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

Author

Henry Bottomley, May 31 2000

Keywords

• base
• easy
• sign

Comments

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

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

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

Examples

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

Links

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

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.

Cf. analogous sequences for bases 2-9: A065359, A065368, A346688, A346689, A346690, A346691, A346731, A346732 and also A373605 (for primorial base).

Programs

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

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

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]

Formula

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

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

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

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

a(n) = n + 11*Sum_{k=10..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, otherwise b(k)=0.

G.f.: (1/(1-x)) * Sum_{k>=1} (1+11*c(k))*x^k, where c(k) = Sum_{j>=2,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>=1} (-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)