cp's OEIS Frontend

Number of unoriented rows of length 6 using up to n colors. For a(0)=0, there are no rows using no colors. For a(1)=1, there is one row using that one color for all positions. For a(2)=36, there are 2^6=64 oriented arrangements of two colors. Of these, 2^3=8 are achiral. That leaves (64-8)/2=28 chiral pairs. Adding achiral and chiral, we get 36. - Robert A. Russell, Nov 14 2018

For n > 0, a(2n+1) is the number of non-isomorphic 8C_m-snakes, where m = 2n+1 or m = 2n (for n>=2). A kC_n-snake is a connected graph in which the k >= 2 blocks are isomorphic to the cycle C_n and the block-cutpoint graph is a path. - Christian Barrientos, May 16 2019

Contribution from Olivier Gérard, Jul 31 2016: (Start)

Let us consider two permutations to be equivalent if they can be obtained from each other by reversal (12345->54321), cyclic rotation (12345->(23451,34512,45123,51234), n+1-complement (31254->35412), or a combination of those three transformations (they commute with each other). a(n) is the number of classes. It is strictly inferior to (n-1)! for n>1.

If rotation is replaced by addition of a constant modulo n, one obtains the same number of classes but not exactly the same permutations starting n=5.

(End)

Original explanation by R. Jerome : Generate all permutations of a string of length n, such as 1234, which has length 4; there are n!=24 of these. Now remove all that have cycles less than 4 characters long; if you only use cyclic notation and not array notation then of the n! possibly only (n-1)! need to be considered. Then calculate the Inverse, Vertical reflection, [VErt reflection inverse], Rotation by 180 degree and [ROt by 180 deg inverse]. If any of these already exist on the list then this permutation is not distinct. Items in []'s are unnecessary since VE(x)=V(I(x))=I(V(x))=R(x) and RO(x)=R(I(x))=I(R(x))=V(x). There are some that are rotationally symmetric and some that are vertically symmetric (only possible for even lengths), but the majority are nonsymmetric.

(-12, a(1)) and (A118675(n), a(n+1)) are solutions (x, y) to the Diophantine equation x^2+(x+47)^2 = y^2.

lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2).

lim_{n -> infinity} a(n)/a(n-1) = (51+14*sqrt(2))/47 for n mod 3 = {0, 2}.

lim_{n -> infinity} a(n)/a(n-1) = (3267+1702*sqrt(2))/47^2 for n mod 3 = 1.

For the generic case x^2+(x+p)^2=y^2 with p= m^2 -2 a prime number in A028871, m>=2, the x values are given by the sequence defined by: a(n)= 6*a(n-3) -a(n-6) +2*p with a(1)=0, a(2)= 2*m +2, a(3)= 3*m^2 -10*m +8, a(4)= 3*p, a(5)= 3*m^2 +10*m +8, a(6)= 20*m^2 -58*m +42. Y values are given by the sequence defined by: b(n)= 6*b(n-3) -b(n-6) with b(1)=p, b(2)= m^2 +2*m +2, b(3)= 5*m^2 -14*m +10, b(4)= 5*p, b(5)= 5*m^2 +14*m +10, b(6)= 29*m^2 -82*m +58. - Mohamed Bouhamida, Sep 09 2009

The sequence of roots of nearest squares begins: 2, 5, 11, 14, 19, 22, 32, 37, 42, 47, 52, 59, 71, 76, 83, 90, 97, 104, 111, 118, 132, ...

The sequence of cube roots of nearest cubes begins: 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 26, ... (Cf. A000037)

The sequence of k-k2 (equals k3-k) begins: 2, 1, 2, 10, -9, 14, -12, -19, -18, -6, 20, -53, -64, 28, -15, -50, -74, -84, -77, -50, ...

If we allow k2=k3 then first missing terms are 0, 1, 64, 729, 4096, ... . - Zak Seidov, Dec 10 2013

For n>1 number of 7-digit positive integers in base n.

Conjecture: a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 3, 4, 6, 9, 13, 16, 20, 21, 24, 25, 44, 50, 51, 65, 84, 189, 290, 484, 616, 664, 680, 917, 1501, 1639, 3013.

Based on our computation, we also formulate the following general conjecture.

General Conjecture: Let T(w) = w*(w+1)/2. We have {P(x,y,z,w): x,y,z,w = 0,1,2,...} = {0,1,2,...} for any of the following polynomials P(x,y,z,w): x^3+y^3+c*z^3+T(w) (c = 2,3,4,6), x^3+y^3+c*z^3+2*T(w) (c = 2,3), x^3+b*y^3+3z^3+3*T(w) (b = 1,2), x^3+2y^3+3z^3+w(5w-1)/2, x^3+2y^3+3z^3+w(5w-3)/2, x^3+2y^3+c*z^3+T(w) (c = 2,3,4,5,6,7,12,20,21,34,35,40), x^3+2y^3+c*z^3+2*T(w) (c = 3,4,5,6,11), x^3+2y^3+c*z^3+w^2 (c = 3,4,5,6), x^3+2y^3+4z^3+w(3w-1)/2, x^3+2y^3+4z^3+w(3w+1)/2, x^3+2y^3+4z^3+w(2w-1), x^3+2y^3+6z^3+w(3w-1)/2, x^3+3y^3+c*z^3+T(w) (c = 3,4,5,6,10,11,13,15,16,18,20), x^3+3y^3+c*z^3+2*T(w) (c = 5,6,11), x^3+4y^3+c*z^3+T(w) (c = 5,10,12,16), x^3+4y^3+5z^3+2*T(w), x^3+5y^3+10z^3+T(w), 2x^3+3y^3+c*z^3+T(w) (c = 4,6), 2x^3+4y^3+8z^3+T(w), x^4+y^3+3z^3+w(3w-1)/2, x^4+y^3+c*z^3+T(w) (c = 2,3,4,5,7,12,13), x^4+y^3+c*z^3+2*T(w) (c = 2,3,4,5), x^4+y^3+2z^3+w^2, x^4+y^3+4z^3+2w^2, x^4+2y^3+c*z^3+T(w) (c = 4,5,12), x^4+2y^3+3z^3+2*T(w), 2x^4+y^3+2z^3+w(3w-1)/2, 2x^4+y^3+c*z^3+T(w) (c = 1,2,3,4,5,6,10,11), 2x^4+y^3+c*z^3+2*T(w) (c = 2,3,4), 2x^4+2y^3+c*z^3+T(w) (c = 3,5), 3x^4+y^3+c*z^3+T(w) (c = 1,2,3,4,5,11), 3x^4+y^3+2z^3+2*T(w), 3x^4+y^3+2z^3+w^2, 3x^4+y^3+2z^3+w(3w-1)/2, 4x^4+y^3+c*z^3+T(w) (c = 2,3,4,6), 4x^4+y^3+2z^3+2*T(w), 5x^4+y^3+c*z^3+T(w) (c = 2,4), a*x^4+y^3+2z^3+T(w) (a = 6,20,28,40), 6x^4+y^3+2z^3+2*T(w), 6x^4+y^3+2z^3+w^2, a*x^4+y^3+3z^3+T(w) (a = 6,8,11), 8x^4+2y^3+4z^3+T(w), x^5+y^3+c*z^3+T(w) (c = 2,3,4), x^5+2y^3+c*z^3+T(w) (c = 3,6,8), 2x^5+y^3+4z^3+T(w), 3x^5+y^3+2z^3+T(w), 5x^5+y^3+c*z^3+T(w) (c = 2,4), x^6+y^3+3z^3+T(w), x^7+y^3+4z^3+T(w), x^4+2y^4+z^3+w^2, x^4+2y^4+2z^3+T(w), x^4+b*y^4+z^3+T(w) (b = 2,3,4), 2x^4+3y^4+z^3+T(w), a*x^5+y^4+z^3+T(w) (a = 1,2), x^5+2y^4+z^3+T(w).

The polynomials listed in the general conjecture should exhaust all those polynomials P(x,y,z,w) = a*x^i+b*y^j+c*z^k+w*(s*w+/-t)/2 with {P(x,y,z,w): x,y,z,w = 0,1,2,...} = {0,1,2,...}, where a,b,c,s > 0, 0 <= t <= s, s == t (mod 2), i >= j >= k >= 3, a <= b if i = j, and b <= c if j = k.

Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 1, 2602.

Note that 1+1+2+3+4+5+7+14 = 37. In 1964 J.-R. Chen proved that any natural number can be written as the sum of 37 fifth powers of nonnegative integers.

For k = 2,3,4,... define s(k) as the smallest positive integer s such that {a(1)*x(1)^k+...+a(s)*x(s)^k: x(1),...,x(s) = 0,1,2,...} = {0,1,2,...} for some positive integers a(1), ..., a(s), and t(k) as the least positive integer t such that {a(1)*x(1)^k+...+a(t)*x(t)^k: x(1),...,x(t) = 0,1,2,...} = {0,1,2,...} for some positive integers a(1), ..., a(t) with a(1)+...+a(t) = g(k), where g(.) is given by A002804. Then s(k) <= t(k) <= g(k). Part (iii) of the conjecture in A271099 implies that t(k) <= 2k-1 for k > 2. It is easy to see that s(2) = t(2) = 4. Our computation suggests that s(3) = t(3) = 5, s(4) = t(4) = 7, s(5) = t(5) = 8 (which is smaller than 2*5-1), and s(6) = t(6) = 10. We conjecture that s(k) = t(k) for any integer k > 1, and that each natural number can be written as x(1)^6+x(2)^6+x(3)^6+2*x(4)^6+3*x(5)^6+5*x(6)^6+6*x(7)^6+10*x(8)^6+18*x(9)^6+26*x(10)^6, where x(1),x(2),...,x(10) are nonnegative integers. Note that 1+1+1+2+3+5+6+10+18+26 = 73 = g(6).

We also conjecture that any natural number can be written as s^5+t^5+2*u^5+3*v^5+4*w^5+6*x^5+8*y^5+12*z^5, with s,t,u,v,w,x,y,z nonnegative integers. Note that 1+1+2+3+4+6+8+12 = 37 = g(5). - Zhi-Wei Sun, Apr 04 2016

Conjecture: a(n) > 0 for all n > 1.

We have verified a(n) > 0 for n = 2..10^7.

Conjecture verified up to 10^10. - Giovanni Resta, Apr 18 2021

6-6-3-3 Conjecture: Each rational number can be written as u^6 - v^6 + x^3 - y^3 with u,v,x,y nonnegative rational numbers. Moreover, a(n) exists for any nonnegative integer n.

As a/b = (a*b^5)/b^6 for any integer a and nonzero integer b, the second assertion in the conjecture implies the first one.

Conjecture 1: Let k be 4 or 5. Then each integer can be written as x^k + y^k - (z^3 + w^3) with x,y,z,w nonnegative integers.

Two examples for k = 5: -4 = 58^5 + 76^5 - (775^3 + 1397^3) and 14 = 40^5 + 67^5 - (125^3 + 1132^3).

Conjecture 2: Let k be among 4, 5, 6 and 7. Then any integer can be written as x^k + y^k - (z^2 + w^2) with x,y,z,w nonnegative integers.

Examples for k = 6, 7: 170 = 9^6 + 15^6 - (2114^2 + 2730^2) and 469 = 7^7 + 8^7 - (1001^2 + 1385^2).

Conjecture 3: For any integer k > 3, there are no nonnegative integers x,y,z,w such that x^k + y^k - (z^k + w^k) = 3.

See also another similar conjecture in A351338.