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 1-10 of 21 results. Next

A019434 Fermat primes: primes of the form 2^(2^k) + 1, for some k >= 0.

Original entry on oeis.org

3, 5, 17, 257, 65537
Offset: 1

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Author

Keywords

Comments

It is conjectured that there are only 5 terms. Currently it has been shown that 2^(2^k) + 1 is composite for 5 <= k <= 32 (see Eric Weisstein's Fermat Primes link). - Dmitry Kamenetsky, Sep 28 2008
No Fermat prime is a Brazilian number. So Fermat primes belong to A220627. For proof see Proposition 3 page 36 in "Les nombres brésiliens" in Links. - Bernard Schott, Dec 29 2012
This sequence and A001220 are disjoint (see "Other theorems about Fermat numbers" in Wikipedia link). - Felix Fröhlich, Sep 07 2014
Numbers n > 1 such that n * 2^(n-2) divides (n-1)! + 2^(n-1). - Thomas Ordowski, Jan 15 2015
From Jaroslav Krizek, Mar 17 2016: (Start)
Primes p such that phi(p) = 2*phi(p-1); primes from A171271.
Primes p such that sigma(p-1) = 2p - 3.
Primes p such that sigma(p-1) = 2*sigma(p) - 5.
For n > 1, a(n) = primes p such that p = 4 * phi((p-1) / 2) + 1.
Subsequence of A256444 and A256439.
Conjectures:
1) primes p such that phi(p) = 2*phi(p-2).
2) primes p such that phi(p) = 2*phi(p-1) = 2*phi(p-2).
3) primes p such that p = sigma(phi(p-2)) + 2.
4) primes p such that phi(p-1) + 1 divides p + 1.
5) numbers n such that sigma(n-1) = 2*sigma(n) - 5. (End)
Odd primes p such that ratio of the form (the number of nonnegative m < p such that m^q == m (mod p))/(the number of nonnegative m < p such that -m^q == m (mod p)) is a divisor of p for all nonnegative q. - Juri-Stepan Gerasimov, Oct 13 2020
Numbers n such that tau(n)*(number of distinct ratio (the number of nonnegative m < n such that m^q == m (mod n))/(the number of nonnegative m < n such that -m^q == m (mod n))) for nonnegative q is equal to 4. - Juri-Stepan Gerasimov, Oct 22 2020
The numbers of primitive roots for the five known terms are 1, 2, 8, 128, 32768. - Gary W. Adamson, Jan 13 2022
Prime numbers such that every residue is either a primitive root or a quadratic residue. - Keith Backman, Jul 11 2022
If there are only 5 Fermat primes, then there are only 31 odd order groups which have a 2-group automorphism group. See the Miles Englezou link for a proof. - Miles Englezou, Mar 10 2025

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 137-141, 197.
  • G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
  • C. F. Gauss, Disquisitiones Arithmeticae, Yale, 1965; see Table 1, p. 458.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §3.2 Prime Numbers, pp. 78-79.
  • Richard K. Guy, Unsolved Problems in Number Theory, A3.
  • Hardy and Wright, An Introduction to the Theory of Numbers, bottom of page 18 in the sixth edition, gives an heuristic argument that this sequence is finite.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 7, 70.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 136-137.

Crossrefs

Subsequence of A147545 and of A334101. Cf. also A333788, A334092.
Cf. A045544.

Programs

Formula

a(n+1) = A180024(A049084(a(n))). - Reinhard Zumkeller, Aug 08 2010
a(n) = 1 + A001146(n-1), if 1 <= n <= 5. - Omar E. Pol, Jun 08 2018

A001317 Sierpiński's triangle (Pascal's triangle mod 2) converted to decimal.

Original entry on oeis.org

1, 3, 5, 15, 17, 51, 85, 255, 257, 771, 1285, 3855, 4369, 13107, 21845, 65535, 65537, 196611, 327685, 983055, 1114129, 3342387, 5570645, 16711935, 16843009, 50529027, 84215045, 252645135, 286331153, 858993459, 1431655765, 4294967295, 4294967297, 12884901891, 21474836485, 64424509455, 73014444049, 219043332147, 365072220245, 1095216660735, 1103806595329, 3311419785987
Offset: 0

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Author

Keywords

Comments

The members are all palindromic in binary, i.e., a subset of A006995. - Ralf Stephan, Sep 28 2004
J. H. Conway writes (in Math Forum): at least the first 31 numbers give odd-sided constructible polygons. See also A047999. - M. Dauchez (mdzzdm(AT)yahoo.fr), Sep 19 2005 [This observation was also made in 1982 by N. L. White (see letter). - N. J. A. Sloane, Jun 15 2015]
Decimal number generated by the binary bits of the n-th generation of the Rule 60 elementary cellular automaton. Thus: 1; 0, 1, 1; 0, 0, 1, 0, 1; 0, 0, 0, 1, 1, 1, 1; 0, 0, 0, 0, 1, 0, 0, 0, 1; ... . - Eric W. Weisstein, Apr 08 2006
Limit_{n->oo} log(a(n))/n = log(2). - Bret Mulvey, May 17 2008
Equals row sums of triangle A166548; e.g., 17 = (2 + 4 + 6 + 4 + 1). - Gary W. Adamson, Oct 16 2009
Equals row sums of triangle A166555. - Gary W. Adamson, Oct 17 2009
For n >= 1, all terms are in A001969. - Vladimir Shevelev, Oct 25 2010
Let n,m >= 0 be such that no carries occur when adding them. Then a(n+m) = a(n)*a(m). - Vladimir Shevelev, Nov 28 2010
Let phi_a(n) be the number of a(k) <= a(n) and respectively prime to a(n) (i.e., totient function over {a(n)}). Then, for n >= 1, phi_a(n) = 2^v(n), where v(n) is the number of 0's in the binary representation of n. - Vladimir Shevelev, Nov 29 2010
Trisection of this sequence gives rows of A008287 mod 2 converted to decimal. See also A177897, A177960. - Vladimir Shevelev, Jan 02 2011
Converting the rows of the powers of the k-nomial (k = 2^e where e >= 1) term-wise to binary and reading the concatenation as binary number gives every (k-1)st term of this sequence. Similarly with powers p^k of any prime. It might be interesting to study how this fails for powers of composites. - Joerg Arndt, Jan 07 2011
This sequence appears in Pascal's triangle mod 2 in another way, too. If we write it as
1111111...
10101010...
11001100...
10001000...
we get (taking the period part in each row):
.(1) (base 2) = 1
.(10) = 2/3
.(1100) = 12/15 = 4/5
.(1000) = 8/15
The k-th row, treated as a binary fraction, seems to be equal to 2^k / a(k). - Katarzyna Matylla, Mar 12 2011
From Daniel Forgues, Jun 16-18 2011: (Start)
Since there are 5 known Fermat primes, there are 32 products of distinct Fermat primes (thus there are 31 constructible odd-sided polygons, since a polygon has at least 3 sides). a(0)=1 (empty product) and a(1) to a(31) are those 31 non-products of distinct Fermat primes.
It can be proved by induction that all terms of this sequence are products of distinct Fermat numbers (A000215):
a(0)=1 (empty product) are products of distinct Fermat numbers in { };
a(2^n+k) = a(k) * (2^(2^n)+1) = a(k) * F_n, n >= 0, 0 <= k <= 2^n - 1.
Thus for n >= 1, 0 <= k <= 2^n - 1, and
a(k) = Product_{i=0..n-1} F_i^(alpha_i), alpha_i in {0, 1},
this implies
a(2^n+k) = Product_{i=0..n-1} F_i^(alpha_i) * F_n, alpha_i in {0, 1}.
(Cf. OEIS Wiki links below.) (End)
The bits in the binary expansion of a(n) give the coefficients of the n-th power of polynomial (X+1) in ring GF(2)[X]. E.g., 3 ("11" in binary) stands for (X+1)^1, 5 ("101" in binary) stands for (X+1)^2 = (X^2 + 1), and so on. - Antti Karttunen, Feb 10 2016

Examples

			Given a(5)=51, a(6)=85 since a(5) XOR 2*a(5) = 51 XOR 102 = 85.
From _Daniel Forgues_, Jun 18 2011: (Start)
  a(0) = 1 (empty product);
  a(1) = 3 = 1 * F_0 = a(2^0+0) = a(0) * F_0;
  a(2) = 5 = 1 * F_1 = a(2^1+0) = a(0) * F_1;
  a(3) = 15 = 3 * 5 = F_0 * F_1 = a(2^1+1) = a(1) * F_1;
  a(4) = 17 = 1 * F_2 = a(2^2+0) = a(0) * F_2;
  a(5) = 51 = 3 * 17 = F_0 * F_2 = a(2^2+1) = a(1) * F_2;
  a(6) = 85 = 5 * 17 = F_1 * F_2 = a(2^2+2) = a(2) * F_2;
  a(7) = 255 = 3 * 5 * 17 = F_0 * F_1 * F_2 = a(2^2+3) = a(3) * F_2;
  ... (End)
		

References

  • Jean-Paul Allouche and Jeffrey Shallit, Automatic sequences, Cambridge University Press, 2003, p. 113.
  • Henry Wadsworth Gould, Exponential Binomial Coefficient Series, Tech. Rep. 4, Math. Dept., West Virginia Univ., Morgantown, WV, Sept. 1961.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 136-137.

Crossrefs

Cf. A038183 (odd bisection, 1D Cellular Automata Rule 90).
Iterates of A048724 (starting from 1).
Row 3 of A048723.
Positions of records in A268389.
Positions of ones in A268669 and A268384 (characteristic function).
Not the same as A045544 nor as A053576.
Cf. A045544.

Programs

  • Haskell
    a001317 = foldr (\u v-> 2*v + u) 0 . map toInteger . a047999_row
    -- Reinhard Zumkeller, Nov 24 2012
    (Scheme, with memoization-macro definec, two variants)
    (definec (A001317 n) (if (zero? n) 1 (A048724 (A001317 (- n 1)))))
    (definec (A001317 n) (if (zero? n) 1 (A048720bi 3 (A001317 (- n 1))))) ;; Where A048720bi implements the dyadic function given in A048720.
    ;; Antti Karttunen, Feb 10 2016
    
  • Magma
    [&+[(Binomial(n, i) mod 2)*2^i: i in [0..n]]: n in [0..41]]; // Vincenzo Librandi, Feb 12 2016
    
  • Maple
    A001317 := proc(n) local k; add((binomial(n,k) mod 2)*2^k, k=0..n); end;
  • Mathematica
    a[n_] := Nest[ BitXor[#, BitShiftLeft[#, 1]] &, 1, n]; Array[a, 42, 0] (* Joel Madigan (dochoncho(AT)gmail.com), Dec 03 2007 *)
    NestList[BitXor[#,2#]&,1,50] (* Harvey P. Dale, Aug 02 2021 *)
  • PARI
    a(n)=sum(i=0,n,(binomial(n,i)%2)*2^i)
    
  • PARI
    a=1; for(n=0, 66, print1(a,", "); a=bitxor(a,a<<1) ); \\ Joerg Arndt, Mar 27 2013
    
  • PARI
    A001317(n,a=1)={for(k=1,n,a=bitxor(a,a<<1));a} \\ M. F. Hasler, Jun 06 2016
    
  • PARI
    a(n) = subst(lift(Mod(1+'x,2)^n), 'x, 2); \\ Gheorghe Coserea, Nov 09 2017
    
  • Python
    from sympy import binomial
    def a(n): return sum([(binomial(n, i)%2)*2**i for i in range(n + 1)]) # Indranil Ghosh, Apr 11 2017
    
  • Python
    def A001317(n): return int(''.join(str(int(not(~n&k))) for k in range(n+1)),2) # Chai Wah Wu, Feb 04 2022

Formula

a(n+1) = a(n) XOR 2*a(n), where XOR is binary exclusive OR operator. - Paul D. Hanna, Apr 27 2003
a(n) = Product_{e(j, n) = 1} (2^(2^j) + 1), where e(j, n) is the j-th least significant digit in the binary representation of n (Roberts: see Allouche & Shallit). - Benoit Cloitre, Jun 08 2004
a(2*n+1) = 3*a(2*n). Proof: Since a(n) = Product_{k in K} (1 + 2^(2^k)), where K is the set of integers such that n = Sum_{k in K} 2^k, clearly K(2*n+1) = K(2*n) union {0}, hence a(2*n+1) = (1+2^(2^0))*a(2*n) = 3*a(2*n). - Emmanuel Ferrand and Ralf Stephan, Sep 28 2004
a(32*n) = 3 ^ (32 * n * log(2) / log(3)) + 1. - Bret Mulvey, May 17 2008
For n >= 1, A000120(a(n)) = 2^A000120(n). - Vladimir Shevelev, Oct 25 2010
a(2^n) = A000215(n); a(2^n-1) = a(2^n)-2; for n >= 1, m >= 0,
a(2^(n-1)-1)*a(2^n*m + 2^(n-1)) = 3*a(2^(n-1))*a(2^n*m + 2^(n-1)-2). - Vladimir Shevelev, Nov 28 2010
Sum_{k>=0} 1/a(k) = Product_{n>=0} (1 + 1/F_n), where F_n=A000215(n);
Sum_{k>=0} (-1)^(m(k))/a(k) = 1/2, where {m(n)} is Thue-Morse sequence (A010060).
If F_n is defined by F_n(z) = z^(2^n) + 1 and a(n) by (1/2)*Sum_{i>=0}(1-(-1)^{binomial(n,i)})*z^i, then, for z > 1, the latter two identities hold as well with the replacement 1/2 in the right hand side of the 2nd one by 1-1/z. - Vladimir Shevelev, Nov 29 2010
G.f.: Product_{k>=0} ( 1 + z^(2^k) + (2*z)^(2^k) ). - conjectured by Shamil Shakirov, proved by Vladimir Shevelev
a(n) = A000225(n+1) - A219843(n). - Reinhard Zumkeller, Nov 30 2012
From Antti Karttunen, Feb 10 2016: (Start)
a(0) = 1, and for n > 1, a(n) = A048720(3, a(n-1)) = A048724(a(n-1)).
a(n) = A048723(3,n).
a(n) = A193231(A000079(n)).
For all n >= 0: A268389(a(n)) = n.
(End)

A053576 Smallest number whose Euler totient is divisible by 2^n.

Original entry on oeis.org

1, 3, 5, 15, 17, 51, 85, 255, 257, 771, 1285, 3855, 4369, 13107, 21845, 65535, 65537, 196611, 327685, 983055, 1114129, 3342387, 5570645, 16711935, 16843009, 50529027, 84215045, 252645135, 286331153, 858993459, 1431655765, 4294967295, 8589934592, 17179869184, 34359738368, 68719476736, 137438953472, 274877906944, 549755813888, 1099511627776
Offset: 0

Views

Author

Labos Elemer, Jan 18 2000

Keywords

Comments

n = 32 is the first place where this differs from A001317, since 2^32 + 1 is not prime. - Mitch Harris, May 02 2007
a(8589934592) is the first unknown term; it is 2^8589934593 if F(33) = 2^(2^33)+1 is composite or F(33) otherwise. - Charles R Greathouse IV, Jul 15 2013
a(n) is the only odd element of the set phi-1(2^n), the totient inverses of 2^n. All other elements are 2*a(n), and the even elements of phi-1(2^(n-1)) * 2. - Torlach Rush, Sep 05 2017

Examples

			1,2,4,8,...,131072 divide phi of 2,3,5,15,...,196611 = 3*65537 respectively.
		

Crossrefs

Programs

  • Mathematica
    With[{s = Array[EulerPhi, 10^6]}, Table[FirstPosition[s, ?(Divisible[#, 2^n] &)][[1]], {n, 0, 19}]] (* _Michael De Vlieger, Sep 05 2017 *)
  • PARI
    a(n)={
      if(n >= 8589934592 && valuation(n>>5,2)>27,
        warning("Result is conjectural on the nonexistence of Fermat primes >= F(33).")
      );
      if(n>31,
        return(2<Charles R Greathouse IV, Jul 15 2013

Extensions

More odd terms from Jud McCranie, Jan 25 2000

A078164 Numbers k such that phi(k) is a perfect biquadrate.

Original entry on oeis.org

1, 2, 17, 32, 34, 40, 48, 60, 257, 512, 514, 544, 640, 680, 768, 816, 960, 1020, 1297, 1387, 1417, 1729, 1971, 2109, 2223, 2289, 2331, 2445, 2457, 2565, 2594, 2608, 2774, 2812, 2834, 2835, 3052, 3260, 3458, 3888, 3912, 3924, 3942, 3996, 4104, 4212, 4218
Offset: 1

Views

Author

Labos Elemer, Nov 27 2002

Keywords

Comments

Corresponding values of phi include 1, 16, 256, 1296, 4096, ... and these arise several times each.
a(3) = A053576(4).
A013776 is a subsequence since phi(2^(4*n+1)) = (2^n)^4. - Bernard Schott, Sep 22 2022
Subsequence of primes is A037896 since in this case: phi(k^4+1) = k^4. - Bernard Schott, Mar 05 2023

Crossrefs

Subsequence of A039770. A037896 is a subsequence.
Sequences where phi(k) is a perfect power: A039770 (square), A039771 (cube), this sequence (4th), A078165 (5th), A078166 (6th), A078167 (7th), A078168 (8th), A078169 (9th), A078170 (10th).

Programs

  • Mathematica
    k=4; Do[s=EulerPhi[n]^(1/k); If[IntegerQ[s], Print[n]], {n, 1, 5000}]
    Select[Range[5000],IntegerQ[Surd[EulerPhi[#],4]]&] (* Harvey P. Dale, Apr 30 2015 *)
  • PARI
    is(n)=ispower(eulerphi(n),4) \\ Charles R Greathouse IV, Apr 24 2020
    
  • Python
    from itertools import count, islice
    from sympy import totient, integer_nthroot
    def A078164_gen(startvalue=1): # generator of terms >= startvalue
        return filter(lambda n:integer_nthroot(totient(n),4)[1], count(max(1,startvalue)))
    A078164_list = list(islice(A078164_gen(),20)) # Chai Wah Wu, Feb 28 2023

A004729 Divisors of 2^32 - 1 (for a(1) to a(31), the 31 regular polygons with an odd number of sides constructible with ruler and compass).

Original entry on oeis.org

1, 3, 5, 15, 17, 51, 85, 255, 257, 771, 1285, 3855, 4369, 13107, 21845, 65535, 65537, 196611, 327685, 983055, 1114129, 3342387, 5570645, 16711935, 16843009, 50529027, 84215045, 252645135, 286331153, 858993459, 1431655765, 4294967295
Offset: 0

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Author

Keywords

Comments

The 32 divisors of the product of the 5 known Fermat primes.
The only known odd numbers whose totient is a power of 2. - Labos Elemer, Dec 06 2000
Equals first 32 members of A001317. Also, equals first 32 members of A053576. - Omar E. Pol, Dec 10 2008
Omitting the first term a(0)=1 gives A045544 (the number of sides of constructible odd-sided regular polygons.)

References

  • J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, New York, 1996; see p. 140.

Crossrefs

Programs

  • Mathematica
    Divisors[2^32-1]
  • PARI
    divisors(1<<32-1)

Extensions

Edited by Daniel Forgues, Jun 17 2011

A235034 Numbers whose prime divisors, when multiplied together without carry-bits (as encodings of GF(2)[X]-polynomials, with A048720), produce the original number; numbers for which A234741(n) = n.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 16, 17, 19, 20, 22, 23, 24, 26, 28, 29, 30, 31, 32, 34, 37, 38, 40, 41, 43, 44, 46, 47, 48, 51, 52, 53, 56, 58, 59, 60, 61, 62, 64, 67, 68, 71, 73, 74, 76, 79, 80, 82, 83, 85, 86, 88, 89, 92, 94, 95, 96, 97, 101
Offset: 1

Views

Author

Antti Karttunen, Jan 02 2014

Keywords

Comments

If n is present, then 2n is present also, as shifting binary representation left never produces any carries.

Examples

			All primes occur in this sequence as no multiplication -> no need to add any intermediate products -> no carry bits produced.
Composite numbers like 15 are also present, as 15 = 3*5, and when these factors (with binary representations '11' and '101') are multiplied as:
   101
  1010
  ----
  1111 = 15
we see that the intermediate products 1*5 and 2*5 can be added together without producing any carry-bits (as they have no 1-bits in the same columns/bit-positions), so A048720(3,5) = 3*5 and thus 15 is included in this sequence.
		

Crossrefs

Gives the positions of zeros in A236378, i.e., n such that A234741(n) = n.
Intersection with A235035 gives A235032.
Other subsequences: A000040 (A091206 and also A091209), A045544 (A004729), A093641, A235040 (gives odd composites in this sequence), A235050, A235490.

A050922 Triangle in which n-th row gives prime factors of n-th Fermat number 2^(2^n)+1.

Original entry on oeis.org

3, 5, 17, 257, 65537, 641, 6700417, 274177, 67280421310721, 59649589127497217, 5704689200685129054721, 1238926361552897, 93461639715357977769163558199606896584051237541638188580280321
Offset: 0

Views

Author

N. J. A. Sloane, Dec 30 1999

Keywords

Comments

Alternatively, list of prime factors of terms of A001317 in order of their first appearance. - Labos Elemer, Jan 21 2002
From T. D. Noe, Jan 29 2009: (Start)
That these two definitions give the same sequence follows from the fact (stated as a formula in A001317) that A001317(n) is the product of Fermat numbers F(i) according to which bits of n are set.
For instance, for n=41, the binary representation of n is 101001, which has bits 0, 3 and 5 set. A001317(n) = 3311419785987 = 3*257*4294967297 = F(0)*F(3)*F(5).
This factorization also explains why the "first 31 numbers give odd-sided constructible polygons". I think Hewgill first noticed this factorization. (End)

Examples

			Triangle begins:
  3;
  5;
  17;
  257;
  65537;
  641,               6700417;
  274177,            67280421310721;
  59649589127497217, 5704689200685129054721;
  1238926361552897,  93461639715357977769163558199606896584051237541638188580280321;
  ...
A001317(127) = 3*5*17*257*65537.641*6700417*274177*6728042130721, A001317(128) = 59649589127497217*5704689200685129054721. See also A050922. Compare with A053576, where 2 and A000215 appear as prime factors. - _Labos Elemer_, Jan 21 2002
		

References

  • M. Aigner and G. M. Ziegler, Proofs from The Book, Springer-Verlag, Berlin, 2nd. ed., 2001; see p. 3.

Crossrefs

Programs

  • Mathematica
    Flatten[Transpose[FactorInteger[#]][[1]]&/@Table[2^(2^n)+1,{n,0,8}]] (* Harvey P. Dale, May 18 2012 *)
  • PARI
    for(n=0, 1e3, f=factor(2^(2^n)+1)[, 1]; for(i=1, #f, print1(f[i], ", "))) \\ Felix Fröhlich, Aug 16 2014

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Apr 13 2000.
Edited by N. J. A. Sloane, Jan 31 2009 at the suggestion of T. D. Noe
Link to Munafo webpage fixed by Robert Munafo, Dec 09 2009

A078165 Numbers k such that phi(k) is a perfect 5th power.

Original entry on oeis.org

1, 2, 51, 64, 68, 80, 96, 102, 120, 1285, 2048, 2056, 2176, 2560, 2570, 2720, 3072, 3084, 3264, 3840, 4080, 7957, 8227, 8279, 9079, 9139, 9709, 9919, 10355, 10595, 11667, 11673, 11691, 12099, 12393, 12483, 12753, 12987, 13797, 14715, 14763
Offset: 1

Views

Author

Labos Elemer, Nov 27 2002

Keywords

Comments

As phi(2^(5*n+1)) = (2^n)^5, A013822 is a subsequence. - Bernard Schott, Sep 26 2022
Numbers of the form u = 2^(5*k)*3^(5*m + 1), k>=1, m>=0, are terms because phi(u) = 2^(5*k)*3^(5*m) = (2^k*3^m)^5. - Marius A. Burtea, Sep 26 2022

Examples

			phi of the sequence includes 1, 32, 1024, 7776, ...; powers arise several times; a(3) = A053576(5) = 51.
		

Crossrefs

A013822 is a subsequence.
Cf. A039770 (square), A039771 (cube), A078164 (4th), A078165 (5th, this sequence), A078166 (6th), A078167 (7th), A078168 (8th), A078169 (9th), A078170 (10th power), A001317, A053576, A045544, A000010.

Programs

  • Mathematica
    k=5; Do[s=EulerPhi[n]^(1/k); If[IntegerQ[s], Print[n]], {n, 1, 5000}]
    Select[Range[15000],IntegerQ[Surd[EulerPhi[#],5]]&] (* Harvey P. Dale, Jul 26 2019 *)
  • PARI
    is(n)=ispower(eulerphi(n),5) \\ Charles R Greathouse IV, Apr 24 2020

A078166 Numbers k such that phi(k) is a perfect sixth power.

Original entry on oeis.org

1, 2, 85, 128, 136, 160, 170, 192, 204, 240, 4369, 8192, 8224, 8704, 8738, 10240, 10280, 10880, 12288, 12336, 13056, 15360, 15420, 16320, 47197, 47239, 47989, 49267, 49589, 50557, 51319, 52429, 52649, 55699, 57589, 57953, 59495, 63973
Offset: 1

Views

Author

Labos Elemer, Nov 27 2002

Keywords

Comments

As phi(2^(6*n+1)) = (2^n)^6, A277757 is a subsequence. - Bernard Schott, Sep 23 2022

Examples

			phi of the sequence includes 1, 64, 4096, 46656,..; powers arise several times; a(3)= A053576(6) = 85; in sequence relatively large jumps are observable when power of new numbers appear.
		

Crossrefs

A277757 is a subsequence.
Numbers k such that phi(k) is a perfect power: A039770 (square), A039771 (cube), A078164 (4th), A078165 (5th), A078166 (6th, this sequence), A078167 (7th), A078168 (8th), A078169 (9th), A078170 (10th power).

Programs

A078167 Numbers k such that phi(k) is a perfect 7th power.

Original entry on oeis.org

1, 2, 255, 256, 272, 320, 340, 384, 408, 480, 510, 21845, 32768, 32896, 34816, 34952, 40960, 41120, 43520, 43690, 49152, 49344, 52224, 52428, 61440, 61680, 65280, 280999, 281587, 282637, 282949, 283897, 294409, 297449, 300409, 302039, 304399
Offset: 1

Views

Author

Labos Elemer, Nov 27 2002

Keywords

Examples

			phi of the sequence includes 1, 128, 16384, 279936, etc..; powers arise several times; a(3) = A053576(7) = 255; in sequence rather large jumps arise when power of new numbers appear.
		

Crossrefs

Cf. A039770 (square), A039771 (cube), A078164 (4th), A078165 (5th), A078166 (6th), A078167 (7th, this sequence), A078168 (8th), A078169 (9th), A078170 (10th power), A001317, A053576, A045544, A000010.

Programs

  • Mathematica
    k=7; Do[s=EulerPhi[n]^(1/k); If[IntegerQ[s], Print[n]], {n, 1, 1000000}]
  • PARI
    is(n)=ispower(eulerphi(n),7) \\ Charles R Greathouse IV, Apr 24 2020
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