A004277 -id:A004277 - cp's OEIS frontend

A171861 Expansion of x(1+x+x^2) / ( (x-1)(x^3+x^2-1) ).

1, 2, 4, 6, 9, 13, 18, 25, 34, 46, 62, 83, 111, 148, 197, 262, 348, 462, 613, 813, 1078, 1429, 1894, 2510, 3326, 4407, 5839, 7736, 10249, 13578, 17988, 23830, 31569, 41821, 55402, 73393, 97226, 128798, 170622, 226027, 299423, 396652, 525453, 696078, 922108

Offset: 1

Ed Pegg Jr, Oct 16 2010

Number of wins in Penney's game if the two players start HHT and TTT and HHT beats TTT.

HHT beats TTT 70% of the time. - Geoffrey Critzer, Mar 01 2014

			a(n) enumerates length n+2 sequences on {H,T} that end in HHT but do not contain the contiguous subsequence TTT.
a(3)=4 because we have: TTHHT, THHHT, HTHHT, HHHHT.
a(4)=6 because we have: TTHHHT, THTHHT, THHHHT, HTTHHT, HTHHHT, HHHHHT. - _Geoffrey Critzer_, Mar 01 2014

G. C. Greubel, Table of n, a(n) for n = 1..1000
Wikipedia, Penney's game
Index entries for linear recurrences with constant coefficients, signature (1,1,0,-1).

Related sequences are A000045 (HHH beats HHT, HTT beats TTH), A006498 (HHH beats HTH), A023434 (HHH beats HTT), A000930 (HHH beats THT, HTH beats HHT), A000931 (HHH beats TTH), A077868 (HHT beats HTH), A002620 (HHT beats HTT), A000012 (HHT beats THH), A004277 (HHT beats THT), A070550 (HTH beats HHH), A000027 (HTH beats HTT), A097333 (HTH beats THH), A040000 (HTH beats TTH), A068921 (HTH beats TTT), A054405 (HTT beats HHH), A008619 (HTT beats HHT), A038718 (HTT beats THT), A128588 (HTT beats TTT).

Cf. A164315 (essentially the same sequence).

A171861 := proc(n) option remember; if n <=4 then op(n,[1,2,4,6]); else procname(n-1)+procname(n-2)-procname(n-4) ; end if; end proc:

nn=44;CoefficientList[Series[x(1+x+x^2)/(1-x-x^2+x^4),{x,0,nn}],x] (* Geoffrey Critzer, Mar 01 2014 *)

a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,0,1,1]^(n-1)*[1;2;4;6])[1,1] \\ Charles R Greathouse IV, Oct 03 2016

a(n) = a(n-1) +a(n-2) -a(n-4) = A000931(n+10)-3 = A134816(n+6)-3 = A078027(n+12)-3.

a(n) = A164315(n-1). - Alois P. Heinz, Oct 12 2017

A054888 Layer counting sequence for hyperbolic tessellation by regular pentagons of angle Pi/2.

Original entry on oeis.org

1, 5, 15, 40, 105, 275, 720, 1885, 4935, 12920, 33825, 88555, 231840, 606965, 1589055, 4160200, 10891545, 28514435, 74651760, 195440845, 511670775, 1339571480, 3507043665, 9181559515, 24037634880, 62931345125

Offset: 0

Paolo Dominici (pl.dm(AT)libero.it), May 23 2000

The layer sequence is the sequence of the cardinalities of the layers accumulating around a (finite-sided) polygon of the tessellation under successive side-reflections.

Reinhard Zumkeller, Table of n, a(n) for n = 0..999 (indices corrected to start at zero by _Sidney Cadot_, Jan 07 2022)
Paolo Dominici, Illustration
Index entries for Coordination Sequences [A layer sequence is a kind of coordination sequence. - _N. J. A. Sloane_, Nov 20 2022]
Index entries for linear recurrences with constant coefficients, signature (3,-1).

Cf. A000045, A001906, A002878, A004277, A203976.

Cf. A054886, A054887, A054889, A054890.

a054888 n = a054888_list !! (n-1)
a054888_list = 1 : zipWith (+) (tail a002878_list) a002878_list
-- Reinhard Zumkeller, Jan 11 2012

[n eq 0 select 1 else 5*Fibonacci(2*n): n in [0..40]]; // G. C. Greubel, Feb 08 2023

LinearRecurrence[{3,-1},{1,5,15},30] (* Harvey P. Dale, Jan 15 2023 *)
Join[{1}, 5*Fibonacci[2*Range[40]]] (* G. C. Greubel, Feb 08 2023 *)

{a(n)=polcoeff(exp(sum(k=1,n,5*fibonacci(k)^2*x^k/k)+x*O(x^n)), n)} /* Paul D. Hanna, Feb 21 2012 */

[5*fibonacci(2*n) + int(n==0) for n in range (41)] # G. C. Greubel, Feb 08 2023

a(n) = 5*A001906(n) + [n=0].
G.f.: (1+x)^2/(1-3*x+x^2).
G.f.: exp( Sum_{n>=1} 5*Fibonacci(n)^2 * x^n/n ). - Paul D. Hanna, Feb 21 2012
a(n) = A001906(n-1) + 2*A001906(n) + A001906(n+1). - R. J. Mathar, Nov 28 2011
a(n) = A203976(A004277(n-1)). - Reinhard Zumkeller, Jan 11 2012
a(n) = 5*A000045(2*n) for n >= 1. - Robert Israel, Jun 01 2015
a(n) = A002878(n-1)+A002878(n). - R. J. Mathar, Jul 09 2024

Offset changed to 0 by N. J. A. Sloane, Jan 03 2022 at the suggestion of Michel Marcus

A103517 Expansion of (1+2*x-x^2)/(1-x)^2.

Original entry on oeis.org

1, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100, 102, 104, 106, 108, 110, 112, 114, 116, 118, 120, 122, 124, 126

Offset: 0

Paul Barry, Feb 09 2005

Row sums of A103516.
Also the number of maximal and maximum cliques in the (n+1) X (n+1) rook graph. - Eric W. Weisstein, Sep 14 2017
Also the number of maximal and maximum independent vertex sets in the (n+1) X (n+1) rook complement graph. - Eric W. Weisstein, Sep 14 2017

G. C. Greubel, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Maximal Clique
Eric Weisstein's World of Mathematics, Maximum Clique
Eric Weisstein's World of Mathematics, Maximal Independent Vertex Set
Eric Weisstein's World of Mathematics, Maximum Independent Vertex Set
Eric Weisstein's World of Mathematics, Rook Complement Graph
Eric Weisstein's World of Mathematics, Rook Graph
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A103516.
Essentially the same as A004277, A005843, A051755, and A076032. - R. J. Mathar, Jul 31 2010
Cf. A272651 (for which this sequence is a conjectured continuation for large n).

[1] cat [2*n+2 : n in [1..60]]; // Wesley Ivan Hurt, Dec 07 2016

a := n -> 2*(n + 1) - 0^n: seq(a(n), n = 0..62); # Peter Luschny, May 12 2023

CoefficientList[Series[(-z^2 + 2*z + 1)/(z - 1)^2, {z, 0, 100}], z] (* Vladimir Joseph Stephan Orlovsky, Jul 10 2011 *)
CoefficientList[Series[2/(x - 1)^2 - 1, {x, 0, 62}], x] (* Robert G. Wilson v, Jan 29 2015 *)
Join[{1}, 2 Range[2, 20]] (* Eric W. Weisstein, Sep 14 2017 *)
Join[{1}, LinearRecurrence[{2, -1}, {4, 6}, 20]] (* Eric W. Weisstein, Sep 14 2017 *)

a(n) = 2*n + 2 - 0^n.
a(n) = Sum_{k=0..n} 0^(k(n-k))*(n+1).
Equals binomial transform of [1, 3, -1, 1, -1, 1, ...]. - Gary W. Adamson, Apr 23 2008
a(n) = 2*a(n-1) - a(n-2) for n > 2. - Eric W. Weisstein, Sep 14 2017
G.f.: (1 + 2*x - x^2)/(-1 + x)^2. - Eric W. Weisstein, Sep 14 2017

A167875 One third of product plus sum of three consecutive nonnegative integers; a(n)=(n+1)(n^2+2n+3)/3.

Original entry on oeis.org

1, 4, 11, 24, 45, 76, 119, 176, 249, 340, 451, 584, 741, 924, 1135, 1376, 1649, 1956, 2299, 2680, 3101, 3564, 4071, 4624, 5225, 5876, 6579, 7336, 8149, 9020, 9951, 10944, 12001, 13124, 14315, 15576, 16909, 18316, 19799, 21360, 23001, 24724, 26531

Offset: 0

Klaus Brockhaus, Nov 14 2009

a(n) = ((n*(n+1)*(n+2))+(n+(n+1)+(n+2)))/3, n >= 0.
Equals A006527 without initial term 0: a(n) = A006527(n+1).
Binomial transform of A167876.
Inverse binomial transform of A080930.
a(n) = A007290(n+2)+n+1.
a(n) = A014820(n)/(n+1) for n > 0.
a(n) = A116731(n+2)-1.
a(n) = A033547(n+1)-n.
a(n) = A054602(n)/3.
a(n) = A086514(n+3)-2.
a(n) = A002061(n+1)+a(n-1) for n > 0.
a(n) = A005894(n)-a(n-1) for n > 0.
First bisection is A057813.
Second differences are in A004277.
a(n) = A177342(n)*(-1)+a(n-1)*5 with n>0. For n=8, a(8)=-A177342(8)+a(7)*5=-631+176*5=249. - Bruno Berselli, May 18 2010

			a(0) = (0*1*2+0+1+2)/3 = (0+3)/3 = 1.
a(1) = (1*2*3+1+2+3)/3 = (6+6)/3 = 4.
a(6)-4*a(5)+6*a(4)-4*a(3)+a(2) = 119-4*76+6*45-4*24+11 = 0. - _Bruno Berselli_, May 26 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A001477 (nonnegative integers),
A006527 ((n^3+2*n)/3),
A167876 (1, 3, 4, 2, 0, 0, 0, 0, ...),
A080930,
A007290 (2*C(n, 3)),
A014820 ((1/3)*(n^2+2*n+3)*(n+1)^2),
A116731,
A033547 (n*(n^2+5)/3),
A054602 (Sum_{d|3} phi(d)*n^(3/d)),
A086514 ((n^3-6*n^2+14*n-6)/3),
A002061 (n^2-n+1),
A005894 (centered tetrahedral numbers),
A057813 ((2*n+1)*(4*n^2+4*n+3)/3),
A004277 (1 and the positive even numbers),
A028387 (n+(n+1)^2),
A166941, A166942, A166943.

[ (&*s + &+s)/3 where s is [n..n+2]: n in [0..42] ];

Select[Table[(n*(n+1)*(n+2)+n+(n+1)+(n+2))/3,{n,0,5!}],IntegerQ[#]&] (* Vladimir Joseph Stephan Orlovsky, Dec 04 2010 *)
(Times@@#+Total[#])/3&/@Partition[Range[0,65],3,1]  (* Harvey P. Dale, Mar 14 2011 *)

a(n)=(n+1)*(n^2+2*n+3)/3 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = (n^3+3*n^2+5*n+3)/3.
a(n) = 3*a(n-1)-3*a(n-2)+a(n-3)+2 for n > 3; a(0)=1, a(1)=4, a(2)=11, a(3)=24.
G.f.: (1+x^2)/(1-x)^4.
a(n) = SUM(A109613(k)*A005408(n-k): 0<=k<=n). - Reinhard Zumkeller, Dec 05 2009
a(n)-4*a(n-1)+6*a(n-2)-4*a(n-3)+a(n-4)=0 for n>3. - Bruno Berselli, May 26 2010

A203976 a(n) = 3*a(n-2) - a(n-4), a(0)=0, a(1)=1, a(2)=5, a(3)=4.

Original entry on oeis.org

0, 1, 5, 4, 15, 11, 40, 29, 105, 76, 275, 199, 720, 521, 1885, 1364, 4935, 3571, 12920, 9349, 33825, 24476, 88555, 64079, 231840, 167761, 606965, 439204, 1589055, 1149851, 4160200, 3010349, 10891545, 7881196, 28514435, 20633239, 74651760, 54018521, 195440845

Offset: 0

Michael Somos, Jan 08 2012

a(n+1) = p(n+2) where p(x) is the unique degree-n polynomial such that p(k) = Lucas(k) for k = 1, ..., n+1.
This is a divisibility sequence; that is, if n divides m, then a(n) divides a(m).
a(n) = row sums of triangle A226377(n), based on differences among Lucas Numbers.  - Richard R. Forberg, Aug 01 2013
A strong divisibility sequence, i.e., gcd(a(n),a(m)) = a(gcd(n,m)) for all natural numbers n and m. The sequence of convergents of the 2-periodic continued fraction [0; 1, -5, 1, -5, ...] = 1/(1 - 1/(5 - 1/(1 - 1/(5 - ...)))) = 1/2*(5 - sqrt(5)) begins [0/1, 1/1, 5/4, 4/3, 15/11, 11/8, 40/29,...]. The present sequence is the sequence of numerators; the sequence of denominators [1, 1, 4, 3, 11, 8, 29,...] is A005013. - Peter Bala, May 19 2014
It appears that the first homology group of the branched n-th cyclic covering of the group of figure-eight knot is the direct sum of cyclic groups of orders a(n) and A005013(n), so the order of that group is the product of these numbers, i. e. A004146(n); see the table on p. 156 of the paper by Fox. - Andrey Zabolotskiy, Mar 16 2023

			a(3) = 4 since p(x) = (-x^2 + 7*x - 4) / 2 interpolates p(1) = 1, p(2) = 3, p(3) = 4, and p(4) = 4.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
R. H. Fox, A quick trip through knot theory, pages 120-167 in: Topology of 3-manifolds and related topics (Proceedings of The University of Georgia Institute, 1961), Prentice-Hall, 1962.
Index to divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-1).

Cf. A000032, A000045, A201157 (bisection), A002878 (bisection). A005013.

a203976 n = a203976_list !! n
a203976_list = 0 : 1 : 5 : 4 : zipWith (-)
   (map (* 3) $ drop 2 a203976_list) a203976_list
-- Reinhard Zumkeller, Jan 10 2012

I:=[0,1,5,4]; [n le 4 select I[n] else 3*Self(n-2)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, Mar 29 2016

LinearRecurrence[{0,3,0,-1},{0,1,5,4},40] (* Harvey P. Dale, Apr 06 2013 *)

{a(n) = if( n%2, fibonacci(n+1) + fibonacci(n-1), 5 * fibonacci(n))}

{a(n) = if( n<0, -a(-n), polcoeff( x * (1 + 5*x + x^2) / (1 - 3*x^2 + x^4) + x * O(x^n), n))}

{a(n) = if( n<0, -a(-n), subst( polinterpolate( vector( n, k, fibonacci(k-1) + fibonacci(k+1) )), x, n + 1))}

a(1) = 1, a(2) = 5, a(3) = 4, a(n) * a(n-3) = a(n-1) * a(n-2) - 5. a(-n) = -a(n).
G.f.: x * (1 + 5*x + x^2) /  ( (x^2+x-1)*(x^2-x-1) ).
a(2*n) = 5 * A000045(2*n) (Fibonacci). a(2*n+1) = A000032(2*n+1) (Lucas).
a(A004277(n)) = A054888(n+1). - Reinhard Zumkeller, Jan 11 2012
a(n) = A000032(n+1) - A061084(n). - R. J. Mathar, Jun 23 2013
a(2n) = a(2n-1) + a(2n+1), for n>0. - Richard R. Forberg, Aug 01 2013
a(n) = (2^(-1-n)*((-5-sqrt(5)+(-1)^n*(-5+sqrt(5)))*((-1+sqrt(5))^n-(1+sqrt(5))^n)))/sqrt(5). - Colin Barker, Mar 28 2016
E.g.f.: exp(-phi*x)*(exp(x) - 1)*(phi*exp(sqrt(5)*x) - 1/phi), where phi = (1 + sqrt(5))/2. - G. C. Greubel, Mar 28 2016

A154325 Triangle with interior all 2's and borders 1.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 2, 2, 1, 1, 2, 2, 2, 2, 1, 1, 2, 2, 2, 2, 2, 1, 1, 2, 2, 2, 2, 2, 2, 1, 1, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1

Offset: 0

Paul Barry, Jan 07 2009

This triangle follows a general construction method as follows: Let a(n) be an integer sequence with a(0)=1, a(1)=1. Then T(n,k,r):=[k<=n](1+r*a(k)*a(n-k)) defines a symmetrical triangle.
Row sums are n + 1 + r*Sum_{k=0..n} a(k)*a(n-k) and central coefficients are 1+r*a(n)^2.
Here a(n)=1-0^n and r=1. Row sums are A004277.
Eigensequence of the triangle = A000129, the Pell sequence. - Gary W. Adamson, Feb 12 2009
Inverse has general element T(n,k)*(-1)^(n-k). - Paul Barry, Oct 06 2010

			Triangle begins
  1;
  1, 1;
  1, 2, 1;
  1, 2, 2, 1;
  1, 2, 2, 2, 1;
  1, 2, 2, 2, 2, 1;
  1, 2, 2, 2, 2, 2, 1;
From _Paul Barry_, Oct 06 2010: (Start)
Production matrix is
  1,  1;
  0,  1, 1;
  0, -1, 0, 1;
  0,  1, 0, 0, 1;
  0, -1, 0, 0, 0, 1;
  0,  1, 0, 0, 0, 0, 1;
  0, -1, 0, 0, 0, 0, 0, 1;
  0,  1, 0, 0, 0, 0, 0, 0, 1; (End)

Cf. A000129, A103451, A129765.

a[n_] :=
 If[Length@
    NestWhileList[# -
       Floor[(Sqrt[8 # + 1] - 1)/2] (Floor[(Sqrt[8 # + 1] - 1)/2] + 1)/
2 &, n, # > 1 &] <= 2, 1, 2] (* David Naccache, Jul 13 2025 *)

row(n) = vector(n+1, k, k--; (2-0^(k*(n-k)))); \\ Michel Marcus, Jul 13 2025

Number triangle T(n,k) = [k<=n](2-0^(n-k)-0^k+0^(n+k)) = [k<=n](2-0^(k*(n-k))).

a(n) = 2 - A103451(n). - Omar E. Pol, Jan 18 2009

More terms from Michel Marcus, Jul 13 2025

A300576 Number of nights required in the worst case to find the princess in a castle with n rooms arranged in a line (Castle and princess puzzle).

Original entry on oeis.org

1, 2, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100, 102, 104, 106, 108, 110, 112, 114, 116, 118, 120, 122

Offset: 1

Dmitry Kamenetsky, Mar 09 2018

This is a logic puzzle. There is a castle with n rooms arranged in a line. The princess living in the castle sleeps in a different room each night, but always one adjacent to the one in which she slept on the previous night. She is free to choose any room in which to sleep on the first night. A prince would like to find the princess, but she will not tell him where she is going to sleep each night. Each night the prince can look in a single room. What strategy should he follow in order to guarantee that he finds the princess in a minimum number of nights?
The strategy to find the princess guaranteed within a(n) nights takes on average k(n) nights until the princess is found with lim_{n->oo} k(n) = n-1.5. For n>4, strategies with lower average numbers of trials exist; A386462 provides this strategy for n=8. See there for more information. - Ruediger Jehn, Aug 05 2025
Christian Perfect (see link) considered the case when the rooms are arranged as a general graph. He showed that the set of solvable graphs is exactly the set of trees not containing the "threesy" subgraph, which is A130131. He also showed that for d-level binary trees with 1 <= d <= 4 the number of required nights is 1, 2, 6, 18. Binary trees with d >= 5 are unsolvable as they contain "threesy".

			For n = 1, there is only one room to search, so a(1) = 1.
For n = 2, the prince searches room 1 on the first night. If the princess is not there that means she was in room 2. If the prince searches room 1 again then he is guaranteed to see the princess as she has to move from room 2 to room 1 (she cannot stay in the same room). So a(2) = 2.
For n = 3, the prince searches room 2 on the first night. If the princess is not there that means she was either in room 1 or 3. On the second night she must go to room 2 and this is where the prince will find her. So a(3) = 2.
For n = 4, a solution that guarantees to find the princess in a(4)=4 nights is to search rooms (2,3,3,2).
For n = 5, a solution that guarantees to find the princess in a(5)=6 nights is to search rooms (2,3,4,4,3,2).
In the general case for n >= 3, a solution guaranteeing success in the minimum number of nights is to search rooms (2,3,...,n-1,n-1,...,3,2), so a(n) = 2*n - 4.

Christian Perfect, Solving the princess on a graph puzzle.
Puzzling Stack Exchange, Why does this solution guarantee that the prince knocks on the right door to find the princess?.
standupmaths, Youtube, The Castle and the Princess Puzzle.
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Essentially the same as A005843, A004277 and A004275.

Cf. A130131, A130132, A331693, A386462.

CoefficientList[ Series[(2x^3 - x^2 + 1)/(x - 1)^2, {x, 0, 62}], x] (* Robert G. Wilson v, Mar 12 2018 *)
Join[{1,2},Range[2,200,2]] (* Harvey P. Dale, Jan 25 2019 *)

For n >= 3, a(n) = 2*n - 4.
From Chai Wah Wu, Apr 14 2024: (Start)
a(n) = 2*a(n-1) - a(n-2) for n > 4.
G.f.: x*(2*x^3 - x^2 + 1)/(x - 1)^2. (End)
E.g.f.: 4 + 2*exp(x)*(x - 2) + 3*x + x^2. - Stefano Spezia, Aug 15 2025

A316723 Square array A(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of Product_{j=1..k} (1+x^j)/(1-x^j).

Original entry on oeis.org

1, 1, 0, 1, 2, 0, 1, 2, 2, 0, 1, 2, 4, 2, 0, 1, 2, 4, 6, 2, 0, 1, 2, 4, 8, 8, 2, 0, 1, 2, 4, 8, 12, 10, 2, 0, 1, 2, 4, 8, 14, 18, 12, 2, 0, 1, 2, 4, 8, 14, 22, 26, 14, 2, 0, 1, 2, 4, 8, 14, 24, 34, 34, 16, 2, 0, 1, 2, 4, 8, 14, 24, 38, 50, 44, 18, 2, 0, 1, 2, 4, 8, 14, 24, 40, 58, 70, 56, 20, 2, 0

Offset: 0

Seiichi Manyama, Jul 11 2018

			Square array begins:
   1, 1,  1,  1,  1,  1,  1, ...
   0, 2,  2,  2,  2,  2,  2, ...
   0, 2,  4,  4,  4,  4,  4, ...
   0, 2,  6,  8,  8,  8,  8, ...
   0, 2,  8, 12, 14, 14, 14, ...
   0, 2, 10, 18, 22, 24, 24, ...
   0, 2, 12, 26, 34, 38, 40, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Columns k=0-10 give: A000007, A040000, A004277, A053799, A053798, A053800, A053801, A053802, A069251, A069252, A069253.

Diagonal gives A015128.

A322419 Number of n-step self-avoiding walks on L-lattice.

Original entry on oeis.org

1, 2, 4, 8, 12, 20, 32, 52, 84, 136, 220, 356, 564, 904, 1448, 2320, 3684, 5872, 9376, 14960, 23688, 37652, 59912, 95316, 150744, 239080, 379528, 602424, 951788, 1507136, 2388252, 3784344, 5973988, 9447880, 14950796, 23658540, 37321752, 58965260, 93206864, 147333080, 232286272

Offset: 0

Robert FERREOL, Dec 07 2018

The L-lattice is an oriented square lattice in which each step must be followed by a step perpendicular to the preceding one.

			a(1) = 2 because there are only two possible directions at each intersection; for the same reason a(2) = 2*2 and a(3) = 2*4 ; but a(4) = 12 (not 16) because four paths return to the starting point and are not self-avoiding. See the 12 paths under "links".

Sean A. Irvine, Table of n, a(n) for n = 0..50
Robert FERREOL, The a(4)=12 walks in L-lattice
Keh-Ying Lin and Yee-Mou Kao, Universal amplitude combinations for self-avoiding walks and polygons on directed lattices, J. Phys. A: Math. Gen. 32 (1999), page 6929.
A. Malakis, Self-avoiding walks on oriented square lattices, Journal of Physics A: Mathematical and General, Volume 8, Number 12 (1975), page 1890.
Wikipedia, Connective constant

Cf. A001411 (square lattice), A117633 (Manhattan lattice), A189722, A004277 (coordination sequence), A151798.

walks:=proc(n)
    option remember;
    local i,father,End,X,walkN,dir,u,x,y;
    if n=1 then [[[0,0]]] else
         father:=walks(n-1):
         walkN:=NULL:
         for i to nops(father) do
            u:=father[i]:End:=u[n-1]:if n mod 2 = 0 then
            dir:=[[1,0], [-1, 0]] else dir := [[0,1], [0, -1]] fi:
            for X in dir do
             if not(member(End+X,u)) then walkN:=walkN,[op(u),End+X] fi;
             od od:
         [walkN] fi end:
n:=5:L:=walks(n):N:=nops(L);
# This program explicitly gives the a(n) walks.

mo = {{1, 0}, {-1, 0}}; moo = {{0, 1}, {0, -1}}; a[0] = 1;
a[tg_, p_: {{0, 0}}] := Module[{e, mv},
If[Mod[tg, 2] == 0, mv = Complement[Last[p] + # & /@ mo, p],
mv = Complement[Last[p] + # & /@ moo, p]];
If[tg == 1, Length@mv, Sum[a[tg - 1, Append[p, e]], {e, mv}]]];
a /@ Range[0, 20] (* after the program from Giovanni Resta at A001411 *)

def add(L, x):
    M = [y for y in L]
    M.append(x)
    return M
plus = lambda L, M: [x + y for x, y in zip(L, M)]
mo = [[1, 0], [-1, 0]]
moo = [[0, 1], [0, -1]]
def a(n, P=[[0, 0]]):
    if n == 0:
        return 1
    if n % 2 == 0:
        mv1 = [plus(P[-1], x) for x in mo]
    else:
        mv1 = [plus(P[-1], x) for x in moo]
    mv2 = [x for x in mv1 if x not in P]
    if n == 1:
        return len(mv2)
    else:
        return sum(a(n - 1, add(P, x)) for x in mv2)
[a(n) for n in range(21)]

a(n) = 4*A189722(n) for n >= 2.

It is proved that a(n)^(1/n) has a limit mu called the "connective constant" of the L-lattice; approximate value of mu: 1.5657. It is only conjectured that a(n + 1) ~ mu * a(n).

A365402 The number of divisors of the largest unitary divisor of n that is an exponentially odd number.

Original entry on oeis.org

1, 2, 2, 1, 2, 4, 2, 4, 1, 4, 2, 2, 2, 4, 4, 1, 2, 2, 2, 2, 4, 4, 2, 8, 1, 4, 4, 2, 2, 8, 2, 6, 4, 4, 4, 1, 2, 4, 4, 8, 2, 8, 2, 2, 2, 4, 2, 2, 1, 2, 4, 2, 2, 8, 4, 8, 4, 4, 2, 4, 2, 4, 2, 1, 4, 8, 2, 2, 4, 8, 2, 4, 2, 4, 2, 2, 4, 8, 2, 2, 1, 4, 2, 4, 4, 4, 4

Offset: 1

Amiram Eldar, Sep 03 2023

The sum of these divisors is A351569(n).

All the terms are either 1 or even (A004277).

Amiram Eldar, Table of n, a(n) for n = 1..10000
Vaclav Kotesovec, Graph - the asymptotic ratio (100000 terms)

Cf. A000005, A000290, A004277, A268335, A350389, A351569, A365401.

f[p_, e_] := If[OddQ[e], e + 1, 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

a(n) = vecprod(apply(x -> if(x%2, x+1, 1), factor(n)[, 2]));

from math import prod
from sympy import factorint
def A365402(n): return prod(e+1 for e in factorint(n).values() if e&1) # Chai Wah Wu, Nov 17 2023

def a(n): return prod((valuation(n,p)+1) for p in prime_divisors(n) if valuation(n,p)%2==1) # Orges Leka, Nov 16 2023

a(n) = A000005(A350389(n)).
a(n) = A000005(n) / A365401(n).
a(n) <= A000005(n) with equality if and only if n is an exponentially odd number (A268335).
a(n) >= 1 with equality if and only if n is a square (A000290).
Multiplicative with a(p^e) = 1 if e is even, and e+1 if e is odd.
Dirichlet g.f.: zeta(2*s)^2 * Product_{p prime} (1 + 2/p^s - 1/p^(2*s)).
From Vaclav Kotesovec, Sep 05 2023: (Start)
Dirichlet g.f.: zeta(s)^2 * zeta(2*s)^2 * Product_{p prime} (1 - 4/p^(2*s) + 4/p^(3*s) - 1/p^(4*s)).
Let f(s) = Product_{p prime} (1 - 4/p^(2*s) + 4/p^(3*s) - 1/p^(4*s)).
Sum_{k=1..n} a(k) ~ f(1) * Pi^4 * n / 36 * (log(n) + 2*gamma - 1 + 24*Zeta'(2)/Pi^2 + f'(1)/f(1)), where
f(1) = Product_{p prime} (1 - 4/p^2 + 4/p^3 - 1/p^4) = 0.2177787166195363783230075141194468131307977550013559376482764035236264911...
f'(1) = f(1) * Sum_{p prime} 4*(2*p - 1) * log(p) / (1 - 3*p + p^2 + p^3) = f(1) * 3.3720882314412399056794495057358594564001229865925330149186567502684770675...
and gamma is the Euler-Mascheroni constant A001620. (End)
a(n) = Sum_{d|n} (-1)^(Sum_{p|gcd(d,n/d)} v_p(d)*v_p(n/d)), where v_p(x) denotes the valuation of x at the prime p. - Orges Leka, Nov 16 2023

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