A154818 Numbers k such that k^4 contains every digit exactly twice.
69636, 70215, 77058, 80892
Offset: 1
Extensions
Keywords fini and full added. - R. J. Mathar, Jan 17 2009
This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
20247 is in the sequence because 20247*74202 = 1502367894 contains ten different digits; 451410 is in the sequence because 451410*14154 = 6389257140 contains ten different digits.
with(numtheory): U:=array(1..50) :c:=0:for i from 5000 to 1000000 do: s1:=0:ll:=length(i):for
q from 0 to ll do:x:=iquo(i, 10^q):y:=irem(x, 10):s1:=s1+y*10^(ll-1-q): od:n:=i*s1:l:=length(n):if l=10 then n0:=n:s:=0:for m from 1 to l do:q:=n0:u:=irem(q,10):v:=iquo(q,10):n0:=v
: U[m]:=u:od: B:={0,1,2,3,4,5,6,7,8,9}: A:=convert(U,set):z:=nops(A):else fi:
if A intersect B = B and z=10 and l=10 then c:=c+1:printf(`%d, `,i): else fi:od:
print(c):
a(2) = A054038(2225) = 588289 = 767^2.
[ n^2: n in [0..2232] | Seqset(Intseq(n^4)) eq {0..9} ];
11826^2 = 139854276 contains all digits from 1 to 9 exactly once. The same is true for all terms up to 30384 whose square is 923187456. These terms are also listed in A071519, they form a subsequence of A054037. The next 3 terms, 32043 (32043^2 = 1026753849), 32286 (32286^2 = 1042385796) and 33144 (33144^2 = 1098524736) contain all of the digits '0' through '9' exactly once: They are the first terms of A054038. The next term, 34273 with 34273^2 = 1174638529, does not have this property, but the next two are again of that type (35172^2 = 1237069584 and 35337^2 = 1248703569).
Select[Range[#, # + 3*10^4] &@ 11111, AllTrue[Most@ DigitCount[#^2], # > 0 &] &] (* Michael De Vlieger, Nov 08 2017 *)
is_A294661(n)=#select(t->t,Set(digits(n^2)))>8 N=100;for(k=10^4,oo,is_A294661(k)||next;print1(k",");N--||break)
fQ[n_] := Length@ Union[ Count[ Sort[ Join[ IntegerDigits[n^2], IntegerDigits[n^3]]], #] & /@ Range[0, 9]] == 1; Select[ Range@ 52000000, fQ] (* Robert G. Wilson v, Jul 01 2023 *)
is(n)={my(v=concat(digits(n^2),digits(n^3)), c=#v); c%10==0 && vecsort(v)==[0..c-1]\(c\10)}
for(n=1,1e6, is(n)&& print1(n","))
309^5 = 2817036000549 contains every digit at least once.
Select[Range@1500, 1+Union@IntegerDigits@(#^5)==Range@10&] (* Hans Rudolf Widmer, Nov 02 2021 *)
isok(m) = #Set(digits(m^5)) == 10; \\ Michel Marcus, Nov 02 2021
def ok(n): return len(set(str(n**5))) == 10 print([m for m in range(1293) if ok(m)]) # Michael S. Branicky, Nov 02 2021
22807116 ^ 4 = 270571148920443982076865351936, which contains exactly 3 times of each digit 0-9.
Select[Range[17824000,31608000],Union[Tally[IntegerDigits[#^4]][[All,2]]]=={3}&] (* Harvey P. Dale, Dec 24 2016 *)
is(n) = my(v=vector(10), d=digits(n^4)); if(#d!=30,return(0)); for(i=1, #d, v[d[i]+1]++; if(v[d[i]+1] > 3, return(0))); 1 \\ David A. Corneth, Aug 19 2025
23 is in the sequence because 23! = 25852016738884976640000 contains every digit at least once.
[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 &]
6534^2 = 42693156, 6534^3 = 278957081304, which together contain each digit 0-9 exactly twice.
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