A346688 -id:A346688 - cp's OEIS frontend

A065359 Alternating bit sum for n: replace 2^k with (-1)^k in binary expansion of n.

0, 1, -1, 0, 1, 2, 0, 1, -1, 0, -2, -1, 0, 1, -1, 0, 1, 2, 0, 1, 2, 3, 1, 2, 0, 1, -1, 0, 1, 2, 0, 1, -1, 0, -2, -1, 0, 1, -1, 0, -2, -1, -3, -2, -1, 0, -2, -1, 0, 1, -1, 0, 1, 2, 0, 1, -1, 0, -2, -1, 0, 1, -1, 0, 1, 2, 0, 1, 2, 3, 1, 2, 0, 1, -1, 0, 1, 2, 0, 1, 2, 3, 1, 2, 3, 4, 2, 3, 1, 2, 0, 1, 2, 3, 1, 2, 0, 1, -1, 0, 1, 2, 0, 1, -1, 0, -2

Offset: 0

Marc LeBrun, Oct 31 2001

Notation: (2)[n](-1)
From David W. Wilson and Ralf Stephan, Jan 09 2007: (Start)
a(n) is even iff n in A001969; a(n) is odd iff n in A000069.
a(n) == 0 (mod 3) iff n == 0 (mod 3).
a(n) == 0 (mod 6) iff (n == 0 (mod 3) and n/3 not in A036556).
a(n) == 3 (mod 6) iff (n == 0 (mod 3) and n/3 in A036556). (End)
a(n) = A030300(n) - A083905(n). - Ralf Stephan, Jul 12 2003
From Robert G. Wilson v, Feb 15 2011: (Start)
First occurrence of k and -k: 0, 1, 2, 5, 10, 21, 42, 85, ..., (A000975); i.e., first 0 occurs for 0, first 1 occurs for 1, first -1 occurs at 2, first 2 occurs for 5, etc.;
  a(n)=-3 only if n mod 3 = 0,
  a(n)=-2 only if n mod 3 = 1,
  a(n)=-1 only if n mod 3 = 2,
  a(n)= 0 only if n mod 3 = 0,
  a(n)= 1 only if n mod 3 = 1,
  a(n)= 2 only if n mod 3 = 2,
  a(n)= 3 only if n mod 3 = 0, ..., . (End)
a(n) modulo 2 is the Prouhet-Thue-Morse sequence A010060. - Philippe Deléham, Oct 20 2011
In the Koch curve, number the segments starting with n=0 for the first segment. The net direction (i.e., the sum of the preceding turns) of segment n is a(n)*60 degrees. This is since in the curve each base-4 digit 0,1,2,3 of n is a sub-curve directed respectively 0, +60, -60, 0 degrees, which is the net 0,+1,-1,0 of two bits in the sum here. - Kevin Ryde, Jan 24 2020

			Alternating bit sum for 11 = 1011 in binary is 1 - 1 + 0 - 1 = -1, so a(11) = -1.

Antti Karttunen, Table of n, a(n) for n = 0..65535 (terms 0..1000 from Harry J. Smith)
J.-P. Allouche and J. Shallit, The ring of k-regular sequences, Theoretical Computer Sci., 98 (1992), 163-197. [Preprint.]
J.-P. Allouche and J. Shallit, The ring of k-regular sequences, II, Theoret. Computer Sci., 307 (2003), 3-29.
H.-K. Hwang, S. Janson and T.-H. Tsai. Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications. Preprint, 2016.
H.-K. Hwang, S. Janson and T.-H. Tsai. Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications. ACM Transactions on Algorithms, 13:4 (2017), #47. DOI:10.1145/3127585.
Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, Identities and periodic oscillations of divide-and-conquer recurrences splitting at half, arXiv:2210.10968 [cs.DS], 2022, p. 45.
Helge von Koch, Une Méthode Géométrique Élémentaire pour l'Étude de Certaines Questions de la Théorie des Courbes Planes, Acta Arithmetica, volume 30, 1906, pages 145-174.
William Paulsen, wpaulsen(AT)csm.astate.edu, Partitioning the [prime] maze
Ralf Stephan, Divide-and-conquer generating functions. I. Elementary sequences, arXiv:math/0307027 [math.CO], 2003.
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions

Cf. A005536 (partial sums), A056832 (abs first differences), A010060 (mod 2), A039004 (indices of 0's).
Cf. A000120, A065360, A065081, A019565, A195017.
Cf. also A004718.
Cf. analogous sequences for bases 3-10: A065368, A346688, A346689, A346690, A346691, A346731, A346732, A055017 and also A373605 (for primorial base).

a065359 0 = 0
a065359 n = - a065359 n' + m where (n', m) = divMod n 2
-- Reinhard Zumkeller, Mar 20 2015

A065359 := proc(n) local dgs ; dgs := convert(n,base,2) ; add( -op(i,dgs)*(-1)^i,i=1..nops(dgs)) ; end proc: # R. J. Mathar, Feb 04 2011

f[0]=0; f[n_] := Plus @@ (-(-1)^Range[ Floor[ Log2@ n + 1]] Reverse@ IntegerDigits[n, 2]); Array[ f, 107, 0]

a(n) = my(s=0, u=1); for(k=0,#binary(n)-1,s+=bittest(n,k)*u;u=-u);s /* Washington Bomfim, Jan 18 2011 */

a(n) = my(b=binary(n)); b*[(-1)^k|k<-[-#b+1..0]]~; \\ Ruud H.G. van Tol, Oct 16 2023

a(n) = if(n==0, 0, 2*hammingweight(bitand(n, ((4<<(2*logint(n,4)))-1)/3)) - hammingweight(n)) \\ Andrew Howroyd, Dec 14 2024

def a(n):
    return sum((-1)**k for k, bi in enumerate(bin(n)[2:][::-1]) if bi=='1')
print([a(n) for n in range(107)]) # Michael S. Branicky, Jul 13 2021

from sympy.ntheory import digits
def A065359(n): return sum((0,1,-1,0)[i] for i in digits(n,4)[1:]) # Chai Wah Wu, Jul 19 2024

G.f.: (1/(1-x)) * Sum_{k>=0} (-1)^k*x^2^k/(1+x^2^k). - Ralf Stephan, Mar 07 2003
a(0) = 0, a(2n) = -a(n), a(2n+1) = 1-a(n). - Ralf Stephan, Mar 07 2003
a(n) = Sum_{k>=0} A030308(n,k)*(-1)^k. - Philippe Deléham, Oct 20 2011
a(n) = -a(floor(n/2)) + n mod 2. - Reinhard Zumkeller, Mar 20 2015
a(n) = A139351(n) - A139352(n). - Kevin Ryde, Jan 24 2020
G.f. A(x) satisfies: A(x) = x / (1 - x^2) - (1 + x) * A(x^2). - Ilya Gutkovskiy, Jul 28 2021
a(n) = A195017(A019565(n)). - Antti Karttunen, Jun 19 2024

More terms from Ralf Stephan, Jul 12 2003

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, ...).

Original entry on oeis.org

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

Henry Bottomley, May 31 2000

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

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

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

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).

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

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

"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]

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)

A346690 Replace 6^k with (-1)^k in base-6 expansion of n.

Original entry on oeis.org

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

Offset: 0

Ilya Gutkovskiy, Jul 29 2021

If n has base-6 expansion abc..xyz with least significant digit z, a(n) = z - y + x - w + ...

			59 = 135_6, 5 - 3 + 1 = 3, so a(59) = 3.

Robert Israel, Table of n, a(n) for n = 0..10000

Cf. A007092, A053827, A055017, A065359, A065368, A346688, A346689, A346691.

f:= proc(n) option remember; (n mod 6) - procname(floor(n/6)) end proc:
f(0):= 0:
map(f, [$1..100]); # Robert Israel, Nov 21 2022

nmax = 104; A[] = 0; Do[A[x] = x (1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4)/(1 - x^6) - (1 + x + x^2 + x^3 + x^4 + x^5) A[x^6] + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
Table[n + 7 Sum[(-1)^k Floor[n/6^k], {k, 1, Floor[Log[6, n]]}], {n, 0, 104}]

a(n) = subst(Pol(digits(n, 6)), 'x, -1); \\ Michel Marcus, Nov 22 2022

from sympy.ntheory.digits import digits
def a(n):
    return sum(bi*(-1)**k for k, bi in enumerate(digits(n, 6)[1:][::-1]))
print([a(n) for n in range(105)]) # Michael S. Branicky, Jul 29 2021

G.f. A(x) satisfies: A(x) = x * (1 + 2*x + 3*x^2 + 4*x^3 + 5*x^4) / (1 - x^6) - (1 + x + x^2 + x^3 + x^4 + x^5) * A(x^6).
a(n) = n + 7 * Sum_{k>=1} (-1)^k * floor(n/6^k).
a(6*n+j) = j - a(n) for 0 <= j <= 5. - Robert Israel, Nov 21 2022

A373605 Sum of the even-indexed digits minus the sum of the odd-indexed digits in the primorial base representation (A049345) of n.

Original entry on oeis.org

0, 1, -1, 0, -2, -1, 1, 2, 0, 1, -1, 0, 2, 3, 1, 2, 0, 1, 3, 4, 2, 3, 1, 2, 4, 5, 3, 4, 2, 3, -1, 0, -2, -1, -3, -2, 0, 1, -1, 0, -2, -1, 1, 2, 0, 1, -1, 0, 2, 3, 1, 2, 0, 1, 3, 4, 2, 3, 1, 2, -2, -1, -3, -2, -4, -3, -1, 0, -2, -1, -3, -2, 0, 1, -1, 0, -2, -1, 1, 2, 0, 1, -1, 0, 2, 3, 1, 2, 0, 1, -3, -2, -4, -3, -5, -4, -2, -1, -3

Offset: 0

Antti Karttunen, Jun 18 2024

Alternating digit sum in primorial base, starting with a positive sign for the rightmost (least significant) digit.

			A049345(85) = 2401, thus the sum of digits at even positions (with the rightmost digit having index 0) is 1+4 = 5, and at the odd positions 0+2 = 2, therefore a(85) = 5-2 = 3.

Antti Karttunen, Table of n, a(n) for n = 0..16170
Index entries for sequences related to primorial base

Cf. A049345, A195017, A276086, A373606, A373607, A373830, A373831 (indices of multiples of 3).

Analogous sequences for bases 2-10: A065359, A065368, A346688, A346689, A346690, A346691, A346731, A346732, A055017.

A373605(n) = { my(p=2, i=1, s=0); while(n, s += i*(n%p); n = n\p; p = nextprime(1+p); i = -i); (s); };

a(n) = A373606(n) - A373607(n).

a(n) = A195017(A276086(n)).

A346689 Replace 5^k with (-1)^k in base-5 expansion of n.

Original entry on oeis.org

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

Offset: 0

Ilya Gutkovskiy, Jul 29 2021

If n has base-5 expansion abc..xyz with least significant digit z, a(n) = z - y + x - w + ...

			48 = 143_5, 3 - 4 + 1 = 0, so a(48) = 0.

Cf. A007091, A053824, A055017, A065359, A065368, A346688, A346690, A346691.

nmax = 104; A[] = 0; Do[A[x] = x (1 + 2 x + 3 x^2 + 4 x^3)/(1 - x^5) - (1 + x + x^2 + x^3 + x^4) A[x^5] + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
Table[n + 6 Sum[(-1)^k Floor[n/5^k], {k, 1, Floor[Log[5, n]]}], {n, 0, 104}]

from sympy.ntheory.digits import digits
def a(n):
    return sum(bi*(-1)**k for k, bi in enumerate(digits(n, 5)[1:][::-1]))
print([a(n) for n in range(105)]) # Michael S. Branicky, Jul 29 2021

G.f. A(x) satisfies: A(x) = x * (1 + 2*x + 3*x^2 + 4*x^3) / (1 - x^5) - (1 + x + x^2 + x^3 + x^4) * A(x^5).

a(n) = n + 6 * Sum_{k>=1} (-1)^k * floor(n/5^k).

A346691 Replace 7^k with (-1)^k in base-7 expansion of n.

Original entry on oeis.org

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

Offset: 0

Ilya Gutkovskiy, Jul 29 2021

If n has base-7 expansion abc..xyz with least significant digit z, a(n) = z - y + x - w + ...

			83 = 146_7, 6 - 4 + 1 = 3, so a(83) = 3.

Cf. A007093, A053828, A055017, A065359, A065368, A346688, A346689, A346690.

nmax = 104; A[] = 0; Do[A[x] = x (1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4 + 6 x^5)/(1 - x^7) - (1 + x + x^2 + x^3 + x^4 + x^5 + x^6) A[x^7] + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
Table[n + 8 Sum[(-1)^k Floor[n/7^k], {k, 1, Floor[Log[7, n]]}], {n, 0, 104}]

from sympy.ntheory.digits import digits
def a(n):
    return sum(bi*(-1)**k for k, bi in enumerate(digits(n, 7)[1:][::-1]))
print([a(n) for n in range(105)]) # Michael S. Branicky, Jul 29 2021

G.f. A(x) satisfies: A(x) = x * (1 + 2*x + 3*x^2 + 4*x^3 + 5*x^4 + 6*x^5) / (1 - x^7) - (1 + x + x^2 + x^3 + x^4 + x^5 + x^6) * A(x^7).

a(n) = n + 8 * Sum_{k>=1} (-1)^k * floor(n/7^k).

A346731 Replace 8^k with (-1)^k in base-8 expansion of n.

Original entry on oeis.org

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

Offset: 0

Ilya Gutkovskiy, Jul 30 2021

If n has base-8 expansion abc..xyz with least significant digit z, a(n) = z - y + x - w + ...

			79 = 117_8, 7 - 1 + 1 = 7, so a(79) = 7.

Cf. A007094, A053829, A055017, A065359, A065368, A346688, A346689, A346690, A346691, A346732.

nmax = 104; A[] = 0; Do[A[x] = x (1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4 + 6 x^5 + 7 x^6)/(1 - x^8) - (1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7) A[x^8] + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
Table[n + 9 Sum[(-1)^k Floor[n/8^k], {k, 1, Floor[Log[8, n]]}], {n, 0, 104}]

from sympy.ntheory.digits import digits
def a(n):
    return sum(bi*(-1)**k for k, bi in enumerate(digits(n, 8)[1:][::-1]))
print([a(n) for n in range(105)]) # Michael S. Branicky, Jul 31 2021

G.f. A(x) satisfies: A(x) = x * (1 + 2*x + 3*x^2 + 4*x^3 + 5*x^4 + 6*x^5 + 7*x^6) / (1 - x^8) - (1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7) * A(x^8).

a(n) = n + 9 * Sum_{k>=1} (-1)^k * floor(n/8^k).

A346732 Replace 9^k with (-1)^k in base-9 expansion of n.

Original entry on oeis.org

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

Offset: 0

Ilya Gutkovskiy, Jul 30 2021

If n has base-9 expansion abc..xyz with least significant digit z, a(n) = z - y + x - w + ...

			89 = 108_9, 8 - 0 + 1 = 9, so a(89) = 9.

Cf. A007095, A053830, A055017, A065359, A065368, A346688, A346689, A346690, A346691, A346731.

nmax = 104; A[] = 0; Do[A[x] = x (1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4 + 6 x^5 + 7 x^6 + 8 x^7)/(1 - x^9) - (1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8) A[x^9] + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
Table[n + 10 Sum[(-1)^k Floor[n/9^k], {k, 1, Floor[Log[9, n]]}], {n, 0, 104}]

from sympy.ntheory.digits import digits
def a(n):
    return sum(bi*(-1)**k for k, bi in enumerate(digits(n, 9)[1:][::-1]))
print([a(n) for n in range(105)]) # Michael S. Branicky, Jul 31 2021

G.f. A(x) satisfies: A(x) = x * (1 + 2*x + 3*x^2 + 4*x^3 + 5*x^4 + 6*x^5 + 7*x^6 + 8*x^7) / (1 - x^9) - (1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8) * A(x^9).

a(n) = n + 10 * Sum_{k>=1} (-1)^k * floor(n/9^k).

cp's OEIS Frontend

A065359 Alternating bit sum for n: replace 2^k with (-1)^k in binary expansion of n.

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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, ...).

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A346690 Replace 6^k with (-1)^k in base-6 expansion of n.

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A373605 Sum of the even-indexed digits minus the sum of the odd-indexed digits in the primorial base representation (A049345) of n.

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A346689 Replace 5^k with (-1)^k in base-5 expansion of n.

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A346691 Replace 7^k with (-1)^k in base-7 expansion of n.

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