A070565 -id:A070565 - cp's OEIS frontend

A062329 a(n) = (sum of digits of n) - (product of digits of n).

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 0, -1, -2, -3, -4, -5, -6, -7, 3, 1, -1, -3, -5, -7, -9, -11, -13, -15, 4, 1, -2, -5, -8, -11, -14, -17, -20, -23, 5, 1, -3, -7, -11, -15, -19, -23, -27, -31, 6, 1, -4, -9, -14, -19, -24, -29, -34, -39, 7, 1, -5, -11, -17, -23, -29, -35, -41, -47, 8, 1, -6, -13, -20, -27

Offset: 0

Amarnath Murthy, Jun 21 2001

			a(23) = 2 + 3 - 2*3 = -1.
a(49) = -(4*9) + (4 + 9) = -36 + 13 = -23.

Harry J. Smith, Table of n, a(n) for n = 0..1000

Cf. A007953, A007954, A034710, A062996, A062997, A062998, A062999, A070565.

A063543 has a similar graph.

a[n_] := (t = IntegerDigits[n]; Plus @@ t - Times @@ t); Table[ a[n], {n, 0, 75}] (* Robert G. Wilson v *)

Corrected and extended by Larry Reeves (larryr(AT)acm.org), Jun 22 2001

Signed version from Henry Bottomley, Jun 29 2001

A178500 a(n) = 10^n * signum(n).

Original entry on oeis.org

0, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000, 1000000000, 10000000000, 100000000000, 1000000000000, 10000000000000, 100000000000000, 1000000000000000, 10000000000000000, 100000000000000000, 1000000000000000000, 10000000000000000000, 100000000000000000000

Offset: 0

Reinhard Zumkeller, May 28 2010

a(n-1) is the minimum difference between an n-digit number (written in base 10, nonzero leading digit) and the product of its digits. For n > 1, it is also a number meeting that bound. See A070565. - Devin Akman, Apr 17 2019

Michael De Vlieger, Table of n, a(n) for n = 0..999
Index entries for linear recurrences with constant coefficients, signature (10).

Cf. A000533, A011557, A057427, A061601, A070565, A109002, A155559, A178501.

Partial sums of A063945.

Array[10^#*Sign[#] &, 20, 0] (* Michael De Vlieger, Apr 21 2019 *)

a(n) = A011557(n)*A057427(n).
For n > 0, a(n) = A011557(n).
a(n) = 10*A178501(n).
a(n) = A000533(n) - 1.
A061601(a(n)) = A109002(n+1).
From Elmo R. Oliveira, Jul 21 2025: (Start)
G.f.: 10*x/(1-10*x).
E.g.f.: 2*exp(5*x)*sinh(5*x).
a(n) = 10*a(n-1) for n > 1. (End)

A229547 Numbers n such that n - product_of_digits(n) is a palindrome.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 29, 34, 46, 57, 61, 78, 82, 93, 101, 129, 143, 187, 202, 218, 226, 244, 247, 252, 269, 294, 297, 303, 319, 336, 348, 357, 361, 364, 386, 404, 412, 419, 437, 453, 462, 488, 505, 514, 519, 524, 534, 539, 544, 554, 564, 574, 584, 594, 597, 606, 613, 615, 617, 619, 625, 635, 638, 645, 655, 663

Offset: 1

Derek Orr, Sep 26 2013

			143 - (1*4*3) = 131 (a palindrome). So, 143 is a member of the sequence.

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Cf. A070565.

f[n_] := Block[{d = n - Times @@ IntegerDigits@ n}, d == FromDigits@ Reverse[IntegerDigits@ d]]; Select[Range[0, 1000], f] (* Michael De Vlieger, Mar 12 2015 *)

for(n=0,10^3,d=digits(n);p=prod(i=1,#d,d[i]);if(Vecrev(digits(n-p))==digits(n-p),print1(n,", "))) \\ Derek Orr, Mar 12 2015

def rev(n):
    return int(''.join(reversed(str(n))))
def DP(n):
    p = 1
    for i in str(n):
        p *= int(i)
    return p
{print(n, end=', ') for n in range(10**3) if n-DP(n)==rev(n-DP(n))}
# Simplified by Derek Orr, Mar 12 2015

More terms from Derek Orr, Mar 12 2015

A124167 a(n) = 10*(10^n-1).

Original entry on oeis.org

0, 90, 990, 9990, 99990, 999990, 9999990, 99999990, 999999990, 9999999990, 99999999990, 999999999990, 9999999999990, 99999999999990, 999999999999990, 9999999999999990, 99999999999999990, 999999999999999990

Offset: 0

Zerinvary Lajos, Dec 02 2006

a(n - 1) is the maximum difference between an n-digit number (written in base 10, nonzero leading digit) and the product of its digits. For n>1, it is also a number meeting that bound. See A070565. - Devin Akman, Apr 17 2019

Michael De Vlieger, Table of n, a(n) for n = 0..999
Index entries for linear recurrences with constant coefficients, signature (11,-10).

Partial sums give 10*A099676.

Cf. A002283, A002275, A070565.

List([0..20], n-> 10*(10^n -1)); # G. C. Greubel, Jun 30 2019

[10*(10^n -1): n in [0..20]]; // G. C. Greubel, Jun 30 2019

a:=n->sum (10^(n-j+2)-10^(n-j+1),j=0..n): seq(a(n),n=0..20);

Array[10 (10^# - 1) &, 20, 0] (* Michael De Vlieger, Apr 21 2019 *)

vector(20, n, n--; 10*(10^n -1)) \\ G. C. Greubel, Jun 30 2019

[10*(10^n -1) for n in (0..20)] # G. C. Greubel, Jun 30 2019

a(n) = 10*A002283(n).
From G. C. Greubel, Jun 30 2019: (Start)
a(n) = 90*A002275(n).
a(n) = 11*a(n-1) - 10*a(n-2).
G.f.: 10*(1/(1-10*x) - 1/(1-x)).
E.g.f.: 10*(exp(10*x) - exp(x)). (End)

A134942 Numbers m such that there exists no number k with k-P(k) = m, where P(k) is the product of digits of k written in base 10.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 21, 23, 27, 29, 32, 33, 36, 39, 41, 43, 44, 47, 48, 49, 51, 53, 54, 56, 57, 61, 62, 63, 65, 67, 68, 69, 71, 72, 75, 76, 77, 78, 79, 81, 83, 84, 85, 86, 87, 88, 89, 91, 92, 93, 94, 95, 96, 97, 98, 99, 121, 123, 127, 129, 132, 133, 136, 139, 141, 143

Offset: 1

Philippe Lallouet (philip.lallouet(AT)orange.fr), Feb 01 2008

Obviously no number containing a zero digit is in the sequence.

			For 0 <= p <= 9, p - P(p) = 0, hence 0 is not in the sequence.
It's easy to see that if p has 2 digits or more the difference p - P(p) has at least 2 digits, hence 1 to 9 are in the sequence.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A070565.

f:= n -> n - convert(convert(n,base,10),`*`):
R:= {}:
S:= {0}:
for d from 1 to 3 do
  S:= map(proc(t) local i; seq(10*t+i,i=1..9) end proc, S);
  V:= S minus map(f, S);
  R:= R union V;
od:
sort(convert(R,list)); # Robert Israel, Jul 22 2025

Corrected by Robert Israel, Jul 22 2025

A336383 a(n) is the smallest number such that, with f(x) = x - (the product of the digits of x), f(a(n)) reaches a fixed point after n iterations.

Original entry on oeis.org

0, 1, 21, 31, 42, 52, 73, 81, 319, 391, 463, 583, 2911, 3667, 6451, 8793, 9927, 237126, 254158, 278393, 2561363, 9398143, 9431623, 9951265, 83543869, 83896381, 83935261, 2843233127, 2847297383, 2853748583, 2885762663, 266998137657, 685718563667, 688373877587

Offset: 0

Alon Ran, Jul 19 2020

A fixed point occurs once the function returns a number that contains the digit 0. After that, the product of the digits will be 0, and so subtracting it from the number will be idempotent.

This sequence is conceptually similar to A003001, though unlike the latter, it is probably infinite.

			a(9) = 391 because:
   1: 391 - 3*9*1 = 364
   2: 364 - 3*6*4 = 292
   3: 292 - 2*9*2 = 256
   4: 256 - 2*5*6 = 196
   5: 196 - 1*9*6 = 142
   6: 142 - 1*4*2 = 134
   7: 134 - 1*3*4 = 122
   8: 122 - 1*2*2 = 118
   9: 118 - 1*1*8 = 110
After iteration 9, the function becomes idempotent:
  10: 110 - 1*1*0 = 110
  11: 110 - 1*1*0 = 110
  12: 110 - 1*1*0 = 110
  ...
Additionally, 391 is the smallest number with this property. Thus, it is a(9).

Cf. A003001, A070565.

nmax = 20; tab = ConstantArray[Null, nmax];
For[k = 0, k <= 1000000, k++,
 l=Length@  NestWhileList[#-Times @@ IntegerDigits[#] &,k,UnsameQ[##] &, 2]-2;
If[tab[[l+1]] == Null, tab[[l+1]] = k]]; tab  (* Robert Price, Sep 13 2020 *)

f(m) = m - vecprod(digits(m)) + (m==0);
lista(nn) = {my(c, m, t); for(k=0, nn, c=0; m=k; while(m!=(m=f(m)), c++); if(c==t, print1(k, ", "); t++)); } \\ Jinyuan Wang, Aug 14 2020

def f(x):
    prod = 1
    for digit in str(x):
        prod *= int(digit)
    return x - prod
def a(n):
    i = 0
    iteration = 0
    while iteration != n:
        i += 1
        j = i
        iteration = 0
        new_j = f(j)
        while j != new_j:
            iteration += 1
            j = new_j
            new_j = f(j)
    return i

a(27)-a(30) from Jinyuan Wang, Aug 14 2020

a(31)-a(33) added by Michael S. Branicky, Aug 29 2020

A336682 a(n) is the number of iterations needed to reach a fixed point starting with n and repeatedly applying f(x) = x - (the product of the digits of x).

Original entry on oeis.org

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

Offset: 0

Robert Price, Sep 13 2020

			a(42) = 4 because:
1: 42 - 4*2 = 34
2: 34 - 3*4 = 22
3: 22 - 2*2 = 18
4: 18 - 1*8 = 10
5: 10 - 1*0 = 10

Robert Price, Table of n, a(n) for n = 0..1000

Cf. A003001, A070565, A336383.

Table[Length@  NestWhileList[# - Times @@ IntegerDigits[#] &, n, UnsameQ[##] &,
2] - 2, {n, 0, 85}];(* Robert Price, Sep 13 2020 *)

A376257 a(n) = (n - c(n))*(-1)^c(n), where c(n) is the product of the digits of n (A007954).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 10, -10, 10, -10, 10, -10, 10, -10, 10, -10, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 30, -28, 26, -24, 22, -20, 18, -16, 14, -12, 40, 37, 34, 31, 28, 25, 22, 19, 16, 13, 50, -46, 42, -38, 34, -30, 26, -22, 18, -14, 60, 55, 50, 45, 40, 35, 30, 25, 20, 15, 70, -64, 58

Offset: 1

Stuart Coe, Sep 17 2024

There is an interesting pattern on the graph of the sequence.

			For n = 129: 1*2*9=18. So a(n) = (129-18)*(-1)^18 = 111.

Paolo Xausa, Table of n, a(n) for n = 1..10000

Cf. A007954, A070565.

A376257[n_] := (n - #)*(-1)^# & [Times @@ IntegerDigits[n]];
Array[A376257, 100] (* Paolo Xausa, Dec 10 2024 *)

a(n) = (n-A007954(n)) * (-1)^A007954(n).

cp's OEIS Frontend

A062329 a(n) = (sum of digits of n) - (product of digits of n).

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A178500 a(n) = 10^n * signum(n).

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A229547 Numbers n such that n - product_of_digits(n) is a palindrome.

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A124167 a(n) = 10*(10^n-1).

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A134942 Numbers m such that there exists no number k with k-P(k) = m, where P(k) is the product of digits of k written in base 10.

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A336383 a(n) is the smallest number such that, with f(x) = x - (the product of the digits of x), f(a(n)) reaches a fixed point after n iterations.

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A336682 a(n) is the number of iterations needed to reach a fixed point starting with n and repeatedly applying f(x) = x - (the product of the digits of x).

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