A090368 -id:A090368 - cp's OEIS frontend

A177966 Indices m for which A177961(m) = 2 + m.

2, 5, 8, 11, 12, 14, 20, 23, 26, 27, 29, 35, 41, 42, 44, 50, 53, 56, 57, 65, 68, 74, 83, 86, 87, 89, 95, 98, 113, 116, 117, 119, 125, 128, 131, 132, 134, 140, 146, 147, 155, 158, 173, 176, 177, 179, 191, 192, 194, 200, 209, 215, 221, 222, 224, 230, 233, 239, 245, 251, 252, 254

Offset: 1

Vladimir Shevelev, May 16 2010

nonn

All m for which 2*m+1 is in A003627 are in the sequence:
This concerns m=2, 5, 8, 11, 14, 20, 23, 26, 29, 35,...
Union of (A003627-1)/2 and (A132235+1)/2. - Robert Israel, Jul 31 2015

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A003627, A132235, A177961, A177964, A177965.

A090368 := proc(n) A020639(2*n-1) ; end proc:
A177961 := proc(n) (A090368(n)+A090368(n+1)) /2 ; end proc:
isA177966 := proc(n) A177961(m) = m+2 ; end proc:
for m from 1 to 800 do if isA177966(m) then printf("%d,",m) ; end if; end do:
# R. J. Mathar, Oct 25 2010
N:= 1000: # to get all terms <= N
A1:= map(t -> (t-1)/2, select(isprime, {seq(6*i-1, i=1..(N+1)/3)})):
A2:= map(t -> (t+1)/2, select(isprime, {seq(23+30*i,i=0..(N-12)/15)})):
sort(convert(A1 union A2,list));
# Robert Israel, Jul 31 2015

M = 1000; (* to get all terms <= M *)
A1 = (Select[Table[6 i - 1, {i, 1, (M + 1)/3}], PrimeQ] - 1)/2;
A2 = (Select[Table[23 + 30 i, {i, 0, (M - 12)/15}], PrimeQ] + 1)/2;
Union[A1, A2] (* Jean-François Alcover, Jul 17 2020, after Robert Israel *)

Corrected (11, 23, 27, etc. inserted) and extended by R. J. Mathar, Oct 25 2010

A195508 Number of iterations in a Draim factorization of 2n+1.

Original entry on oeis.org

1, 2, 3, 1, 5, 6, 1, 8, 9, 1, 11, 2, 1, 14, 15, 1, 2, 18, 1, 20, 21, 1, 23, 3, 1, 26, 2, 1, 29, 30, 1, 2, 33, 1, 35, 36, 1, 3, 39, 1, 41, 2, 1, 44, 3, 1, 2, 48, 1, 50, 51, 1, 53, 54, 1, 56, 2, 1, 3, 5, 1, 2, 63, 1, 65, 3, 1, 68, 69, 1, 5, 2, 1, 74, 75, 1, 2, 78, 1, 3, 81, 1, 83, 6, 1, 86, 2, 1, 89, 90, 1, 2, 5, 1, 95, 96, 1, 98, 99

Offset: 1

Frank Ellermann, Sep 19 2011

A Draim factorization determines the smallest divisor d of 2n+1 with simple operations (integer division, remainder, +, -, *) and needs a(n)=(d-1)/2 steps.

Least m>0 for which gcd(n+1+m, n-m) > 1. [Clark Kimberling, Jul 18 2012]

			a(12)=2 because the Draim algorithm needs 2 steps to find the smallest divisor of 25=2*12+1; any a(n)=2 indicates a smallest divisor 5 of 2n+1.

H. Davenport, The Higher Arithmetics, 7th ed. 1999, Cambridge University Press, pp. 32-35.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A090368.

a[n_] := Module[{m = 1}, While[GCD[n + m + 1, n - m] == 1, m++]; m]; Array[a, 100] (* Amiram Eldar, Nov 06 2019 *)

SEQ = '' ;  do N = 1 to 50 ;  X = 2 * N + 1  ;  M = X
   do Y = 3 by 2 until R = 0
      Q = X % Y   ;  R = X // Y  ;  M = M - 2 * Q  ;  X = M + R
   end Y ;  SEQ = SEQ (( Y - 1 ) / 2 ) ;  end N ;  say SEQ

a(n) = (A090368(n+1)-1)/2.

A227198 Odd terms in A227197.

Original entry on oeis.org

9, 15, 25, 33, 39, 49, 57, 63, 81, 87, 95, 105, 111, 119, 121, 129, 135, 145, 153, 159, 169, 177, 183, 201, 207, 215, 225, 231, 249, 255, 265, 273, 279, 289, 297, 303, 321, 327, 335, 345, 351, 361, 369, 375, 385, 393, 399, 417, 423, 441, 447, 455, 465, 471, 489

Offset: 1

Antti Karttunen, Jul 06 2013

nonn

Gives all terms n for which A090368((n+1)/2) = A112046((n-1)/2).

Contains all odd squares. What else?

Cf. A227196, A227197, A090368, A112046.

A227196(n) = for(k=1, n, if(kronecker(k, n)<1, return(k)))
for(n=2,1000,if((0==kronecker(A227196(n),n)&&1==(n%2)),print1(n,", ")))

A117368 a(n) = largest prime less than the smallest prime dividing (2n-1).

Original entry on oeis.org

2, 3, 5, 2, 7, 11, 2, 13, 17, 2, 19, 3, 2, 23, 29, 2, 3, 31, 2, 37, 41, 2, 43, 5, 2, 47, 3, 2, 53, 59, 2, 3, 61, 2, 67, 71, 2, 5, 73, 2, 79, 3, 2, 83, 5, 2, 3, 89, 2, 97, 101, 2, 103, 107, 2, 109, 3, 2, 5, 7, 2, 3, 113, 2, 127, 5, 2, 131, 137, 2, 7, 3, 2, 139, 149, 2, 3, 151, 2, 5, 157, 2

Offset: 2

Leroy Quet, Mar 10 2006

nonn

Placing a 1 between each term of this sequence gets sequence A117365.

			a(13) = 3 because smallest prime dividing 25 is 5 and largest prime less than 5 is 3.

Robert Israel, Table of n, a(n) for n = 2..10000

Cf. A006512, A090368, A117365, A151799.

with(numtheory): a:=proc(n): prevprime(factorset(2*n-1)[1]): end: seq(a(n),n=2..90); # Emeric Deutsch, Apr 22 2006

prs=Prime[Range[50]];
f[n_]:=NextPrime[First[Select[prs,Divisible[2 n-1,#]&]],-1]
f/@Range[2,90]  (* Harvey P. Dale, Jan 23 2011 *)

From Robert Israel, Apr 14 2019: (Start)
a(n) = A151799(A090368(n)).
a(n) = 2*n-3 if 2*n-1 is in A006512. (End)

More terms from Emeric Deutsch, Apr 22 2006

A177983 a(1)=3. Otherwise the average of the least prime divisors of 2n-1 and 2n+3.

Original entry on oeis.org

3, 5, 4, 9, 8, 7, 15, 11, 10, 21, 4, 13, 17, 17, 16, 18, 20, 4, 39, 23, 22, 45, 5, 25, 30, 4, 28, 32, 32, 31, 33, 35, 4, 69, 38, 37, 40, 41, 5, 81, 4, 43, 47, 5, 46, 6, 50, 4, 99, 53, 52, 105, 56, 55, 111, 4, 58, 6, 7, 5, 8, 65, 4, 129, 5, 67, 72, 71, 70, 75, 4, 7, 77, 77, 76, 78, 80, 4, 82, 83

Offset: 1

Vladimir Shevelev, May 16 2010

Lim inf (n->infinity) (a(n)/n)=0.

If there exist infinitely many cousin primes (A023200), then lim sup (n->infinity) (a(n)/ n)=2.

Cf. A020639, A177961.

A090368 := proc(n) A020639(2*n-1) ; end proc:
A177983 := proc(n) (A090368(n)+A090368(n+2)) /2 ; end proc:
seq(A177983(n),n=1..80) ; # R. J. Mathar, Oct 25 2010

Table[Mean[{FactorInteger[2n-1][[1,1]],FactorInteger[2n+3][[1,1]]}],{n,80}] (* Harvey P. Dale, Nov 08 2022 *)

a(n) = if (n==1, 3, (factor(2*n-1)[1,1] + factor(2*n+3)[1,1])/2); \\ Michel Marcus, Feb 08 2016

a(n) = (A090368(n)+A090368(n+2))/2 . [R. J. Mathar, Oct 25 2010]

I corrected a(25). It should be 30 (not 31) Vladimir Shevelev, May 22 2010

More terms from R. J. Mathar, Oct 25 2010

A090370 Least m > 3 such that gcd(n-1, m*n - 1) = m-1.

Original entry on oeis.org

4, 5, 6, 4, 8, 5, 4, 6, 12, 4, 14, 8, 4, 5, 18, 4, 20, 5, 4, 12, 24, 4, 6, 14, 4, 5, 30, 4, 32, 5, 4, 18, 6, 4, 38, 20, 4, 5, 42, 4, 44, 5, 4, 24, 48, 4, 8, 6, 4, 5, 54, 4, 6, 5, 4, 30, 60, 4, 62, 32, 4, 5, 6, 4

Offset: 4

Lekraj Beedassy, Nov 27 2003

nonn

Choosing a pair (m, n) so as to redefine 1 hour = m*n minutes and 1 minute = m*n seconds, then the three hands of a fictitious n-hour clock coincide in exactly m-1 equally spaced positions, including that of the n o'clock position. For instance, in the cases where we select (m, n) as (6, 11), (8, 15), (4, 25), with m*n respectively equal to 66, 120, 100 (implying 1 hour = 66 minutes, 1 minute = 66 seconds; 1 hour = 120 minutes, 1 minute = 120 seconds; 1 hour = 100 minutes, 1 minute = 100 seconds), the hands coincide in exactly 6-1=5, 8-1=7, 4-1=3 equally spaced positions on a 11-hour, 15-hour, 25-hour clock respectively.

			We have a(50)=8 because 50*8 = 400 is the least multiple of 50 such that gcd(50-1, 400-1) = 8 - 1 = 7.

Giovanni Resta, Table of n, a(n) for n = 4..10000

Cf. A090368, A090369.

A090370:=proc(n) local m; m:=4; while  (gcd(n-1, m*n - 1) <> m-1) do m:=m+1; end;  return m; end; # Søren Eilers, Aug 09 2018

a[n_] := Block[{m=4}, While[GCD[n-1, n*m-1] != m-1, m++]; m]; Table[a[k], {k, 4, 67}] (* Giovanni Resta, Aug 09 2018 *)

a(n) = {m = 4; while (gcd(n-1,m*n - 1) != m-1, m++); return (m);} \\ Michel Marcus, Jul 27 2013

a(n) = 1 + A090368(k) for n=2k. [corrected by Søren Eilers, Aug 09 2018]

a(n) = 1 + A090369(k) for n=2k+1.

a(46) and a(49) corrected by Søren Eilers, Aug 09 2018

A133346 a(n) = smallest k such that prime(n+2) = prime(n) + (prime(n) mod k), or 0 if no such k exists.

Original entry on oeis.org

0, 0, 0, 0, 0, 7, 11, 0, 15, 21, 21, 31, 7, 11, 35, 9, 17, 17, 61, 9, 21, 23, 23, 77, 7, 19, 97, 101, 91, 19, 13, 41, 25, 127, 47, 139, 21, 17, 31, 11, 167, 13, 37, 11, 61, 25, 39, 7, 13, 73, 9, 227, 25, 239, 35, 15, 9, 29, 271, 269, 37, 25, 7, 61, 59, 27, 21, 13, 11, 113, 113

Offset: 1

Rémi Eismann, Oct 20 2007

nonn

			For n = 1 we have prime(n) = 2, prime(n+2) = 5; there is no k such that 5 - 2 = 3 = (2 mod k), hence a(1) = 0.
For n = 6 we have prime(n) = 13, prime(n+2) = 19; 7 is the smallest k such that 19 - 13 = 6 = (13 mod k), hence a(6) = 7.
For n = 30 we have prime(n) = 113, prime(n+2) = 131; 19 is the smallest k such that 131 - 113 = 18 = (113 mod k), hence a(30) = 19.

Remi Eismann, Table of n, a(n) for n = 1..10000

Cf. A000040, A117078, A117563, A001223, A118534, A020639, A090369, A090368, A130533, A130650, A130703, A130889, A130882.

A133347 a(n) = smallest k such that prime(n+3) = prime(n) + (prime(n) mod k), or 0 if no such k exists.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 17, 19, 27, 29, 27, 33, 39, 47, 49, 55, 59, 19, 61, 65, 15, 29, 31, 31, 29, 29, 89, 23, 113, 41, 121, 15, 27, 47, 21, 17, 31, 15, 33, 61, 25, 57, 57, 193, 71, 43, 31, 43, 221, 73, 233, 27, 83, 257, 37, 29, 51, 51, 21, 11, 97, 289, 41, 313, 107, 67

Offset: 1

Rémi Eismann, Oct 20 2007

nonn

			For n = 1 we have prime(n) = 2, prime(n+3) = 7; there is no k such that 7 - 2 = 5 = (2 mod k), hence a(1) = 0.
For n = 10 we have prime(n) = 29, prime(n+3) = 41; 17 is the smallest k such that 41 - 29 = 12 = (29 mod k), hence a(10) = 17.
For n = 53 we have prime(n) = 241, prime(n+3) = 263; 73 is the smallest k such that 263 - 241 = 22 = (241 mod k), hence a(53) = 73.

Remi Eismann, Table of n, a(n) for n = 1..10000

Cf. A000040, A117078, A117563, A001223, A118534, A020639, A090369, A090368, A130533, A130650, A130703, A130889, A130882.

A306536 The smallest integer k such that floor((2*n)^k/k) is an odd number.

Original entry on oeis.org

12, 3, 5, 5, 3, 7, 10, 3, 5, 11, 3, 7, 5, 3, 17, 12, 3, 5, 5, 3, 11, 10, 3, 5, 7, 3, 7, 5, 3, 11, 11, 3, 5, 5, 3, 13, 10, 3, 5, 7, 3, 10, 5, 3, 17, 7, 3, 5, 5, 3, 11, 10, 3, 5, 7, 3, 10, 5, 3, 7, 7, 3, 5, 5, 3, 15, 7, 3, 5, 12, 3, 10, 5, 3, 7

Offset: 1

Jinyuan Wang, Feb 22 2019

nonn

For n > 0, there are infinitely many numbers k such that floor(n^k/k) is an odd number. For odd number n, it's obvious because k can be n^j for any j >= 0. For n = 2, k can be 3*2^j for any j >= 1.

For even number n > 2, k can be A090368(n/2)^j for any j >= 1. Proof: if odd number k makes n^k == 1 (mod k), then n^k = 1 + k*(odd number t), so floor(n^k/k) = t is an odd number. A090368(n/2)^j (j>0) is such k.

Cf. A090368.

a(n) = {k=1; while((2*n)^k\k%2==0, k++); k; }

a(2^j) <= 12. (This is because floor((2*2^j)^12/12) = floor(2^(12j+10)/3) = (2^(12j+10) - 1)/3 is an odd integer for all j >= 0. - Jianing Song, Feb 24 2019)

For n > 1, a(n) <= A090368(n).

cp's OEIS Frontend

A177966 Indices m for which A177961(m) = 2 + m.

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A195508 Number of iterations in a Draim factorization of 2n+1.

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A227198 Odd terms in A227197.

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A117368 a(n) = largest prime less than the smallest prime dividing (2n-1).

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A177983 a(1)=3. Otherwise the average of the least prime divisors of 2n-1 and 2n+3.

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A090370 Least m > 3 such that gcd(n-1, m*n - 1) = m-1.

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A133346 a(n) = smallest k such that prime(n+2) = prime(n) + (prime(n) mod k), or 0 if no such k exists.

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A133347 a(n) = smallest k such that prime(n+3) = prime(n) + (prime(n) mod k), or 0 if no such k exists.