Loading [MathJax]/jax/output/SVG/jax.js
 
+0  
 
0
1259
5
avatar

Please give the next 4 terms of this sequence. Thanks for any help:

2, 4, 7, 10, 15, 18, 23, 26............640 (is 100th term)

 Jan 14, 2016

Best Answer 

 #5
avatar+26396 
+15

Please give the next 4 terms of this sequence. Thanks for any help:

2, 4, 7, 10, 15, 18, 23, 26............640 (is 100th term)

 

Thanks heureka: brilliant work!. In my book it is stated slightly different: It is the sequence of natual or counting numbers added to prime numbers - 1. So, we have:

Counting numbers: 1, 2, 3, 4, 5, 6............

Prime numbers    + 2, 3, 5, 7, 11, 13...............

Subtract 1  from      2, 4, 7, 10, 15, 18..........etc.

each term

 

The formula is an=an1+p(n)p(n1)+1n2a1=2 a1=2a2=a1+p(2)p(1)+1=2+p(2)2+1=p(2)+1a3=a2+p(3)p(2)+1=( p(2)+1 )+p(3)p(2)+1=p(3)+2a4=a3+p(4)p(3)+1=( p(3)+2 )+p(4)p(3)+1=p(4)+3a5=a4+p(5)p(4)+1=( p(4)+3 )+p(5)p(4)+1=p(5)+4an=an1+p(n)p(n1)+1=( p(n1)+n2 )+p(n)p(n1)+1=p(n)+n1 an=n+p(n)1 

 

laugh

 Jan 15, 2016
edited by heureka  Jan 15, 2016
edited by heureka  Jan 15, 2016
 #1
avatar+26396 
+15

Please give the next 4 terms of this sequence. Thanks for any help:

2, 4, 7, 10, 15, 18, 23, 26............640 (is 100th term)

 

We need the prime number table

 

The prime numbers. np(n)np(n)np(n)np(n)np(n)np(n)1221734117961283814191015472322794218162293824213523834319163307834314724894419364311844335112597451976531385439613261014619966317864437172710347211673318744981928107482236833788457923291094922769347894611029301135022970349904631131311275123371353914671237321315223972359924791341331375324173367934871443341395425174373944911547351495525775379954991653361515626376383965031759371575726977389975091861381635827178397985211967391675927779401995232071401736028180409100541

 

The formula is  an=an1+p(n)p(n1)+1n2a1=2 

 

a1=2a2=a1+p(2)p(1)+1=2+32+1=4a3=a2+p(3)p(2)+1=4+53+1=7a4=a3+p(4)p(3)+1=7+75+1=10a5=a4+p(5)p(4)+1=10+117+1=15a6=a5+p(6)p(5)+1=15+1311+1=18a7=a6+p(7)p(6)+1=18+1713+1=23a8=a7+p(8)p(7)+1=23+1917+1=26a9=a8+p(9)p(8)+1=26+2319+1=31a10=a9+p(10)p(9)+1=31+2923+1=38a11=a10+p(11)p(10)+1=38+3129+1=41a12=a11+p(12)p(11)+1=41+3731+1=48a100=a99+p(100)p(99)+1=621+541523+1=640

 

a1a50na(n)na(n)na(n)na(n)na(n)1211412193311574121924124822100321624222237135323105331694323341014562411234172442365151561251213518345241618166826126361864624472317752712937193472578261878281343820048270931198529137392054927510382090301424021250278

 

a51a101na(n)na(n)na(n)na(n)na(n)na(n)51283613437142381499915571016475229062354724308250292570532936336973439835139357954304643747444684516945845531165377754538552395593563186638276458865289659857325673977746587535976055832868404784748854498618593356941579479895499962160340704188048890552100640

 

laugh

 Jan 14, 2016
 #2
avatar+118696 
+5

Thanks Heureka,

Here is a site that you will like guest :)

It is the

Online Encyclopeadia of Integer Sequences  :)

 

https://oeis.org/search?q=2%2C+4%2C+7%2C+10%2C+15%2C+18%2C+23%2C+26&sort=&language=english&go=Search

 Jan 14, 2016
 #3
avatar
+5

Thanks heureka: brilliant work!. In my book it is stated slightly different: It is the sequence of natual or counting numbers added to prime numbers - 1. So, we have:

Counting numbers: 1, 2, 3, 4, 5, 6............

Prime numbers    + 2, 3, 5, 7, 11, 13...............

Subtract 1  from      2, 4, 7, 10, 15, 18..........etc.

each term

 Jan 14, 2016
 #4
avatar+130466 
0

That really IS brilliant, Heureka....!!!!!

 

 

cool cool cool

 Jan 14, 2016
 #5
avatar+26396 
+15
Best Answer

Please give the next 4 terms of this sequence. Thanks for any help:

2, 4, 7, 10, 15, 18, 23, 26............640 (is 100th term)

 

Thanks heureka: brilliant work!. In my book it is stated slightly different: It is the sequence of natual or counting numbers added to prime numbers - 1. So, we have:

Counting numbers: 1, 2, 3, 4, 5, 6............

Prime numbers    + 2, 3, 5, 7, 11, 13...............

Subtract 1  from      2, 4, 7, 10, 15, 18..........etc.

each term

 

The formula is an=an1+p(n)p(n1)+1n2a1=2 a1=2a2=a1+p(2)p(1)+1=2+p(2)2+1=p(2)+1a3=a2+p(3)p(2)+1=( p(2)+1 )+p(3)p(2)+1=p(3)+2a4=a3+p(4)p(3)+1=( p(3)+2 )+p(4)p(3)+1=p(4)+3a5=a4+p(5)p(4)+1=( p(4)+3 )+p(5)p(4)+1=p(5)+4an=an1+p(n)p(n1)+1=( p(n1)+n2 )+p(n)p(n1)+1=p(n)+n1 an=n+p(n)1 

 

laugh

heureka Jan 15, 2016
edited by heureka  Jan 15, 2016
edited by heureka  Jan 15, 2016

3 Online Users

avatar