A055017 Difference between sums of alternate digits of n starting with the last, i.e., (Sum of ultimate digit of n, antepenultimate digit of n,...)-(sum of penultimate digit of n, preantepenultimate digit of n,...).

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, -8, -7, -6, -5, -4, -3

Offset

0,3

Comments

n is divisible by 11 iff a(n) is a multiple of 11

Digital sum with alternating signs starting with a positive sign for the rightmost digit. - Hieronymus Fischer, Jun 18 2007

For n<100 equal to (n mod 10 - floor(n/10)) = -A076313(n). - Hieronymus Fischer, Jun 18 2007

Links

Hieronymus Fischer, Table of n, a(n) for n = 0..10000

Formula

From Hieronymus Fischer, Jun 18 2007, Jun 25 2007, Mar 23 2014: (Start)

a(n) = n + 11*Sum_{k>0} (-1)^k*floor(n/10^k).

a(10n+k) = k-a(n), 0<=k<10.

G.f.: Sum_{k>0} (x^k-x^(k+10^k)+(-1)^k*11*x^(10^k))/(1-x^(10^k)) / (1-x).

a(n) = n+11*sum{10<=k<=n, sum{j|k,j>=10, (-1)^floor(log_10(j))*(floor(log_10(j))-floor(log_10(j-1)))}}.

G.f. expressed in terms of Lambert series: g(x)=(x/(1-x)+11*L[b(k)](x))/(1-x) where L[b(k)](x)=sum{k>=0, b(k)*x^k/(1-x^k)} is a Lambert series with b(k)=(-1)^floor(log_10(k)), if k>1 is a power of 10, else b(k)=0.

G.f.: sum{k>0, (1+11*c(k))*x^k}/(1-x), where c(k)=sum{j>1,j|k, (-1)^floor(log_10(j))*(floor(log_10(j))-floor(log_10(j-1)))}.

Formulas for general bases b > 1 (b = 10 for this sequence).

a(n) = sum_{k>=0} (-1)^k*(floor(n/b^k) mod b).

a(n) = n + (b+1)*sum_{k>0} (-1)^k*floor(n/b^k). Both sums are finite with floor(log_b(n)) as the highest index.

a(n) = a(n mod b^k) + (-1)^k*a(floor(n/b^k)), for all k>=0.

a(n) = a(n mod b) - a(floor(n/b)).

a(n) = a(n mod b^2) + a(floor(n/b^2)).

a(n) = (-1)^m*A225693(n), where m = floor(log_b(n)).

a(n) = (-1)^k*A225693(A004086(n)), where k = is the number of trailing 0's of n, formally, k = max(j | n == 0 mod 10^j).

a(A004086(A004086(n))) = (-1)^k*a(n), where k = is the number of trailing 0's in the decimal representation of n. (End)

Example

a(123)=3-2+1=2, a(9875)=5-7+8-9=-3.

Maple

sumodigs := proc(n) local dg; dg := convert(n, base, 10) ; add(op(1+2*i, dg), i=0..floor(nops(dg)-1)/2) ; end proc:
sumedigs := proc(n) local dg; dg := convert(n, base, 10) ; add(op(2+2*i, dg), i=0..floor(nops(dg)-2)/2) ; end proc:
A055017 := proc(n) sumodigs(n)-sumedigs(n) ; end proc: # R. J. Mathar, Aug 26 2011

Prog

(Smalltalk)
"Recursive version for general bases"
"Set base = 10 for this sequence"
altDigitalSumRight: base
| s |
base = 1 ifTrue: [^self \\ 2].
(s := self // base) > 0
  ifTrue: [^(self - (s * base) - (s altDigitalSumRight: base))]
  ifFalse: [^self]
[by Hieronymus Fischer, Mar 23 2014]
(Python)
def A055017(n): return sum((-1 if i % 2 else 1)*int(j) for i, j in enumerate(str(n)[::-1])) # Chai Wah Wu, May 11 2022

Crossrefs

Cf. A225693 (alternating sum of digits).

Unsigned version differs from A040114 and A040115 when n=100 and from A040997 when n=101.

Cf. A076313, A076314, A007953, A003132.

Cf. A004086.

Sequence in context: A274580 A241494 A076313 * A225693 A040997 A256851

Adjacent sequences: A055014 A055015 A055016 * A055018 A055019 A055020

Keyword

base,easy,sign

Author

Henry Bottomley, May 31 2000

Status

approved