Question #267832

 For each list of integers, provide a simple formula or rule that generates the terms of an integer sequence that begins with the given list. Assume the sequence starts with . Do not forget initial conditions if required.

a)     7, 11, 15, 19, 23, 27, 31, 35, 39, 43

b)    -1, 2, 1, 3, 4, 7, 11, 18, 29, 47


1
Expert's answer
2021-11-22T14:36:20-0500

a)


a2a1=117=4,a3a2=1511=4,a_2-a_1=11-7=4,a_3-a_2=15-11=4,

a4a3=1915=4,a5a4=2319=4,a_4-a_3=19-15=4,a_5-a_4=23-19=4,

a6a5=2723=4,a7a6=3127=4,a_6-a_5=27-23=4,a_7-a_6=31-27=4,

The sequence

a8a7=3531=4,a9a8=3935=4,a_8-a_7=35-31=4,a_9-a_8=39-35=4,

a10a9=4339=4a_{10}-a_9=43-39=4

The sequence {an}\{a_n\} is arithmetic progression with a=7,d=4.a=7, d=4.


b)

a0=1,a1=2,a_0=-1, a_1=2,

a3=1=1+2=a1+a2,a_3=1=-1+2=a_1+a_2,

a4=3=2+1=a2+a3,a_4=3=2+1=a_2+a_3,

a5=4=1+3=a3+a4,a_5=4=1+3=a_3+a_4,

a6=7=3+4=a4+a5,a_6=7=3+4=a_4+a_5,

a7=11=4+7=a5+a6,a_7=11=4+7=a_5+a_6,

a8=18=7+11=a6+a7,a_8=18=7+11=a_6+a_7,

a9=29=11+18=a7+a8,a_9=29=11+18=a_7+a_8,

a10=47=18+29=a8+a9a_{10}=47=18+29=a_8+a_9

The sequence {an}\{a_n\} satisfies the recurrence relation an=an2+an1,n=2,3,...a_n=a_{n-2}+a_{n-1}, n=2, 3, ... with a0=1,a1=2.a_0=-1, a_1=2.


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