cp's OEIS Frontend

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

%I A081352 #44 Sep 08 2022 08:45:09
%S A081352 1,7,11,21,29,43,55,73,89,111,131,157,181,211,239,273,305,343,379,421,
%T A081352 461,507,551,601,649,703,755,813,869,931,991,1057,1121,1191,1259,1333,
%U A081352 1405,1483,1559,1641,1721,1807,1891,1981,2069,2163,2255,2353,2449,2551
%N A081352 Main diagonal of square maze arrangement of natural numbers A081349.
%C A081352 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
%H A081352 B. Berselli, <a href="/A081352/b081352.txt">Table of n, a(n) for n = 0..10000</a>
%H A081352 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,-2,1).
%F A081352 a(n) = (n + 1)*(n + 2) - (-1)^n = 2*C(n+2, 2) - (-1)^n.
%F A081352 G.f.: (1 +5*x -3*x^2 +x^3) / ((1+x)*(1-x)^3). [_Bruno Berselli_, Aug 01 2010]
%F A081352 a(n) -2*a(n-1) +2*a(n-3) -a(n-4) = 0 with n>3. [_Bruno Berselli_, Aug 01 2010]
%F A081352 a(n) = 3*A000982(n + 2) - A000982(n + 3). - _Miko Labalan_, Mar 26 2016
%F A081352 a(n) = A116940(n) + A236283(n + 1). - _Miko Labalan_, Dec 04 2016
%F A081352 a(n) = (2*n^2 + 6*n - 2*(-1)^n + (-1)^(2*n) + 3)/2. - _Kritsada Moomuang_, Oct 24 2019
%p A081352 A081352:=n->(n + 1)*(n + 2) - (-1)^n; seq(A081352(n), n=0..50); # _Wesley Ivan Hurt_, Feb 26 2014
%t A081352 CoefficientList[Series[(1 + 5 x - 3 x^2 + x^3) / ((1 + x) (1 - x)^3), {x, 0, 60}], x] (* _Vincenzo Librandi_, Aug 08 2013 *)
%o A081352 (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
%o A081352 (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
%Y A081352 Cf. A081350, A081351, A054569.
%K A081352 nonn,easy
%O A081352 0,2
%A A081352 _Paul Barry_, Mar 19 2003