Closed form representation of recurrence relation 4
I want to find a closed form of the following recurrence relation:
$$
a_0=0; \
a_{n+1}=begin{cases}0, & exists kin mathbb{N}_0: n=frac12 k(k+1); \
a_n+1, & nexists k in mathbb{N}_0:n=frac12 k(k+1). end{cases}
$$
(In particular, this recurrence relation describes the sequence
$$
(0,0,1,0,1,2,0,1,2,3,0,1,2,ldots)
$$
.)
I was not successful in finding an adequate generating function as normal.
How can I proceed?
real-analysis sequences-and-series limits recurrence-relations
asked Nov 26 '18 at 19:25
Jean Paul
11
add a comment |
I want to find a closed form of the following recurrence relation:
$$
a_0=0; \
a_{n+1}=begin{cases}0, & exists kin mathbb{N}_0: n=frac12 k(k+1); \
a_n+1, & nexists k in mathbb{N}_0:n=frac12 k(k+1). end{cases}
$$
(In particular, this recurrence relation describes the sequence
$$
(0,0,1,0,1,2,0,1,2,3,0,1,2,ldots)
$$
.)
I was not successful in finding an adequate generating function as normal.
How can I proceed?
real-analysis sequences-and-series limits recurrence-relations
asked Nov 26 '18 at 19:25
Jean Paul
11
I think you intended $a_n=a_{n-1}+1$ when $n$ is not a triangular number.
– John Wayland Bales
Nov 26 '18 at 19:48
@JohnWaylandBales You're correct! I'll edit my post.
– Jean Paul
Nov 26 '18 at 19:56
Look at the Mathematica section of oeis.org/A002262
– John Wayland Bales
Nov 26 '18 at 19:59
@JohnWaylandBales Could you elaborate on it further? I'm not sure about how to decipher that entry...
– Jean Paul
Nov 26 '18 at 20:25
I tried to decipher it but must have done it wrong, because it is not producing the right numbers. Perhaps someone who knows how to code in Mathematica can help.
– John Wayland Bales
Nov 26 '18 at 21:14
add a comment |
I want to find a closed form of the following recurrence relation:
$$
a_0=0; \
a_{n+1}=begin{cases}0, & exists kin mathbb{N}_0: n=frac12 k(k+1); \
a_n+1, & nexists k in mathbb{N}_0:n=frac12 k(k+1). end{cases}
$$
(In particular, this recurrence relation describes the sequence
$$
(0,0,1,0,1,2,0,1,2,3,0,1,2,ldots)
$$
.)
I was not successful in finding an adequate generating function as normal.
How can I proceed?
real-analysis sequences-and-series limits recurrence-relations
asked Nov 26 '18 at 19:25
Jean Paul
11
I want to find a closed form of the following recurrence relation:
$$
a_0=0; \
a_{n+1}=begin{cases}0, & exists kin mathbb{N}_0: n=frac12 k(k+1); \
a_n+1, & nexists k in mathbb{N}_0:n=frac12 k(k+1). end{cases}
$$
(In particular, this recurrence relation describes the sequence
$$
(0,0,1,0,1,2,0,1,2,3,0,1,2,ldots)
$$
.)
I was not successful in finding an adequate generating function as normal.
How can I proceed?
real-analysis sequences-and-series limits recurrence-relations
real-analysis sequences-and-series limits recurrence-relations
asked Nov 26 '18 at 19:25
Jean Paul
11
asked Nov 26 '18 at 19:25
Jean Paul
11
edited Nov 26 '18 at 19:57
asked Nov 26 '18 at 19:25
Jean Paul
11
asked Nov 26 '18 at 19:25
Jean Paul
11
asked Nov 26 '18 at 19:25
Jean Paul
11
11
I think you intended $a_n=a_{n-1}+1$ when $n$ is not a triangular number.
– John Wayland Bales
Nov 26 '18 at 19:48
@JohnWaylandBales You're correct! I'll edit my post.
– Jean Paul
Nov 26 '18 at 19:56
Look at the Mathematica section of oeis.org/A002262
– John Wayland Bales
Nov 26 '18 at 19:59
@JohnWaylandBales Could you elaborate on it further? I'm not sure about how to decipher that entry...
– Jean Paul
Nov 26 '18 at 20:25
I tried to decipher it but must have done it wrong, because it is not producing the right numbers. Perhaps someone who knows how to code in Mathematica can help.
– John Wayland Bales
Nov 26 '18 at 21:14
add a comment |
I think you intended $a_n=a_{n-1}+1$ when $n$ is not a triangular number.
– John Wayland Bales
Nov 26 '18 at 19:48
@JohnWaylandBales You're correct! I'll edit my post.
– Jean Paul
Nov 26 '18 at 19:56
Look at the Mathematica section of oeis.org/A002262
– John Wayland Bales
Nov 26 '18 at 19:59
@JohnWaylandBales Could you elaborate on it further? I'm not sure about how to decipher that entry...
– Jean Paul
Nov 26 '18 at 20:25
I tried to decipher it but must have done it wrong, because it is not producing the right numbers. Perhaps someone who knows how to code in Mathematica can help.
– John Wayland Bales
Nov 26 '18 at 21:14
I think you intended $a_n=a_{n-1}+1$ when $n$ is not a triangular number.
– John Wayland Bales
Nov 26 '18 at 19:48
I think you intended $a_n=a_{n-1}+1$ when $n$ is not a triangular number.
– John Wayland Bales
Nov 26 '18 at 19:48
@JohnWaylandBales You're correct! I'll edit my post.
– Jean Paul
Nov 26 '18 at 19:56
@JohnWaylandBales You're correct! I'll edit my post.
– Jean Paul
Nov 26 '18 at 19:56
Look at the Mathematica section of oeis.org/A002262
– John Wayland Bales
Nov 26 '18 at 19:59
Look at the Mathematica section of oeis.org/A002262
– John Wayland Bales
Nov 26 '18 at 19:59
@JohnWaylandBales Could you elaborate on it further? I'm not sure about how to decipher that entry...
– Jean Paul
Nov 26 '18 at 20:25
@JohnWaylandBales Could you elaborate on it further? I'm not sure about how to decipher that entry...
– Jean Paul
Nov 26 '18 at 20:25
I tried to decipher it but must have done it wrong, because it is not producing the right numbers. Perhaps someone who knows how to code in Mathematica can help.
– John Wayland Bales
Nov 26 '18 at 21:14
I tried to decipher it but must have done it wrong, because it is not producing the right numbers. Perhaps someone who knows how to code in Mathematica can help.
– John Wayland Bales
Nov 26 '18 at 21:14
add a comment |
1 Answer
1
active
oldest
votes
Adapting the Mathematica results from https://oeis.org/A002262, if we define
$$ m(k)=leftlfloor frac{sqrt{8k-7}-1}{2}rightrfloorquadtext{ for }kge1 $$
then
$$ a_k= k-m(k+1)cdotfrac{m(k+1)+1}{2}$$
This produces the sequence
$$ (0,0,1,0,1,2,0,1,2,3,0,1,2,3,4,0,1,2,3,4,5,0,ldots) $$
answered Nov 26 '18 at 22:03
John Wayland Bales
13.9k21237
add a comment |
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1 Answer
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oldest
votes
1 Answer
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active
oldest
votes
active
oldest
votes
active
oldest
votes
Adapting the Mathematica results from https://oeis.org/A002262, if we define
$$ m(k)=leftlfloor frac{sqrt{8k-7}-1}{2}rightrfloorquadtext{ for }kge1 $$
then
$$ a_k= k-m(k+1)cdotfrac{m(k+1)+1}{2}$$
This produces the sequence
$$ (0,0,1,0,1,2,0,1,2,3,0,1,2,3,4,0,1,2,3,4,5,0,ldots) $$
answered Nov 26 '18 at 22:03
John Wayland Bales
13.9k21237
add a comment |
Adapting the Mathematica results from https://oeis.org/A002262, if we define
$$ m(k)=leftlfloor frac{sqrt{8k-7}-1}{2}rightrfloorquadtext{ for }kge1 $$
then
$$ a_k= k-m(k+1)cdotfrac{m(k+1)+1}{2}$$
This produces the sequence
$$ (0,0,1,0,1,2,0,1,2,3,0,1,2,3,4,0,1,2,3,4,5,0,ldots) $$
answered Nov 26 '18 at 22:03
John Wayland Bales
13.9k21237
add a comment |
Adapting the Mathematica results from https://oeis.org/A002262, if we define
$$ m(k)=leftlfloor frac{sqrt{8k-7}-1}{2}rightrfloorquadtext{ for }kge1 $$
then
$$ a_k= k-m(k+1)cdotfrac{m(k+1)+1}{2}$$
This produces the sequence
$$ (0,0,1,0,1,2,0,1,2,3,0,1,2,3,4,0,1,2,3,4,5,0,ldots) $$
answered Nov 26 '18 at 22:03
John Wayland Bales
13.9k21237
Adapting the Mathematica results from https://oeis.org/A002262, if we define
$$ m(k)=leftlfloor frac{sqrt{8k-7}-1}{2}rightrfloorquadtext{ for }kge1 $$
then
$$ a_k= k-m(k+1)cdotfrac{m(k+1)+1}{2}$$
This produces the sequence
$$ (0,0,1,0,1,2,0,1,2,3,0,1,2,3,4,0,1,2,3,4,5,0,ldots) $$
answered Nov 26 '18 at 22:03
John Wayland Bales
13.9k21237
answered Nov 26 '18 at 22:03
John Wayland Bales
13.9k21237
answered Nov 26 '18 at 22:03
John Wayland Bales
13.9k21237
answered Nov 26 '18 at 22:03
John Wayland Bales
13.9k21237
13.9k21237
add a comment |
add a comment |
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I think you intended $a_n=a_{n-1}+1$ when $n$ is not a triangular number.
– John Wayland Bales
Nov 26 '18 at 19:48
@JohnWaylandBales You're correct! I'll edit my post.
– Jean Paul
Nov 26 '18 at 19:56
Look at the Mathematica section of oeis.org/A002262
– John Wayland Bales
Nov 26 '18 at 19:59
@JohnWaylandBales Could you elaborate on it further? I'm not sure about how to decipher that entry...
– Jean Paul
Nov 26 '18 at 20:25
I tried to decipher it but must have done it wrong, because it is not producing the right numbers. Perhaps someone who knows how to code in Mathematica can help.
– John Wayland Bales
Nov 26 '18 at 21:14