A125169 -id:A125169 - cp's OEIS frontend

A127576 Primes of the form 16n+15.

31, 47, 79, 127, 191, 223, 239, 271, 367, 383, 431, 463, 479, 607, 719, 751, 863, 911, 991, 1039, 1087, 1103, 1151, 1231, 1279, 1327, 1423, 1439, 1471, 1487, 1567, 1583, 1663, 1759, 1823, 1871, 1951, 1999, 2063, 2111, 2143, 2207, 2239, 2287, 2351, 2383

Offset: 1

Artur Jasinski, Jan 19 2007

nonn

Subsequence of A007522. - R. J. Mathar, Jan 07 2009

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A125169, A127575.

a = {}; Do[If[PrimeQ[16n + 15], AppendTo[a, 16n + 15]], {n, 1, 200}]; a
Select[16*Range[400]+15,PrimeQ] (* Harvey P. Dale, Feb 06 2013 *)

select(n->n%16==15, primes(100)) \\ Charles R Greathouse IV, Apr 28 2015

A127575 Numbers n such that 16n+15 is prime.

Original entry on oeis.org

1, 2, 4, 7, 11, 13, 14, 16, 22, 23, 26, 28, 29, 37, 44, 46, 53, 56, 61, 64, 67, 68, 71, 76, 79, 82, 88, 89, 91, 92, 97, 98, 103, 109, 113, 116, 121, 124, 128, 131, 133, 137, 139, 142, 146, 148, 149, 152, 158, 161, 166, 167, 169, 172, 179, 182, 188, 194, 197

Offset: 1

Artur Jasinski, Jan 19 2007

nonn

Daniel Starodubtsev, Table of n, a(n) for n = 1..10000

Cf. A125169 (16n+15), A127576 (resulting primes).

a = {}; Do[If[PrimeQ[16n + 15], AppendTo[a, n]], {n, 1, 200}]; a

is(n)=isprime(16*n+15) \\ Charles R Greathouse IV, Apr 29 2015

A239126 Rectangular array showing the starting values M(n, k), k >= 1, for the Collatz operation (ud)^n, n >= 1, ending in an odd number, read by antidiagonals.

Original entry on oeis.org

3, 7, 7, 11, 15, 15, 15, 23, 31, 31, 19, 31, 47, 63, 63, 23, 39, 63, 95, 127, 127, 27, 47, 79, 127, 191, 255, 255, 31, 55, 95, 159, 255, 383, 511, 511, 35, 63, 111, 191, 319, 511, 767, 1023, 1023, 39, 71, 127, 223, 383, 639, 1023, 1535, 2047, 2047

Offset: 1

Wolfdieter Lang, Mar 13 2014

The companion array and triangle for the odd end numbers N(n, k) is given in A239127.
The two operations on natural numbers m used in the Collatz 3x+1 conjecture are here (following the M. Trümper paper given in the link) denoted by u for 'up' and d for 'down': u m = 3*m+1, if m is odd, and d m = m/2 if m is even. The present array gives all start numbers M(n, k) for the Collatz word (ud)^n = s^n (s = ud is useful because, except for the one letter word u, at least one d follows a letter u), with n >= 1, and k >= 1. Such Collatz sequences have the maximal number of u's (grow fastest).
This rectangular array is M of Example 2.2. with x=y = n, n >= 1, of the M. Trümper reference, pp. 7-8, written as a triangle by taking NE-SW diagonals. The Collatz sequence starting with M(n, k) has length 2*n+1 for each k and it ends in the odd number N(n, k) given in A239127.
The first row sequences of the array M (columns of triangle TM) are A004767, A004771, A125169, A239128, ...

			The rectangular array M(n, k) begins:
n\k     1    2    3    4     5     6     7     8     9    10 ...
1:      3    7   11   15    19    23    27    31    35    39
2:      7   15   23   31    39    47    55    63    71    79
3:     15   31   47   63    79    95   111   127   143   159
4:     31   63   95  127   159   191   223   255   287   319
5:     63  127  191  255   319   383   447   511   575   639
6:    127  255  383  511   639   767   895  1023  1151  1279
7:    255  511  767 1023  1279  1535  1791  2047  2303  2559
8:    511 1023 1535 2047  2559  3071  3583  4095  4607  5119
9:   1023 2047 3071 4095  5119  6143  7167  8191  9215 10239
10:  2047 4095 6143 8191 10239 12287 14335 16383 18431 20479
...
The triangle TM(m, n) begins (zeros are not shown):
m\n   1    2     3     4     5     6      7      8      9    10 ...
1:    3
2:    7    7
3:   11   15    15
4:   15   23    31    31
5:   19   31    47    63    63
6:   23   39    63    95   127   127
7:   27   47    79   127   191   255    255
8:   31   55    95   159   255   383    511    511
9:   35   63   111   191   319   511    767   1023   1023
10:  39   71   127   223   383   639   1023   1535   2047  2047
...
---------------------------------------------------------------------
n=1, ud, k=1: M(1, 1) = 3 = TM(1, 1), N(1,1) = 5 with the Collatz sequence  [3, 10, 5] of length 3.
n=1, ud, k=2: M(1, 2) = 7 = TM(2, 1), N(1,2) = 11 with the Collatz sequence  [7, 22, 11] of length 3.
n=4, (ud)^4, k=2: M(4, 2) = 63 = TM(5, 4), N(4,2) = 323 with the Collatz sequence  [63, 190, 95, 286, 143, 430, 215, 646, 323] of length 9.
n=5, (ud)^5, k=1: M(5, 1) = 63 =  TM(5, 5), N(5,1) = 485 with the Collatz sequence  [63, 190, 95, 286, 143, 430, 215, 646, 323, 970, 485] of length 11.

Wolfdieter Lang, On Collatz' Words, Sequences, and Trees, J. of Integer Sequences, Vol. 17 (2014), Article 14.11.7.
Manfred Trümper, The Collatz Problem in the Light of an Infinite Free Semigroup, Chinese Journal of Mathematics, Vol. 2014, Article ID 756917, 21 pages.
Eric Weisstein's World of Mathematics, Collatz Problem.
Wikipedia, Collatz Conjecture.

Cf. A006577, A139399, A112695, A238475, A238476, A004767, A004771, A125169, A239127, A239128.

The array: M(n, k) = 2^(n+1)*k - 1 for n >= 1 and k >= 1.
The triangle: TM(m, n) = M(n, m-n+1) = 2^(n+1)*(m-n+1) - 1 for m >= n >= 1 and 0 for m < n.
a(n) = 4*A087808(A130328(n-1)) - 1 (conjectured). - Christian Krause, Jun 15 2021

A319282 Numbers of the form 16^i(16j + 15).

Original entry on oeis.org

15, 31, 47, 63, 79, 95, 111, 127, 143, 159, 175, 191, 207, 223, 239, 240, 255, 271, 287, 303, 319, 335, 351, 367, 383, 399, 415, 431, 447, 463, 479, 495, 496, 511, 527, 543, 559, 575, 591, 607, 623, 639, 655, 671, 687, 703, 719, 735, 751, 752, 767, 783, 799, 815

Offset: 1

Jianing Song, Sep 16 2018

nonn

{-a(n)} gives all negative fourth powers modulo all powers of 2, that is, negative fourth powers over 2-adic integers.

Jianing Song, Table of n, a(n) for n = 1..9999 (all terms <= 150000)

A125169 is a proper subsequence.
Perfect powers over 2-adic integers:
Squares: positive: A234000; negative: A004215 (negated);
Cubes: A191257;
Fourth powers: positive: A319281; negative: this sequence (negated).

isA319282(n)= n\16^valuation(n, 16)%16==15

def A319282(n):
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        kmin = kmax >> 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    def f(x): return n+x-sum((((x>>(i<<2))-15)>>4)+1 for i in range(x.bit_length()>>2))
    return bisection(f,n,n) # Chai Wah Wu, Feb 17 2025

a(n) = 15*n + O(log(n)).

A158058 a(n) = 16n^2 - 2n.

Original entry on oeis.org

14, 60, 138, 248, 390, 564, 770, 1008, 1278, 1580, 1914, 2280, 2678, 3108, 3570, 4064, 4590, 5148, 5738, 6360, 7014, 7700, 8418, 9168, 9950, 10764, 11610, 12488, 13398, 14340, 15314, 16320, 17358, 18428, 19530, 20664, 21830, 23028, 24258, 25520

Offset: 1

Vincenzo Librandi, Mar 12 2009

The identity (16*(n-1) + 15)^2 - (16*n^2 - 2*n)*4^2 = 1 can be written as A125169(n-1)^2 - a(n)*4^2 = 1. - Vincenzo Librandi, Feb 01 2012
Sequence found by reading the line from 14, in the direction 14, 60, ... in the square spiral whose vertices are the generalized decagonal numbers A074377. - Omar E. Pol, Nov 02 2012
The continued fraction expansion of sqrt(a(n)) is [4n-1; {1, 2, 1, 8n-2}]. - Magus K. Chu, Nov 08 2022

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
E. J. Barbeau, Polynomial Excursions, Chapter 10: Diophantine equations (2010), pages 84-85 (row 15 in the first table at p. 85, case d(t) = t*(4^2*t-2)).
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A125169.

Magma
```
[16*n^2-2*n: n in [1..40]]
```

seq(16*n^2-2*n,n=1..40); # Nathaniel Johnston, Jun 26 2011

LinearRecurrence[{3,-3,1},{14,60,138},40]

PARI
```
a(n) = 16*n^2-2*n.
```

G.f.: x*(-14 - 18*x)/(x-1)^3.

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

A177065 a(n) = (8n+3)(8*n+5).

Original entry on oeis.org

15, 143, 399, 783, 1295, 1935, 2703, 3599, 4623, 5775, 7055, 8463, 9999, 11663, 13455, 15375, 17423, 19599, 21903, 24335, 26895, 29583, 32399, 35343, 38415, 41615, 44943, 48399, 51983, 55695, 59535, 63503, 67599, 71823, 76175, 80655, 85263, 89999, 94863, 99855

Offset: 0

Vincenzo Librandi, May 31 2010

Cf. comment of Reinhard Zumkeller in A177059: in general, (h*n+h-k)*(h*n+k) = h^2*A002061(n+1) + (h-k)*k - h^2; therefore a(n) = 64*A002061(n+1) - 49. - Bruno Berselli, Aug 24 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A002061, A004770, A016754, A017101, A125169, A177059.

[(8*n+3)*(8*n+5): n in [0..50]]; // Vincenzo Librandi, Apr 08 2013

A177065:=n->(8*n+3)*(8*n+5): seq(A177065(n), n=0..100); # Wesley Ivan Hurt, Apr 24 2017

Table[(8n+3)(8n+5),{n,0,40}] (* or *) LinearRecurrence[{3,-3,1},{15,143,399},40] (* Harvey P. Dale, Mar 13 2013 *)
CoefficientList[Series[(15 + 98 x + 15 x^2)/(1-x)^3, {x, 0, 50}], x] (* Vincenzo Librandi, Apr 08 2013 *)

a(n)=(8*n+3)*(8*n+5) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 128*n + a(n-1) with n > 0, a(0)=15.
a(n) = A125169(A016754(n) - 1). - Reinhard Zumkeller, Jul 05 2010
a(0)=15, a(1)=143, a(2)=399, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Mar 13 2013
G.f.: (15+98*x+15*x^2)/(1-x)^3. - Vincenzo Librandi, Apr 08 2013
From Amiram Eldar, Feb 19 2023: (Start)
a(n) = A017101(n)*A004770(n).
Sum_{n>=0} 1/a(n) = (sqrt(2)-1)*Pi/16.
Sum_{n>=0} (-1)^n/a(n) = (cos(Pi/8) * log(tan(3*Pi/16)) + sin(Pi/8) * log(cot(Pi/16)))/4.
Product_{n>=0} (1 - 1/a(n)) = sec(Pi/8)*cos(Pi/(4*sqrt(2))).
Product_{n>=0} (1 + 1/a(n)) = sec(Pi/8). (End)
E.g.f.: exp(x)*(15 + 64*x*(2 + x)). - Elmo R. Oliveira, Oct 25 2024

Edited by N. J. A. Sloane, Jun 22 2010

A214863 Numbers n such that n XOR 11 = n - 11.

Original entry on oeis.org

11, 15, 27, 31, 43, 47, 59, 63, 75, 79, 91, 95, 107, 111, 123, 127, 139, 143, 155, 159, 171, 175, 187, 191, 203, 207, 219, 223, 235, 239, 251, 255, 267, 271, 283, 287, 299, 303, 315, 319, 331, 335, 347, 351, 363

Offset: 1

Brad Clardy, Mar 09 2013

Links to sequences of the form n XOR m = n - m are found below with the value of m specified.

Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A005408 (m=1), A042964 (m=2), A131098 (m=3), A047566 (m=4), A047550 (m=5), A047589 (m=6), A004771 (m=7), A115419 (m=8), A214865 (m=9), A214864 (m=10), A133894 (m=12), A125169 (m=15).

Cf. also A016825, A168392.

XOR := func;
m:=11;
for n in [1 .. 500] do
      if (XOR(n, m) eq n-m) then n; end if;
end for;

Select[Range[400],BitXor[#,11]==#-11&] (* or *) LinearRecurrence[{1,1,-1},{11,15,27},50] (* Harvey P. Dale, Jun 05 2021 *)

a(n)= 1+8*n-2*(-1)^n.
a(n)=A016825(n) + A168392(n) +  for n>0.
G.f. x*(11+4*x+x^2) / ( (1+x)*(x-1)^2 ). - R. J. Mathar, Mar 10 2013

A239128 a(n) = 32*n - 1, n >= 1. Fourth column of triangle A239126, related to the Collatz problem.

Original entry on oeis.org

31, 63, 95, 127, 159, 191, 223, 255, 287, 319, 351, 383, 415, 447, 479, 511, 543, 575, 607, 639, 671, 703, 735, 767, 799, 831, 863, 895, 927, 959, 991, 1023, 1055, 1087, 1119, 1151, 1183, 1215, 1247, 1279, 1311, 1343, 1375, 1407, 1439, 1471, 1503, 1535, 1567, 1599

Offset: 1

Wolfdieter Lang, Mar 13 2014

This sequence gives all starting values a(n) (in increasing order) of Collatz sequences of length 9 following the pattern (ud)^4, with u (for `up'), mapping an odd number m to 3*m+1, and d (for `down'), mapping an even number m to m/2. The last entry of this sequence is required to be odd and it is given by 162*n-1.

This appears in Example 2.2. for x=y = 4 in the M. Trümper paper on p. 7, given as a link below.

			a(1) = 31 because the Collatz sequence following the pattern udududud is [31, 94, 47, 142, 71, 214, 107, 322, 161], with length 9, ending in the odd number N(4,1) = 161 = 162*1 - 1 from the array A239127, and 31 is the smallest positive number whose Collatz sequence follows this pattern and ends in an odd number.
a(4) = 127 with the Collatz sequence [127, 382, 191, 574, 287, 862, 431, 1294, 647] ending in N(4,4) = 647 = 32*4 - 1. 127 is the fourth smallest positive number following this pattern with odd end number.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Wolfdieter Lang, On Collatz' Words, Sequences, and Trees, J. of Integer Sequences, Vol. 17 (2014), Article 14.11.7.
Manfred Trümper, The Collatz Problem in the Light of an Infinite Free Semigroup, Chinese Journal of Mathematics, Vol. 2014, Article ID 756917, 21 pages.
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A125169 (third column), A239126, A239127.

CoefficientList[Series[(31 + x)/(1 - x)^2, {x, 0, 40}], x] (* Vincenzo Librandi, Mar 16 2014 *)
32*Range[50]-1 (* Harvey P. Dale, Jan 25 2021 *)

O.g.f.: x*(31+x)/(1-x)^2.
From Elmo R. Oliveira, Apr 04 2025: (Start)
E.g.f.: exp(x)*(32*x - 1) + 1.
a(n) = 2*a(n-1) - a(n-2) for n > 2. (End)

cp's OEIS Frontend

A127576 Primes of the form 16n+15.

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A127575 Numbers n such that 16n+15 is prime.

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A239126 Rectangular array showing the starting values M(n, k), k >= 1, for the Collatz operation (ud)^n, n >= 1, ending in an odd number, read by antidiagonals.

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A319282 Numbers of the form 16^i*(16*j + 15).

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A158058 a(n) = 16*n^2 - 2*n.

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A177065 a(n) = (8*n+3)*(8*n+5).

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A214863 Numbers n such that n XOR 11 = n - 11.

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A239128 a(n) = 32*n - 1, n >= 1. Fourth column of triangle A239126, related to the Collatz problem.

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A319282 Numbers of the form 16^i(16j + 15).

A158058 a(n) = 16n^2 - 2n.

A177065 a(n) = (8n+3)(8*n+5).