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 1-4 of 4 results.

A072496 k-morphic numbers for any k such that (k-1)/10 is an odd integer not divisible by 5.

Original entry on oeis.org

0, 1, 4, 5, 6, 9, 11, 16, 19, 21, 24, 25, 29, 31, 36, 39, 41, 44, 49, 51, 56, 59, 61, 64, 69, 71, 75, 76, 79, 81, 84, 89, 91, 96, 99, 101, 125, 149, 151, 176, 199, 201, 224, 249, 251, 299, 301, 349, 351, 375, 376, 399, 401, 424, 449, 451, 499, 501, 549, 551
Offset: 1

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Author

Benoit Cloitre, Oct 19 2002

Keywords

Comments

Definition: n is a k-morphic number if n^k ends with n.

Crossrefs

Cf. A072495.
3-morphic numbers = 7-morphic numbers, see A033819; 5-morphic numbers = 13-morphic numbers, see A068407.

Programs

  • PARI
    k=11; for(n=1,10000,if((n^k)%(10^ceil(log(n)/log(10)))==n, print1(n,","))); \\ starting with 4, 5, ...
    
  • Sage
    def automorphic(maxdigits, pow, base=10) :
        morphs = [[0]]
        for i in range(maxdigits):
            T=[d*base^i+x for x in morphs[-1] for d in range(base)]
            morphs.append([x for x in T if x^pow % base^(i+1) == x])
        res = list(set(sum(morphs, []))); res.sort()
        return res
    # (call with pow=11 for this sequence), Eric M. Schmidt, Jul 30 2013

Extensions

Sequence corrected by Eric M. Schmidt, Jul 30 2013

A227071 Let s(m) = the set of k > 0 such that k^m ends with k. Then a(n) = least m such that s(m) = s(n).

Original entry on oeis.org

1, 2, 3, 2, 5, 6, 3, 2, 9, 2, 11, 2, 5, 2, 3, 6, 17, 2, 3, 2, 21, 2, 3, 2, 9, 26, 3, 2, 5, 2, 11, 2, 33, 2, 3, 6, 5, 2, 3, 2, 41, 2, 3, 2, 5, 6, 3, 2, 17, 2, 51, 2, 5, 2, 3, 6, 9, 2, 3, 2, 21, 2, 3, 2, 65, 6, 3, 2, 5, 2, 11, 2, 9, 2, 3, 26, 5, 2, 3, 2, 81, 2
Offset: 1

Views

Author

T. D. Noe, Jul 29 2013

Keywords

Comments

See A227070 for more details and for the numbers n such that n = a(n).
The entries in the b-file have been tentatively obtained by comparing the terms < 10^30 in the sets s(n). - Giovanni Resta, Jul 30 2013

Crossrefs

Cf. A003226 (n=2), A033819 (n=3), A068407 (n=5), A068408 (n=6).
Cf. A072496 (n=11), A072495 (n=21), A076650 (n=26).
Cf. A227070 (n such that n = a(n)).

Programs

  • Mathematica
    ts = {{}}; t2 = {1}; te = {1}; Do[s = Select[Range[0, 10^7], PowerMod[#, n, 10^IntegerLength[#]] == # &]; If[MemberQ[ts, s], AppendTo[t2, te[[Position[ts, s, 1, 1][[1, 1]]]]], AppendTo[ts, s]; AppendTo[te, n]; AppendTo[t2, n]], {n, 2, 82}]; t2

Formula

Conjecture: a(n+1) = A132741(n) + 1. - Eric M. Schmidt, Jul 30 2013

Extensions

Mathematica program and some entries corrected by Giovanni Resta, Jul 30 2013

A227070 Powers n such that the set s(n) = {k > 0 such that k^n ends with k} does not occur for smaller n.

Original entry on oeis.org

1, 2, 3, 5, 6, 9, 11, 17, 21, 26, 33, 41, 51, 65, 81, 101, 126, 129, 161, 201, 251, 257, 321, 401, 501, 513, 626, 641, 801, 1001, 1025, 1251, 1281, 1601, 2001, 2049, 2501, 2561, 3126, 3201, 4001, 4097, 5001, 5121, 6251, 6401, 8001, 8193, 10001
Offset: 1

Views

Author

T. D. Noe, Jul 29 2013

Keywords

Comments

These numbers might be called automorphic powers because the sets s(n) are called automorphic numbers. It appears that all numbers of the form 1 + 5^i are here. In fact, these appear to produce the only even numbers here. The set s(4) equals s(2). The set s(7) equals s(3). The set s(9) does not differ from s(5) until k = 10443. The set s(17) does not differ from s(9) until k = 108307. The sequence also has 126, 201, 251, 501, and 626, but there may be missing numbers.
Entries a(17)-a(49) have been tentatively obtained by comparing the terms < 10^30 in the sets s(n), for 2 <= n <= 10001. - Giovanni Resta, Jul 30 2013

Crossrefs

Cf. A003226 (n=2), A033819 (n=3), A068407 (n=5), A068408 (n=6).
Cf. A072496 (n=11), A072495 (n=21), A076650 (n=26).
Cf. A227071.

Programs

  • Mathematica
    ts = {}; t = {}; Do[s = Select[Range[11000000], PowerMod[#, n, 10^IntegerLength[#]] == # &]; If[! MemberQ[ts, s], Print[n]; AppendTo[ts, s]; AppendTo[t, n]], {n, 2, 101}]; t = Join[{1}, t]

Formula

Conjecture: a(n+1) = A003592(n) + 1. - Eric M. Schmidt, Jul 30 2013

Extensions

a(17)-a(49) from Giovanni Resta, Jul 30 2013

A255577 Numbers n not coprime to 10 such that there exists an integer k > 1 where n^k contains n as its last digits in base 10.

Original entry on oeis.org

2, 4, 5, 6, 8, 12, 16, 24, 25, 28, 32, 36, 44, 48, 52, 56, 64, 68, 72, 75, 76, 84, 88, 92, 96, 104, 112, 125, 128, 136, 144, 152, 168, 176, 184, 192, 208, 216, 224, 232, 248, 256, 264, 272, 288, 296, 304, 312, 328
Offset: 1

Views

Author

Keywords

Comments

Intersection of A065502 (numbers not coprime with 10) and A072495 (k-morphic numbers).
Also defined as all n not coprime with 10 where there exists k > 1 such that n^k mod 10^floor(log_10(n)) = n.
For n > 1, a(n) <= a(n-1) + 2^(ceiling(log_10(a(n))) + 1) (conjectured).
For a(n) >= 10^k where k >= 1, there exists a(m) = a(n) mod 10^j where m < n and j < k.
From Robert Israel, May 14 2015: (Start)
n with d digits is in the sequence if and only if n is either divisible by 2^d but not by 5, or divisible by 5^d but not by 2.
For d >= 2 the number of terms with d digits is 4*5^(d-1) + 2^(d-1) - 4*floor(5^d/50) - floor(2^d/20) - x(d) where x(d) = 3 if d == 2 or 3 mod 4, 2 otherwise.
(End)

Examples

			For n = 2, we have n^5 = 2^5 = 32, whose last digit is 2 = n, so 2 is in the sequence.
For n = 3, we have n^5 = 3^5 = 243, so 3 is in the sequence.
For n = 4, we have n^3 = 4^3 = 64, so 4 is in the sequence.
...
As a counterexample, n = 41 is not in the sequence because it is coprime with 10, even though we have 41^6 = 4750104241, whose last 2 digits are 41.
		

Crossrefs

Programs

  • Maple
    F:= d -> (seq(seq(2^d*(5*j+i),i=1..4),j=0..5^(d-1)-1), seq(5^d*(2*j+1),j=0..2^(d-1)-1)):
    sort(convert({seq(F(d),d=1..4)},list)); # Robert Israel, May 14 2015
  • Tcl
    See a255577.tcl in LINKS.

Formula

For n > 6, a(n) > 2n since no term is divisible by 10 (but all are divisible by either 2 or 5). - Charles R Greathouse IV, May 13 2015
Showing 1-4 of 4 results.