Ben Paul Thurston - cp's OEIS frontend

A334264 Numbers k > 1 such that (3/2)^k sets a new record for closest fractional part to 1/2.

2, 3, 5, 9, 11, 69, 420, 2361, 12432, 21565, 28226, 128389, 274555, 497269, 836000, 1151341, 1973112, 2202332, 2458844, 5402520

Offset: 1

Ben Paul Thurston, Apr 20 2020

A081464 is also related to (3/2) to a power being a record distance from a value of an integer.

Cf. A002379, A034082, A179523.

off := 1; for i from 2 to 1000 do t := (1+1/2)^i-floor((1+1/2)^i); d := abs(1/2-t); if d < off then off := d; print(i) end if end do

dm = 1; r = 3/2; s = {}; Do[r *= 3/2; If[(d = Abs[r - Floor[r] - 1/2]) < dm, dm = d; AppendTo[s, n + 1]], {n, 1, 10^7}]; s (* Amiram Eldar, Jun 08 2020 *)

a(8)-a(13) from Amiram Eldar, Jun 08 2020
a(14)-a(16) from Chai Wah Wu, Jul 02 2020
a(17)-a(20) from Bert Dobbelaere, Apr 25 2021

A309815 a(n) is the smallest positive integer x such that sqrt(2) + sqrt(x) is closer to an integer than any other value already in the sequence.

Original entry on oeis.org

1, 2, 3, 6, 7, 13, 21, 112, 243, 275, 466, 761, 1128, 4704, 9523, 10730, 17579, 28085, 41041, 165312, 331299, 372815, 607754, 967441, 1410360, 5648160, 11300259, 12713402, 20707831, 32942845, 48005301, 192060400, 384143763, 432165299, 703818922, 1119543881, 1631318640

Offset: 1

Ben Paul Thurston, Aug 18 2019

nonn

If b(n) = round(sqrt(2) + sqrt(a(n))), then (b(n)^2 + 2 - a(n))/(2*b(n)) is an approximation for sqrt(2). Conjecture: all convergents of the continued fraction of sqrt(2) except 1 arise in this way. - Robert Israel, Aug 18 2019

			a(6) = 13 because sqrt(2)+sqrt(13) is closer to an integer than any of the previous 5 terms.

R:= 1: delta:= sqrt(2)-1:
for r from 2 to 10000 do
   x0:= ceil((r - sqrt(2)-delta)^2);
   x1:= floor((r-sqrt(2)+delta)^2);
   for x from x0 to x1 do
     dx:= abs(sqrt(2)+sqrt(x)-r);
     if is(dx < delta) then
       delta:= dx;
       R:= R, x;
     fi
   od
od:
R; # Robert Israel, Aug 18 2019

d[x_] := Abs[x - Round[x]]; dm = 1; s = {}; Do[If[(d1 = d[Sqrt[2] + Sqrt[n]]) < dm, dm = d1; AppendTo[s, n]], {n, 1, 10^5}]; s (* Amiram Eldar, Aug 18 2019 *)

import math
a = 2**(1/2)
l = []
closest = 1.0
for i in range(1, 100000000):
    b = i**(1/2)
    c = abs(a+b - round(a+b))
    if c < closest:
        print(i, c)
        closest = c
        l.append(i)
print(l)

More terms from Giovanni Resta, Aug 19 2019

A317778 Starting with 1,2,3,4,5,6: a(n) is the next smallest number greater than a(n-1) such that a[i] + a[j] + a[k] != a[x] + a[y] + a[z] for 1 <= i,j,k,x,y,z <= n all distinct.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 13, 22, 39, 72, 131, 229, 386, 641, 896, 1164, 1419, 1855, 2831, 3545, 5036, 5750, 8034, 10022, 12227, 14377, 17455, 19951, 24701, 27197, 36455, 42303, 49751, 57232, 65684, 83879, 94391, 110073, 124015, 137442, 156835, 175130, 209215, 229396, 242692

Offset: 1

Ben Paul Thurston, Aug 06 2018

nonn

a(n) <= a(n-1) + a(n-2) + a(n-3) - 2. - Charlie Neder, Feb 09 2019

			After 1,2,3,4,5,6: 7 cannot be the next term because 1+3+7 = 2+4+5.

Charlie Neder, Table of n, a(n) for n = 1..70

Cf. A011185.

def u(series):
    for i in range(0, len(series)):
        for j in range(i+1, len(series)):
            for k in range(j+1, len(series)):
                for l in range(0, len(series)):
                    for m in range(l+1, len(series)):
                        for n in range(m+1, len(series)):
                            if len(set([i,j,k,l,m,n]))==6:
                                if series[i]+series[j]+series[k]==series[l]+series[m]+series[n]:
                                    return False
    return True
def a(series, n):
    a = []
    for i in range(0, len(series)):
        a.append(series[i])
    a.append(n)
    return a
series = [1, 2, 3,4,5,6]
for i in range(7, 1000):
    print(i)
    nseries = a(series, i)
    if u(nseries):
        series.append(i)
        print(series)
print(series)

a(24)-a(45) from Charlie Neder, Feb 09 2019

A303610 Circle aliasing numbers with 1/n size steps.

Original entry on oeis.org

10, 1010, 110100, 11010100, 1101100100, 110110100100, 11101010101000, 1110110101001000, 111011010101001000, 11101101101001001000, 1110111010101010001000, 111011101010101010001000, 11110110110101010010010000, 1111011011010101010010010000, 111101110101101010010100010000

Offset: 1

Ben Paul Thurston, May 06 2018

nonn

Starting from [-1,0] taking 2*n steps of length 1/n each either up or right, follow the path staying as close to the unit circle as possible. Every step up is considered a 1, every step right is considered a 0.

			For n=3, we have 110100, meaning if we were to start at [-1, 0] and take 2*n=6 steps of length 1/n = 1/6 which can either be up or to the right, to follow the path of the unit circle the closest we would move up 1, up 1 again, then right, then up again, then right two more times, which we translate to the binary number 110100.

Subsequence of A035928 in binary.

def closer(pos1, pos2):
    dpos1 = (pos1[0]**2.0+pos1[1]**2.0)**.5
    dpos2 = (pos2[0]**2.0+pos2[1]**2.0)**.5
    if (1.0-dpos1)**2.0 < (1.0-dpos2)**2.0:
        return True
    else:
        return False
def converts(path):
    return ''.join(path)
l = []
for steps in range(1, 20):
    stepsize = 1.0/steps
    pos = [-1.0, 0.0]
    paths = []
    for i in range(0, 2*steps):
        if closer([pos[0]+stepsize, pos[1]], [pos[0], pos[1]+stepsize]):
            pos = [pos[0]+stepsize, pos[1]]
            paths.append(str(0))
        else:
            pos = [pos[0], pos[1]+stepsize]
            paths.append(str(1))
    l.append(int(converts(paths)))
print(l)

A266238 a(n+1) = 2^(2n - 1) + (-1)^n a(n), a(1) = 1.

Original entry on oeis.org

1, 1, 9, 23, 151, 361, 2409, 5783, 38551, 92521, 616809, 1480343, 9868951, 23685481, 157903209, 378967703, 2526451351, 6063483241, 40423221609, 97015731863, 646771545751, 1552251709801, 10348344732009, 24836027356823, 165573515712151, 397376437709161

Offset: 1

Ben Paul Thurston, Dec 25 2015

			a(4) = 2^(2*3 - 1) + (-1)^3 * 9 = 23.

Iain Fox, Table of n, a(n) for n = 1..1662
Iain Fox, Proof for generating function and recurrence relation
Index entries for linear recurrences with constant coefficients, signature (0,15,0,16)

f:= gfun:-rectoproc({a(n) = 15*a(n-2) + 16*a(n-4),a(1)=1,a(2)=1,a(3)=9,a(4)=23},a(n),remember):
map(f, [$1..50]); # Robert Israel, Dec 25 2017

RecurrenceTable[{a[1]==1,a[n+1]==2^(2n-1)+(-1)^n a[n]},a,{n,30}] (* Harvey P. Dale, Dec 20 2017 *)
f[n_]:= ((7 +7I)(-I)^n + (7 -7I)*I^n +(-1)^(1 +n) 2^(2n) +2^(2 +2n))/34; Array[f, 26] (* or *)
CoefficientList[ Series[ -(8x^3 -6x^2 +x +1)/(16x^4 +15x^2 -1), {x, 0, 25}], x] (* or *)
LinearRecurrence[{0, 15, 0, 16}, {1, 1, 9, 23}, 26] (* Robert G. Wilson v, Dec 24 2017 *)

a=vector(10^3); a[1]=1; for(n=2, #a, a[n] = 2^(2*n-3)-(-1)^n*a[n-1]); a \\ Altug Alkan, Dec 20 2017

first(n) = Vec(x*(1 + x - 6*x^2 + 8*x^3)/((1 - 4*x)*(1 + 4*x)*(1 + x^2)) + O(x^(n+1))) \\ Iain Fox, Dec 21 2017

From Colin Barker, Dec 21 2017: (Start)
G.f.: x*(1 + x - 6*x^2 + 8*x^3) / ((1 - 4*x)*(1 + 4*x)*(1 + x^2)). [Proved by Iain Fox, Dec 21 2017]
a(n) = ((7+7*i)*(-i)^n + (7-7*i)*i^n + (-1)^(1+n)*4^n + 4^(1+n)) / 34 where i=sqrt(-1).
a(n) = 15*a(n-2) + 16*a(n-4) for n > 4. [Proved by Iain Fox, Dec 21 2017] (End)

Corrected by Harvey P. Dale, Dec 20 2017

A265500 Balance scale sequence.

Original entry on oeis.org

1, 2, 7, 21, 52, 167, 501, 1503, 4509, 13527, 40581

Offset: 1

Ben Paul Thurston, Dec 09 2015

Every natural number up to sum(a[1], a[i]) can be written as adding or subtracting unique entries up to a(i).

			17 = 21-7+1+2, 18=21-1-2, 19=17+2.

The Ben Paul Thurston Blog, Balance scale series

a(i+1)= 1+2*sum(a[1], a[i]) (proved).

A265945 n-th derivative of x^(2*x) at x=1.

Original entry on oeis.org

1, 2, 6, 18, 64, 220, 888, 3192, 15104, 48096, 338400, 285120, 13728768, -75484032, 1327431168, -15621137280, 232048220160, -3468200017920, 56208508250112, -959557688285184, 17344153658234880, -330228975428997120, 6611419866845122560, -138844103855232061440

Offset: 0

Ben Paul Thurston, Dec 23 2015

sign

Robert Israel, Table of n, a(n) for n = 0..450

Cf. A005727.

S:= series((x+1)^(2*x+2), x, 51):
seq(coeff(S,x,j)*j!, j=0..50); # Robert Israel, Dec 23 2015

With[{nn=30},CoefficientList[Series[(x+1)^(2x+2),{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, May 20 2018 *)

E.g.f.: (x+1)^(2*x+2).
From Robert Israel, Dec 23 2015: (Start)
a(n) = Sum_{k=0..n} binomial(n,j)*A005727(j)*A005727(n-j).
a(n+1) = 2*(a(n) + Sum_{k=0..n-1} (-1)^(n-1-k)*n!*a(k)/((n-k)*k!)). (End)

More terms from Alois P. Heinz, Dec 23 2015

A235915 a(1) = 1, a(n) = a(n-1) + (digsum(a(n-1)) mod 5) + 1, digsum = A007953.

Original entry on oeis.org

1, 3, 7, 10, 12, 16, 19, 20, 23, 24, 26, 30, 34, 37, 38, 40, 45, 50, 51, 53, 57, 60, 62, 66, 69, 70, 73, 74, 76, 80, 84, 87, 88, 90, 95, 100, 102, 106, 109, 110, 113, 114, 116, 120, 124, 127, 128, 130, 135, 140, 141

Offset: 1

Ben Paul Thurston, Jan 16 2014

			For n = 7, a(6) is 16, where the sum of the digits is 7, of which the remainder when divided by 5 is 2, then plus 1 is 3. Thus a(7) is a(6) + 3 or 19.

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Ben Paul Thurston, Low Kolmorogov complexity but never repeating series?

Cf. A007953.

a:= proc(n) a(n):= `if`(n=1, 1, a(n-1) +1 +irem(
            add(i, i=convert(a(n-1), base, 10)), 5)) end:
seq(a(n), n=1..100);  # Alois P. Heinz, Feb 15 2014

NestList[#+Mod[Total[IntegerDigits[#]],5]+1&,1,50] (* Harvey P. Dale, Nov 23 2023 *)

digsum(n)=d=eval(Vec(Str(n))); sum(i=1, #d, d[i])
a=vector(1000); a[1]=1; for(n=2, #a, a[n]=a[n-1]+digsum(a[n-1])%5+1); a \\ Colin Barker, Feb 14 2014

def adddigits(i):
  s = str(i)
  t=0
  for j in s:
    t = t+int(j)
  return t
n = 1
a = [1]
for i in range(0, 100):
  r = adddigits(n)%5+1
  n = n+r
  a.append(n)
print(a)

A179332 a(1)=1; for each n > 1, a(n) is the smallest number such that Sum_{i=1..n} 1/a(i)^2 < sqrt(2).

Original entry on oeis.org

1, 2, 3, 5, 9, 37, 195, 8584, 1281621, 1325419784, 40182098746967, 203448501599750774078, 4275655952199444141114482835180, 10920781877316031992615629928696178128586477545

Offset: 1

Ben Paul Thurston, Jul 10 2010

In other words, the sequence is the lexicographically first infinite sequence of positive integers whose squared reciprocals sum to less than sqrt(2). After a(1)=1, each term is the smallest number that will not cause the sum of the squares of the reciprocals to exceed the square root of 2.

			a(1)=1; 1/1^2 = 1;
a(2)=2; 1 + 1/2^2 = 5/4 = 1.25;
a(3)=3; 5/4 + 1/3^2 = 49/36 = 1.3611111111...;
a(4)=5; 49/36 + 1/5^2 = 1261/900 = 1.4011111111...;
a(5)=9; 1261/900 + 1/9^2 = 11449/8100 = 1.4134567901...;
(sums approach sqrt(2) = 1.4142135623...).

Cf. A216245.

Digits := 200 : A179332 := proc(n) option remember; if n = 1 then 1; else sqrt(2)-add( 1/procname(i)^2,i=1..n-1) ; ceil( 1/sqrt(%)) ; end if; end proc: seq(A179332(n),n=1..14) ; # R. J. Mathar, Jul 11 2010

a(n+1) = ceiling(1/sqrt(sqrt(2) - Sum_{i=1..n} 1/a(i)^2)). - R. J. Mathar, Jul 11 2010

More terms from R. J. Mathar, Jul 11 2010

Name changed, comments expanded, and example corrected and expanded by Jon E. Schoenfield, Feb 28 2014

A160785 Even squarefree numbers plus 1.

Original entry on oeis.org

3, 7, 11, 15, 23, 27, 31, 35, 39, 43, 47, 59, 63, 67, 71, 75, 79, 83, 87, 95, 103, 107, 111, 115, 119, 123, 131, 135, 139, 143, 147, 155, 159, 167, 171, 175, 179, 183, 187, 191, 195, 203, 207, 211, 215, 219, 223, 227, 231, 239, 247, 255, 259, 263, 267, 275, 279

Offset: 1

Ben Paul Thurston, May 26 2009

			10 is an even squarefree number, so 11 is in this list.

Cf. A005117.

with(numtheory): a := proc (n) if issqrfree(2*n) = true then 2*n+1 else end if end proc: seq(a(n), n = 1 .. 150); # Emeric Deutsch, Jun 20 2009

a(n) = A039956(n) + 1.

More terms from Emeric Deutsch, Jun 20 2009

cp's OEIS Frontend

User: Ben Paul Thurston

A334264 Numbers k > 1 such that (3/2)^k sets a new record for closest fractional part to 1/2.

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A309815 a(n) is the smallest positive integer x such that sqrt(2) + sqrt(x) is closer to an integer than any other value already in the sequence.

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A317778 Starting with 1,2,3,4,5,6: a(n) is the next smallest number greater than a(n-1) such that a[i] + a[j] + a[k] != a[x] + a[y] + a[z] for 1 <= i,j,k,x,y,z <= n all distinct.

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Python

Extensions

A303610 Circle aliasing numbers with 1/n size steps.

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Python

A266238 a(n+1) = 2^(2*n - 1) + (-1)^n * a(n), a(1) = 1.

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A265500 Balance scale sequence.

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A265945 n-th derivative of x^(2*x) at x=1.

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A235915 a(1) = 1, a(n) = a(n-1) + (digsum(a(n-1)) mod 5) + 1, digsum = A007953.

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A266238 a(n+1) = 2^(2n - 1) + (-1)^n a(n), a(1) = 1.