%I #35 Aug 14 2024 08:34:28
%S 0,1,4,10,10,15,16,24,30,35,40,46,54,55,60,60,66,69,70,70,70,71,74,80,
%T 80,85,86,94,100,105,110,116,124,125,130,130,136,139,140,140,140,141,
%U 144,150,150,155,156,164,170,175,180,186,194,195,200,200,206,209,210
%N Write the digit string 0123456789, repeated infinitely many times. Then, starting from the first "0" digit at the left end, move to the right by one digit (to the "1"), then two digits (to the "3"), then three digits (to the "6"), four digits ("0"), five digits ("5"), and so on. Partial sums of the digits thus reached are 0, 1, 4, 10, 10, 15, ...
%C The first differences 0, 1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, 0, etc. are in A008954.
%D Giorgio Balzarotti and Paolo P. Lava, Le sequenze di numeri interi, Hoepli, 2008, p. 62.
%H Michael De Vlieger, <a href="/A111072/b111072.txt">Table of n, a(n) for n = 0..10000</a>
%H J. Bokowski & N. J. A. Sloane, <a href="/A006248/a006248.pdf">Emails, June 1994</a>.
%F a(n+1) = a(n) + (a(n) - a(n-1) + (n+1) mod 10) mod 10, with a(0)=0, a(1)=1.
%F G.f.: x*(x^12+3*x^11+6*x^10+5*x^8+5*x^6+5*x^4+6*x^2+3*x+1) / (x^16 -x^15 -x^11 +x^10 +x^6 -x^5 -x +1). - _Alois P. Heinz_, Jan 23 2021
%e a(9) = 35 because a(8) - a(7) + (9 mod 10) = 30 - 24 + 9 = 15 and a(8) + (15 mod 10) = 30 + 5 = 35.
%e Jumping we move to the numbers 0, 1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, 0, 0, 1, 3, 6, 0, 5, 1, 8, 6, etc. Summing the numbers we obtain 0, 0+1 = 1, 1+3 = 4, 4+6 = 10, 10+0 = 10, 10+5 = 16, etc.
%p a:= proc(n) option remember; `if`(n=0, 0, a(n-1)+
%p [0,1,3,6,0,5,1,8,6,5,5,6,8,1,5,0,6,3,1,0,0]
%p [1+irem(n, 20)])
%p end:
%p seq(a(n), n=0..60); # _Alois P. Heinz_, Jan 23 2021
%t Fold[Append[#1, #1[[-1]] + Mod[(#1[[-1]] - #1[[-2]] + Mod[#2, 10]), 10]] &, {0, 1}, Range[2, 58]] (* _Michael De Vlieger_, Nov 05 2017 *)
%Y Cf. A008954.
%K nonn,base,easy
%O 0,3
%A _Giorgio Balzarotti_ and _Paolo P. Lava_, Oct 10 2005