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A111072
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, ...
4
0, 1, 4, 10, 10, 15, 16, 24, 30, 35, 40, 46, 54, 55, 60, 60, 66, 69, 70, 70, 70, 71, 74, 80, 80, 85, 86, 94, 100, 105, 110, 116, 124, 125, 130, 130, 136, 139, 140, 140, 140, 141, 144, 150, 150, 155, 156, 164, 170, 175, 180, 186, 194, 195, 200, 200, 206, 209, 210
OFFSET
0,3
COMMENTS
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.
REFERENCES
Giorgio Balzarotti and Paolo P. Lava, Le sequenze di numeri interi, Hoepli, 2008, p. 62.
LINKS
J. Bokowski & N. J. A. Sloane, Emails, June 1994.
FORMULA
a(n+1) = a(n) + (a(n) - a(n-1) + (n+1) mod 10) mod 10, with a(0)=0, a(1)=1.
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
EXAMPLE
a(9) = 35 because a(8) - a(7) + (9 mod 10) = 30 - 24 + 9 = 15 and a(8) + (15 mod 10) = 30 + 5 = 35.
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.
MAPLE
a:= proc(n) option remember; `if`(n=0, 0, a(n-1)+
[0, 1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, 0, 0]
[1+irem(n, 20)])
end:
seq(a(n), n=0..60); # Alois P. Heinz, Jan 23 2021
MATHEMATICA
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 *)
CROSSREFS
Cf. A008954.
Sequence in context: A081547 A264272 A264257 * A189895 A310333 A180862
KEYWORD
nonn,base,easy
AUTHOR
STATUS
approved