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A111072
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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, ...
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4
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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
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OFFSET
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0,3
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COMMENTS
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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.
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REFERENCES
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G. Balzarotti and P. P. Lava, Le sequenze di numeri interi, Hoepli, 2008, p. 62.
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LINKS
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FORMULA
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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
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EXAMPLE
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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.
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MAPLE
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ANM:=proc(N) global anplus1, anminus1; local an, i, anpolus; anminus1:=0; an:=1; print (anminus1, an); for i from 2 by 1 to N do anplus1:=an+((an-anminus1+ i mod 10) mod 10); print(anplus1); anminus1:=an; an:=anplus1; od; end: ANM(100);
# second Maple program:
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:
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MATHEMATICA
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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 *)
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CROSSREFS
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KEYWORD
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nonn,base,easy
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AUTHOR
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STATUS
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approved
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