Where to find a list of the Padovan numbers? [closed]

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0












$begingroup$


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?










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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
















0












$begingroup$


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?










share|cite|improve this question











$endgroup$



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














0












0








0





$begingroup$


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?










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 1:40









Travis

61.9k767148




61.9k767148










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














  • 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










1 Answer
1






active

oldest

votes


















5












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















5












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














5












5








5





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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



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k8WgWRBgH,TA4goB1I9,L5TAye,Ed4Vn9COb tYTnQtZDbUA1LEN,422iblkJoq 8n oLWQS2,3FuD0kF,tnzh

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