A000215 -id:A000215 - cp's OEIS frontend

A037096 Periodic vertical binary vectors computed for powers of 3: a(n) = Sum_{k=0 .. (2^n)-1} (floor((3^k)/(2^n)) mod 2) * 2^k.

1, 2, 0, 204, 30840, 3743473440, 400814250895866480, 192435610587299441243182587501623263200, 2911899996313975217187797869354128351340558818020188112521784134070351919360

Offset: 0

Antti Karttunen, Jan 29 1999

This sequence can be also computed with a recurrence that does not explicitly refer to 3^n. See the C program.

Conjecture: For n >= 3, each term a(n), when considered as a GF(2)[X] polynomial, is divisible by the GF(2)[X] polynomial (x + 1) ^ A055010(n-1). If this holds, then for n >= 3, a(n) = A048720(A136386(n), A048723(3,A055010(n-1))).

			When powers of 3 are written in binary (see A004656), under each other as:
  000000000001 (1)
  000000000011 (3)
  000000001001 (9)
  000000011011 (27)
  000001010001 (81)
  000011110011 (243)
  001011011001 (729)
  100010001011 (2187)
it can be seen that the bits in the n-th column from the right can be arranged in periods of 2^n: 1, 2, 4, 8, ... This sequence is formed from those bits: 1, is binary for 1, thus a(0) = 1. 01, reversed is 10, which is binary for 2, thus a(1) = 2, 0000 is binary for 0, thus a(2)=0, 000110011, reversed is 11001100 = A007088(204), thus a(3) = 204.

S. Wolfram, A New Kind of Science, Wolfram Media Inc., (2002), p. 119.

Antti Karttunen, Table of n, a(n) for n = 0..11
Antti Karttunen, C program for computing this sequence.
S. Wolfram, A New Kind of Science, Wolfram Media Inc., (2002), p. 119.

Cf. A036284, A037095, A037097, A136386 for related sequences.
Cf. A000079, A000215, A000225, A000244, A004656, A007088, A048720, A048723, A055010.
Cf. also A004642, A265209, A265210 (for 2^n written in base 3).

a(n) := sum( 'bit_n(3^i, n)*(2^i)', 'i'=0..(2^(n))-1);
bit_n := (x, n) -> `mod`(floor(x/(2^n)), 2);

a(n) = Sum_{k=0 .. A000225(n)} (floor(A000244(k)/(2^n)) mod 2) * 2^k.
Other identities and observations:
For n >= 2, a(n) = A000215(n-1)*A037097(n) = A048720(A037097(n), A048723(3, A000079(n-1))).

Entry revised by Antti Karttunen, Dec 29 2007

Name changed and the example corrected by Antti Karttunen, Dec 05 2015

A051158 Decimal expansion of Sum_{n >= 0} 1/(2^2^n+1).

Original entry on oeis.org

5, 9, 6, 0, 6, 3, 1, 7, 2, 1, 1, 7, 8, 2, 1, 6, 7, 9, 4, 2, 3, 7, 9, 3, 9, 2, 5, 8, 6, 2, 7, 9, 0, 6, 4, 5, 4, 6, 2, 3, 6, 1, 2, 3, 8, 4, 7, 8, 1, 0, 9, 9, 3, 2, 6, 2, 1, 4, 4, 2, 4, 5, 9, 9, 6, 0, 9, 1, 0, 8, 9, 9, 7, 7, 4, 8, 8, 6, 0, 8, 8, 8, 9, 9, 3, 6, 1, 9, 1, 8, 4, 6, 4, 6, 4, 4, 0, 7, 4

Offset: 0

Robert Lozyniak (11(AT)onna.com)

			0.59606317211782167942...

Joerg Arndt, Matters Computational (The Fxtbook), section 38.7, p.740 (gives method for divisionless computation corresponding to PARI/GP code below).
S. Audinarayana Moorthy, Problem E2455, The American Mathematical Monthly, Vol. 81, No. 1 (1974), p. 85, solution, ibid., Vol. 82, No. 2 (1975), pp. 173-174.
Michael Coons, On the rational approximation of the sum of the reciprocals of the Fermat numbers, Raman. J., Vol. 28 (2013), pp. 39-65.
Michael Coons, Addendum to: On the rational approximation of the sum of the reciprocals of the Fermat numbers, arXiv:1511.08147 [math.NT], 2015.
Steven R. Finch, Mathematical Constants II, Encyclopedia of Mathematics and Its Applications, Cambridge University Press, Cambridge, 2018, p. 247.
Solomon W. Golomb, On the sum of the reciprocals of the Fermat numbers and related irrationalities, Canad. J. Math., Vol. 15 (1963), pp. 475-478.

A048649 + A051158 = 2.
Terms in continued fraction: A159243. - Enrique Pérez Herrero, Nov 17 2009
Cf. A000215, A000120, A007814.

RealDigits[Sum[1/(2^2^n + 1), {n, 0, 10}], 10, 111][[1]] (* Robert G. Wilson v, Jul 03 2014 *)

/* divisionless routine from fxtbook */
s2(y, N=7)=
{ local(in, y2, A); /* as powerseries correct to order = 2^N-1 */
    in = 1; /* 1+y+y^2+y^3+...+y^(2^k-1) */
    A = y; for(k=2, N, in *= (1+y); y *= y; A += y*(in + A); );
    return( A ); }
a=0.5*s2(0.5) /* computation of the constant 0.596063172117821... */
/* Joerg Arndt, Apr 15 2010 */

suminf(n=0, 1/(2^2^n+1)) \\ Michel Marcus, May 15 2020

Equals (1/2) * Sum_{k>=1} A000120(k)/2^k (S. Audinarayana Moorthy, 1974). - Amiram Eldar, May 15 2020

Equals 1 - Sum_{n>=1} A007814(n)/2^n = 2/3 - Sum_{n>=1} A007814(n)/4^n = 3/5 - Sum_{n>=1} A007814(n)/16^n. - Amiram Eldar, Nov 06 2020

A087046 Algebraic order of r_n, the value of r in the logistic map that corresponding to the onset of the period 2^n-cycle.

Original entry on oeis.org

1, 2, 12, 240, 65280, 4294901760, 18446744069414584320, 340282366920938463444927863358058659840, 115792089237316195423570985008687907852929702298719625575994209400481361428480

Offset: 1

Eric W. Weisstein, Aug 04 2003

Ilias S. Kotsireas and Kostas Karamanos, Exact computation of the bifurcation point B4 of the logistic map and the Bailey-Broadhurst conjectures, Internat. J. Bifur. Chaos Appl. Sci. Engrg., Vol. 14, No. 7 (2004), pp. 2417-2423; alternative link.
Eric Weisstein's World of Mathematics, Logistic Map.

Cf. A086180, A086181, A087089, A118454, A000215, A346192.

Cf. A051179 (partial sums).

Table[If[n <= 1, 1, 2^(2^(n - 1)) - 2^(2^(n - 2))], {n, 1, 10}] (* Cheng Zhang, Apr 02 2012 *)

a(n) = 1<<(1<<(n-1)) - 1<<(1<<(n-2)); \\ Kevin Ryde, Jan 18 2024

a(n) = 2^(2^(n - 1)) - 2^(2^(n - 2)) with n>1, a(1)=1. - Cheng Zhang, Apr 02 2012

Sum_{n>=1} 1/a(n) = 1 + A346192. - Amiram Eldar, Jul 18 2021

More terms from Cheng Zhang, Apr 03 2012

A119563 Define F(n) = 2^(2^n)+1 = n-th Fermat number, M(n) = 2^n-1 = the n-th Mersenne number. Then a(n) = F(n)+M(n)-1 = 2^(2^n) + 2^n - 1.

Original entry on oeis.org

2, 5, 19, 263, 65551, 4294967327, 18446744073709551679, 340282366920938463463374607431768211583, 115792089237316195423570985008687907853269984665640564039457584007913129640191

Offset: 0

Cino Hilliard, May 31 2006

nonn

The first 5 entries are primes. Are there infinitely many primes in this sequence?

			F(2) = 2^(2^2)+1 = 17, M(2) = 2^2-1 = 3, F(2)+ M(2) - 1 = 19

fm3(n) = for(x=0,n,y=2^(2^x)+2^x-1;print1(y","))

a(n) = A119561(n)-2=A000215(n)+A000225(n)-1. - R. J. Mathar, Apr 22 2007

Edited by N. J. A. Sloane, Jun 03 2006

A129802 Possible bases for Pepin's primality test for Fermat numbers.

Original entry on oeis.org

3, 5, 6, 7, 10, 12, 14, 20, 24, 27, 28, 39, 40, 41, 45, 48, 51, 54, 56, 63, 65, 75, 78, 80, 82, 85, 90, 91, 96, 102, 105, 108, 112, 119, 125, 126, 130, 147, 150, 156, 160, 164, 170, 175, 180, 182, 192, 204, 210, 216, 224, 238, 243, 245, 250, 252, 260, 291, 294, 300

Offset: 1

Max Alekseyev, Jun 14 2007, corrected Dec 29 2007. Thanks to Ant King for pointing out an error in the earlier version of this sequence

nonn

Prime elements of this sequence are given by A102742.
From Jianing Song, May 15 2024: (Start)
Let m be an odd number and ord(2,m) = 2^r*d be the multiplicative order of 2 modulo m, where d is odd, then 2^2^n + 1 is congruent to one of 2^2^r + 1, 2^2^(r+1) + 1, ..., 2^2^(r+ord(2,d)-1) + 1 modulo m, so it suffices to check these ord(2,d) numbers.
Note that if m > 1, then m does not divide 2^2^n + 1 for n >= r, otherwise we would have 2^(2^n*d) = (2^ord(2,m))^2^(n-r) == 1 (mod m) and 2^(2^n*d) = (2^2^n)^d == (-1)^d == -1 (mod m). As a result, m is a term if and only if the Jacobi symbol ((2^2^n + 1)/m) is equal to -1 for m = r, r+1, ..., r+ord(2,d)-1.
By definition, a squarefree number that is a product of elite primes (A102742) or anti-elite primes (A128852) is a term if and only if its number of elite factors is odd. But a squarefree term can have factors that are neither elite nor anti-elite, the smallest being 551 = 19*29. (End)

			For n >= 2, we have 2^2^n + 1 == 170, 461, 17, 257, 519, 539 (mod 551) respectively for n == 0, 1, 2, 3, 4, 5 (mod 6). As we have (170/551) = (461/551) = (17/551) = (257/551) = (519/551) = (539/551) = -1, 551 is a term. - _Jianing Song_, May 19 2024

Arkadiusz Wesolowski, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Pepin's Test.

Cf. A000215, A019434, A060377, A102742, A128852, A372894.

{ isPepin(n) = local(s,S=Set(),t); n\=2^valuation(n,2); s=Mod(3,n); while( !setsearch(S,s), S=setunion(S,[s]); s=(s-1)^2+1); t=s; until( t==s, if( kronecker(lift(t),n)==1, return(0)); t=(t-1)^2+1);1 }
for(n=2,1000,if(isPepin(n),print1(n,", ")))

for(b=2, 300, k=b/2^valuation(b, 2); if(k>1, i=logint(k, 2); m=Mod(2, k); z=znorder(m); e=znorder(Mod(2, z/2^valuation(z, 2))); t=0; for(c=1, e, if(kronecker(lift(m^2^(i+c))+1, k)==-1, t++, break)); if(t==e, print1(b, ", ")))); \\ Arkadiusz Wesolowski, Sep 22 2021

isA129802(n) = n = (n >> valuation(n,2)); my(d = znorder(Mod(2, n)), StartPoint = valuation(d, 2), LengthTest = znorder(Mod(2, d >> StartPoint))); for(i = StartPoint, StartPoint + LengthTest - 1, if(kronecker(lift(Mod(2, n)^2^i + 1), n) == 1, return(0))); 1 \\ Jianing Song, May 19 2024

A positive integer 2^k*m, where m is odd and k >= 0, belongs to this sequence iff the Jacobi symbol (F_n/m) = 1 for only a finite number of Fermat numbers F_n = A000215(n).

A135590 Numbers k such that k^2 + 1 is a Sarrus number (pseudoprime to base 2).

Original entry on oeis.org

216, 948, 1560, 4872, 8208, 9828, 18200, 29640, 37024, 65536, 89550, 283800, 535920, 592956, 649800, 825930, 1042320, 1382400, 1536220, 3688230, 4215120, 4321800, 5103210, 19078930, 21415680, 24471720, 214067490, 435457620, 535019100

Offset: 1

Jason Earls, Feb 25 2008

nonn

Note that A000215(5) corresponds to a(10), and A000215(6) corresponds to a(33), and in general when A000215(n) is composite, this sequence has corresponding entry. - Jeppe Stig Nielsen, Mar 26 2016

Amiram Eldar, Table of n, a(n) for n = 1..40 (terms 1..33 from Jeppe Stig Nielsen)

Cf. A000215, A001567, A002522, A005574.

fQ[n_] := ( !PrimeQ[n^2 + 1] && PowerMod[2, n^2, n^2 + 1] == 1); lst = {}; Do[ If[ fQ@ n, AppendTo[lst, n]], {n, 2, 440000000, 2}]; lst (* Robert G. Wilson v, Apr 18 2008 *)

is(n) = {Mod(2, n)^(n-1)==1 && !ispseudoprime(n) && n > 1};
for(n=1, 1e10, if(is(n^2+1), print1(n, ", "))); \\ Altug Alkan, Mar 26 2016

More terms from Robert G. Wilson v, Apr 18 2008

A152155 Minimal residues of Pepin's Test for Fermat Numbers using the base 3.

Original entry on oeis.org

0, -1, -1, -1, -1, 10324303, -6586524273069171148, 110780954395540516579111562860048860420, 5864545399742183862578018016183410025465491904722516203269973267547486512819

Offset: 0

Dennis Martin (dennis.martin(AT)dptechnology.com), Nov 27 2008

sign

For n>=1 the Fermat Number F(n) is prime if and only if 3^((F(n) - 1)/2) is congruent to -1 (mod F(n)).

Any positive integer k for which the Jacobi symbol (k|F(n)) is -1 can be used as the base instead.

			a(4) = 3^(32768) (mod 65537) = 65536 = -1 (mod F(4)), therefore F(4) is prime.
a(5) = 3^(2147483648) (mod 4294967297) = 10324303 (mod F(5)), therefore F(5) is composite.

M. Krizek, F. Luca & L. Somer, 17 Lectures on Fermat Numbers, Springer-Verlag NY 2001, pp. 42-43.

Dennis Martin, Table of n, a(n) for n = 0..11
Chris Caldwell, The Prime Pages: Pepin's Test.

A000215, A019434, A152153, A152154, A152156.

f:= proc(n) local F;
   F:= 2^(2^n) + 1;
   `mods`(3 &^ ((F-1)/2), F)
end proc:
seq(f(n), n=0..10); # Robert Israel, Dec 19 2016

a(n)=centerlift(Mod(3,2^(2^n)+1)^(2^(2^n-1))) \\ Jeppe Stig Nielsen, Dec 19 2016

a(n) = 3^((F(n) - 1)/2) (mod F(n)), where F(n) is the n-th Fermat Number, using the symmetry mod (so (-F(n)-1)/2 < a(n) < (F(n)-1)/2).

A185121 Smallest prime factor of 10^(2^n) + 1.

Original entry on oeis.org

11, 101, 73, 17, 353, 19841, 1265011073, 257, 10753, 1514497, 1856104284667693057, 106907803649, 458924033, 3635898263938497962802538435084289

Offset: 0

Sergio Pimentel, Jan 22 2012

10^k+1 can only be prime if k is a power of 2. So far the only known primes of this form are a(0) = 11 and a(1) = 101. [Edited by M. F. Hasler, Aug 03 2019]
a(n) >= 2^(n+1)+1; we have a(n) = 2^(n+1)+1 for n=3, n=7, and n=15.
a(14) > 10^16. - Max Alekseyev, Jun 28 2013
From the Keller link a(15)-a(20) = 65537, 8257537, 175636481, 639631361, 70254593, 167772161. - Ray Chandler, Dec 27 2013

			For n=2, a(2)=73 since 10^(2^2) + 1 = 10001 = 73 * 137.

FactorDB, Factorizations of 10^(2^n)+1
Makoto Kamada, Factorizations of 100...001
Wilfrid Keller, Prime factors of generalized Fermat numbers Fm(10) and complete factoring status

Cf. A038371, A000533, A000215, A093179, A102050

Essentially the same as A102050. - Sean A. Irvine, Feb 17 2013

Table[With[{k = 2^n}, FactorInteger[10^k + 1]][[1, 1]], {n, 0, 13, 1}] (* Vincenzo Librandi, Jul 23 2013 *)

a(n) = factor(10^(2^n)+1)[1, 1] \\ Michel Marcus, May 30 2013

a(n) = A038371(2^n). - M. F. Hasler, Jul 30 2019

A063486 a(n) = 2^(2^n) + 5.

Original entry on oeis.org

7, 9, 21, 261, 65541, 4294967301, 18446744073709551621, 340282366920938463463374607431768211461, 115792089237316195423570985008687907853269984665640564039457584007913129639941

Offset: 0

Jason Earls, Jul 28 2001

D. M. Burton, Elementary Number Theory, Allyn and Bacon Inc., Boston, MA, 1976, p. 238.
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, Example 5.1 on page 153.

Harry J. Smith, Table of n, a(n) for n=0..11
Tigran Hakobyan, On the unboundedness of common divisors of distinct terms of the sequence a(n)=2^2^n+d for d>1, arXiv:1601.04946 [math.NT], 2016.

Cf. A000215, A001146, A130729, A130730.

2^2^Range[0, 10] + 5 (* Paolo Xausa, Apr 17 2024 *)

PARI
```
for(n=0,8,print1(2^(2^n)+5, ", "))
```

{ for (n=0, 11, write("b063486.txt", n, " ", 2^(2^n) + 5) ) } \\ Harry J. Smith, Aug 23 2009

[2**2**n + 5 for n in (0..8)] # Stefano Spezia, Jul 20 2025

cp's OEIS Frontend

A109022 Integers with mutual residues of 2 or more.

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A037096 Periodic vertical binary vectors computed for powers of 3: a(n) = Sum_{k=0 .. (2^n)-1} (floor((3^k)/(2^n)) mod 2) * 2^k.

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A051158 Decimal expansion of Sum_{n >= 0} 1/(2^2^n+1).

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A087046 Algebraic order of r_n, the value of r in the logistic map that corresponding to the onset of the period 2^n-cycle.

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Mathematica

PARI

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A119563 Define F(n) = 2^(2^n)+1 = n-th Fermat number, M(n) = 2^n-1 = the n-th Mersenne number. Then a(n) = F(n)+M(n)-1 = 2^(2^n) + 2^n - 1.

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PARI

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A129802 Possible bases for Pepin's primality test for Fermat numbers.

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PARI

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A135590 Numbers k such that k^2 + 1 is a Sarrus number (pseudoprime to base 2).

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