A002450 -id:A002450 - cp's OEIS frontend

A061547 Number of 132 and 213-avoiding derangements of {1,2,...,n}.

1, 0, 1, 2, 6, 10, 26, 42, 106, 170, 426, 682, 1706, 2730, 6826, 10922, 27306, 43690, 109226, 174762, 436906, 699050, 1747626, 2796202, 6990506, 11184810, 27962026, 44739242, 111848106, 178956970, 447392426, 715827882, 1789569706, 2863311530, 7158278826

Offset: 0

Emeric Deutsch, May 16 2001

Or, number of permutations with no fixed points avoiding 213 and 132.
Number of derangements of {1,2,...,n} having ascending runs consisting of consecutive integers. Example: a(4)=6 because we have 234/1, 34/12, 34/2/1, 4/123, 4/3/12, 4/3/2/1, the ascending runs being as indicated. - Emeric Deutsch, Dec 08 2004
Let c be twice the sequence A002450 interlaced with itself (from the second term), i.e., c = 2*(0, 1, 1, 5, 5, 21, 21, 85, 85, 341, 341, ...). Let d be powers of 4 interlaced with the zero sequence: d = (1, 0, 4, 0, 16, 0, 64, 0, 256, 0, ...). Then a(n+1) = c(n) + d(n). - Creighton Dement, May 09 2005
Inverse binomial transform of A094705 (0, 1, 4, 15). - Paul Curtz, Jun 15 2008
Equals row sums of triangle A177993. - Gary W. Adamson, May 16 2010
a(n-1) is also the number of order preserving partial isometries (of an n-chain) of fix 1 (fix of alpha equals the number of fixed points of alpha). - Abdullahi Umar, Dec 28 2010
a(n+1) <= A218553(n) is also the Moore lower bound on the order of a (5,n)-cage. - Jason Kimberley, Oct 31 2011
For n > 0, a(n) is the location of the n-th new number to make a first appearance in A087230. E.g., the 17th number to make its first appearance in A087230 is 18 and this occurs at A087230(43690) and a(17)=43690. - K D Pegrume, Jan 26 2022
Position in A002487 of 2 adjacent terms of A000045. E.g., 3/5 at 10, 5/8 at 26, 8/13 at 42, ... - Ed Pegg Jr, Dec 27 2022

			a(4)=6 because the only 132 and 213-avoiding permutations of {1,2,3,4} without fixed points are: 2341, 3412, 3421, 4123, 4312 and 4321.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
F. Al-Kharousi, R. Kehinde and A. Umar, Combinatorial results for certain semigroups of partial isometries of a finite chain, The Australasian Journal of Combinatorics, Volume 58 (3) (2014), 363-375.
J. Brillhart and P. Morton, A case study in mathematical research: the Golay-Rudin-Shapiro sequence, Amer. Math. Monthly, 103 (1996) 854-869 (contains the sequence of the odd-subscripted terms and that of the even-subscripted terms).
Emeric Deutsch, Derangements That Don't Rise Too Fast: 10902, Amer. Math. Monthly, Vol. 110, No. 7 (2003), pp. 639-640.
K. Dilcher and K. B. Stolarsky, Stern polynomials and double-limit continued fractions, Acta Arithmetica 140 (2009), 119-134
R. Kehinde and A. Umar, On the semigroup of partial isometries of a finite chain, arXiv:1101.0049 [math.GR], 2010.
T. Mansour and A. Robertson, Refined restricted permutations avoiding subsets of patterns of length three, arXiv:math/0204005 [math.CO], 2002
Index entries for linear recurrences with constant coefficients, signature (1,4,-4).

Cf. A000166, A020988, A020989, A068018.
Cf. A177993. - Gary W. Adamson, May 16 2010
Cf. A183158, A183159. - Abdullahi Umar, Dec 28 2010
Moore lower bound on the order of a (k,g) cage: A198300 (square); rows: A000027 (k=2), A027383 (k=3), A062318 (k=4), this sequence (k=5), A198306 (k=6), A198307 (k=7), A198308 (k=8), A198309 (k=9), A198310 (k=10), A094626 (k=11); columns: A020725 (g=3), A005843 (g=4), A002522 (g=5), A051890 (g=6), A188377 (g=7). - Jason Kimberley, Oct 31 2011

[(3/8)*2^n +(1/24)*(-2)^n - 2/3: n in [1..35]]; // Vincenzo Librandi, Aug 13 2011

A061547:=n->ceil(abs((3/8)*2^n +(1/24)*(-2)^n - 2/3)); seq(A061547(n), n=1..30); # Wesley Ivan Hurt, Apr 03 2014

f[n_] := (9*2^(n-3) - (-2)^(n-3) - 2)/3; Array[f, 32] (* Robert G. Wilson v, Aug 13 2011 *)

a(n)=(3/8)*2^n+(1/24)*(-2)^n-2/3 \\ Charles R Greathouse IV, Sep 24 2015

a(n) = (3/8)*2^n + (1/24)*(-2)^n - 2/3 for n>=1.
a(n) = 4*a(n-2) + 2, a(0)=1, a(1)=0, a(2)=1.
G.f: (5*z^3-3*z^2-z+1)/((z-1)*(4*z^2-1)).
a(n) = A020989((n-2)/2) for n=2, 4, 6, ... and A020988((n-3)/2) for n=3, 5, 7, ... .
a(n+1)-2*a(n) = A078008 signed. Differences: doubled A000302. - Paul Curtz, Jun 15 2008
a(2i+1) = 2*Sum_{j=0..i-1} 4^j = string "2"^i read in base 4.
a(2i+2) = 4^i + 2*Sum_{j=0..i-1} 4^j = string "1"*"2"^i read in base 4.
a(n+2) = Sum_{k=0..n} A144464(n,k)^2 = Sum_{k=0..n} A152716(n,k). - Philippe Deléham and Michel Marcus, Feb 26 2014
a(2*n-1) = A176965(2*n), a(2*n) = A176965(2*n-1) for n>0. - Yosu Yurramendi, Dec 23 2016
a(2*n-1) = A020988(k-1), a(2*n)= A020989(n-1) for n>0. - Yosu Yurramendi, Jan 03 2017
a(n+2) = 2*A086893(n), n > 0. - Yosu Yurramendi, Mar 07 2017
E.g.f.: (15 - 8*cosh(x) + 5*cosh(2*x) - 8*sinh(x) + 4*sinh(2*x))/12. - Stefano Spezia, Apr 07 2022

a(0)=1 prepended by Alois P. Heinz, Jan 27 2022

A218736 a(n) = (33^n - 1)/32.

Original entry on oeis.org

0, 1, 34, 1123, 37060, 1222981, 40358374, 1331826343, 43950269320, 1450358887561, 47861843289514, 1579440828553963, 52121547342280780, 1720011062295265741, 56760365055743769454, 1873092046839544391983, 61812037545704964935440, 2039797239008263842869521

Offset: 0

M. F. Hasler, Nov 04 2012

Partial sums of powers of 33 (A009977).

Vincenzo Librandi, Table of n, a(n) for n = 0..600
Index entries related to partial sums.
Index entries related to q-numbers.
Index entries for linear recurrences with constant coefficients, signature (34,-33).

Cf. similar sequences of the form (k^n-1)/(k-1): A000225, A003462, A002450, A003463, A003464, A023000, A023001, A002452, A002275, A016123, A016125, A091030, A135519, A135518, A131865, A091045, A218721, A218722, A064108, A218724-A218734, A132469, A218737-A218753, A133853, A094028, A218723.

[n le 2 select n-1 else 34*Self(n-1)-33*Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 07 2012

LinearRecurrence[{34, -33}, {0, 1}, 30] (* Vincenzo Librandi, Nov 07 2012 *)

A218736(n):=(33^n-1)/32$
makelist(A218736(n),n,0,30); /* Martin Ettl, Nov 05 2012 */

PARI
```
A218736(n)=33^n>>5
```

From Vincenzo Librandi, Nov 07 2012: (Start)
G.f.: x/((1 - x)*(1 - 33*x)).
a(n) = 34*a(n-1) - 33*a(n-2).
a(n) = floor(33^n/32). (End)
E.g.f.: exp(x)*(exp(32*x) - 1)/32. - Stefano Spezia, Mar 24 2023

A086893 a(n) is the index of F(n+1) at the unique occurrence of the ordered pair of reversed consecutive terms (F(n+1),F(n)) in Stern's diatomic sequence A002487, where F(k) denotes the k-th term of the Fibonacci sequence A000045.

Original entry on oeis.org

1, 3, 5, 13, 21, 53, 85, 213, 341, 853, 1365, 3413, 5461, 13653, 21845, 54613, 87381, 218453, 349525, 873813, 1398101, 3495253, 5592405, 13981013, 22369621, 55924053, 89478485, 223696213, 357913941, 894784853, 1431655765, 3579139413

Offset: 1

John W. Layman, Sep 18 2003

If the Fibonacci pairs are kept in the natural order (F(n),F(n+1)), it appears that the first term of the pair occurs in A002487 at the index given by A061547(n).
Equals row sums of triangle A177954. - Gary W. Adamson, May 15 2010
Starting at n=3, begin subtracting from (2^(n-1)-1)/2^(n-1): 3/4 - 1/2 = 1/4 with 1+4=5=a(3); 7/8 - 1/4 = 5/8 with 5+8=13=a(4); 15/16 - 5/8 = 5/16 with 5+16=21= a(5); 31/32 - 5/16 = 21/32 with 21+32=53=a(6); 63/64 - 21/32 = 21/64 with 21+64=85=a(7) and so on. For n odd in the first fraction (2^(n-1)-1)/2^(n-1), the result approaches 1/3, and for n even in the first fraction, the result approaches 2/3. - J. M. Bergot, May 08 2015
Also, the decimal representation of the x-axis, from the left edge to the origin, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 678", based on the 5-celled von Neumann neighborhood, initialized with a single black (ON) cell at stage zero. See A283641. - Robert Price, Mar 12 2017

			A002487 begins 0,1,1,2,1,3,2,... with offset 0. Thus a(1)=1 since (F(2),F(1)) = (1,1) occurs at term 1 of A002487. Similarly, a(2)=3 and a(3)=5, since (F(3),F(2))=(2,1) occurs at term 3 and (F(4),F(3))=(3,2) at term 5 of A002487.

A. J. Macfarlane, Generating functions for integer sequences defined by the evolution of cellular automata..., Fig. 12.

Cf. A000045, A002487, A061547, A077954, A283641, A372286, A372352.
Interleaving of A002450\{0} and A072197.
Positive terms of A096773 in ascending order.
Partial sums of A158302.

[2^(n-1)*(3-(-1)^n/3)-1/3: n in [0..35]]; // Vincenzo Librandi, May 09 2015

f[n_] := Module[{a = 1, b = 0, m = n}, While[m > 0, If[OddQ@ m, b = a + b, a = a + b]; m = Floor[m/2]]; b]; a = Table[f[n], {n, 0, 10^6}]; b = Reverse /@ Partition[Map[Fibonacci, Range[Ceiling@ Log[GoldenRatio, Max@ a] + 1]], 2, 1]; Map[If[Length@ # > 0, #[[1, 1]] - 1, 0] &@ SequencePosition[a, #] &, b] (* Michael De Vlieger, Mar 15 2017, Version 10.1, after Jean-François Alcover at A002487 *)

a(n)=if(n%2,2^(n+1),2^(n+1)+2^(n-1))\3 \\ Charles R Greathouse IV, May 08 2015

def A086893(n): return (1<Chai Wah Wu, Apr 29 2024

It appears that a(n)=(4^((n+1)/2)-1)/3 if n is odd and a(n)=(a(n-1)+a(n+1))/2 if n is even.
G.f.: (1+2*x-2*x^2)/((1-x)*(1-4*x^2)); a(n) = 2^(n-1)(3-(-1)^n/3)-1/3 (offset 0); a(n) = Sum{k=0..n+1, 4^floor(k/2)/2} (offset 0); a(2n) = A002450(n+1) (offset 0); a(2n+1) = A072197(n) (offset 0). - Paul Barry, May 21 2004
a(n+2) = 4*a(n) + 1, a(1) = 1, a(2) = 3, n > 0. - Yosu Yurramendi, Mar 07 2017
a(n+1) = a(n) + A158302(n), a(1) = 1, n > 0. - Yosu Yurramendi, Mar 07 2017

More terms from Paul Barry, May 21 2004

A072197 a(n) = 4*a(n-1) + 1 with a(0) = 3.

Original entry on oeis.org

3, 13, 53, 213, 853, 3413, 13653, 54613, 218453, 873813, 3495253, 13981013, 55924053, 223696213, 894784853, 3579139413, 14316557653, 57266230613, 229064922453, 916259689813, 3665038759253, 14660155037013, 58640620148053, 234562480592213, 938249922368853, 3752999689475413

Offset: 0

N. Rathankar (rathankar(AT)yahoo.com), Jul 03 2002

Let A be the Hessenberg matrix of order n, defined by: A[1, j] = 1, A[i, i] := 2, (i > 1), A[i, i - 1] = -1, and A[i, j] = 0 otherwise. Then, for n >= 1, a(n - 1) = (-1)^n*charpoly(A, -2). - Milan Janjic, Jan 26 2010
Numbers whose binary representation is 11 together with n times 01. For example, 213 = 11010101 (2). - Omar E. Pol, Nov 22 2012
The Collatz-function starting with a(n) will terminate at 1 after 2*n + 7 steps. This is because 3*a(n) + 1 = 5*2^(2n + 1), and the Collatz-function starting with 5 terminates at 1 after 5 additional steps.  So for example, a(2) = 53; Collatz sequence starting with 53 follows: 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1 (11 steps). - Bob Selcoe, Apr 03 2015
a(n) is also the sum of the numerator and denominator of the binary fractions 0.1, 0.101, 0.10101, 0.1010101... Thus 0.1 = 1/2 with 1 + 2 = 3, 0.101 = 1/2 + 1/8 = 5/8 with 5 + 8 = 13; 0.10101 = 1/2 + 1/8 + 1/32 = 21/32 with 21 + 32 = 53. - J. M. Bergot, Sep 28 2016
a(n), for n >= 2, is also the smallest odd number congruent to 5 modulo 8 for which the modified reduced Collatz map given in A324036 has n consecutive extra steps compared to the reduced Collatz map given in A075677. - Nicolas Vaillant, Philippe Delarue, Wolfdieter Lang, May 09 2019

			a(1) = 13 because a(0) = 3 and 4 * 3 + 1 = 13.
a(2) = 53 because a(1) = 13 and 4 * 13 + 1 = 53.
a(3) = 213 because a(2) = 53 and 4 * 53 + 1 = 213.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Petro Kosobutskyy and Volodymyr Karkulovskyy, Recurrence and structuring of sequences of transformations 3n+ 1 as arguments for confirmation of the Collatz hypothesis, Comp. Des. Sys., Theor. Practice (2023) Vol. 5, No. 1, pp. 28-33. See p. 29.
Petro Kosobutskyy, Anastasiia Yedyharova, and Taras Slobodzyan, From Newton's binomial and Pascal's triangle to Collatz's problem, Comp. Des. Sys., Theor. Practice (2023) Vol. 5, No. 1, pp. 121-127.
Index entries for linear recurrences with constant coefficients, signature (5,-4).
Index entries for sequences related to 3x+1 (or Collatz) problem.

Cf. A000302, A002066 (first differences), A002450, A007583, A020988, A075677, A178415, A324036, A347834.

[(10*4^n-1)/3: n in [0..30]]; // Vincenzo Librandi, Oct 31 2011

A072197:=n->(10*4^n - 1)/3: seq(A072197(n), n=0..30); # Wesley Ivan Hurt, Sep 29 2016

Table[(10(4^n) - 1)/3, {n, 0, 19}] (* Alonso del Arte, Nov 22 2012 *)
NestList[4#+1&,3,30] (* Harvey P. Dale, Mar 09 2019 *)

a(n)=10*4^n\3 \\ Charles R Greathouse IV, Apr 06 2016

Vec((3-2*x)/((1-x)*(1-4*x)) + O(x^30)) \\ Colin Barker, Sep 28 2016

a(n) = (10*4^n - 1)/3 = 10*A002450(n) + 3. - Henry Bottomley, Dec 02 2002
a(n) = 5*a(n-1) - 4*a(n-2), n > 1. - Vincenzo Librandi, Oct 31 2011
a(n) = 2^(2*(n + 1)) - (2^(2*n + 1) + 1)/3 = A000302(n + 1) - A007583(n). - Vladimir Pletser, Apr 12 2014
a(n) = (5*2^(2*n + 1) - 1)/3. - Bob Selcoe, Apr 03 2015
G.f.: (3-2*x)/((1-x)*(1-4*x)). - Colin Barker, Sep 28 2016
a(n) = A020988(n) + A020988(n+1) + 1 = 2*(A002450(n) + A002450(n+1)) + 1. - Yosu Yurramendi, Jan 24 2017
a(n) = A002450(n+1) + 2^(2*n+1). - Adam Michael Bere, May 13 2021
a(n) = a(n-1) + 5*2^(2*n-1), for n >= 1, with a(0) = 3. - Wolfdieter Lang, Aug 16 2021
a(n) = A178415(2,n+1) = A347834(2,n), arrays, for n >= 0. - Wolfdieter Lang, Nov 29 2021
E.g.f.: exp(x)*(10*exp(3*x) - 1)/3. - Elmo R. Oliveira, Apr 02 2025

More terms from Henry Bottomley, Dec 02 2002

A020989 a(n) = (5*4^n - 2)/3.

Original entry on oeis.org

1, 6, 26, 106, 426, 1706, 6826, 27306, 109226, 436906, 1747626, 6990506, 27962026, 111848106, 447392426, 1789569706, 7158278826, 28633115306, 114532461226, 458129844906, 1832519379626, 7330077518506, 29320310074026, 117281240296106, 469124961184426

Offset: 0

N. J. A. Sloane

Let Zb[n](x) = polynomial in x whose coefficients are the corresponding digits of index n in base b. Then Z2[(5*4^k-2)/3](1/tau) = 1. - Marc LeBrun, Mar 01 2001
a(n)=number of derangements of [2n+2] with runs consisting of consecutive integers. E.g., a(1)=6 because the derangements of {1,2,3,4} with runs consisting of consecutive integers are 4|123, 34|12, 4|3|12, 4|3|2|1, 234|1 and 34|2|1 (the bars delimit the runs). - Emeric Deutsch, May 26 2003
Sum of n-th row of triangle of powers of 4: 1; 1 4 1; 1 4 16 4 1; 1 4 16 64 16 4 1; ... - Philippe Deléham, Feb 22 2014

			a(0) = 1;
a(1) = 1 + 4 + 1 = 6;
a(2) = 1 + 4 + 16 + 4 + 1 = 26;
a(3) = 1 + 4 + 16 + 64 + 16 + 4 + 1 = 106; etc. - _Philippe Deléham_, Feb 22 2014

Clifford A. Pickover, A Passion for Mathematics, John Wiley & Sons, Inc., 2005, at pp. 104 and 311 (for "Mr. Zanti's ants").

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
John Brillhart and Patrick Morton, Über Summen von Rudin-Shapiroschen Koeffizienten, Illinois Journal of Mathematics, volume 22, issue 1, 1978, pages 126-148. See Satz 9(a) page 132 and Satz 21 page 144 m_k = a(k).
John Brillhart and Patrick Morton, A case study in mathematical research: the Golay-Rudin-Shapiro sequence, Amer. Math. Monthly, 103 (1996) 854-869, see page 858 m_k = a(k).
Kevin Ryde, Iterations of the Alternate Paperfolding Curve, see index "m_k".
Index entries for linear recurrences with constant coefficients, signature (5,-4).

Cf. A002450, A020988, A112468.

A column of A119726.

[(5*4^n-2)/3: n in [0..25]]; // Vincenzo Librandi, Jul 24 2011

NestList[4#+2&,1,25] (* Harvey P. Dale, Jul 23 2011 *)

a(n)=(5*4^n-2)/3 \\ Charles R Greathouse IV, Jul 02 2013

a(0) = 1, a(n) = 4*a(n-1) + 2; a(n) = a(n-1)+ 5*{4^(n-1)}; - Amarnath Murthy, May 27 2001
G.f.: (1+x)/((1-4*x)*(1-x)). - Zerinvary Lajos, Jan 11 2009; adapted to offset by Philippe Deléham, Feb 22 2014
a(n) = 5*a(n-1) - 4*a(n-2), a(0) = 1, a(1) = 6. - Philippe Deléham, Feb 22 2014
a(n) = Sum_{k=0..n} A112468(n,k)*5^k. - Philippe Deléham, Feb 22 2014
a(n) = (A020988(n) + A020988(n+1))/2. - Yosu Yurramendi, Jan 23 2017
a(n) = A002450(n) + A002450(n+1). - Yosu Yurramendi, Jan 24 2017
a(n) = 10*A020988(n-1) + 6. - Yosu Yurramendi, Feb 19 2017
E.g.f.: exp(x)*(5*exp(3*x) - 2)/3. - Stefano Spezia, Apr 10 2022

A175046 Write n in binary, then increase each run of 0's by one 0, and increase each run of 1's by one 1. a(n) is the decimal equivalent of the result.

Original entry on oeis.org

3, 12, 7, 24, 51, 28, 15, 48, 99, 204, 103, 56, 115, 60, 31, 96, 195, 396, 199, 408, 819, 412, 207, 112, 227, 460, 231, 120, 243, 124, 63, 192, 387, 780, 391, 792, 1587, 796, 399, 816, 1635, 3276, 1639, 824, 1651, 828, 415, 224, 451, 908, 455, 920, 1843, 924

Offset: 1

Leroy Quet, Dec 02 2009

A318921 expands the runs in a similar way, and A318921(a(n)) = A001477(n). - Andrew Weimholt, Sep 08 2018
From Chai Wah Wu, Nov 18 2018: (Start)
Let f(k) = Sum_{i=2^k..2^(k+1)-1} a(i), i.e., the sum ranges over all numbers with a (k+1)-bit binary expansion. Thus f(0) = a(1) = 3 and f(1) = a(2) + a(3) = 19.
Then f(k) = 20*6^(k-1) - 2^(k-1) for k > 0.
Proof: by summing over the recurrence relations for a(n) (see formula section), we get f(k+2) = Sum_{i=2^k..2^(k+1)-1} (f(4i) + f(4i+1) + f(4i+2) + f(4i+3)) = Sum_{i=2^k..2^(k+1)-1} (6*a(2i) + 6*a(2i+1) + 4) = 6*f(k+1) + 2^(k+2). Solving this first-order recurrence relation with the initial condition f(1) = 19 shows that f(k) = 20*6^(k-1)-2^(k-1) for k > 0.
(End)

			6 in binary is 110. Increase each run by one digit to get 11100, which is 28 in decimal. So a(6) = 28.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
N. J. A. Sloane, Coordination Sequences, Planing Numbers, and Other Recent Sequences (II), Experimental Mathematics Seminar, Rutgers University, Jan 31 2019, Part I, Part 2, Slides. (Mentions this sequence)
Chai Wah Wu, Record values in appending and prepending bitstrings to runs of binary digits, arXiv:1810.02293 [math.NT], 2018.

Cf. A175047, A175048, A324127 (partial sums).
Cf. also A030308, A007088, A182512, A318921.
For records see A319422, A319423, A319424.

import Data.List (group)
a175046 = foldr (\b v -> 2 * v + b) 0 .
          concatMap (\bs@(b:_) -> b : bs) . group . a030308_row
-- Reinhard Zumkeller, Jul 05 2013

a[n_] := (Append[#, #[[1]]]& /@ Split[IntegerDigits[n, 2]]) // Flatten // FromDigits[#, 2]&;
Array[a, 60] (* Jean-François Alcover, Nov 12 2018 *)

A175046(n)={for(i=2,#n=binary (n*2+bittest (n,0)),n[i]!=n[i-1]&&n[i-1]*=[1,1]);fromdigits(concat(n),2)} \\ M. F. Hasler, Sep 08 2018

from re import split
def A175046(n):
    return int(''.join(d+'1' if '1' in d else d+'0' for d in split('(0+)|(1+)',bin(n)[2:]) if d != '' and d != None),2) # Chai Wah Wu, Sep 24 2018

def a(n):
    b = bin(n)[2:]
    return int(b.replace("01", "001").replace("10", "110") + b[-1], 2)
print([a(n) for n in range(1, 55)]) # Michael S. Branicky, Dec 07 2021

2n+1 <= a(n) < 2*(n+1/n)^2; a(n) mod 4 = 3*(n mod 2). - M. F. Hasler, Sep 08 2018
a(n) <= (9*n^2 + 12*n)/5, with equality iff n = (2/3)*(4^k-1) = A182512(k) for some k, i.e., n = 10101...10 in binary. - Conjectured by N. J. A. Sloane, Sep 09 2018, proved by M. F. Hasler, Sep 12 2018
From M. F. Hasler, Sep 12 2018: (Start)
Proof of N. J. A. Sloane's formula: For given (binary) length L(n) = floor(log_2(n)+1), the length of a(n) is maximal, L(a(n)) = 2*L(n), if and only if n's bits are alternating, i.e., n in A020988 (if even) or in A002450 (if odd).
For n = A020988(k) (= k times '10' in base 2) = (4^k - 1)*2/3, one has a(n) = A108020(k) (= k times '1100' in base 2) = (16^k - 1)*4/5. This yields a(n)/n = (4^k + 1)*6/5 = (n*9 + 12)/5, i.e., the given upper bound.
For n = A002450(k) = (4^k - 1)/3, one gets a(n) = A182512(k) = (16^k - 1)/5, whence a(n)/n = (4^k + 1)*3/5 = (n*9 + 6)/5, smaller than the bound.
If L(a(n)) < 2 L(n) - 1, then log_2(a(n)) < floor(log_2(a(n))+1) = L(a(n)) <= 2*L(n) - 2 = 2*floor(log_2(n)+1)-2 = 2*floor(log_2(n)) <= 2*log_2(n), whence a(n) < n^2.
It remains to consider the case L(a(n)) = 2 L(n) - 1. There are two possibilities:
If n = 10..._2, then n >= 2^(L(n)-1) and a(n) = 1100..._2 < 1101_2 * 2^(L(a(n))-4) = 13*2^(2*L(n)-5), so a(n)/n^2 < 13*2^(-5+2) = 13/8 = 1.625 < 9/5 = 1.8.
If n = 11..._2, then n >= 3*2^(L(n)-2) and a(n) = 111..._2 < 2^L(a(n)) = 2^(2*L(n)-1), so a(n)/n^2 < 2^(-1+4)/9 = 8/9 < 1 < 9/5.
This shows that a(n)/n^2 <= 9/5 + 12/(5*n) always holds, with equality iff n is in A020988; and a(n)/n^2 < 13/8 if n is not in A020988 or A002450. (End)
From M. F. Hasler, Sep 10 2018: (Start)
Right inverse of A318921: A318921 o A175046 = id (= A001477).
a(A020988(k)) = A108020(k); a(A002450(k)) = A182512(k); a(A000225(k)) = A000225(k+1) (achieves the lower bound a(n) >= 2n + 1) for all k >= 0. (End)
From David A. Corneth, Sep 20 2018: (Start)
a(4*k)   = 2*a(2*k).
a(4*k+1) = 4*a(2*k) + 3.
a(4*k+2) = 4*a(2*k+1).
a(4*k+3) = 2*a(2*k+1) + 1. (End)

Extended by Ray Chandler, Dec 18 2009

A022168 Triangle of Gaussian binomial coefficients [ n,k ] for q = 4.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 21, 21, 1, 1, 85, 357, 85, 1, 1, 341, 5797, 5797, 341, 1, 1, 1365, 93093, 376805, 93093, 1365, 1, 1, 5461, 1490853, 24208613, 24208613, 1490853, 5461, 1, 1, 21845, 23859109, 1550842085, 6221613541

Offset: 0

N. J. A. Sloane

The coefficients of the matrix inverse are apparently given by T^(-1)(n,k) = (-1)^n*A157784(n,k). - R. J. Mathar, Mar 12 2013

			Triangle begins:
  1;
  1,    1;
  1,    5,       1;
  1,   21,      21,        1;
  1,   85,     357,       85,        1;
  1,  341,    5797,     5797,      341,       1;
  1, 1365,   93093,   376805,    93093,    1365,    1;
  1, 5461, 1490853, 24208613, 24208613, 1490853, 5461, 1;

F. J. MacWilliams and N. J. A. Sloane, The Theory of Error-Correcting Codes, Elsevier-North Holland, 1978, p. 698.

T. D. Noe, Rows n=0..50 of triangle, flattened
R. Mestrovic, Lucas' theorem: its generalizations, extensions and applications (1878--2014), arXiv preprint arXiv:1409.3820 [math.NT], 2014.
Kent E. Morrison, Integer Sequences and Matrices Over Finite Fields, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.1.
M. Sved, Gaussians and binomials, Ars. Combinatoria, 17A (1984), 325-351. (Annotated scanned copy)
Index entries for sequences related to Gaussian binomial coefficients

Cf. A006118 (row sums), A002450 (k=1), A006105 (k=2), A006106 (k=3).

A022168 := proc(n,m)
        A027637(n)/A027637(n-m)/A027637(m) ;
end proc: # R. J. Mathar, Nov 14 2011

gaussianBinom[n_, k_, q_] := Product[q^i - 1, {i, n}]/Product[q^j - 1, {j, n - k}]/Product[q^l - 1, {l, k}]; Column[Table[gaussianBinom[n, k, 4], {n, 0, 8}, {k, 0, n}], Center] (* Alonso del Arte, Nov 14 2011 *)
Table[QBinomial[n,k,4], {n,0,10}, {k,0,n}]//Flatten (* or *) q:= 4; T[n_, 0]:= 1; T[n_,n_]:= 1; T[n_,k_]:= T[n,k] = If[k < 0 || n < k, 0, T[n-1, k -1] +q^k*T[n-1,k]]; Table[T[n,k], {n,0,10}, {k,0,n}] // Flatten  (* G. C. Greubel, May 27 2018 *)

{q=4; T(n,k) = if(k==0,1, if (k==n, 1, if (k<0 || nG. C. Greubel, May 27 2018

T(n,k) = T(n-1,k-1) + q^k * T(n-1,k). - Peter A. Lawrence, Jul 13 2017

G.f. of column k: x^k * exp( Sum_{j>=1} f((k+1)*j)/f(j) * x^j/j ), where f(j) = 4^j - 1. - Seiichi Manyama, May 09 2025

A069294 Number of n X 3 binary arrays with a path of adjacent 1's from upper left corner to anywhere in right hand column.

Original entry on oeis.org

12, 110, 926, 7556, 60920, 488860, 3915640, 31340216, 250769592, 2006308480, 16050948896, 128409116176, 1027277763840, 8218237436320, 65745948074080, 525967738606656, 4207742397091072, 33661940724484800, 269295530702399616

Offset: 2

R. H. Hardin, Mar 14 2002

G. C. Greubel, Table of n, a(n) for n = 2..1001
Index entries for linear recurrences with constant coefficients, signature (12,-34,14,16).

Cf. n X 2 A002450, n X 4 A069295, n X 5 A069296, n X 6 A069297, n X 7 A069298, n X 8 A069299, n X 9 A069300, n X 10 A069301, n X 11 A069302, n X 12 A069303, n X 13 A069304, n X 14 A069305, read by rows A069306-A069320.

I:=[12, 110, 926, 7556]; [n le 4 select I[n] else 12*Self(n-1) - 34*Self(n-2) +14*Self(n-3) + 16*Self(n-4): n in [1..30]]; // G. C. Greubel, Apr 22 2018

LinearRecurrence[{12, -34, 14, 16}, {12, 110, 926, 7556}, 50] (* G. C. Greubel, Apr 22 2018 *)

a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; 16,14,-34,12]^(n-2)*[12;110;926;7556])[1,1] \\ Charles R Greathouse IV, May 10 2016

x='x+O('x^30); Vec(2*x^2*(6-17*x+7*x^2+8*x^3)/((1-8*x)*(2*x^3 +2*x^2-4*x+1))) \\ G. C. Greubel, Apr 22 2018

G.f.: 2*x^2*(6-17*x+7*x^2+8*x^3)/(1-8*x)/(2*x^3+2*x^2-4*x+1). - Vladeta Jovovic, Jul 02 2003

A069305 Number of n X 14 binary arrays with a path of adjacent 1's from upper left corner to anywhere in right hand column.

Original entry on oeis.org

195025, 9893561055, 306680449108553, 7502586497280931603, 160836875185955529578497, 3180073668375520890944381755, 59642420242861089993013318912175, 1078941052043395111216099344387994533, 19026742282555500292841393785480166859877

Offset: 2

R. H. Hardin, Mar 14 2002

nonn

Cf. n X 2 A002450, n X 3 A069294, n X 4 A069295, n X 5 A069296, n X 6 A069297, n X 7 A069298, n X 8 A069299, n X 9 A069300, n X 10 A069301, n X 11 A069302, n X 12 A069303, n X 13 A069304, read by rows A069306-A069320.

a(6)-a(10) from Sean A. Irvine, Apr 26 2024

A169707 Total number of ON cells at stage n of two-dimensional cellular automaton defined by "Rule 750" using the von Neumann neighborhood.

Original entry on oeis.org

1, 5, 9, 21, 25, 37, 57, 85, 89, 101, 121, 149, 169, 213, 281, 341, 345, 357, 377, 405, 425, 469, 537, 597, 617, 661, 729, 805, 889, 1045, 1241, 1365, 1369, 1381, 1401, 1429, 1449, 1493, 1561, 1621, 1641, 1685, 1753, 1829, 1913, 2069, 2265, 2389, 2409, 2453, 2521

Offset: 1

N. J. A. Sloane, Apr 17 2010

Square grid, 4 neighbors per cell (N, E, S, W cells), turn ON iff exactly 1 or 3 neighbors are ON; once ON, cells stay ON.
The terms agree with those of A246335 for n <= 11, although the configurations are different starting at n = 7. - N. J. A. Sloane, Sep 21 2014
Offset 1 is best for giving a formula for a(n), although the Maple and Mathematica programs index the states starting at state 0.
It appears that this shares infinitely many terms with both A162795 and A147562, see Formula section and Example section. - Omar E. Pol, Feb 19 2015

			Divides naturally into blocks of sizes 1,2,4,8,16,...:
1,
5, 9,
21, 25, 37, 57,
85, 89, 101, 121, 149, 169, 213, 281, <- terms 8 through 15
341, 345, 357, 377, 405, 425, 469, 537, 597, 617, 661, 729, 805, 889, 1045, 1241,
1365, 1369, 1381, 1401, 1429, 1449, 1493, 1561, 1621, 1641, 1685, 1753, 1829, 1913, 2069, 2265, 2389, 2409, 2453, 2521, ...
From _Omar E. Pol_, Feb 18 2015: (Start)
Also, written as an irregular triangle T(j,k), k>=1, in which the row lengths are the terms of A011782:
1;
5;
9,   21;
25,  37,   57,  85;
89,  101, 121, 149, 169, 213, 281, 341;
345, 357, 377, 405, 425, 469, 537, 597, 617, 661, 729, 805, 889, 1045, 1241, 1365;
The right border gives the positive terms of A002450.
It appears that T(j,k) = A162795(j,k) = A147562(j,k), if k is a power of 2, for example: it appears that the three mentioned triangles only share the elements from the columns 1, 2, 4, 8, 16, ...
(End)

S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 928.

N. J. A. Sloane and Vincenzo Librandi, Table of n, a(n) for n = 1..513, computed from the definition, not using the conjectured formula. [The first 200 terms were found by _Vincenzo Librandi_]
David Applegate, The movie version
David Applegate, Omar E. Pol and N. J. A. Sloane, The Toothpick Sequence and Other Sequences from Cellular Automata, Congressus Numerantium, Vol. 206 (2010), 157-191. [There is a typo in Theorem 6: (13) should read u(n) = 4.3^(wt(n-1)-1) for n >= 2.], arXiv:1004.3036 [math.CO].
N. J. A. Sloane, Table of n, a(n) for n=1..8192 assuming the recurrence is correct. Note that these terms are merely conjectures.
N. J. A. Sloane, Illustration of first 24 generations
N. J. A. Sloane, Illustration of first 24 generations of the NW-NE-SE-SW version of this CA
N. J. A. Sloane, Illustration of generation 7 of the NW-NE-SE-SW version of this CA (containing a(7) = 57 ON cells)
N. J. A. Sloane, On the Number of ON Cells in Cellular Automata, arXiv:1503.01168 [math.CO], 2015.
N. J. A. Sloane, Catalog of Toothpick and Cellular Automata Sequences in the OEIS
Index entries for sequences related to cellular automata

Cf. A169708 (first differences), A147562, A147582, A169648, A169649, A169709, A169710, A246333, A246334, A246335, A246336, A253098 (partial sums).
Cf. also A162795, A150552, A160104, A170903, A048645, A187200.
See A253088 for the analogous CA using Rule 750 and a 9-celled neighborhood.

(Maple program that uses the actual definition of the automaton, rather than the (conjectured) formula, from N. J. A. Sloane, Feb 15 2015):
# Count terms in a polynomial:
C := f->`if`(type(f, `+`), nops(f), 1);
# Replace all nonzero coeffts by 1:
bool := proc(f) local ix, iy, f2, i, t1, t2, A;
f2:=expand(f);
if whattype(f) = `+` then
t1:=nops(f2); A:=0;
for i from 1 to t1 do t2:=op(i, f2); ix:=degree(t2, x); iy:=degree(t2, y);
A:=A+x^ix*y^iy; od: A;
else ix:=degree(f2, x); iy:=degree(f2, y); x^ix*y^iy;
fi;
end;
# a loop that produces M steps of A169707 and A169708:
M:=20;
F:=x*y+x/y+1/x*y+1/x/y mod 2;
GG[0]:=1;
for n from 1 to M do dd[n]:=expand(F*GG[n-1]) mod 2;
GG[n]:=bool(GG[n-1]+dd[n]);
lprint(n,C(GG[n]), C(GG[n]-GG[n-1])); od:

Map[Function[Apply[Plus,Flatten[ #1]]], CellularAutomaton[{ 750, {2,{{0,2,0},{2,1,2},{0,2,0}}},{1,1}},{{{1}},0},100]]
ArrayPlot /@ CellularAutomaton[{750, {2, {{0, 2, 0}, {2, 1, 2}, {0, 2, 0}}}, {1, 1}}, {{{1}}, 0}, 23]
(* The next two lines deal with the equivalent CA based on neighbors NW, NE, SE, SW. This is to facilitate the comparison with A246333 and A246335 *)
Map[Function[Apply[Plus, Flatten[ #1]]], CellularAutomaton[{ 750, {2, {{2, 0, 2}, {0, 1, 0}, {2, 0, 2}}}, {1, 1}}, {{{1}}, 0}, 100]]
ArrayPlot /@ CellularAutomaton[{750, {2, {{2, 0, 2}, {0, 1, 0}, {2, 0, 2}}}, {1, 1}}, {{{1}}, 0}, 23]

a(2^k + i) = (4^(k+1)-1)/3 + 4*A246336(i), for k >= 0, 0 <= i < 2^k. For example, if n = 15 = 2^3 + 7, so k=3, i=7, we have a(15) = (4^4-1)/3 + 4*A246336(7) = 85 + 4*49 = 281.
a(n) = 1 + 2*(A139250(n) - A160552(n)) = A160164(n) - A170903(n) = A187220(n) + 2*(A160552(n-1)). - Omar E. Pol, Feb 18 2015
It appears that a(n) = A162795(n) = A147562(n), if n is a member of A048645, otherwise a(n) > A162795(n) > A147562(n). - Omar E. Pol, Feb 19 2015
It appears that a(n) = 1 + 4*A255747(n-1). - Omar E. Pol, Mar 05 2015
It appears that a(n) = 1 + 4*(A139250(n-1) - (a(n-1) - 1)/4), n > 1. - Omar E. Pol, Jul 24 2015
It appears that a(2n) = 1 + 4*A162795(n). - Omar E. Pol, Jul 04 2017

Edited (added formula, illustration, etc.) by N. J. A. Sloane, Aug 30 2014

Offset changed to 1 by N. J. A. Sloane, Feb 09 2015

cp's OEIS Frontend

A061547 Number of 132 and 213-avoiding derangements of {1,2,...,n}.

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A218736 a(n) = (33^n - 1)/32.

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A086893 a(n) is the index of F(n+1) at the unique occurrence of the ordered pair of reversed consecutive terms (F(n+1),F(n)) in Stern's diatomic sequence A002487, where F(k) denotes the k-th term of the Fibonacci sequence A000045.

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A072197 a(n) = 4*a(n-1) + 1 with a(0) = 3.

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A020989 a(n) = (5*4^n - 2)/3.

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A175046 Write n in binary, then increase each run of 0's by one 0, and increase each run of 1's by one 1. a(n) is the decimal equivalent of the result.

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