A020666 -id:A020666 - cp's OEIS frontend

A378771 a(n) is the least k such that the last k digits of m = A020666(n)^n contain all 10 possible digits (0 through 9).

10, 10, 10, 11, 13, 11, 13, 17, 15, 16, 15, 15, 16, 18, 17, 15, 17, 15, 13, 16, 17, 15, 24, 17, 23, 16, 19, 20, 20, 22, 22, 25, 32, 17, 20, 23, 20, 19, 19, 23, 19, 14, 21, 19, 17, 25, 22, 17, 27, 19, 24, 13, 20, 28, 18, 33, 26, 18, 33, 22, 23, 20, 25, 25, 25, 22

Offset: 1

David A. Corneth, Dec 06 2024

A conjecture over at A020666 says that A020666(n) = 2 for n >= 189. Perhaps looking at last digits and some modular arithmetic leads to a proof.

A020666(189) = 2 and a(189) = 33 so the last 33 digits of 2^189 contain all 10 digits. Therefore for k = 189 + 4*5^(33-1) the last 33 digits of 2^k contain all 10 possible digits.

			a(4) = 11 since m = A020666(4)^4 = 763^4 = 338920744561. The last 11 digits contain all 10 possible digits (0 through 9) but the last 10 digits do not; 3 is missing in the last 10 digits.

David A. Corneth, Table of n, a(n) for n = 1..10000

Cf. A005054, A020666.

a(n) = {
	if(n == 1, return(10));
	my(i, todo = 10, v = vector(10), d);
	for(i = 2, oo,
		d = digits(i^n);
		if(#Set(d) == 10,
			forstep(i = #d, 1, -1,
				if(v[d[i]+1] == 0,
					v[d[i]+1] = 1;
					todo--;
					if(todo == 0,
						return(#d - i + 1))))));
}

A020667 Least n-th power containing every digit.

Original entry on oeis.org

1023456789, 1026753849, 12584301976, 338920744561, 2817036000549, 16157819263041, 1727094849536, 13685690504052736, 1628413597910449, 3656158440062976, 2384185791015625, 129746337890625, 1490116119384765625, 168377826559400929, 2862423051509815793

Offset: 1

David W. Wilson

It is extremely probable that a(n) = 2^n for all n >= 169.

Seiichi Manyama, Table of n, a(n) for n = 1..3321

Cf. A020666, A137214.

def a(n):
    if n == 1: return 1023456789
    a020667n = 2
    while not(len(set(str(a020667n**n))) == 10): a020667n += 1
    return a020667n**n
print([a(n) for n in range(1, 16)]) # Michael S. Branicky, Jul 04 2021

a(n) = A020666(n)^n. - Seiichi Manyama, Dec 05 2024

A217368 Smallest number having a power that in decimal has exactly n copies of all ten digits.

Original entry on oeis.org

32043, 69636, 643905, 421359, 320127, 3976581, 47745831, 15763347, 31064268, 44626422, 248967789, 85810806, 458764971, 500282265, 2068553967, 711974055, 2652652791, 901992825, 175536645, 3048377607, 3322858521, 1427472867, 3730866429, 9793730157

Offset: 1

James G. Merickel, Oct 01 2012

The exponents that produce the number with a fixed number of copies of each digit are listed in sequence A217378. See there for further comments.
Since we allow A217378(n)=1, the sequence is well defined, with the upper bound a(n) <= 100...99 ~ 10^(10n-1) (n copies of each digit, sorted in increasing order, except for one "1" permuted to the first position). - M. F. Hasler, Oct 05 2012
What is the minimum value of a(n)? Can it be proved that a(n) > 2 for all n? - Charles R Greathouse IV, Oct 16 2012

			The third term raised to the fifth power (A217378(3)=5), 643905^5 = 110690152879433875483274690625, has three copies of each digit (in its decimal representation), and no number smaller than 643905 has a power with this feature.

Cf. A020666, A054038, A054039, A066825, A067635, A074205, A156977, A217378.

f[n_] := Block[{k = 2, t = Table[n, {10}], r = Range[0, 9]}, While[c = Count[ IntegerDigits[k^Floor[ Log[k, 10^(10 n)]]], #] & /@ r; c != t, k++]; k] (* Robert G. Wilson v, Nov 28 2012 *)

is(n,k)=my(v);for(e=ceil((10*n-1)*log(10)/log(k)), 10*n*log(10)/log(k), v=vecsort(digits(k^e)); for(i=1,9,if(v[i*n]!=i-1 || v[i*n+1]!=i, return(0))); return(1)); 0
a(n)=my(k=2); while(!is(n,k),k++); k \\ Charles R Greathouse IV, Oct 16 2012

a(13)-a(14) from James G. Merickel, Oct 06 2012 and Oct 08 2012
a(15)-a(16) from Charles R Greathouse IV, Oct 17 2012
a(17)-a(19) from Charles R Greathouse IV, Oct 18 2012
a(20) from Charles R Greathouse IV, Oct 22 2012
a(21)-a(24) from Giovanni Resta, May 05 2017

A112388 a(n) is the smallest prime p such that p^n contains every digit.

Original entry on oeis.org

10123457689, 101723, 5437, 2339, 1009, 257, 139, 173, 83, 67, 31, 29, 37, 17, 17, 47, 19, 7, 5, 23, 23, 5, 11, 11, 17, 5, 5, 5, 5, 11, 5, 11, 11, 5, 5, 7, 5, 7, 3, 5, 5, 7, 7, 7, 3, 7, 3, 3, 5, 5, 5, 5, 3, 7, 7, 5, 3, 7, 5, 3, 3, 3, 3, 3, 3, 5, 3, 2, 3, 2, 3, 3, 3, 3, 5, 3, 3, 3, 2, 3, 5, 2

Offset: 1

Tanya Khovanova, Dec 05 2005

Conjecture: a(n)=2 for all n>168. Checked up to n = 20000. - Robert Israel, Aug 28 2020

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

Cf. A020666, A020665, A007377, A137214.

f:= proc(n) local k;
  k:= 1:
  do k:= nextprime(k);
    if convert(convert(k^n,base,10),set) = {$0..9} then return k fi
  od
end proc:
f(1):= 10123457689:
map(f, [$1..100]); # Robert Israel, Aug 28 2020

f[n_] := Block[{k = 1}, While[ Union@IntegerDigits[ Prime[k]^n] != {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, k++ ]; Prime[k]]; Array[f, 82] (* Robert G. Wilson v, Dec 06 2005 *)

from sympy import nextprime
def a(n):
    if n == 1: return 10123457689
    p = 2
    while not(len(set(str(p**n))) == 10): p = nextprime(p)
    return p
print([a(n) for n in range(1, 83)]) # Michael S. Branicky, Jul 04 2021

More terms from Robert G. Wilson v, Dec 06 2005

A178960 Numbers n such that n! contains every digit at least once.

Original entry on oeis.org

23, 27, 31, 33, 34, 35, 36, 37, 39, 40, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101

Offset: 1

Michel Lagneau, Dec 31 2010

			23 is in the sequence because 23! = 25852016738884976640000 contains every
  digit at least once.

Cf. A178929, A054038, A156977, A054038, A074205, A020666.

[n: n in [0..101] | Seqset(Intseq(Factorial(n))) eq {0..9}]; // Bruno Berselli, May 17 2011

with(numtheory):Digits:=200:B:={0,1,2,3,4,5,6,7,8,9}: T:=array(1..250) : for
  p from 1 to 200 do:ind:=0:n:=p!:l:=length(n):n0:=n:s:=0:for m from 1 to l do:q:=n0:u:=irem(q,10):v:=iquo(q,10):n0:=v : T[m]:=u:od: A:=convert(T,set):z:=nops(A):if A intersect B = B and ind=0 then ind:=1: printf(`%d, `,p):else fi:od:

Select[Range[101], Length[Union[IntegerDigits[#!]]] == 10 &]

A217535 Least number having in its decimal representation each digit n times.

Original entry on oeis.org

1023456789, 10012233445566778899, 100011222333444555666777888999, 1000011122223333444455556666777788889999, 10000011112222233333444445555566666777778888899999, 100000011111222222333333444444555555666666777777888888999999

Offset: 1

M. F. Hasler, Oct 05 2012

Alois P. Heinz, Table of n, a(n) for n = 1..100

Cf. A050278, A051018, A049363, A074205, A154566, A204047, A020666, A020667, A061604, A180489.

A subsequence of A171102.

a:= n-> parse(cat(1,0$n,1$(n-1),seq(i$n, i=2..9))):
seq(a(n), n=1..10);  # Alois P. Heinz, Jun 25 2017

A217535(n)=sum(d=1,9,10^(n-(d==1))\9*d*10^(n*(9-d)))+10^(10*n-1)

a(n) ~ 10^(10n-1). See PARI code for an exact formula.

cp's OEIS Frontend

A378771 a(n) is the least k such that the last k digits of m = A020666(n)^n contain all 10 possible digits (0 through 9).

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A020667 Least n-th power containing every digit.

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A217368 Smallest number having a power that in decimal has exactly n copies of all ten digits.

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A112388 a(n) is the smallest prime p such that p^n contains every digit.

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Maple

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A178960 Numbers n such that n! contains every digit at least once.

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Magma

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Mathematica

A217535 Least number having in its decimal representation each digit n times.

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Maple

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Formula