Want 3 consecutive integers?

Take 4 consecutive terms!

Hello SeqFans,

I must have missed the correct entry in the OEIS because I cannot find this easy (core?) seq.

"To find 3 consecutive naturals in S, you have to take 4 consecutive terms of S -- no less":

S = 1,3,0,2,4,1,3,5,2,4,6,3,5,7,4,6,8,5,7,9,6,8,10,7,9,11,8,10,12,9,11,13,10,12,14,11,13,15,12,14,16,13,15,17,14,16,18,15,...

Ex: taking the first 3 terms doesn’t allow you to handle 3 consecutive natural numbers as they are 0,1 and... 3.

But if you take the fourth term (2), you’ll have in hand 0,1,2 [and even another triplet of consecutive naturals, which is (1,2,3)].

Formula is easy to compute.

Best,

E.

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Alex M.

Simplest formula I could get: a(3n)=n-1; a(3n+1)=n+1; a(3n+2)=n+3. Or, a(n)=1+a(n-3).

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Richard Mathar:

Confirming Alex we see that this falls into the pattern

a(n)= +a(n-1) +a(n-3) -a(n-4) = (n+3+5*A049347(n-1))/3  (assuming offset 0)

also known as the ocean of

<a href="Sindx_Rea.html#recLCC">Index to sequences with linear recurrences with constant coefficients</a>, signature (1,0,1,-1).

Generating function

(1+2*x+x^3-3*x^2)/(1+x+x^2)/(x-1)^2

These almost-no-growth sequences are bad for the economy.

RJM

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Alexander P-sky:

PURRS Demo Results

Verified exact solution for x(n) = 1+x(-3+n) for the initial conditions

x(0) = 1

x(1) = 3

x(2) = 0

Verified solution x(n) =

1+1/3*n-(1/18*I)*sqrt(3)*(-1/2+(1/2*I)*sqrt(3))^n+1/2*(-1/2-(1/2*I)*sqrt(3))^n+1

/2*(-1/2+(1/2*I)*sqrt(3))^n+(1/18*I)*sqrt(3)*(-1/2-(1/2*I)*sqrt(3))^n+(-1/2-(1/2

*I)*sqrt(3))^(-1)*(-1/2-(1/2*I)*sqrt(3))^n+(-1/2+(1/2*I)*sqrt(3))^(-1)*(-1/2+(1

/2*I)*sqrt(3))^n

for each n >= 0

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PURRS Demo Results

Verified exact solution for x(n) = x(-1+n)-x(-4+n)+x(-3+n) for the initial conditions

x(0) = 1

x(1) = 3

x(2) = 0

x(3) = 2

Verified solution

x(n) = 1+1/3*n-(5/9*I)*sqrt(3)*(-1/2+(1/2*I)*sqrt(3))^n+(5/9*I)*sqrt(3)*(-1/2-(1/2*I)*sqrt(3))^n

for each n >= 0

Computing the exact solution took about 61 ms of CPU time; verifying it took about 19 ms of CPU time.

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S building method is easy:

S = 1,3,0,2,4,1,3,5,2,4,6,3,5,7,4,6,8,5,7,9,6,8,10,7,9,11,8,10,12,9,11,13,10,12,14,11,13,15,12,14,16,13,15,17,14,16,18,15,...

S = 1,3,0,2,4,1,3,5,2,4,6,3,5,7,4,6,8,5,7,9,6,8,10,7,9,11,8,10,12,9,11,13,10,12,14,11,13,15,12,14,16,13,15,17,14,16,18,15,...

S = 1,3,0,2,4,1,3,5,2,4,6,3,5,7,4,6,8,5,7,9,6,8,10,7,9,11,8,10,12,9,11,13,10,12,14,11,13,15,12,14,16,13,15,17,14,16,18,15,...

Best,

Thanks to all,

É.