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A081352
Main diagonal of square maze arrangement of natural numbers A081349.
10
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
0,2
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
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
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 *)
PROG
(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
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Paul Barry, Mar 19 2003
STATUS
approved