A089186 -id:A089186 - cp's OEIS frontend

A372082 Primes p such that the 10's complement A089186(p) and the concatenations of p and A089186(p) and of A089186(p) and p are all prime.

3, 7, 17, 29, 71, 83, 281, 719, 1637, 2309, 3701, 4493, 5507, 6299, 7691, 8363, 9029, 11003, 13163, 17117, 18371, 20807, 31181, 31793, 32693, 32843, 33617, 33893, 34211, 34673, 37277, 38453, 49409, 50591, 61547, 62723, 65327, 65789, 66107, 66383, 67157, 67307, 68207, 68819, 79193, 81629, 82883

Offset: 1

Robert Israel, Jul 03 2024

If p is a term and starts with 1 to 8, then its 10's complement A089186(p) is also a term. This is not the case if p starts with 9, as then A089186(A089186(p)) <> p. For example, 9029 is a term but its 10's complement 971 is not a term.

			a(3) = 17 is a term because 17 is a prime, its 10's complement 83 is a prime, and the concatenations 1783 and 8317 are primes.

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

Cf. A089186. Subset of A083989.

filter:= proc(n) local d,c;
if not isprime(n) then return false fi;
d:= 10^(1+ilog10(n)); c:= d-n;
isprime(c) and isprime(c*d+n) and isprime(n*10^(1+ilog10(c))+c)
end proc:
select(filter, [seq(i,i=3..10000,2)]);

A110396 10's complement factorial of n: a(n) = (10's complement of n)(10's complement of n-1)...(10's complement of 2)(10's complement of 1).

Original entry on oeis.org

1, 9, 72, 504, 3024, 15120, 60480, 181440, 362880, 362880, 32659200, 2906668800, 255786854400, 22253456332800, 1913797244620800, 162672765792768000, 13664512326592512000, 1134154523107178496000, 93000670894788636672000, 7533054342477879570432000

Offset: 0

Amarnath Murthy, Jul 29 2005

			a(3) = (10-3)*(10-2)*(10-1) = 7*8*9 = 504.

Alois P. Heinz, Table of n, a(n) for n = 0..400

Cf. A089186, A109631, A110394, A110395.

s:=proc(m) nops(convert(m,base,10)) end: for q from 1 to 120 do c[q]:=10^s(q)-q od: a:=n->product(c[i],i=1..n): seq(a(n),n=0..20); # Emeric Deutsch, Jul 31 2005
# second Maple program:
a:= proc(n) option remember; `if`(n=0, 1,
       (10^length(n)-n)*a(n-1))
    end:
seq(a(n), n=0..30);  # Alois P. Heinz, Sep 22 2015

a(n) = prod(i=1, n, 10^(1+logint(i, 10))-i); \\ Jinyuan Wang, Aug 09 2025

from functools import cache
def a(n): return 1 if n == 0 else (10**len(str(n))-n)*a(n-1)
print([a(n) for n in range(21)]) # Michael S. Branicky, Aug 13 2025

a(n) = Product_{i=1..n} c(i), where c(i) = A089186(i) is the difference between i and the next power of 10 (for example, c(13) = 100 - 13 = 87; c(100) = 1000 - 100 = 900). - Emeric Deutsch, Jul 31 2005

More terms from Emeric Deutsch, Jul 31 2005

a(0)=1 prepended by Alois P. Heinz, Aug 13 2025

A068822 a(n) = gcd(n,c(n)), where c(n) is the 10's complement of n.

Original entry on oeis.org

1, 2, 1, 2, 5, 2, 1, 2, 1, 10, 1, 4, 1, 2, 5, 4, 1, 2, 1, 20, 1, 2, 1, 4, 25, 2, 1, 4, 1, 10, 1, 4, 1, 2, 5, 4, 1, 2, 1, 20, 1, 2, 1, 4, 5, 2, 1, 4, 1, 50, 1, 4, 1, 2, 5, 4, 1, 2, 1, 20, 1, 2, 1, 4, 5, 2, 1, 4, 1, 10, 1, 4, 1, 2, 25, 4, 1, 2, 1, 20, 1, 2

Offset: 1

Amarnath Murthy, Mar 08 2002

			a(45) = 5 as 10's complement of 45 is 100-45 = 55 and (45,55) = 5.

Alois P. Heinz, Table of n, a(n) for n = 1..9999

Cf. A089186, A178914.

a:=n-> igcd((10^length(n)-n), n):
seq(a(n), n=1..100);  # Alois P. Heinz, Sep 22 2015

GCD[#, 10^(IntegerLength[#]) - #] & /@ Range[82] (* Jayanta Basu, Aug 07 2013 *)

a(n) = gcd(n,A089186(n)) = gcd(n,A178914(n)). - Alois P. Heinz, Nov 05 2018

A083989 Concatenation of prime k and its 10's complement is a prime.

Original entry on oeis.org

3, 7, 17, 23, 29, 31, 41, 43, 53, 67, 71, 83, 109, 127, 139, 151, 173, 179, 197, 211, 229, 263, 271, 281, 307, 359, 463, 547, 557, 569, 587, 593, 673, 677, 683, 691, 701, 719, 727, 757, 769, 823, 839, 881, 883, 887, 907, 937, 983, 997, 1087, 1103, 1171, 1181

Offset: 1

Amarnath Murthy, May 23 2003

			43 belongs to this sequence as 4357 is also a prime.

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

Cf. A083990, A089186.

g:= proc(n) local c; c:= 10^(1+ilog10(n))-n; n*10^(1+ilog10(c))+c end proc:
select(t -> isprime(t) and isprime(g(t)), [seq(i,i=3..10000,2)]); # Robert Israel, Jul 03 2024

More terms from Jason Earls, Jun 01 2003

A097327 Least positive integer m such that m*n has greater decimal digit length than n.

Original entry on oeis.org

10, 5, 4, 3, 2, 2, 2, 2, 2, 10, 10, 9, 8, 8, 7, 7, 6, 6, 6, 5, 5, 5, 5, 5, 4, 4, 4, 4, 4, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 10, 10, 10

Offset: 1

Rick L. Shepherd, Aug 04 2004

For any positive base B >= 2 the corresponding sequence contains only terms from 2 to B inclusive so the corresponding sequence for binary is all 2s (A007395).

			a(12) = 9 since 12 has two decimal digits and 9*12 = 108 has three (but 8*12 = 96 has only two).

Michael S. Branicky, Table of n, a(n) for n = 1..10000

Cf. A089186 (analog for decimal m+n), A080079 (analog for binary m+n), A097326.

Cf. A055642.

Table[Ceiling[10^IntegerLength[n]/n], {n, 100}] (* Paolo Xausa, Nov 02 2024 *)

a(n) = my(m=1, sn=#Str(n)); while (#Str(m*n) <= sn, m++); m; \\ Michel Marcus, Oct 05 2021

def a(n): return (10**len(str(n))-1)//n + 1
print([a(n) for n in range(1, 103)]) # Michael S. Branicky, Oct 05 2021

a(n) = A097326(n) + 1.

a(n) = ceiling(10^A055642(n)/n). - Michael S. Branicky, Oct 05 2021

A087325 Numbers k such that k and its 10's complement both have the same prime signature.

Original entry on oeis.org

3, 5, 7, 11, 14, 15, 17, 26, 29, 30, 35, 38, 41, 47, 50, 53, 59, 62, 65, 70, 71, 74, 83, 85, 86, 89, 94, 97, 110, 111, 113, 122, 129, 132, 134, 137, 140, 150, 153, 158, 170, 173, 174, 179, 183, 185, 186, 187, 191, 195, 201, 206, 209, 212, 215, 219, 221, 227, 236

Offset: 1

Amarnath Murthy, Sep 04 2003

Conjecture: (1) Sequence is infinite. (2) For every prime signature there corresponds a term in this sequence.
From Robert Israel, Jul 02 2024: (Start)
Conjecture (2) is false: k and its 10's complement can't both have prime signature p^m where m is even.
If k is a term, then so is 10 * k.
It appears that the first term with m prime factors, counted with multiplicity, is 3 * 10^((m-1)/2) if m is odd and 132 * 10^((m-4)/2) if m >= 4 is even. (End)

			35 is a member as 35= 5*7 and its 10's complement (100-35) = 65 = 13*5 both have the prime signature p*q.
35 is a member as 35 = 5*7 and its 10's complement (100-35) = 65 = 13*5 both have the prime signature p*q.

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

Cf. A087324, A089186. Contains A068811.

ps:= n -> sort(ifactors(n)[2][..,2]):
tc:= n -> 10^(1+ilog10(n))-n:
select(n -> ps(n) = ps(tc(n)), [$1..1000]); # Robert Israel, Jul 02 2024

More terms from David Wasserman, May 06 2005

A109640 Indices of records in A109631.

Original entry on oeis.org

1, 2, 5, 11, 13, 19, 23, 31, 47, 97, 101, 113, 131, 151, 181, 227, 307, 457, 907, 1009, 1129, 1289, 1511, 1801, 2251, 3001, 4507, 9001, 10007, 11251, 12889, 15013, 18013, 22501, 30011, 45007, 90001, 100003, 112501, 128591, 150001, 180001, 225023, 300007, 450001

Offset: 1

Jason Earls, Aug 04 2005

Previous name was: Values of k which are incrementally the largest values of the function: Smallest number m such that k divides (10's complement factorial of m).

Cf. A002034, A109631, A110396.

A089186 := proc(n) 10^max(1,ilog10(n)+1)-n ; end: A110396 := proc(n) mul( A089186(i),i=1..n) ; end: A109631 := proc(n) local a; for a from 1 do if A110396(a) mod n = 0 then RETURN(a) ; fi ; od: end: A109640 := proc(n) option remember ; local nprev,aprev,a ; if n = 1 then RETURN(1); else nprev := A109640(n-1) ; aprev := A109631(nprev) ; for a from nprev+1 do if A109631(a) > aprev then RETURN(a) ; fi ; od; fi ; end: for n from 1 do printf("%d, ",A109640(n)) ; od: # R. J. Mathar, Feb 12 2008

More terms from R. J. Mathar, Feb 12 2008
a(23)-a(45) from Jinyuan Wang, Aug 09 2025
New name (using comment from R. J. Mathar) from Joerg Arndt, Aug 09 2025

A178914 10's complement of nonnegative numbers.

Original entry on oeis.org

10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 90, 89, 88, 87, 86, 85, 84, 83, 82, 81, 80, 79, 78, 77, 76, 75, 74, 73, 72, 71, 70, 69, 68, 67, 66, 65, 64, 63, 62, 61, 60, 59, 58, 57, 56, 55, 54, 53, 52, 51, 50, 49, 48, 47, 46, 45, 44, 43, 42, 41, 40, 39, 38, 37, 36, 35, 34, 33, 32, 31, 30, 29, 28

Offset: 0

Amarnath Murthy, Jun 23 2010

Apart from the initial a(0) a duplicate of A089186. - R. J. Mathar, Jun 25 2010

			a(11) = 10's complement of 11 = 89

Cf. A089186.

Join[{10},Table[10^IntegerLength[n]-n,{n,80}]] (* Harvey P. Dale, Feb 06 2015 *)

a(n) = 10^k - n where k is the number of digits in n.

A279913 Largest n-digit number ending in n.

Original entry on oeis.org

1, 92, 993, 9994, 99995, 999996, 9999997, 99999998, 999999999, 9999999910, 99999999911, 999999999912, 9999999999913, 99999999999914, 999999999999915, 9999999999999916, 99999999999999917, 999999999999999918, 9999999999999999919, 99999999999999999920

Offset: 1

Wesley Ivan Hurt, Dec 22 2016

Cf. A266959.

Cf. A011557 (10^n), A089186.

Table[n + 10^n - 10^(1 + Floor[Log10[n]]), {n, 100}] (* Wesley Ivan Hurt, Mar 30 2020 *)

a(n) = 10^n + n - 10^(#Str(n)); \\ Michel Marcus, Jul 25 2022

def A279913(n): return 10**n+n-10**(len(str(n))) # Chai Wah Wu, Jul 25 2022

a(n) = n + 10^n - 10^(1 + floor(log_10(n))). - Wesley Ivan Hurt, Mar 30 2020

a(n) = 10^n - A089186(n). - Michel Marcus, Jul 25 2022

A039690 Ambitious numbers: numbers n with the property that if a number ends in n then it is divisible by n.

Original entry on oeis.org

1, 2, 5, 10, 20, 25, 50, 100, 125, 200, 250, 500, 1000, 1250, 2000, 2500, 5000, 10000, 12500, 20000, 25000, 50000, 100000, 125000, 200000, 250000, 500000, 1000000, 1250000, 2000000, 2500000, 5000000, 10000000, 12500000, 20000000, 25000000

Offset: 1

Erich Friedman

Number whose 10's complement (A089186) is a multiple of it. 125 is a member as its 10's complement is 1000-125 = 875 = 125*7. - Amarnath Murthy, Mar 08 2002

			If a number ends in 2 then it is even and so is divisible by 2, so 2 is in the sequence.

P. J. Davis and R. Hersh, The Mathematical Experience, Birkhäuser, Boston and Basel, 1981; see pp. 293-298.

Problem of the week, Web site - problem 881
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,10).

Cf. A039690, A089186.

LinearRecurrence[{0,0,0,0,10},{1,2,5,10,20,25,50,100,125},40] (* Harvey P. Dale, Sep 08 2018 *)

Consists of the numbers 1, 2, 5, 25 or 125 times a power of 10.

a(n) = 10*a(n-5). - Wesley Ivan Hurt, May 03 2023

Entry revised by N. J. A. Sloane, Aug 03 2004

cp's OEIS Frontend

A372082 Primes p such that the 10's complement A089186(p) and the concatenations of p and A089186(p) and of A089186(p) and p are all prime.

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A110396 10's complement factorial of n: a(n) = (10's complement of n)*(10's complement of n-1)*...*(10's complement of 2)*(10's complement of 1).

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A068822 a(n) = gcd(n,c(n)), where c(n) is the 10's complement of n.

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A083989 Concatenation of prime k and its 10's complement is a prime.

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Maple

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A097327 Least positive integer m such that m*n has greater decimal digit length than n.

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Mathematica

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A087325 Numbers k such that k and its 10's complement both have the same prime signature.

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A109640 Indices of records in A109631.

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A178914 10's complement of nonnegative numbers.

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A110396 10's complement factorial of n: a(n) = (10's complement of n)(10's complement of n-1)...(10's complement of 2)(10's complement of 1).