A108981 -id:A108981 - cp's OEIS frontend

A126473 Number of strings over a 5 symbol alphabet with adjacent symbols differing by three or less.

1, 5, 23, 107, 497, 2309, 10727, 49835, 231521, 1075589, 4996919, 23214443, 107848529, 501037445, 2327695367, 10813893803, 50238661313, 233396326661, 1084301290583, 5037394142315, 23402480441009, 108722104190981, 505095858086951, 2346549744920747

Offset: 0

R. H. Hardin, Dec 27 2006

[Empirical] a(base,n) = a(base-1,n) + 7^(n-1) for base >= 3n-2; a(base,n) = a(base-1,n) + 7^(n-1)-2 when base = 3n-3.
From Johannes W. Meijer, Aug 01 2010: (Start)
The a(n) represent the number of n-move routes of a fairy chess piece starting in a given side square (m = 2, 4, 6 or 8) on a 3 X 3 chessboard. This fairy chess piece behaves like a king on the eight side and corner squares but on the central square the king goes crazy and turns into a red king, see A179596.
For the side squares the 512 red kings lead to 47 different red king sequences, see the cross-references for some examples.
The sequence above corresponds to four A[5] vectors with the decimal [binary] values 367 [1,0,1,1,0,1,1,1,1], 463 [1,1,1,0,0,1,1,1,1], 487 [1,1,1,1,0,0,1,1,1] and 493 [1,1,1,1,0,1,1,0,1]. These vectors lead for the corner squares to A179596 and for the central square to A179597.
This sequence belongs to a family of sequences with g.f. (1+x)/(1-4*x-k*x^2). Red king sequences that are members of this family are A003947 (k=0), A015448 (k=1), A123347 (k=2), A126473 (k=3; this sequence) and A086347 (k=4). Other members of this family are A000351 (k=5), A001834 (k=-1), A111567 (k=-2), A048473 (k=-3) and A053220 (k=-4)
Inverse binomial transform of A154244. (End)
Equals the INVERT transform of A055099: (1, 4, 14, 50, 178, ...). - Gary W. Adamson, Aug 14 2010
Number of one-sided n-step walks taking steps from {E, W, N, NE, NW}. - Shanzhen Gao, May 10 2011
For n>=1, a(n) equals the numbers of words of length n-1 on alphabet {0,1,2,3,4} containing no subwords 00 and 11. - Milan Janjic, Jan 31 2015

Shanzhen Gao and Keh-Hsun Chen, Tackling Sequences From Prudent Self-Avoiding Walks, FCS'14, The 2014 International Conference on Foundations of Computer Science.
S. Gao and H. Niederhausen, Sequences Arising From Prudent Self-Avoiding Walks, 2010.
Index entries for linear recurrences with constant coefficients, signature (4,3).

Cf. 5 symbol differing by two or less A126392, one or less A057960.
Cf. Red king sequences side squares [numerical value A[5]]: A086347 [495], A179598 [239], A126473 [367], A123347 [335], A179602 [95], A154964 [31], A015448 [327], A152187 [27], A003947 [325], A108981 [11], A007483 [2]. - Johannes W. Meijer, Aug 01 2010
Cf. A055099.

with(LinearAlgebra): nmax:=19; m:=2; A[5]:= [1,0,1,1,0,1,1,1,1]: A:=Matrix([[0,1,0,1,1,0,0,0,0],[1,0,1,1,1,1,0,0,0],[0,1,0,0,1,1,0,0,0],[1,1,0,0,1,0,1,1,0],A[5],[0,1,1,0,1,0,0,1,1],[0,0,0,1,1,0,0,1,0],[0,0,0,1,1,1,1,0,1],[0,0,0,0,1,1,0,1,0]]): 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); # Johannes W. Meijer, Aug 01 2010
# second Maple program:
a:= n-> (M-> M[1,2]+M[2,2])(<<0|1>, <3|4>>^n):
seq(a(n), n=0..24);  # Alois P. Heinz, Jun 28 2021

LinearRecurrence[{4, 3}, {1, 5}, 24] (* Jean-François Alcover, Dec 10 2024 *)

a(n)=([0,1; 3,4]^n*[1;5])[1,1] \\ Charles R Greathouse IV, May 10 2016

From Johannes W. Meijer, Aug 01 2010: (Start)
G.f.: (1+x)/(1-4*x-3*x^2).
a(n) = 4*a(n-1) + 3*a(n-2) with a(0) = 1 and a(1) = 5.
a(n) = ((1+3/sqrt(7))/2)*(A)^(-n) + ((1-3/sqrt(7))/2)*(B)^(-n) with A = (-2 + sqrt(7))/3 and B = (-2-sqrt(7))/3.
Lim_{k->oo} a(n+k)/a(k) = (-1)^(n+1)*A000244(n)/(A015530(n)*sqrt(7)-A108851(n))
(End)
a(n) = A015330(n)+A015330(n+1). - R. J. Mathar, May 09 2023

Edited by Johannes W. Meijer, Aug 10 2010

A122117 a(n) = 3a(n-1) + 4a(n-2), with a(0)=1, a(1)=2.

Original entry on oeis.org

1, 2, 10, 38, 154, 614, 2458, 9830, 39322, 157286, 629146, 2516582, 10066330, 40265318, 161061274, 644245094, 2576980378, 10307921510, 41231686042, 164926744166, 659706976666, 2638827906662, 10555311626650, 42221246506598

Offset: 0

Philippe Deléham, Oct 19 2006

Inverse binomial transform of A005053. Binomial transform of [1, 1, 7, 13, 55, ...] = A015441(n+1).
Convolved with [1, 2, 2, 2, ...] = powers of 4: [1, 4, 16, 64, ...]. - Gary W. Adamson, Jun 02 2009
a(n) is the number of compositions of n when there are 2 types of 1 and 6 types of other natural numbers. - Milan Janjic, Aug 13 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,4).

Cf. A055380, A100088, A108981, A122016, A201455.

a:=[1,2];; for n in [3..30] do a[n]:=3*a[n-1]+4*a[n-2]; od; a; # G. C. Greubel, May 18 2019

I:=[1, 2]; [n le 2 select I[n] else 3*Self(n-1)+4*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jul 06 2012

CoefficientList[Series[(1-x)/(1-3*x-4*x^2),{x,0,30}],x] (* Vincenzo Librandi, Jul 06 2012 *)

Vec((1-x)/(1-3*x-4*x^2)+O(x^30)) \\ Charles R Greathouse IV, Jan 11 2012

def A122117(n): return ((4<<(m:=n<<1))|2)//5-((1<Chai Wah Wu, Apr 22 2025

from sage.combinat.sloane_functions import recur_gen2b; it = recur_gen2b(1,2,3,4, lambda n: 0); [next(it) for i in range(24)] # Zerinvary Lajos, Jul 03 2008

((1-x)/(1-3*x-4*x^2)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, May 18 2019

a(n) = 2*A108981(n-1) for n > 0, with a(0) = 1.
a(2*n) = 4*a(2*n-1) + 2, a(2*n+1) = 4*a(2*n) - 2.
a(n) = Sum_{k=0..n} 2^(n-k)*A055380(n,k).
G.f.: (1-x)/(1-3*x-4*x^2).
Lim_{n->infinity} a(n+1)/a(n) = 4.
a(n) = Sum_{k=0..n} A122016(n,k)*2^k. - Philippe Deléham, Nov 05 2008
a(n) = A100088(2*n). - Chai Wah Wu, Apr 22 2025

Corrected by T. D. Noe, Nov 07 2006

A152187 a(n) = 3a(n-1) + 5a(n-2), with a(0)=1, a(1)=5.

Original entry on oeis.org

1, 5, 20, 85, 355, 1490, 6245, 26185, 109780, 460265, 1929695, 8090410, 33919705, 142211165, 596232020, 2499751885, 10480415755, 43940006690, 184222098845, 772366329985, 3238209484180, 13576460102465, 56920427728295

Offset: 0

Philippe Deléham, Nov 28 2008

Unsigned version of A152185.
From Johannes W. Meijer, Aug 01 2010: (Start)
The a(n) represent the number of n-move routes of a fairy chess piece starting in a given side square (m = 2, 4, 6 and 8) on a 3 X 3 chessboard. This fairy chess piece behaves like a king on the eight side and corner squares but on the central square the king goes crazy and turns into a red king, see A179596.
The sequence above corresponds to 24 red king vectors, i.e., A[5] vectors, with decimal values 27, 30, 51, 54, 57, 60, 90, 114, 120, 147, 150, 153, 156, 177, 180, 210, 216, 240, 282, 306, 312, 402, 408 and 432. These vectors lead for the corner squares to A015523 and for the central square to A179606.
This sequence belongs to a family of sequences with g.f. (1+2*x)/(1 - 3*x - k*x^2). Red king sequences that are members of this family are A007483 (k=2), A108981 (k=4), A152187 (k=5; this sequence), A154964 (k=6), A179602 (k=7) and A179598 (k=8). We observe that there is no red king sequence for k=3. Other members of this family are A036563 (k=-2), A054486 (k=-1), A084244 (k=0), A108300 (k=1) and A000351 (k=10).
Inverse binomial transform of A015449 (without the first leading 1).
(End)

Index entries for linear recurrences with constant coefficients, signature (3, 5).

Cf. A015523, A072263, A072264, A179606, A197189.

LinearRecurrence[{3,5},{1,5},40] (* Harvey P. Dale, May 03 2013 *)

G.f.: (1+2*x)/(1 - 3*x - 5*x^2).
Lim_{k->infinity} a(n+k)/a(k) = (A072263(n) + A015523(n)*sqrt(29))/2. - Johannes W. Meijer, Aug 01 2010
G.f.: G(0)*(1+2*x)/(2-3*x), where G(k) = 1 + 1/(1 - x*(29*k-9)/(x*(29*k+20) - 6/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 17 2013

A193729 Mirror of the triangle A193728.

Original entry on oeis.org

1, 1, 2, 3, 10, 8, 9, 42, 64, 32, 27, 162, 360, 352, 128, 81, 594, 1728, 2496, 1792, 512, 243, 2106, 7560, 14400, 15360, 8704, 2048, 729, 7290, 31104, 73440, 103680, 87552, 40960, 8192, 2187, 24786, 122472, 344736, 604800, 677376, 473088, 188416, 32768

Offset: 0

Clark Kimberling, Aug 04 2011

T(n, k) is obtained by reversing the rows of the triangle A193728.

Triangle T(n,k), read by rows, given by [1,2,0,0,0,0,...] DELTA [2,2,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 05 2011

			First six rows:
   1;
   1,   2;
   3,  10,    8;
   9,  42,   64,   32;
  27, 162,  360,  352,  128;
  81, 594, 1728, 2496, 1792, 512;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000420, A033999, A081038, A081294, A084938.

Cf. A108981, A133494, A193722, A193728, A247560.

function T(n, k) // T = A193729
  if k lt 0 or k gt n then return 0;
  elif n lt 2 then return k+1;
  else return 3*T(n-1, k) + 4*T(n-1, k-1);
  end if;
end function;
[T(n, k): k in [0..n], n in [0..12]]; // G. C. Greubel, Nov 28 2023

(* First program *)
z = 8; a = 1; b = 2; c = 2; d = 1;
p[n_, x_] := (a*x + b)^n ; q[n_, x_] := (c*x + d)^n
t[n_, k_] := Coefficient[p[n, x], x^k]; t[n_, 0] := p[n, x] /. x -> 0;
w[n_, x_] := Sum[t[n, k]*q[n + 1 - k, x], {k, 0, n}]; w[-1, x_] := 1
g[n_] := CoefficientList[w[n, x], {x}]
TableForm[Table[Reverse[g[n]], {n, -1, z}]]
Flatten[Table[Reverse[g[n]], {n, -1, z}]]  (* A193728 *)
TableForm[Table[g[n], {n, -1, z}]]
Flatten[Table[g[n], {n, -1, z}]]   (* A193729 *)
(* Second program *)
T[n_, k_]:= T[n, k]= If[k<0 || k>n, 0, If[n<2, k+1, 3*T[n-1,k] + 4*T[n -1, k-1]]];
Table[T[n, k], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Nov 28 2023 *)

def T(n, k): # T = A193729
    if (k<0 or k>n): return 0
    elif (n<2): return k+1
    else: return 3*T(n-1, k) + 4*T(n-1, k-1)
flatten([[T(n, k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Nov 28 2023

Let w(n,k) be the triangle of A193728, then the triangle in this sequence is given by w(n,n-k).
T(n,k) = 4*T(n-1,k-1) + 3*T(n-1,k) with T(0,0)=T(1,0)=1 and T(1,1)=2. - Philippe Deléham, Oct 05 2011
G.f.: (1-2*x-2*x*y)/(1-3*x-4*x*y). - R. J. Mathar, Aug 11 2015
From G. C. Greubel, Nov 28 2023: (Start)
T(n, 0) = A133494(n).
T(n, 1) = 2*A081038(n-1).
T(n, n) = A081294(n).
Sum_{k=0..n} T(n, k) = (1/7)*(4*[n=0] + 3*A000420(n)).
Sum_{k=0..n} (-1)^k * T(n, k) = A033999(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = (1/2)*[n=0] + A108981(n-1).
Sum_{k=0..floor(n/2)} (-1)^k * T(n-k, k) = (1/2)*[n=0] + A247560(n-1).
(End)

A124793 Numbers in a perpendicular plane intersecting a 3D clockwise spiral produced by powers of 2.

Original entry on oeis.org

1, 5, 35, 157, 1123, 5021, 35939, 160669, 1150051, 5141405, 36801635, 164524957, 1177652323, 5264798621, 37684874339, 168473555869, 1205915978851, 5391153787805, 38589311323235, 172516921209757, 1234857962343523

Offset: 1

Paolo P. Lava and Giorgio Balzarotti, Jun 27 2007

The general formula for powers of k integer is a(n) = k^((1/4)*(10*n - 7 - (-1)^n)) + k^((1/4)*(10*n - 1 + (-1)^n)) - a(n-1), with a(0)=1 and where k is an integer value. If we replace k with "i" or "-i" where i=sqrt(-1), we get a periodic complex sequence (period 8).

			Write powers of 2 in a sort of 3D clockwise spiral. After the initial 1 (2^0) move right till 2^1=2 (practically only one step); then move down till 2^2=4 (3,4); then left till 2^3=8 (5,6,7,8). When writing number 5 we are in the same column of 1 so 5 is the second number of the sequence. Then move up till 2^4=16. Then move up perpendicularly to the plane till 2^5=32 and again right till 2^6=64. The number 35 is in the sequence because it lies in the same line as 1 and 5. The process continues down, left, up, perpendicular, right and so on.

Charles R Greathouse IV, Table of n, a(n) for n = 1..1329
Index entries for linear recurrences with constant coefficients, signature (-1,32,32).

Cf. A108981, A001107.

P:=proc(n) local a,i,x,y; a:=1; print(a); for i from 1 by 1 to n do x:=1/4*(10*i-7-(-1)^i); y:=1/4*(10*i-1+(-1)^i); a:=2^x+2^y-a; print(a); od; end: P(100);

LinearRecurrence[{-1, 32, 32}, {1, 5, 35}, 25] (* Paolo Xausa, Feb 23 2024 *)

a(n) = 2^((1/4)*(10*n - 7 - (-1)^n)) + 2^((1/4)*(10*n - 1 + (-1)^n)) - a(n-1), with a(0)=1.
From Colin Barker, Jul 07 2012: (Start)
a(n) = -a(n-1) + 32*a(n-2) + 32*a(n-3).
G.f.: x*(1+2*x)*(1+4*x)/((1+x)*(1-32*x^2)). (End)
a(2n) = 3/31 + 19*32^n/124, a(2n+1) = -3/31 + 136*32^n/124. [R. J. Mathar, Jul 10 2012]

A140724 Period 10: 1, 5, 9, 7, 7, 9, 5, 1, 3, 3 repeated.

Original entry on oeis.org

1, 5, 9, 7, 7, 9, 5, 1, 3, 3, 1, 5, 9, 7, 7, 9, 5, 1, 3, 3, 1, 5, 9, 7, 7, 9, 5, 1, 3, 3, 1, 5, 9, 7, 7, 9, 5, 1, 3, 3, 1, 5, 9, 7, 7, 9, 5, 1, 3, 3, 1, 5, 9, 7, 7, 9, 5, 1, 3, 3, 1, 5, 9, 7, 7, 9, 5, 1, 3, 3, 1, 5, 9, 7, 7, 9, 5, 1, 3, 3, 1, 5, 9, 7, 7, 9, 5, 1, 3, 3, 1, 5, 9, 7, 7, 9, 5, 1, 3, 3, 1, 5, 9, 7, 7

Offset: 0

Paul Curtz, Jul 12 2008

The last digit of A108981(n).
Also the continued fraction of (290003+sqrt(240183699293))/652402.
Also the decimal expansion of 13073/81819.
The period contains each of the 5 odd digits twice.

Index entries for linear recurrences with constant coefficients, signature (2,-2,2,-2,1).

PadRight[{},120,{1,5,9,7,7,9,5,1,3,3}] (* Harvey P. Dale, Apr 23 2020 *)

13073/81819. \\ Charles R Greathouse IV, Jul 21 2015

a(n)+a(n+5) = 10 = A010692(n).
a(n) = a(n+10) .
a(10*k+9+i) = a(10*k+18-i) (palindromic).
a(n) = 2*a(n-1)-2*a(n-2)+2*a(n-3)-2*a(n-4)+a(n-5). G.f.: -(1+3*x+x^2-3*x^3+3*x^4)/ ((x-1) * (x^4-x^3+x^2-x+1)).

Edited by R. J. Mathar, Sep 07 2009

cp's OEIS Frontend

A126473 Number of strings over a 5 symbol alphabet with adjacent symbols differing by three or less.

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A122117 a(n) = 3*a(n-1) + 4*a(n-2), with a(0)=1, a(1)=2.

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A152187 a(n) = 3*a(n-1) + 5*a(n-2), with a(0)=1, a(1)=5.

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A193729 Mirror of the triangle A193728.

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A124793 Numbers in a perpendicular plane intersecting a 3D clockwise spiral produced by powers of 2.

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A140724 Period 10: 1, 5, 9, 7, 7, 9, 5, 1, 3, 3 repeated.

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A122117 a(n) = 3a(n-1) + 4a(n-2), with a(0)=1, a(1)=2.

A152187 a(n) = 3a(n-1) + 5a(n-2), with a(0)=1, a(1)=5.