A256379 Cumulative sum for n in base 10 when alternately adding and subtracting each digit of a particular value.

1, 3, 6, 10, 15, 21, 28, 36, 45, 44, 44, 43, 39, 36, 30, 25, 17, 10, 0, 2, 1, 1, 6, 8, 15, 19, 28, 34, 45, 42, 44, 39, 39, 38, 30, 27, 17, 12, 0, 4, 1, 7, 6, 6, 15, 17, 28, 32, 45, 40, 44, 37, 39, 30, 30, 29, 17, 14, 0, 6, 1, 9, 6, 16, 15, 15, 28, 30, 45, 38, 44, 35, 39, 28, 30, 17, 17, 16, 0, 8, 1, 11, 6, 18, 15, 29, 28, 28, 45, 36, 44, 33, 39, 26, 30

Offset: 1

Author

Anthony Sand, Mar 27 2015

Keywords

• nonn
• base

Comments

For n = 1..9, the function is encountering each digit for the first time, therefore each is added to the cumulative sum. At n = 9, the sum is 45. At n = 10, the sum is 44, because the digit 1 is encountered for the second time and is therefore subtracted. At n = 11, the sum is again 44, because 1 is added and then subtracted. At n = 12, the sum is 43, because 1 is added and 2, encountered for the second time, is subtracted.

0 <= a(n) <= 45 for all n; a(n) = a(n mod 20) for odd n. - Danny Rorabaugh, Mar 31 2015

Links

• Anthony Sand, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A037123, A167232, A256100, A256851 (first differences).

Programs

Mathematica

f[n_] := Block[{g, r = PadRight[Range@ 9, 10]}, g[x_] := Boole[OddQ /@ DigitCount[x]]; Total[r Boole[OddQ /@ Total[g /@ Range@ n]]]]; Array[f, 120] (* Michael De Vlieger, Mar 29 2015 *)

PARI

{ nmx=1000; b=10; dig=vector(b); for(i=1,b,dig[i]=1); n=0; s=0; while(n
				

Formula

a(n) = Sum_{k=1..n} M(k), with M(k) := Sum_{m=1..r(k)} (-1)^(a(k,m) + 1)*digit(k,m), where a(k,m) = A256100(k,m) read as an array with row length r(k) (number of digits of k), and digit(k,m) is the m-th digit of k. - Wolfdieter Lang, Apr 08 2015