cp's OEIS Frontend

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.

Showing 1-3 of 3 results.

A034089 Numbers that are proper divisors of the number you get by rotating digits right once.

Original entry on oeis.org

102564, 128205, 142857, 153846, 179487, 205128, 230769, 102564102564, 128205128205, 142857142857, 153846153846, 179487179487, 205128205128, 230769230769, 1012658227848, 1139240506329, 102564102564102564
Offset: 1

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Author

Keywords

Comments

Let p(q) denote the period of the fraction q; then sequence is generated by p(i / (10k-1)), k=2,3,4,5,6,7,8,9; k <= i <= 9 and the concatenations of those periods, e.g., p(7/39)=a(5) p(2/19)=a(17).
Example if k=5: p((5+2)/49)=142857 which is in the sequence as the concatenations 142857142857, 142857142857142857, 142857142857142857142857, etc. - Benoit Cloitre, Feb 02 2002
The i in p(i / (10k-1)) is the last digit of the period, while k is equal to the ratio (right-rotated of p)/p. Thus no concatenation of any different such p's can be in the sequence. There are 8*9/2 = 36 terms which are not concatenation of previous terms, the last one being a(124) = 1525423728813559322033898305084745762711864406779661016949 with 58 digits. The term a(3)=p(7/49) is the only period of length (6) different from the length (42) of the other terms corresponding to the same value of k. - M. F. Hasler, Nov 18 2007
Numbers comprising multiple copies of a single digit, e.g., 111111, are not permitted. - Harvey P. Dale, Mar 08 2013
From Emmanuel Vantieghem, Oct 25 2015: (Start)
Subsequence of A245680.
Every element of the sequence is a multiple of 3.
The leading digit of every element is < 5.
(End)

Crossrefs

Subsequences of this sequence (with quotient k): A146088 (k=2), A146561 (k=3), A146569 (k=4), A146754 (k=5), A291354 (k=6), A291215 (k=7), A291321 (k=8), A291353 (k=9).

Programs

  • PARI
    period(p,q,S=[])=until(setsearch(S,p),S=setunion(S,[p]);p=10*p%q);S=[];until(p==S[1],S=concat(S,p);p=10*p%q);S*10\q /* print list of periods, right-rotated and ratio */ rotquo(n,d)={d=divrem(n,10);d[1]+=d[2]*10^#Str(d[1]);[n,d[1],d[1]/n]} for(k=2,9,for(i=k,9,print1( i/(10*k-1),"\t",rotquo(sum(j=1,#p=period(i,k*10-1),p[j]*10^(#p-j))))) /* build the sequence up to the greatest period */ A034089()={local(S=[],p); for(k=2,9,for(i=k,9,S=concat(S,sum(j=1,#p=period(i,k*10-1),p[j]*10^(#p-j))))); S=vecsort(S); for(i=1,#S, for(c=2,58\p=#Str(S[i]), S=concat(S,S[i]*(10^(c*p)-1)/(10^p-1)) )); vecsort(S)} \\ M. F. Hasler, Nov 18 2007

Extensions

Edited, corrected and extended by M. F. Hasler, Nov 18 2007

A245682 Numbers x whose digits can be permuted to produce more than a single multiple of x.

Original entry on oeis.org

123876, 142857, 153846, 230769, 285714, 1028574, 1218753, 1238760, 1239876, 1246878, 1294857, 1402857, 1420785, 1425897, 1428507, 1428570, 1428597, 1428705, 1429857, 1485792, 1492857, 1538460, 1539846, 1570284, 1584297, 2300769, 2307690, 2307699, 2309769, 2857014, 2857140, 2859714, 2985714, 10028574, 10178649
Offset: 1

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Author

Paolo P. Lava, Jul 29 2014

Keywords

Comments

It is a subset of A245680.
If x < 10^d is in the sequence, then so are x*10^j*(1+10^d+...+10^k*d) for all nonnegative integers j and k. - Robert Israel, Jul 29 2014

Examples

			Two permutations of 123876 are 371628, 867132  and  371628 / 123876 = 3, 867132  / 123876 = 7.
Five permutations of 142857 are 285714, 428571, 571428, 714285, 857142 and 285714 / 142857 = 2, 428571 / 142857 = 3, 571428 / 142857 = 4, 714285 / 142857 = 5, 857142 / 142857 = 6.
		

Crossrefs

Programs

  • Maple
    P:=proc(q) local a,b,c,i,j,k,n,t; for n from 1 to q do a:=n; b:=[];
    while a>0 do b:=[a mod 10,op(b)]; a:=trunc(a/10); od;
    t:=0; for i from 2 to 9 do a:=i*n; c:=[];
    while a>0 do c:=[a mod 10,op(c)]; a:=trunc(a/10); od;
    if sort(b)=sort(c) then t:=t+1; fi; if t>1 then print(n); break;
    fi; od; od; end: P(10^10);
    # Alternative
    N:= 10: # get a(1) to a(N)
    count:= 0:
    for x from 10 while count < N do
      M:= 10^(ilog10(x)+1)-1;
      L:= sort(convert(x,base,10));
      mults:= 0;
      for i from 2 to floor(M/x) do
        Lp:= sort(convert(i*x,base,10));
        if Lp = L then
          mults:= mults+1;
          if mults = 2 then
            count:= count+1;
            A[count]:= x;
            print(x);
            break;
          fi
        fi
       od
    od:
    seq(A[i],i=1..count); # Robert Israel, Jul 29 2014
  • PARI
    for(n=1,10^8,d=vecsort(digits(n));p=0;for(k=2,9,dd=vecsort(digits(n*k));if(d==dd,p++));if(p>1,print1(n,", "))) \\ faster program Derek Orr, Jul 29 2014
  • Python
    import itertools
    from itertools import permutations
    for n in range(1,10**8):
      plist = list(permutations(str(n)))
      count = 0
      lst = []
      for i in plist:
        num = ''
        for j in range(len(i)):
          num += i[j]
        if int(num)%n==0 and int(num)/n > 1:
          if int(num) not in lst:
            lst.append(int(num))
            count += 1
      if count > 1:
        print(n,end=', ') # Derek Orr, Jul 29 2014
    

Extensions

a(7) to a(10) from Robert Israel, Jul 29 2014
a(11) - a(35) from Derek Orr, Jul 29 2014

A373407 Smallest positive integer k such that no more than n numbers (formed by multiplying k by a digit) are anagrams of k, or -1 if no such number exists.

Original entry on oeis.org

1, 1035, 123876, 1402857, 1037520684, 142857
Offset: 1

Views

Author

Jean-Marc Rebert, Jun 04 2024

Keywords

Comments

For n = 2..6 all terms are divisible by 9.
For n >= 4, a(n) must be divisible by 9, or a(n) = -1, because all anagrams d*k of k for d = 2, 3, 5, 6, 8 and 9 are divisible by 9. Thus there are only 3 values of d, i.e., 1, 4 and 7, for which k*d must not be divisible by 9.
If a(n) exists for n > 1 then 9|a(n). Holds for n = 2 and n = 3 by inspection. Proof for n >= 4: if k*d is an anagram of k where 2 <= d <= 9 then k*d - k = k*(d-1) is a multiple of 9. For this to be true, k must be a multiple of 9 as d is not of the form 1 (mod 3) for all d. - David A. Corneth, Jun 04 2024
From Michael S. Branicky, Jun 07 2024: (Start)
The following were constructed from multiples of cyclic numbers (cf. A180340, Wikipedia):
a(6) = 142857 = (10^6 - 1) / 7;
a(7) <= 1304347826086956521739 = 3*(10^22 - 1) / 23;
a(8) <= 1176470588235294 = 2*(10^16 - 1) / 17;
a(9) <= 105263157894736842 = 2*(10^18 - 1) / 19. (End)

Examples

			a(2) = 1035, because 1035 * 1 = 1035 and 1035 * 3 = 3105 are anagrams of 1035, and no other number 1035 * i with digit i is an anagram of 1035, and no lesser number verifies this property.
Table n, k, set of multipliers.
  1   1          [1]
  2   1035       [1, 3]
  3   123876     [1, 3, 7]
  4   1402857    [1, 2, 3, 5]
  5   1037520684 [1, 2, 4, 5, 8]
  6   142857     [1, 2, 3, 4, 5, 6]
		

Crossrefs

Programs

  • PARI
    isok(k, n) = my(d=vecsort(digits(k))); sum(i=1, 9, vecsort(digits(k*i)) == d) == n; \\ Michel Marcus, Jun 04 2024
Showing 1-3 of 3 results.