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.

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A245386 Numbers in A245385 where P, Q, R, and S are all distinct.

Original entry on oeis.org

164, 195, 265, 498, 1664, 1995, 2665, 4847, 4998, 6545, 7424, 16664, 19995, 21775, 24996, 26665, 43243, 49998, 86486, 148480, 166664, 175150, 199995, 217775, 249996, 266665, 368180, 484847, 499998, 654545, 742424, 1001001, 1081075, 1216216, 1249992, 1297290, 1451850, 1471468, 1481477
Offset: 1

Views

Author

Derek Orr, Jul 20 2014

Keywords

Comments

This sequence does not contain any repdigits, unlike A245385.

Examples

			4*84847 = 48484*7 = 339388. Thus 484847 is a member of this sequence.
		

Crossrefs

Cf. A262743 (predestined numbers).

Programs

  • Python
    for n in range(1,10**7):
      s = str(n)
      count = 0
      for i in range(1,len(s)):
        if i != len(s) - i:
          if int(s[:i]) != int(s[len(s)-i:]):
            num = int(s[:i])*int(s[i:])
            if num != 0:
              if num == int(s[:len(s)-i])*int(s[len(s)-i:]):
                count += 1
                break
      if count > 0:
        print(n,end=', ')

A245385 Numbers N such that N = P//Q = R//S, where // is the concatenation function, satisfying the following properties: P and S are m-digit integers, Q and R are k-digit integers, k and m are distinct positive integers, and P*Q = R*S.

Original entry on oeis.org

111, 164, 195, 222, 265, 333, 444, 498, 555, 666, 777, 888, 999, 1111, 1664, 1995, 2222, 2665, 3333, 4444, 4847, 4998, 5555, 6545, 6666, 7424, 7777, 8888, 9999, 11111, 16664, 19995, 21775, 22222, 24996, 26665, 33333, 43243, 44444, 49998, 55555, 66666, 77777, 86486, 88888, 99999
Offset: 1

Views

Author

Derek Orr, Jul 20 2014

Keywords

Examples

			Take the number 21775.
2*1775 != 2177*5.
21*775 == 217*75 = 16275. Thus 21775 is a member of this sequence.
		

Crossrefs

Programs

  • Python
    for n in range(1,10**5):
      s = str(n)
      count = 0
      for i in range(1,len(s)):
        num = int(s[:i])*int(s[i:])
        if i != len(s) - i:
          if num != 0:
            if num == int(s[:len(s)-i])*int(s[len(s)-i:]):
              count += 1
              break
      if count > 0:
        print(n,end=', ')
Showing 1-2 of 2 results.