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 1-5 of 5 results.

A181129 Smallest primes of the form (i+1)(i+2)...(h-1)(h)1234...(i-1)(i). These elements, by definition, belong to A001292.

Original entry on oeis.org

2341, 89101234567, 45678910111213123, 23456789101112131415161718192021222324251, 30313233341234567891011121314151617181920212223242526272829, 20212223242526272829303132333435363738394041424344454612345678910111213141516171819, 42434445461234567891011121314151617181920212223242526272829303132333435363738394041, 14151617181920212223242526272829303132333435363738394041424344454647484950515212345678910111213
Offset: 1

Views

Author

Marco Ripà, Jan 23 2011

Keywords

Comments

If we indicate by p(j) the j-th term of A001292, the sequence above can be synthesized as:
p(8), p(53), p(82), p(302), p(591), p(1055), p(1077), p(1340), p(1499), p(1890), p(2231), p(3109), p(3145), p(3620), p(3878), p(4405), p(6248), p(8878), p(8888), p(11329), p(11439), p(12310), p(12344), p(13323), p(13747), p(15883), p(17471), p(17985), p(19815), p(20335), p(21676).
The first 30 terms of the sequence contain fewer than 500 digits. Among the first 22155 terms of A001292 only 31 are primes.

References

  • Marco Ripà, "Rudimatematici", Bookshelf, October 2010.
  • M. Vassilev-Missana and K. Atanassov, "Some Smarandache problems", Hexis, 2004.

Crossrefs

Programs

Extensions

Edited by N. J. A. Sloane, Jan 25 2011

A263106 Semiprimes such that the leftward cyclic permutation of its decimal digits is also semiprime.

Original entry on oeis.org

4, 6, 9, 15, 22, 26, 33, 39, 49, 51, 55, 58, 62, 77, 85, 93, 94, 111, 122, 129, 134, 141, 145, 155, 158, 159, 161, 177, 178, 183, 185, 187, 202, 206, 214, 226, 254, 262, 298, 303, 309, 314, 321, 339, 341, 355, 358, 362, 371, 381, 391, 393, 394, 403, 407, 413
Offset: 1

Views

Author

Zak Seidov, Oct 09 2015

Keywords

Comments

First 18 terms are also in A085751.

Examples

			15 = 3 * 5, 51 = 3 * 17; 889 = 7 * 17, 898 = 2 * 449.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[4, 1000], 2 == PrimeOmega[#] == PrimeOmega[FromDigits[RotateLeft[IntegerDigits[#]]]] &]
  • PARI
    shl(n)=if(n<10,return(n)); my(d=digits(n)); fromdigits(concat(d[2..#d], d[1]))
    is(n)=bigomega(n)==2 && bigomega(shl(n))==2 \\ Charles R Greathouse IV, Oct 12 2015

A352991 Concatenation of all the distinct permutations of the first 1, 2, 3, ... (strictly) positive integers, arranged in ascending numerical order.

Original entry on oeis.org

1, 12, 21, 123, 132, 213, 231, 312, 321, 1234, 1243, 1324, 1342, 1423, 1432, 2134, 2143, 2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 3412, 3421, 4123, 4132, 4213, 4231, 4312, 4321, 12345, 12354, 12435, 12453, 12534, 12543, 13245, 13254, 13425
Offset: 1

Views

Author

Marco Ripà, Apr 16 2022

Keywords

Comments

This sequence differs from A030299 starting from a(409114) = 10123456789. All the permutations are listed once and only once (e.g., the concatenation of the permutations of the elements of the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} originates the number 1112345678910 which is a unique element of this sequence and appears only once, since 1_11_23456789 = 11_1_23456789 = 1112345678910).
A001292 is a subsequence of the present sequence. An open problem, published by Kenichiro Kashihara in 1996 (see References, p. 25, #30, Problem 2), is to find how many terms of A001292 (which is a subsequence of A030299) are powers of integers; Kashihara conjectured that there are none (even if, clearly, A001292(1) = 1 should be disregarded in order to keep the conjecture alive). Currently, only the terms up to the prime a(409120) = 10123457689 have been directly checked by the author of this sequence and no nontrivial perfect power has been found. On the other hand, many (maybe infinitely many) terms of the present sequence are nontrivial powers of integers (e.g., A352329(2) to A352329(36) are squares of integers and belong to this sequence).
Although A181129 is a subsequence of the present one, so that A181129(1) = a(19) = 2341, a(14) is the smallest prime in this sequence.
The number of digits of a(n) comes from A058183. There are exactly k! (Cf. A000142) terms having A058183(k) digits. - David A. Corneth, Apr 17 2022

Examples

			a(3) = 21, since the number of permutations of {1, 2} is 2! = 2 and the concatenation 1_2 is smaller than 2_1 (while {1} originates only a(1) = 1, so that a(2) = 21).
		

References

  • Kenichiro Kashihara, Comments and Topics on Smarandache Notions and Problems, 25. Erhus University Press, Arizona, 1996. ISBN: 1-879585-55-3.

Crossrefs

Programs

  • Python
    from itertools import count, islice, permutations
    def agen(): # generator of terms
        for k in count(1):
            s = (int("".join(map(str, p))) for p in permutations(range(1, k+1)))
            yield from sorted(set(s))
    print(list(islice(agen(), 42))) # Michael S. Branicky, Apr 16 2022

A353025 Terms of A352991 which are perfect powers.

Original entry on oeis.org

1, 13527684, 34857216, 65318724, 73256481, 81432576, 139854276, 152843769, 157326849, 215384976, 245893761, 254817369, 326597184, 361874529, 375468129, 382945761, 385297641, 412739856, 523814769, 529874361, 537219684, 549386721, 587432169, 589324176, 597362481, 615387249, 627953481, 653927184
Offset: 1

Views

Author

Marco Ripà, Apr 17 2022

Keywords

Comments

It appears that all terms are terms of A062503.
We note that a(n)=A352329(n) up to a(36)=A352329(36)=923187456, while the mentioned match does not hold starting from a(37)=14102987536 (since A352329(37)=1234608769).
There are no perfect powers among terms t which are permutations of 123_...(m - 1)_m for m == {2, 3, 5, 6} (mod 9). This is since 10 == 1 (mod 9) and also (1 + 0) == 1 (mod 9), so digit position has no effect. Hence, t == A134804(m) (mod 9). Now, if m is such that A134804(m) = {3, 6}, there is a lone factor of 3, which is not a perfect power (indeed).
Therefore, all terms are necessarily congruent modulo 9 to 0 or 1 (see Marco Ripà link).
All terms up to 10^34 are squares (in particular, there are 67 squares with no more than 17 digits). - Aldo Roberto Pessolano, May 12 2022

Examples

			75910168324 is a term since 75910168324 = 275518^2.
		

Crossrefs

Programs

  • Mathematica
    z = 1; Do[r = Range[k];
    n = ToExpression[StringJoin[ToString[#] & /@ r]];
    If[And[Mod[n, 9] != 3, Mod[n, 9] != 6], d = DigitCount[n];
      s = IntegerPart[Sqrt[10^(IntegerLength[n] - 1)]];
      f = IntegerPart[Sqrt[10^(IntegerLength[n])]];
      Do[y = x^2;
       If[DigitCount[y] == d, c = True;
        Do[If[Not[StringContainsQ[ToString[y], ToString[i]]],
          c = False], {i, 10, k}]; If[c, Print[z, " ", y]; z++]], {x, s,
        f}]], {k, 1, 10}] (* Aldo Roberto Pessolano, May 12 2022 *)

Formula

Digit sum of a(n) is always congruent to 0 or 1 modulo 9.
a(n) = m^2, where the integer m := m(n) is not a perfect power itself (conjectured).

A263108 Semiprimes m such that the leftward cyclic permutation of its decimal digits is a larger semiprime.

Original entry on oeis.org

15, 26, 39, 49, 58, 122, 129, 134, 141, 145, 155, 158, 159, 161, 177, 178, 183, 185, 187, 226, 254, 262, 298, 339, 341, 355, 358, 362, 371, 381, 391, 393, 394, 445, 451, 469, 473, 493, 497, 565, 581, 583, 586, 589, 674, 781, 791, 889, 895, 899, 1114, 1119
Offset: 1

Views

Author

Zak Seidov, Oct 09 2015

Keywords

Comments

Subsequence of A263106.

Examples

			Permute the digits of 15 = 3 * 5 to get 51 = 3 * 17.
Permute the digits of 26 = 2 * 13 to get 62 = 2 * 31.
Permute the digits of 122 = 2 * 61 to get 221 = 13 * 17.
Permute the digits of 129 = 3 * 43 to get 291 = 3 * 97.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[4, 1000000], 2 == PrimeOmega[#] == PrimeOmega[fd = FromDigits[RotateLeft[IntegerDigits[#]]]] && fd > # &] (* for b-file *)
Showing 1-5 of 5 results.