what is next three numbers in sequence 5, 11,9, 18, 23

Question

What is the pattern?

Answer 1

Let f(n)=k+an+bn²+cn³+dn⁴.

f(0)=5=k

f(1)=a+b+c+d=6

f(2)=2a+4b+8c+16d=4

f(3)=3a+9b+27c+81d=13

f(4)=4a+16b+64c+256d=18

f(2)-2f(1)=2b+6c+14d=-8 ①

f(3)-3f(1)=6b+24c+78d=-5 ②

f(4)-4f(1)=12b+60c+252d=-6, 2b+10c+42d=-1 ③

③-①: 4c+28d=7 ④

②-3①: 6c+36d=19 ⑤

3④-2⑤: 12d=-17, d=-17/12⇒c=35/3.

From ①, b+3c+7d=-4, b+35-119/12=-4, b=-349/12.

f(1)=a+b+c+d=a-349/12+35/3-17/12=6, a=149/6.

f(n)=5+149n/6-349n²/12+35n³/3-17n⁴/12. This is a pattern for all terms in the series, t_n=f(n) where t_n is the nth term and integer n≥0. f(n) can be written f(n)=(60+298n-349n²+140n³-17n⁴)/12.

This gives the numeric series 5, 11, 9, 18, 23, -25, -209, -646, -1487, -2917, -5155, ... for n≥0.

This is only one solution. There are many others (probably simpler!).

Another possibility is 5, 11, 9, 18, 23, 35, 47, 62 where the pattern is two interlaced series:

5, 9, 23, 47, 81 where the differences between the terms are 4, 14, 24, 34...;

11, 18, 35, 62 where the differences are 7, 17, 27...