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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A248802 Smallest prime factor of 2^(2^n+2) + 3.

Original entry on oeis.org

11, 19, 67, 13, 262147, 13, 1669, 13, 255127, 13, 2383, 13, 67, 13, 32544331, 13, 271, 13, 4057, 13
Offset: 0

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Author

Arkadiusz Wesolowski, Oct 14 2014

Keywords

Comments

These numbers do not occur in A023394 (prime factors of Fermat numbers A000215).
From Chai Wah Wu, Oct 21 2019: (Start)
a(22) = 67, a(26) = 1399, a(28) = 10957, a(30) = 117127, a(32) = 67, a(36) = 12781849, a(38) = 262147, a(42) = 67, a(48) = 6391117, a(50) = 1265347, a(52) = 67, a(54) = 2383, a(58) = 26833, a(62) = 67, a(64) = 517261, a(68) = 2251, a(72) = 67, a(74) = 137077, a(78) = 562273, a(82) = 67, a(84) = 1399, a(86) = 3253, a(88) = 271, a(92) = 67, a(94) = 2203, a(96) = 329347, a(98) = 2383, a(100) = 5323, a(110) = 2759137, a(114) = 122653, a(116) = 659941, a(126) = 48337, a(130) = 2403229, a(134) = 2534659, a(140) = 41257.
Theorem: a(n) >= 13 for n > 0.
Proof. 2^(2^n+2) + 3 is odd and not a multiple of 3, so a(n) > 3. For all primes 3 < p < 14, p-3 is a power of 2. For p = 5, 2^4 == 1 mod 5, so for n = 1, 2^(2^n+2) + 3 == 4 mod 5 and for n > 1, 2^(2^n+2) + 3 == 7 == 2 mod 5. For p = 7, 2^3 == 1 mod 7. Since 2^n+2 <> 2 mod 3, 2^(2^n+2) <> 4 mod 7 and thus 2^(2^n+2) + 3 <> 0 mod 7.
For p = 11, 2^10 == 1 mod 11. Since 2^n+2 is even for n > 0, 2^n+2 <> 3 mod 10 and thus 2^(2^n+2) <> 2^3 mod 11 and 2^(2^n+2) + 3 <> 0 mod 11. End of proof.
Theorem: a(2n+1) = 13 for n >= 1.
Proof by induction. a(3) = 13 since 2^(2^3+2) + 3 = 1027 = 13*79.
Suppose a(2n+1) = 13, this implies that 2^(2^(2n+1)+2) == 10 mod 13.
Then 2^(2^(2n+3)+2) = 2^(3*2^(2n+1)) * 2^(2^(2n+1)+2). For n >= 1, 2^(2n+1) is a multiple of 4, and thus 2^(3*2^(2n+1)) == 2^12 == 1 mod 13.
This implies that 2^(2^(2n+3)+2) == 2^(2^(2n+1)+2) == 10 mod 13 and thus a(2n+3) <= 13. By the first result above, a(2n+3) = 13.
End of proof.
Conjecture 1: a(10n+2) = 67 for n >= 0.
Conjecture 2: a(36n+16) = 271 for n >= 0 and n <> 1 mod 5.
Conjecture 3: a(84n+22) = 523 for n >= 0 and n <> 0 mod 5.
Conjecture 4: a(58n+26) = 1399 for n >= 0 and when it is not covered by Conjectures 1-3.
Conjecture 5: a(138n+6) = 1669 for n >= 0 and n <> 2 mod 5.
Conjecture 6: a(44n+10) = 2383 for n >= 0 and when it is not covered by Conjectures 1-5.
(End)

Crossrefs

Cf. A000215, A023394, A057733.

Programs

  • Mathematica
    PrimeFactors[n_]:= Flatten[Table[#[[1]], {1}]&/@FactorInteger[n]]; Table[PrimeFactors[2^(2^n + 2) + 3] [[1]], {n, 0, 7}] (* Vincenzo Librandi, Oct 15 2014 *)
  • PARI
    a(n) = factor(2^(2^n+2) + 3)[1, 1]; \\ Michel Marcus, Oct 15 2014
  • PARI
    for(n=1,19,my(x=2^(2^n+2)+3);forprime(k=3,oo,if(x%k==0,print1(k,", ");break))) \\ Hugo Pfoertner, Aug 08 2019

Formula

Smallest prime factor of 4*A000215(n) - 1, with the Fermat numbers A000215. - Wolfdieter Lang, Nov 05 2014

A291880 Numbers n such that phi(n) - 1 | sigma(n).

Original entry on oeis.org

3, 4, 5, 6, 8, 10, 20, 22, 40, 76, 80, 108, 160, 204, 320, 640, 1072, 1280, 2560, 4192, 5120, 10240, 20480, 40960, 49344, 81920, 163840, 327680, 655360, 1310720, 2621440, 4197376, 5242880, 10485760, 20971520, 41943040, 83886080, 167772160, 268460032, 335544320, 671088640, 1073790976, 1342177280, 2684354560, 5368709120
Offset: 1

Views

Author

Paolo P. Lava, Sep 05 2017

Keywords

Comments

Numbers n such that A109606(n) | A000203(n).
All numbers of the form 5*2^x, with x >= 0, are part of the sequence (A020714).
Values of the ratio sigma(n)/(phi(n)-1) are 4, 7, 2, 12, 5, 6, 6, 4, 6, 4, 6, 8, 6, 8, 6, 6, 4, 6, 6, 4, 6, 6, 6, 6, 8, 6, 6, 6, 6, 6, 6, 4, 6, ...
Sequence contains also terms of the form 2^(n-2)*(2^n+3) where 2^n+3 is a prime and n > 3, like 22, 76, 1072, 4192, 4197376, 268460032. See A057733 for primes of the form 2^n+3. - Michel Marcus, Sep 17 2017

Examples

			sigma(1072) = 2108, phi(1072) = 528 and 2108/(528 - 1) = 4.

Crossrefs

Cf. A000010, A000203, A020714, A056097, A057733, A109606.

Programs

  • Maple
    with(numtheory): P:=proc(q) local n; for n from 3 to q do
if type(sigma(n)/(phi(n)-1),integer) then  print(n); fi; od;  end: P(10^7);
  • Mathematica
    Select[Range[3, 10^6], Divisible[DivisorSigma[1, #], EulerPhi[#] - 1] &] (* Michael De Vlieger, Sep 06 2017 *)
  • PARI
    isok(n) = denominator(sigma(n)/(eulerphi(n)-1)) == 1; \\ Michel Marcus, Sep 06 2017

Extensions

a(34)-a(41) from Michel Marcus, Sep 15 2017
a(42)-a(45) from Michel Marcus, Sep 21 2017

A363464 Numbers k in A052294 with arithmetic derivative k' (A003415) in A052294.

Original entry on oeis.org

6, 9, 10, 14, 18, 20, 21, 22, 24, 25, 33, 34, 35, 38, 40, 42, 44, 48, 49, 52, 62, 65, 66, 68, 69, 70, 76, 80, 84, 88, 91, 93, 94, 96, 100, 104, 110, 115, 117, 118, 121, 132, 133, 134, 138, 140, 143, 144, 145, 148, 152, 155, 158, 164, 174, 182, 185, 186, 188, 192
Offset: 1

Views

Author

Marius A. Burtea, Jul 08 2023

Keywords

Comments

If p > 2 is in A092506 then m = 2*p and u = 4*p are terms. Indeed, if p = 2^k + 1, k >= 1, m = 2*(2^k + 1) = 2^(k+1) + 2^1 has two 1's in its binary expansion, and m' = p+2 = 2^k + 3 = 2^k + 2^1 + 1 has three 1's in its binary expansion. Similarly u = 4*(2^k + 1) = 2^(k+2) + 2^2 and u' = 4*p + 4 = 2^(k+2) + 2^3.
If p is in A057733 then the number m = 2*p is a term. Indeed, if p = 2^k + 3, k >= 1, m = 2*(2^k + 3) = 2^(k+1) + 2^2 + 2 has three 1's in its binary expansion, and m' = p+2 = 2^k + 5 = 2^k + 2^2 + 1 has three 1's in its binary expansion.
If p > 7 is in A057733 then the number m = 4*p is a term. Indeed, if p = 2^k + 3, k >= 3, m = 4*(2^k + 3) = 2^(k+2) + 2^3 + 2 has three 1's in its binary expansion, and m' = 4*(p + 1) = 4*(2^k + 4) = 2^(k+2) + 2^4 has two 1's in its binary expansion.
If p is in A123250 then the number m = 4*p is a term. Indeed, if p = 2^k + 5, k >= 1, m = 4*(2^k + 5) = 2^(k+2) + 2^4 + 2^2 has three 1's its binary expansion, and m' = 4*(p+1) = 4*(2^k + 6) = 2^(k+2) + 2^4 + 2^2 has three 1's in its binary expansion.
If p is in A104070 then the number m = 4*p is a term. Indeed, if p = 2^k + 9, k >= 1, m = 4*(2^k + 9) = 2^(k+2) + 2^5 + 2^2 has three 1's its binary expansion, and m' = 4*(p+1) = 4*(2^k + 10) = 2^(k+2) + 2^5 + 2^3 has three 1's in its binary expansion.

Examples

			6 = 110_2 has two 1's, 6' = 5 = 101_2 has two 1's, so 6 is a term.
9 = 101_2 has two 1's, 9' = 6 = 110_2 has two 1's, so 9 is a term.
10 = 1010_2 has two 1's, 10' = 7 = 111_2 has three 1's, so 10 is a term.
18 = 10010_2 has two 1's, 18' = 21 = 10101_2 has three 1's, so 18 is a term.

Crossrefs

Cf. A000120, A003415, A052294, A057733, A081092, A092506, A104070, A123250.

Programs

  • Magma
    fp:=func; f:=func; [n:n in [1..200]| fp(n) and fp(Floor(f(n)))];
  • Mathematica
    pernQ[n_] := PrimeQ[DigitCount[n, 2, 1]]; d[0] = d[1] = 0; d[n_] := n*Plus @@ ((Last[#]/First[#]) & /@ FactorInteger[n]); Select[Range[200], And @@ pernQ[{#, d[#]}] &] (* Amiram Eldar, Jul 10 2023 *)
Previous Showing 31-33 of 33 results.

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