A034804 Consider the sequence of 4-tuples {0,a,b,c} (c>=a+b; a,b,c>0) which have the smallest integer 'c' required to reach {k,k,k,k} in n steps under map {r,s,t,u}->{|r-s|,|s-t|,|t-u|,|u-r|}. This sequence gives the third term 'b' of these quadruples.
0, 1, 0, 2, 1, 2, 4, 5, 6, 14, 17, 20, 48, 57, 68, 162, 193, 230, 548, 653, 778, 1854, 2209, 2632, 6272, 7473, 8904, 21218, 25281, 30122, 71780, 85525, 101902, 242830, 289329, 344732, 821488, 978793, 1166220, 2779074, 3311233, 3945294, 9401540
Offset: 1
Keywords
Examples
a(10)=14 because {0, 5, 14, 31}->{5, 9, 17, 31}->{4, 8, 14, 26}->{4, 6, 12, 22}->{2, 6, 10, 18}->{4, 4, 8, 16}->{0, 4, 8, 12}->{4, 4, 4, 12}->{0, 0, 8, 8}->{0, 8, 0, 8}->{8, 8, 8, 8} ('b'=14 in the first 4-tuple and there is no quadruple with a+b<=c<=31 and 10 steps).
Formula
a(n)= 2*Trib(2*q) if r=0; Trib(2*q-1)+Trib(2*q+1) if r=1; Trib(2*q)+Trib(2*q+1) if r=2 where q=[(n-1)/3], r=n-1 (mod 3) and Trib denotes the tribonacci sequence (A000073) with Trib(-1)=1. G.f.: (-x^7+2*x^6-2*x^5+2*x^4-2*x^3-x)/(x^9+x^6+3*x^3-1). Recurrence: a(n)=3*a(n-3)+a(n-6)+a(n-9), n >= 10.
Extensions
Better description, more terms, formula, etc. from Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Jun 24 2001