A081352 Main diagonal of square maze arrangement of natural numbers A081349.

1, 7, 11, 21, 29, 43, 55, 73, 89, 111, 131, 157, 181, 211, 239, 273, 305, 343, 379, 421, 461, 507, 551, 601, 649, 703, 755, 813, 869, 931, 991, 1057, 1121, 1191, 1259, 1333, 1405, 1483, 1559, 1641, 1721, 1807, 1891, 1981, 2069, 2163, 2255, 2353, 2449, 2551

Offset: 0

Author

Paul Barry, Mar 19 2003

Keywords

• nonn
• easy

Comments

Conjecture: let a and b be integers such that 0 < a < b so that 0 < a/b is a proper fraction. Define the map f(a,b,D) = a/b + gcd(a,b)/D. Of course, all such a/b can be partially ordered by value, i.e., 1/2 = 0.5 < 2/3 = 4/6 = 6/9 = 0.6666... < 3/4 = 6/8 = 0.75 < 4/5 = 0.8 etc. The map f appears to specify a total strict order on the co-domain for all a/b that is consistent with the given partial order of the domain, i.e., the partial order remains intact, while equivalent fractions are given a total strict order themselves. Moreover, equivalent fractions are strictly ordered by numerator (or denominator), e.g., 1/2 < 2/4 < 3/6 etc. The conditions are that for n >= 4 all of the fractions with denominator b <= n are listed and the minimum integer value of D to achieve the total strict order of the co-domain is 2*C(n-1,2) - (-1)^(n-1). So, a(n-3) = D for n >= 4. Example: given n = 4, we have D = 2*(4-1,2) - (-1)^(4-1) = 2*3 + 1 = 7 = a(4-3) = a(1). Partial order of domain. 1/4 < 1/3 < 1/2 = 2/4 < 2/3 < 3/4. Total order of co-domain. f(1,4,7) = 1/4 + 1/7 = 33/84 < f(1,3,7) = 1/3 + 1/7 = 40/84 < f(1,2,7) = 1/2 + 1/7 = 54/84 < f(2,4,7) = 2/4 + 2/7 = 66/84 < f(2,3,7) = 2/3 + 2/7 = 68/84 < f(3,4,7) = 3/4 + 1/7 = 75/84. Observe that if D = 6, then f(2,4,6) = 2/4 + 2/6 = 10/12 = f(2,3,6) = 2/3 + 1/6. Computation shows the same failure to achieve total strict order of the co-domain for D = 2..5. (As a >= 1, then b >=2, from the above). Computation also shows that the conjecture holds for n = 4..17. - Ross La Haye, Oct 02 2016

Links

• B. Berselli, Table of n, a(n) for n = 0..10000
• Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Crossrefs

Cf. A081350, A081351, A054569.

Programs

Magma

I:=[1,7,11,21]; [n le 4 select I[n] else 2*Self(n-1)-2*Self(n-3)+Self(n-4): n in [1..50]]; // Vincenzo Librandi, Aug 08 2013

Maple

A081352:=n->(n + 1)*(n + 2) - (-1)^n; seq(A081352(n), n=0..50); # Wesley Ivan Hurt, Feb 26 2014

Mathematica

CoefficientList[Series[(1 + 5 x - 3 x^2 + x^3) / ((1 + x) (1 - x)^3), {x, 0, 60}], x] (* Vincenzo Librandi, Aug 08 2013 *)

PARI

x='x+O('x^99); Vec((1+5*x-3*x^2+x^3)/((1+x)*(1-x)^3)) \\ Altug Alkan, Mar 26 2016

Formula

a(n) = (n + 1)*(n + 2) - (-1)^n = 2*C(n+2, 2) - (-1)^n.

G.f.: (1 +5*x -3*x^2 +x^3) / ((1+x)*(1-x)^3). [Bruno Berselli, Aug 01 2010]

a(n) -2*a(n-1) +2*a(n-3) -a(n-4) = 0 with n>3. [Bruno Berselli, Aug 01 2010]

a(n) = 3*A000982(n + 2) - A000982(n + 3). - Miko Labalan, Mar 26 2016

a(n) = A116940(n) + A236283(n + 1). - Miko Labalan, Dec 04 2016

a(n) = (2*n^2 + 6*n - 2*(-1)^n + (-1)^(2*n) + 3)/2. - Kritsada Moomuang, Oct 24 2019