A062709 -id:A062709 - cp's OEIS frontend

A057732 Numbers k such that 2^k + 3 is prime.

1, 2, 3, 4, 6, 7, 12, 15, 16, 18, 28, 30, 55, 67, 84, 228, 390, 784, 1110, 1704, 2008, 2139, 2191, 2367, 2370, 4002, 4060, 4062, 4552, 5547, 8739, 17187, 17220, 17934, 20724, 22732, 25927, 31854, 33028, 35754, 38244, 39796, 40347, 55456, 58312, 122550, 205962, 235326, 363120, 479844, 685578, 742452, 1213815, 1434400, 1594947, 1875552, 1940812, 2205444

Offset: 1

G. L. Honaker, Jr., Oct 29 2000

Some of the larger entries may only correspond to probable primes.

A number k is in this sequence iff A062709(k) is in A057733; this is the case iff A257273(k) is in A125246. - M. F. Hasler, Apr 27 2015

			For k = 6, 2^6 + 3 = 67 is prime.
For k = 28, 2^28 + 3 = 268435459 is prime.

Mike Oakes, posting to primenumbers(AT)yahoogroups.com on Jul 08 2001

Keith Conrad, Square patterns and infinitude of primes, University of Connecticut, 2019.
Henri Lifchitz and Renaud Lifchitz (Editors), Search for 2^n+3, PRP Top Records.

Cf. A050414, A057733, A062709, A094076, A125246, A257273.

Cf. A019434 (primes 2^k+1), this sequence (2^k+3), A059242 (2^k+5), A057195 (2^k+7), A057196(2^k+9), A102633 (2^k+11), A102634 (2^k+13), A057197 (2^k+15), A057200 (2^k+17), A057221 (2^k+19), A057201 (2^k+21), A057203 (2^k+23).

[n: n in [0..1000] | IsPrime(2^n+3)]; // Vincenzo Librandi, Apr 27 2015

Select[Range[10000], PrimeQ[2^# + 3] &] (* Vincenzo Librandi, Apr 27 2015 *)

for(n=1, 2200, if(isprime(2^n+3), print1(n, ", ")));

for (n=1, 2, if (isprime(2^n+3), print1(n, ", "))); for(n=3, 100000, N=2^n+3 ; S=(N-5)/2 ; x=S ; for(j=1, n-1, x=Mod(x^2-2, N)) ; if(x==S , print1(n, ", "))) \\ produces terms corresponding to probable primes, see formula; Tony Reix, Aug 27 2015

Here is an LLT-like algorithm, using a cycle of the digraph x^2-2 modulo N, that finds terms of this sequence generating a PRP (PRobable Prime) of A057733 numbers: N=2^k+3; S0=(N-5)/2; s(0)=S0; s(i+1)=s(i)^2-2 modulo N; if s(k-1) == S0 then N is prime. - Tony Reix, Aug 27 2015

More terms from Jason Earls, Jul 18 2001 and Mike Oakes, Jul 28 2001
a(47)-a(50) from Donovan Johnson 2006, verified by Paul Bourdelais, Mar 22 2012
a(51) is a probable prime based on trial factoring to 1E9 and PRP testing base 3,5,7 (PFGW v3.3.1). Discovered by Paul Bourdelais, Apr 09 2012
a(52)-a(54) from Paul Bourdelais, Jun 18 2019
a(55) from Paul Bourdelais, Jul 16 2019
a(56) from Paul Bourdelais, Apr 22 2020
a(57) from Paul Bourdelais, Jun 12 2020
a(58) from Paul Bourdelais, Aug 04 2020

A264977 a(0) = 0, a(1) = 1, a(2n) = 2a(n), a(2*n+1) = a(n) XOR a(n+1).

Original entry on oeis.org

0, 1, 2, 3, 4, 1, 6, 7, 8, 5, 2, 7, 12, 1, 14, 15, 16, 13, 10, 7, 4, 5, 14, 11, 24, 13, 2, 15, 28, 1, 30, 31, 32, 29, 26, 7, 20, 13, 14, 3, 8, 1, 10, 11, 28, 5, 22, 19, 48, 21, 26, 15, 4, 13, 30, 19, 56, 29, 2, 31, 60, 1, 62, 63, 64, 61, 58, 7, 52, 29, 14, 19, 40, 25, 26, 3, 28, 13, 6, 11, 16, 9, 2, 11, 20, 1, 22

Offset: 0

Antti Karttunen, Dec 10 2015

a(n) is the n-th Stern polynomial (cf. A260443, A125184) evaluated at X = 2 over the field GF(2).

For n >= 1, a(n) gives the index of the row where n occurs in array A277710.

			In this example, binary numbers are given zero-padded to four bits.
a(2) = 2a(1) = 2 * 1 = 2.
a(3) = a(1) XOR a(2) = 1 XOR 2 = 0001 XOR 0010 = 0011 = 3.
a(4) = 2a(2) = 2 * 2 = 4.
a(5) = a(2) XOR a(3) = 2 XOR 3 = 0010 XOR 0011 = 0001 = 1.
a(6) = 2a(3) = 2 * 3 = 6.
a(7) = a(3) XOR a(4) = 3 XOR 4 = 0011 XOR 0100 = 0111 = 7.

Antti Karttunen, Table of n, a(n) for n = 0..16384

Cf. A000225, A001511, A002487, A003987, A010060, A011655, A048675, A055396, A125184, A248663, A260443, A265397, A277330.
Cf. A023758 (the fixed points).
Cf. also A068156, A168081, A265407.
Cf. A277700 (binary weight of terms).
Cf. A277701, A277712, A277713 (positions of 1's, 2's and 3's in this sequence).
Cf. A277711 (position of the first occurrence of each n in this sequence).
Cf. also arrays A277710, A099884.

recurXOR[0] = 0; recurXOR[1] = 1; recurXOR[n_] := recurXOR[n] = If[EvenQ[n], 2recurXOR[n/2], BitXor[recurXOR[(n - 1)/2 + 1], recurXOR[(n - 1)/2]]]; Table[recurXOR[n], {n, 0, 100}] (* Jean-François Alcover, Oct 23 2016 *)

class Memoize:
    def _init_(self, func):
        self.func=func
        self.cache={}
    def _call_(self, arg):
        if arg not in self.cache:
            self.cache[arg] = self.func(arg)
        return self.cache[arg]
@Memoize
def a(n): return n if n<2 else 2*a(n//2) if n%2==0 else a((n - 1)//2)^a((n + 1)//2)
print([a(n) for n in range(51)]) # Indranil Ghosh, Jul 27 2017

a(0) = 0, a(1) = 1, a(2*n) = 2*a(n), a(2*n+1) = a(n) XOR a(n+1).
a(n) = A248663(A260443(n)).
a(n) = A048675(A277330(n)). - Antti Karttunen, Oct 27 2016
Other identities. For all n >= 0:
a(n) = n - A265397(n).
From Antti Karttunen, Oct 28 2016: (Start)
A000035(a(n)) = A000035(n). [Preserves the parity of n.]
A010873(a(n)) = A010873(n). [a(n) mod 4 = n mod 4.]
A001511(a(n)) = A001511(n) = A055396(A277330(n)). [In general, the 2-adic valuation of n is preserved.]
A010060(a(n)) = A011655(n).
a(n) <= n.
For n >= 2, a((2^n)+1) = (2^n) - 3.
The following two identities are so far unproved:
For n >= 2, a(3*A000225(n)) = a(A068156(n)) = 5.
For n >= 2, a(A068156(n)-2) = A062709(n) = 2^n + 3.
(End)

A254364 a(n) = 34^n + 102^n + 6*3^n + 5^n + 15.

Original entry on oeis.org

35, 70, 182, 574, 2054, 7990, 32942, 141694, 629174, 2862790, 13275902, 62494414, 297701894, 1431677590, 6937683662, 33825224734, 165731728214, 815255212390, 4023182840222, 19905098860654, 98686897716134, 490094080827190, 2437150677449582, 12132600130570174, 60450764450513654

Offset: 0

Luciano Ancora, Jan 29 2015

This is the sequence of fifth terms of "third partial sums of m-th powers".

Colin Barker, Table of n, a(n) for n = 0..1000
Luciano Ancora, Demonstration of formulas, page 2.
Luciano Ancora, Recurrence relations for partial sums of m-th powers.
Index entries for linear recurrences with constant coefficients, signature (15,-85,225,-274,120).

Cf. A062709, A254362, A254363.

Table[3 4^n + 10 2^n + 6 3^n + 5^n + 15, {n, 0, 30}] (* Bruno Berselli, Jan 30 2015 *)
LinearRecurrence[{15,-85,225,-274,120},{35,70,182,574,2054},30] (* Harvey P. Dale, Aug 11 2016 *)

vector(30, n, n--; 3*4^n + 10*2^n + 6*3^n + 5^n + 15) \\ Colin Barker, Jan 30 2015

From Colin Barker, Jan 30 2015: (Start)
G.f.: -(2754*x^4 - 4081*x^3 + 2107*x^2 - 455*x + 35)/((x-1)*(2*x-1)*(3*x-1)*(4*x-1)*(5*x-1)).
a(n) = 15*a(n-1) - 85*a(n-2) + 225*a(n-3) - 274*a(n-4) + 120*a(n-5). (End)
E.g.f.: exp(x)*(exp(4*x) + 3*exp(3*x) + 6*exp(2*x) + 10*exp(x) + 15). - Elmo R. Oliveira, Sep 16 2024

A131426 a(n) = 2*prime(n) - 3.

Original entry on oeis.org

1, 3, 7, 11, 19, 23, 31, 35, 43, 55, 59, 71, 79, 83, 91, 103, 115, 119, 131, 139, 143, 155, 163, 175, 191, 199, 203, 211, 215, 223, 251, 259, 271, 275, 295, 299, 311, 323, 331, 343, 355, 359, 379, 383, 391, 395, 419, 443, 451, 455, 463, 475, 479, 499, 511, 523, 535, 539, 551, 559, 563, 583, 611

Offset: 1

Gary W. Adamson, Jul 10 2007

Right border of triangle A131424.

			a(4) = 11 = 2*p(4) - 3 = 2*7 - 3.

Cf. A000040, A062709, A131424.

[2*NthPrime(n) - 3: n in [1..65]]; // Vincenzo Librandi, Mar 30 2015

A131426:=n->2*ithprime(n)-3: seq(A131426(n), n=1..70); # Wesley Ivan Hurt, Mar 30 2015

2#-3&/@Prime[Range[80]] (* Harvey P. Dale, Jul 24 2011 *)

a(n) = 2*prime(n) - 3 \\ Charles R Greathouse IV, Jul 12 2016

a(n) = 2*A000040(n) - 3.

A116623 a(0)=1, a(2n) = a(n)+A000079(A000523(2n)), a(2n+1) = 3*a(n) + A000079(A000523(2n+1)+1).

Original entry on oeis.org

1, 5, 7, 19, 11, 29, 23, 65, 19, 49, 37, 103, 31, 85, 73, 211, 35, 89, 65, 179, 53, 143, 119, 341, 47, 125, 101, 287, 89, 251, 227, 665, 67, 169, 121, 331, 97, 259, 211, 601, 85, 223, 175, 493, 151, 421, 373, 1087, 79, 205, 157, 439, 133, 367, 319, 925, 121

Offset: 0

Antti Karttunen, Feb 20 2006. Proposed by Pierre Lamothe (plamothe(AT)aei.ca), May 21 2004

Viewed as a binary tree, this is (1); 5; 7,19; 11,29,23,65; ... Related to the parity vectors of Collatz and Terras trajectories.

Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. a(n) = A116640(A059893(n)). a(A000225(n)) = A001047(n+1). For n>= 1 a(A000079(n)) = A062709(n+1). A116641 gives the terms in ascending order and without duplicates.

A116623 := proc(n)
    option remember;
    if n = 0 then
        1;
    elif type(n,'even') then
        procname(n/2)+2^A000523(n) ;
    else
        3*procname(floor(n/2))+2^(1+A000523(n)) ;
    end if;
end proc: # R. J. Mathar, Nov 28 2016

a[n_] := a[n] = Which[n == 0, 1, EvenQ[n], a[n/2] + 2^Floor@Log2[n], True, 3a[Floor[n/2]] + 2^(1 + Floor@Log2[n])];
Table[a[n], {n, 0, 56}] (* Jean-François Alcover, Sep 01 2023 *)

A116640 a(n) = A116623(A059893(n)).

Original entry on oeis.org

1, 5, 7, 19, 11, 23, 29, 65, 19, 31, 37, 73, 49, 85, 103, 211, 35, 47, 53, 89, 65, 101, 119, 227, 89, 125, 143, 251, 179, 287, 341, 665, 67, 79, 85, 121, 97, 133, 151, 259, 121, 157, 175, 283, 211, 319, 373, 697, 169, 205, 223, 331, 259, 367, 421, 745, 331

Offset: 0

Antti Karttunen, Feb 20 2006. Proposed by Pierre Lamothe (plamothe(AT)aei.ca), May 21 2004

Viewed as a binary tree, this is (1); 5; 7,19; 11,23,29,65; ... Cf. A116623.

If we treat (2n+1) as a binary number with the nonzero bits numbered (highest bit first) from 0..k and the regular binary place value of each nonzero bit numbered from b(0) to b(k) then a(n) = 3^0 * b(0) + 3^1 * b(1) + .. + 3^k. For instance, if n=6 then 2n+1 = 13, which is equal to 8+4+1 or 1101 base(2); and a(n)=29 which is 8*1 + 4*3 + 1*9. - Joe Slater, Jan 23 2016

Cf. A000079, A000225, A001047, A062709.

a:= proc(n) option remember; piecewise(
    n mod 4 = 0, 3*procname(n/2) - 2*procname(n/4),
  n mod 4 = 1, 6*procname((n-1)/4) - procname((n-1)/2),
  n mod 4 = 2, procname(n/2) + 2*procname((n-2)/4),
  5*procname((n-1)/2) - 6*procname((n-3)/4))
end proc:
a(0):= 1:
map(a, [$0..100]); # Robert Israel, Jan 19 2016

a[n_] := a[n] = Switch[Mod[n, 4], 0, 3a[Floor[n/2]] - 2a[Floor[n/4]], 1, 6a[Floor[n/4]] - a[Floor[n/2]], 2, a[Floor[n/2]] + 2a[Floor[n/4]], 3, 5a[Floor[n/2]] - 6a[Floor[n/4]]]; a[0]=1; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Feb 28 2016 *)

a(n) = if(n==0, return(1)); 2*a(n\2) - (-1)^n * 3^hammingweight(n) \\ Charles R Greathouse IV, Jan 21 2016

a(n) = my(p=2*n+1,v=vecextract(vector(#binary(p),j,2^(j-1)),p));sum(i=0,#v-1,3^i*v[#v-i]) \\ Joe Slater, May 09 2017

a(A000225(n)) = A001047(n+1).
For n>= 1 a(A000079(n)) = A062709(n+1).
From Joe Slater, Jan 19 2016: (Start)
a(0) = 1,
a(n) = 3*a(floor(n/2)) - 2*a(floor(n/4)) for n=0 (mod 4) and n>0,
a(n) = 6*a(floor(n/4)) - a(floor(n/2)) for n=1 (mod 4),
a(n) = a(floor(n/2)) + 2*a(floor(n/4)) for n=2 (mod 4),
a(n) = 5*a(floor(n/2)) - 6*a(floor(n/4)) for n=3 (mod 4)
(End)
a(0) = 1, a(n) = 2*a(floor(n/2)) - A033999(n) * A048883(n) for n>0. -
Joe Slater, Jan 22 2016

A173921 Sums of rows of the triangle in A173920.

Original entry on oeis.org

0, 1, 1, 4, 2, 5, 6, 12, 4, 7, 9, 14, 12, 19, 21, 32, 8, 11, 17, 22, 19, 27, 30, 40, 24, 35, 37, 52, 41, 57, 60, 80, 16, 19, 33, 38, 34, 41, 51, 60, 40, 51, 57, 70, 64, 81, 85, 104, 48, 67, 69, 92, 76, 99, 101, 128, 84, 111, 113, 144, 118, 151, 155, 192, 32, 35, 65, 70, 66, 73

Offset: 0

Reinhard Zumkeller, Mar 04 2010

nonn

a(n) = SUM(A173920(n,k): 0<=k<=n);
a(A000225(n)) = A001787(n);
a(A000079(n)) = A011782(n);
a(A000051(n)) = A062709(n) for n>0;
A000120(n)<=a(n)<=A000788(n); a(n)=A000788(n) iff n=2^k-1.

R. Zumkeller, Table of n, a(n) for n = 0..320

Cf. A000225, A001787.

A257273 a(n) = 2^(n-1)*(2^n+3).

Original entry on oeis.org

2, 5, 14, 44, 152, 560, 2144, 8384, 33152, 131840, 525824, 2100224, 8394752, 33566720, 134242304, 536920064, 2147581952, 8590131200, 34360131584, 137439739904, 549757386752, 2199026401280, 8796099313664, 35184384671744, 140737513521152, 562950003752960, 2251799914348544, 9007199456067584

Offset: 0

M. F. Hasler, Apr 27 2015

a(n) is in A125246 <=> n is in A057732 <=> A062709(n) is in A057733.

These are also the row sum of the triangle A146769: For n>=1, a(n-1) is the sum of row n of A146769.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-8).

Cf. A000079, A007582, A028403, A256873, A256871, A257272.

Cf. A062709, A057732, A057733.

[2^(n-1)*(2^n+3): n in [0..35]]; // Vincenzo Librandi, Apr 27 2015

Table[2^(n - 1) (2^n + 3), {n, 0, 30}] (* Bruno Berselli, Apr 27 2015 *)
CoefficientList[Series[(2 - 7 x)/((1 - 4 x) (1 - 2 x)), {x, 0, 30}], x] (* Vincenzo Librandi, Apr 27 2015 *)
LinearRecurrence[{6,-8},{2,5},30] (* Harvey P. Dale, Dec 21 2024 *)

PARI
```
a(n)=2^(n-1)*(2^n+3)
```

Vec((2-7*x)/((1-4*x)*(1-2*x)) + O(x^100)) \\ Colin Barker, Apr 27 2015

G.f.: (2-7*x)/((1-4*x)*(1-2*x)). - Vincenzo Librandi, Apr 27 2015

a(n) = 6*a(n-1)-8*a(n-2). - Colin Barker, Apr 27 2015

A194455 a(n) = 2^n + 3n + 1.

Original entry on oeis.org

2, 6, 11, 18, 29, 48, 83, 150, 281, 540, 1055, 2082, 4133, 8232, 16427, 32814, 65585, 131124, 262199, 524346, 1048637, 2097216, 4194371, 8388678, 16777289, 33554508, 67108943, 134217810, 268435541, 536871000, 1073741915, 2147483742, 4294967393, 8589934692, 17179869287

Offset: 0

Bruno Berselli, Sep 01 2011

Inverse binomial transform of this sequence: 2,4,1,1 (1 continued).

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-5,2).

Cf. A000051, A005126, A176691; A120845.
Cf. A062709 (first differences), A000079 (second and successive differences).
Cf. A146529 (differences between alternate terms, for n>2).
Cf. A086653, A115067.

Magma
```
[2^n+3*n+1: n in [0..31]];
```

Table[2^n + 3 n + 1, {n, 0, 40}] (* Vincenzo Librandi, Mar 26 2013 *)
LinearRecurrence[{4,-5,2},{2,6,11},40] (* Harvey P. Dale, Oct 01 2014 *)

PARI
```
for(n=0, 31, print1(2^n+3*n+1", "));
```

G.f.: (2 - 2*x - 3*x^2)/((1 - 2*x)*(1 - x)^2).
a(n) = A086653(n) - 1  for n > 0.
Sum_{i=0..n} a(i) = A115067(n+1) + 2^(n+1).
a(n) = 3*a(n-1) - 2*a(n-2) - 3  for n > 1.
a(n)^2 = 2^(n+1)*(a(n-1) + 3) + (3*n + 1)^2  for n > 2.
E.g.f.: exp(x)*(1 + exp(x) + 3*x). - Stefano Spezia, May 06 2023

A254362 a(n) = 3*2^n + 3^n + 6.

Original entry on oeis.org

10, 15, 27, 57, 135, 345, 927, 2577, 7335, 21225, 62127, 183297, 543735, 1618905, 4832127, 14447217, 43243335, 129533385, 388206927, 1163834337, 3489930135, 10466644665, 31393642527, 94168344657, 282479868135, 847389272745, 2542067154927, 7626000138177

Offset: 0

Luciano Ancora, Jan 29 2015

This is the sequence of third terms of "third partial sums of m-th powers".

Is this 10 followed by A087719? [Bruno Berselli, Jan 30 2015]

Colin Barker, Table of n, a(n) for n = 0..1000
Luciano Ancora, Demonstration of formulas, page 1.
Luciano Ancora, Recurrence relations for partial sums of m-th powers
Index entries for linear recurrences with constant coefficients, signature (6,-11,6).

Cf. A062709, A254363, A254364.

Table[3 * 2^n + 3^n + 6, {n, 0, 29}] (* Alonso del Arte, Jan 29 2015 *)
LinearRecurrence[{6,-11,6},{10,15,27},30] (* Harvey P. Dale, Oct 11 2024 *)

vector(30, n, n--; 3*2^n + 3^n + 6) \\ Colin Barker, Jan 30 2015

G.f.: -(47*x^2-45*x+10) / ((x-1)*(2*x-1)*(3*x-1)). - Colin Barker, Jan 30 2015

a(n) = 6*a(n-1) -11*a(n-2) +6*a(n-3). - Colin Barker, Jan 30 2015

cp's OEIS Frontend

A057732 Numbers k such that 2^k + 3 is prime.

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A264977 a(0) = 0, a(1) = 1, a(2*n) = 2*a(n), a(2*n+1) = a(n) XOR a(n+1).

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A254364 a(n) = 3*4^n + 10*2^n + 6*3^n + 5^n + 15.

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A131426 a(n) = 2*prime(n) - 3.

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A116623 a(0)=1, a(2n) = a(n)+A000079(A000523(2n)), a(2n+1) = 3*a(n) + A000079(A000523(2n+1)+1).

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A116640 a(n) = A116623(A059893(n)).

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A173921 Sums of rows of the triangle in A173920.

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A264977 a(0) = 0, a(1) = 1, a(2n) = 2a(n), a(2*n+1) = a(n) XOR a(n+1).

A254364 a(n) = 34^n + 102^n + 6*3^n + 5^n + 15.