Where to find a list of the Padovan numbers? [closed]
$begingroup$
I am writing some code that calculates the Padovan sequence.
I want to test my code to check that the $73$rd term is correct (without using my program to calculate it, as that would defeat the purpose of the test).
Does anyone know where I can find a list of the first 100+ (the more the merrier) Padovan numbers?
sequences-and-series
edited Jan 2 at 1:40
Travis
61.9k767148
asked Jan 2 at 1:07
danday74danday74
1296
$endgroup$
closed as off-topic by Saad, Misha Lavrov, metamorphy, Paul Frost, user91500 Jan 2 at 11:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Misha Lavrov, metamorphy, Paul Frost, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I am writing some code that calculates the Padovan sequence.
I want to test my code to check that the $73$rd term is correct (without using my program to calculate it, as that would defeat the purpose of the test).
Does anyone know where I can find a list of the first 100+ (the more the merrier) Padovan numbers?
sequences-and-series
edited Jan 2 at 1:40
Travis
61.9k767148
asked Jan 2 at 1:07
danday74danday74
1296
$endgroup$
closed as off-topic by Saad, Misha Lavrov, metamorphy, Paul Frost, user91500 Jan 2 at 11:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Misha Lavrov, metamorphy, Paul Frost, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
It's a recursion, no? Just generate the terms on a spreadsheet. Usual is, I think, $a_0=1,a_1=0=a_2$ and $a_n=a_{n-2}+a_{n-3}$, no? If so, I see $a_{73}=145547525$ but maybe there are different definitions.
$endgroup$
– lulu
Jan 2 at 1:10
$begingroup$
thanks P(0) = P(1) = P(2) = 1
$endgroup$
– danday74
Jan 2 at 1:13
$begingroup$
with $a_0=a_1=a_2=1$ I see $a_{73}=593775046$
$endgroup$
– lulu
Jan 2 at 1:14
$begingroup$
thanks test passed :)
$endgroup$
– danday74
Jan 2 at 1:15
add a comment |
$begingroup$
I am writing some code that calculates the Padovan sequence.
I want to test my code to check that the $73$rd term is correct (without using my program to calculate it, as that would defeat the purpose of the test).
Does anyone know where I can find a list of the first 100+ (the more the merrier) Padovan numbers?
sequences-and-series
edited Jan 2 at 1:40
Travis
61.9k767148
asked Jan 2 at 1:07
danday74danday74
1296
$endgroup$
I am writing some code that calculates the Padovan sequence.
I want to test my code to check that the $73$rd term is correct (without using my program to calculate it, as that would defeat the purpose of the test).
Does anyone know where I can find a list of the first 100+ (the more the merrier) Padovan numbers?
sequences-and-series
sequences-and-series
edited Jan 2 at 1:40
Travis
61.9k767148
asked Jan 2 at 1:07
danday74danday74
1296
edited Jan 2 at 1:40
Travis
61.9k767148
asked Jan 2 at 1:07
danday74danday74
1296
edited Jan 2 at 1:40
Travis
61.9k767148
edited Jan 2 at 1:40
Travis
61.9k767148
edited Jan 2 at 1:40
Travis
61.9k767148
61.9k767148
asked Jan 2 at 1:07
danday74danday74
1296
asked Jan 2 at 1:07
danday74danday74
1296
asked Jan 2 at 1:07
danday74danday74
1296
1296
closed as off-topic by Saad, Misha Lavrov, metamorphy, Paul Frost, user91500 Jan 2 at 11:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Misha Lavrov, metamorphy, Paul Frost, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Misha Lavrov, metamorphy, Paul Frost, user91500 Jan 2 at 11:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Misha Lavrov, metamorphy, Paul Frost, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
It's a recursion, no? Just generate the terms on a spreadsheet. Usual is, I think, $a_0=1,a_1=0=a_2$ and $a_n=a_{n-2}+a_{n-3}$, no? If so, I see $a_{73}=145547525$ but maybe there are different definitions.
$endgroup$
– lulu
Jan 2 at 1:10
$begingroup$
thanks P(0) = P(1) = P(2) = 1
$endgroup$
– danday74
Jan 2 at 1:13
$begingroup$
with $a_0=a_1=a_2=1$ I see $a_{73}=593775046$
$endgroup$
– lulu
Jan 2 at 1:14
$begingroup$
thanks test passed :)
$endgroup$
– danday74
Jan 2 at 1:15
add a comment |
1
$begingroup$
It's a recursion, no? Just generate the terms on a spreadsheet. Usual is, I think, $a_0=1,a_1=0=a_2$ and $a_n=a_{n-2}+a_{n-3}$, no? If so, I see $a_{73}=145547525$ but maybe there are different definitions.
$endgroup$
– lulu
Jan 2 at 1:10
$begingroup$
thanks P(0) = P(1) = P(2) = 1
$endgroup$
– danday74
Jan 2 at 1:13
$begingroup$
with $a_0=a_1=a_2=1$ I see $a_{73}=593775046$
$endgroup$
– lulu
Jan 2 at 1:14
$begingroup$
thanks test passed :)
$endgroup$
– danday74
Jan 2 at 1:15
1
1
$begingroup$
It's a recursion, no? Just generate the terms on a spreadsheet. Usual is, I think, $a_0=1,a_1=0=a_2$ and $a_n=a_{n-2}+a_{n-3}$, no? If so, I see $a_{73}=145547525$ but maybe there are different definitions.
$endgroup$
– lulu
Jan 2 at 1:10
$begingroup$
It's a recursion, no? Just generate the terms on a spreadsheet. Usual is, I think, $a_0=1,a_1=0=a_2$ and $a_n=a_{n-2}+a_{n-3}$, no? If so, I see $a_{73}=145547525$ but maybe there are different definitions.
$endgroup$
– lulu
Jan 2 at 1:10
$begingroup$
thanks P(0) = P(1) = P(2) = 1
$endgroup$
– danday74
Jan 2 at 1:13
$begingroup$
thanks P(0) = P(1) = P(2) = 1
$endgroup$
– danday74
Jan 2 at 1:13
$begingroup$
with $a_0=a_1=a_2=1$ I see $a_{73}=593775046$
$endgroup$
– lulu
Jan 2 at 1:14
$begingroup$
with $a_0=a_1=a_2=1$ I see $a_{73}=593775046$
$endgroup$
– lulu
Jan 2 at 1:14
$begingroup$
thanks test passed :)
$endgroup$
– danday74
Jan 2 at 1:15
$begingroup$
thanks test passed :)
$endgroup$
– danday74
Jan 2 at 1:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The standard reference for integer sequences is the OEIS.
A quick search shows that the Padovan numbers are sequence A000931. NB for many sequences including this one there are varying conventions about the indexing of the sequence. (From the comments I see that OP is using convention $P(0) = P(1) = P(2) = 1$, and the entry's convention has a relative offset of $-5$ from this one.) The first entry in the $texttt{LINKS}$ subsection of the latter link is a list of the first $sim 8000$ entries of the sequence.
answered Jan 2 at 1:19
TravisTravis
61.9k767148
$endgroup$
$begingroup$
exactly what I needed many thanks
$endgroup$
– danday74
Jan 2 at 1:29
1
$begingroup$
Cheers, I'm glad you found it useful.
$endgroup$
– Travis
Jan 2 at 1:39
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The standard reference for integer sequences is the OEIS.
A quick search shows that the Padovan numbers are sequence A000931. NB for many sequences including this one there are varying conventions about the indexing of the sequence. (From the comments I see that OP is using convention $P(0) = P(1) = P(2) = 1$, and the entry's convention has a relative offset of $-5$ from this one.) The first entry in the $texttt{LINKS}$ subsection of the latter link is a list of the first $sim 8000$ entries of the sequence.
answered Jan 2 at 1:19
TravisTravis
61.9k767148
$endgroup$
$begingroup$
exactly what I needed many thanks
$endgroup$
– danday74
Jan 2 at 1:29
1
$begingroup$
Cheers, I'm glad you found it useful.
$endgroup$
– Travis
Jan 2 at 1:39
add a comment |
$begingroup$
The standard reference for integer sequences is the OEIS.
A quick search shows that the Padovan numbers are sequence A000931. NB for many sequences including this one there are varying conventions about the indexing of the sequence. (From the comments I see that OP is using convention $P(0) = P(1) = P(2) = 1$, and the entry's convention has a relative offset of $-5$ from this one.) The first entry in the $texttt{LINKS}$ subsection of the latter link is a list of the first $sim 8000$ entries of the sequence.
answered Jan 2 at 1:19
TravisTravis
61.9k767148
$endgroup$
$begingroup$
exactly what I needed many thanks
$endgroup$
– danday74
Jan 2 at 1:29
1
$begingroup$
Cheers, I'm glad you found it useful.
$endgroup$
– Travis
Jan 2 at 1:39
add a comment |
$begingroup$
The standard reference for integer sequences is the OEIS.
A quick search shows that the Padovan numbers are sequence A000931. NB for many sequences including this one there are varying conventions about the indexing of the sequence. (From the comments I see that OP is using convention $P(0) = P(1) = P(2) = 1$, and the entry's convention has a relative offset of $-5$ from this one.) The first entry in the $texttt{LINKS}$ subsection of the latter link is a list of the first $sim 8000$ entries of the sequence.
answered Jan 2 at 1:19
TravisTravis
61.9k767148
$endgroup$
The standard reference for integer sequences is the OEIS.
A quick search shows that the Padovan numbers are sequence A000931. NB for many sequences including this one there are varying conventions about the indexing of the sequence. (From the comments I see that OP is using convention $P(0) = P(1) = P(2) = 1$, and the entry's convention has a relative offset of $-5$ from this one.) The first entry in the $texttt{LINKS}$ subsection of the latter link is a list of the first $sim 8000$ entries of the sequence.
answered Jan 2 at 1:19
TravisTravis
61.9k767148
answered Jan 2 at 1:19
TravisTravis
61.9k767148
answered Jan 2 at 1:19
TravisTravis
61.9k767148
answered Jan 2 at 1:19
TravisTravis
61.9k767148
61.9k767148
$begingroup$
exactly what I needed many thanks
$endgroup$
– danday74
Jan 2 at 1:29
1
$begingroup$
Cheers, I'm glad you found it useful.
$endgroup$
– Travis
Jan 2 at 1:39
add a comment |
$begingroup$
exactly what I needed many thanks
$endgroup$
– danday74
Jan 2 at 1:29
1
$begingroup$
Cheers, I'm glad you found it useful.
$endgroup$
– Travis
Jan 2 at 1:39
$begingroup$
exactly what I needed many thanks
$endgroup$
– danday74
Jan 2 at 1:29
$begingroup$
exactly what I needed many thanks
$endgroup$
– danday74
Jan 2 at 1:29
1
1
$begingroup$
Cheers, I'm glad you found it useful.
$endgroup$
– Travis
Jan 2 at 1:39
$begingroup$
Cheers, I'm glad you found it useful.
$endgroup$
– Travis
Jan 2 at 1:39
add a comment |
1
$begingroup$
It's a recursion, no? Just generate the terms on a spreadsheet. Usual is, I think, $a_0=1,a_1=0=a_2$ and $a_n=a_{n-2}+a_{n-3}$, no? If so, I see $a_{73}=145547525$ but maybe there are different definitions.
$endgroup$
– lulu
Jan 2 at 1:10
$begingroup$
thanks P(0) = P(1) = P(2) = 1
$endgroup$
– danday74
Jan 2 at 1:13
$begingroup$
with $a_0=a_1=a_2=1$ I see $a_{73}=593775046$
$endgroup$
– lulu
Jan 2 at 1:14
$begingroup$
thanks test passed :)
$endgroup$
– danday74
Jan 2 at 1:15