumkeller, Mar 04 2011
E.g.f.: (cosh(sqrt(5)*x) + sqrt(5)*sinh(sqrt(5)*x) - 1)/5. - Stefano Spezia, May 24 2024

A032122 Number of reversible strings with n beads of 5 colors.

Original entry on oeis.org

1, 5, 15, 75, 325, 1625, 7875, 39375, 195625, 978125, 4884375, 24421875, 122078125, 610390625, 3051796875, 15258984375, 76294140625, 381470703125, 1907349609375, 9536748046875, 47683720703125

Offset: 0

Christian G. Bower

			For a(2)=15, the five achiral strings are AA, BB, CC, DD, and EE; the 10 (equivalent) chiral pairs are AB-BA, AC-CA, AD-DA, AE-EA, BC-CB, BD-DB, BE-EB, CD-DC, CE-EC, and DE-ED.

Vincenzo Librandi, Table of n, a(n) for n = 0..500
C. G. Bower, Transforms (2)
Index entries for linear recurrences with constant coefficients, signature (5,5,-25).

Column 5 of A277504.

Cf. A000351 (oriented), A032088(n>1) (chiral), A056451 (achiral).

I:=[1, 5, 15]; [n le 3 select I[n] else 5*Self(n-1)+ 5*Self(n-2)-25*Self(n-3): n in [1..30]]; // Vincenzo Librandi, Jan 31 2012

LinearRecurrence[{5, 5, -25}, {1, 5, 15}, 31] (* Vincenzo Librandi, Jan 31 2012 *)
k=5; Table[(k^n+k^Ceiling[n/2])/2,{n,0,30}] (*Robert A. Russell, Nov 25 2017*)

a(n)=(5^((n+1)\2)+5^n)/2 \\ Charles R Greathouse IV, Jan 31 2012

"BIK" (reversible, indistinct, unlabeled) transform of 5, 0, 0, 0...
a(n) = 1/2 * (5^n + 5^floor((n+1)/2)) = 5*A001447(n+1). - Ralf Stephan, Jul 07 2003
G.f.: (1-15*x^2) / ((1-5*x)*(1-5*x^2)). - Colin Barker, Jan 30 2012 [Adapted to offset 0 by Robert A. Russell, Nov 10 2018]
a(n) = 5*a(n-1) + 5*a(n-2) - 25*a(n-3). - Vincenzo Librandi, Jan 31 2012
a(n) = (A000351(n) + A056451(n)) / 2. - Robert A. Russell, Nov 10 2018

a(0)=1 prepended by Robert A. Russell, Nov 10 2018

A056456 Number of palindromes of length n using exactly five different symbols.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 120, 120, 1800, 1800, 16800, 16800, 126000, 126000, 834120, 834120, 5103000, 5103000, 29607600, 29607600, 165528000, 165528000, 901020120, 901020120, 4809004200, 4809004200, 25292030400

Offset: 1

Marks R. Nester

M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2.]

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,14,-14,-71,71,154,-154,-120,120).

Cf. A056451, A001118.

k=5; Table[k! StirlingS2[Ceiling[n/2],k],{n,1,30}] (* Robert A. Russell, Sep 25 2018 *)
LinearRecurrence[{1, 14, -14, -71, 71, 154, -154, -120, 120}, {0, 0, 0, 0, 0, 0, 0, 0, 120}, 30] (* Vincenzo Librandi, Sep 29 2018 *)

a(n) = 5!*stirling((n+1)\2, 5, 2); \\ Altug Alkan, Sep 25 2018

a(n) = 5! * Stirling2( [(n+1)/2], 5).
G.f.: -120*x^9/((x-1)*(2*x-1)*(2*x+1)*(2*x^2-1)*(3*x^2-1)*(5*x^2-1)). - Colin Barker, Sep 03 2012
G.f.: k!(x^(2k-1)+x^(2k))/Product_{i=1..k}(1-ix^2), where k=5 is the number of symbols. - Robert A. Russell, Sep 25 2018

A321391 Array read by antidiagonals: T(n,k) is the number of achiral rows of n colors using up to k colors.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 2, 1, 0, 1, 4, 3, 4, 1, 0, 1, 5, 4, 9, 4, 1, 0, 1, 6, 5, 16, 9, 8, 1, 0, 1, 7, 6, 25, 16, 27, 8, 1, 0, 1, 8, 7, 36, 25, 64, 27, 16, 1, 0, 1, 9, 8, 49, 36, 125, 64, 81, 16, 1, 0, 1, 10, 9, 64, 49, 216, 125, 256, 81, 32, 1, 0

Offset: 0

Robert A. Russell, Nov 08 2018

The antidiagonals go from top-right to bottom-left.

			The array begins with T(0,0):
1 1  1   1    1     1     1      1      1      1       1       1 ...
0 1  2   3    4     5     6      7      8      9      10      11 ...
0 1  2   3    4     5     6      7      8      9      10      11 ...
0 1  4   9   16    25    36     49     64     81     100     121 ...
0 1  4   9   16    25    36     49     64     81     100     121 ...
0 1  8  27   64   125   216    343    512    729    1000    1331 ...
0 1  8  27   64   125   216    343    512    729    1000    1331 ...
0 1 16  81  256   625  1296   2401   4096   6561   10000   14641 ...
0 1 16  81  256   625  1296   2401   4096   6561   10000   14641 ...
0 1 32 243 1024  3125  7776  16807  32768  59049  100000  161051 ...
0 1 32 243 1024  3125  7776  16807  32768  59049  100000  161051 ...
0 1 64 729 4096 15625 46656 117649 262144 531441 1000000 1771561 ...
For T(3,3)=9, the rows are AAA, ABA, ACA, BAB, BBB, BCB, CAC, CBC, and CCC.

Columns 0-6 are A000007, A000012, A060546, A056449, A056450, A056451, A056452.

Cf. A003992 (oriented), A277504 (unoriented), A293500 (chiral).

Table[If[n>0, (n-k)^Ceiling[k/2], 1], {n, 0, 12}, {k, 0, n}] // Flatten

T(n,k) = [n==0] + [n>0] * k^ceiling(n/2).

The generating function for column k is (1+k*x) / (1-k*x^2).

A032088 Number of reversible strings with n beads of 5 colors. If more than 1 bead, not palindromic.

Original entry on oeis.org

5, 10, 50, 300, 1500, 7750, 38750, 195000, 975000, 4881250, 24406250, 122062500, 610312500, 3051718750, 15258593750, 76293750000, 381468750000, 1907347656250, 9536738281250, 47683710937500

Offset: 1

Christian G. Bower

Andrew Howroyd, Table of n, a(n) for n = 1..200
C. G. Bower, Transforms (2)
Index entries for linear recurrences with constant coefficients, signature (5, 5, -25).

Column 5 of A293500 for n>1.
Cf. A032122.
Equals (A000351 - A056451) / 2 for n>1.

Join[{5}, LinearRecurrence[{5, 5, -25}, {10, 50, 300}, 19]] (* Jean-François Alcover, Oct 11 2017 *)

a(n) = if(n<2, [5][n], (5^n - 5^(ceil(n/2)))/2); \\ Andrew Howroyd, Oct 10 2017

"BHK" (reversible, identity, unlabeled) transform of 5, 0, 0, 0, ...
Conjectures from Colin Barker, Jul 07 2012: (Start)
a(n) = 5*a(n-1) + 5*a(n-2) - 25*a(n-3) for n > 4.
G.f.: 5*x*(1 - 3*x - 5*x^2 + 25*x^3)/((1 - 5*x)*(1 - 5*x^2)).
(End)
Conjectures from Colin Barker, Mar 09 2017: (Start)
a(n) = 5^(n/2)*(5^(n/2) - 1) / 2 for n > 1 and even.
a(n) = -5*(5^(n/2-1/2) - 5^(n-1)) / 2 for n > 1 and odd.
(End)
The above conjectures are true: The second set follows from the definition and the first set can be derived from that. - Andrew Howroyd, Oct 10 2017
a(n) = (5^n - 5^(ceiling(n/2))) / 2 = (A000351(n) - A056451(n)) / 2 for n>1. - Robert A. Russell and Danny Rorabaugh, Jun 22 2018

A056470 Number of palindromic structures using a maximum of five different symbols.

Original entry on oeis.org

1, 1, 2, 2, 5, 5, 15, 15, 52, 52, 202, 202, 855, 855, 3845, 3845, 18002, 18002, 86472, 86472, 422005, 422005, 2079475, 2079475, 10306752, 10306752, 51263942, 51263942, 255514355, 255514355, 1275163905

Offset: 1

Marks R. Nester

Permuting the symbols will not change the structure.

M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2]

Index entries for linear recurrences with constant coefficients, signature (1,10,-10,-31,31,30,-30).

Cf. A056451.

LinearRecurrence[{1,10,-10,-31,31,30,-30},{1,1,2,2,5,5,15},40] (* Harvey P. Dale, Dec 16 2017 *)

a(2n-1) = a(2n) = A056272(n). [R. J. Mathar, Nov 24 2010]

G.f.: -x*(19*x^6-24*x^4+9*x^2-1) / ((x-1)*(2*x^2-1)*(3*x^2-1)*(5*x^2-1)). [Colin Barker, Dec 05 2012]

A056461 Number of primitive (aperiodic) palindromes using a maximum of five different symbols.

Original entry on oeis.org

5, 0, 20, 20, 120, 100, 620, 600, 3100, 3000, 15620, 15480, 78120, 77500, 390480, 390000, 1953120, 1949900, 9765620, 9762480, 48827480, 48812500, 244140620, 244124400, 1220703000, 1220625000, 6103512500, 6103437480, 30517578120, 30517184400, 152587890620, 152587500000

Offset: 1

Marks R. Nester

nonn

M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2]

Cf. A056391, A056451.

Column 5 of A284823.

a(n) = sumdiv(n, d, moebius(d)*5^((1 + n/d)\2));
for(n=1, 40, print1(a(n), ", ")); \\ Petros Hadjicostas, Apr 24 2020

Sum_{d|n} mu(d)*A056451(n/d).

More terms from Petros Hadjicostas, Apr 24 2020

A240437 Number of non-palindromic n-tuples of 5 distinct elements.

Original entry on oeis.org

0, 20, 100, 600, 3000, 15500, 77500, 390000, 1950000, 9762500, 48812500, 244125000, 1220625000, 6103437500, 30517187500, 152587500000, 762937500000, 3814695312500, 19073476562500, 95367421875000, 476837109375000, 2384185742187500, 11920928710937500, 59604644531250000, 298023222656250000

Offset: 1

Mikk Heidemaa, Aug 17 2014

			For n=3 a(3)=100 solutions are:
{0,0,1}, {0,0,2}, {0,0,3}, {0,0,4}, {0,1,1}, {0,1,2}, {0,1,3}, {0,1,4},
{0,2,1}, {0,2,2}, {0,2,3}, {0,2,4}, {0,3,1}, {0,3,2}, {0,3,3}, {0,3,4},
{0,4,1}, {0,4,2}, {0,4,3}, {0,4,4}, {1,0,0}, {1,0,2}, {1,0,3}, {1,0,4},
{1,1,0}, {1,1,2}, {1,1,3}, {1,1,4}, {1,2,0}, {1,2,2}, {1,2,3}, {1,2,4},
{1,3,0}, {1,3,2}, {1,3,3}, {1,3,4}, {1,4,0}, {1,4,2}, {1,4,3}, {1,4,4},
{2,0,0}, {2,0,1}, {2,0,3}, {2,0,4}, {2,1,0}, {2,1,1}, {2,1,3}, {2,1,4},
{2,2,0}, {2,2,1}, {2,2,3}, {2,2,4}, {2,3,0}, {2,3,1}, {2,3,3}, {2,3,4},
{2,4,0}, {2,4,1}, {2,4,3}, {2,4,4}, {3,0,0}, {3,0,1}, {3,0,2}, {3,0,4},
{3,1,0}, {3,1,1}, {3,1,2}, {3,1,4}, {3,2,0}, {3,2,1}, {3,2,2}, {3,2,4},
{3,3,0}, {3,3,1}, {3,3,2}, {3,3,4}, {3,4,0}, {3,4,1}, {3,4,2}, {3,4,4},
{4,0,0}, {4,0,1}, {4,0,2}, {4,0,3}, {4,1,0}, {4,1,1}, {4,1,2}, {4,1,3},
{4,2,0}, {4,2,1}, {4,2,2}, {4,2,3}, {4,3,0}, {4,3,1}, {4,3,2}, {4,3,3},
{4,4,0}, {4,4,1}, {4,4,2}, {4,4,3}.

Index entries for linear recurrences with constant coefficients, signature (5, 5, -25).

Cf. A233411, A242026, A242278.

gf := (20*x^2) / (1 - 5*x - 5*x^2 + 25*x^3): ser := series(gf, x, 26):
seq(coeff(ser,x,n), n=1..25); # Peter Luschny, May 13 2019

Table[1/2 * 5^(n/2) * ((Sqrt[5]-1) * (-1)^n - Sqrt[5]-1) + 5^n, {n, 25}]

concat([0], Vec( ( (20*x^2) / (1 - 5*x - 5*x^2 + 25*x^3) + O(x^30) ) ) ) \\ Joerg Arndt, Aug 18 2014

a(n) = 1/2 * 5^(n/2) * ((sqrt(5)-1) * (-1)^n - sqrt(5)-1) + 5^n.
a(n) = 5^n - 5^ceiling(n/2).
a(n) = A000351(n) - A056451(n).
G.f.: (20*x^2) / (1 - 5*x - 5*x^2 + 25*x^3). [corrected by Peter Luschny, May 13 2019]

A162962 a(n) = 5*a(n-2) for n > 2; a(1) = 1, a(2) = 5.

Original entry on oeis.org

1, 5, 5, 25, 25, 125, 125, 625, 625, 3125, 3125, 15625, 15625, 78125, 78125, 390625, 390625, 1953125, 1953125, 9765625, 9765625, 48828125, 48828125, 244140625, 244140625, 1220703125, 1220703125, 6103515625, 6103515625, 30517578125

Offset: 1

Klaus Brockhaus, Jul 19 2009

Apparently a(n) = A074872(n+1), a(n) = A056451(n-1) for n > 1.

Binomial transform is A084057 without initial 1, second binomial transform is A048876, third binomial transform is A082762, fourth binomial transform is A162769, fifth binomial transform is A093145 without initial 0.

Index entries for linear recurrences with constant coefficients, signature (0,5).

Cf. A000351 (powers of 5), A074872 (powers of 5 repeated), A056451 (5^floor((n+1)/2)), A084057, A048876, A082762, A162769, A093145.

[ n le 2 select 4*n-3 else 5*Self(n-2): n in [1..30] ];

LinearRecurrence[{0,5},{1,5},30] (* Harvey P. Dale, Mar 18 2023 *)

a(n) = 5^((1/4)*(2*n-1+(-1)^n)).

G.f.: x*(1+5*x)/(1-5*x^2).

cp's OEIS Frontend

A056487 a(n) = 5^(n/2) for n even, a(n) = 3*5^((n-1)/2) for n odd.

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A074872 Inverse BinomialMean transform of the Fibonacci sequence A000045 (with the initial 0 omitted).

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A032122 Number of reversible strings with n beads of 5 colors.

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A056456 Number of palindromes of length n using exactly five different symbols.

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A321391 Array read by antidiagonals: T(n,k) is the number of achiral rows of n colors using up to k colors.

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A032088 Number of reversible strings with n beads of 5 colors. If more than 1 bead, not palindromic.

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A056470 Number of palindromic structures using a maximum of five different symbols.

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