Rok Cestnik - cp's OEIS frontend

A366397 Decimal expansion of the number whose continued fraction terms are one larger than those of Pi.

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

Offset: 1

Rok Cestnik, Oct 08 2023

			4.12406010228786539167585... = 4 + 1/(8 + 1/(16 + 1/(2 + 1/(293 + ...)))).
Pi = 3.141592653589793238... = 3 + 1/(7 + 1/(15 + 1/(1 + 1/(292 + ...)))).

Sean A. Irvine, Java program (github)

Cf. A000796, A001203, A059833.

N = 25;
cf(v) = my(m=contfracpnqn(v)); m[1, 1]/m[2, 1];
summand(k) = (-1)^k/2^(10*k)*(-2^5/(4*k+1)-1/(4*k+3)+2^8/(10*k+1)-2^6/(10*k+3)-2^2/(10*k+5)-2^2/(10*k+7)+1/(10*k+9));
pi1 = contfrac(1/2^6*sum(k=0,N,summand(k)));
pi2 = contfrac(1/2^6*sum(k=0,N+1,summand(k)));
n = 0; while(pi1[1..n+1] == pi2[1..n+1], n++);
ap1 = cf(apply(x->x+1, pi1[1..n-1]));
ap2 = cf(apply(x->x+1, pi1[1..n]));
n = 0; while(digits(floor(10^(n+1)*ap1)) == digits(floor(10^(n+1)*ap2)), n++);
A366397 = digits(floor(10^n*ap1));

A367120 Decimal expansion of continued fraction 2+1/(4+3/(6+5/(8+7/(...)))).

Original entry on oeis.org

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

Offset: 1

Rok Cestnik, Nov 13 2023

			2.224412437956340467163837541384021939...

Cf. A113014, A113011, A194807, A365307.

First[RealDigits[2/HypergeometricPFQ[{1, 1}, {3/2, 3}, -1/2], 10, 100]] (* or *)
First[RealDigits[2/Sum[(-1)^k/Binomial[k+2, 2]/(2*k+1)!!, {k, 0, Infinity}], 10, 100]] (* Paolo Xausa, Nov 18 2024 *)

N=50;
doblfac(n) = if(n<0, 0, n<2, 1, n*doblfac(n-2));
ap1 = 2 / sum(k=0,N, (-1)^k/binomial(k+2,2)/doblfac(2*k+1));
ap2 = 2 / sum(k=0,N+1, (-1)^k/binomial(k+2,2)/doblfac(2*k+1));
n = 0; while(digits(floor(10^(n+1)*ap1)) == digits(floor(10^(n+1)*ap2)), n++);
A367120 = digits(floor(10^n*ap1));

Equals 2 / pFq(1,1; 3/2,3; -1/2) where pFq() is the generalized hypergeometric function.

Equals 2 / Sum_{k>=0} (-1)^k/binomial(k+2,2)/(2*k+1)!! = 2 / (1 - 1/9 + 1/90 - 1/1050 + 1/14175 - 1/218295 + ... ).

A365125 Put a positive charge at 0 and a negative charge at 1, then keep adding alternating charges at points of zero potential; this is the decimal expansion of the limit.

Original entry on oeis.org

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

Offset: 0

Rok Cestnik, Aug 22 2023

The potential is inversely proportional to the distance: a positive charge at p has potential 1/abs(x-p). Starting from a positive charge at 0 and a negative charge at 1, the potential is zero at the midpoint 1/2, so another positive charge is added there. Then the potential is zero at 0.7886... and a negative charge is added there, and so on. The first few zero potential points are: 0.5, 0.7886..., 0.6325..., 0.7179..., 0.6714..., ... The limit is the constant of this sequence.

After initial points p_0 = 0 and p_1 = 1, each new zero point fits between previous two points and is the solution to Sum_{n>=0} 1/(x-p_n) = 0.

			0.68784181032838926327134404409883348611583979...

Rok Cestnik, Visualization

p={0,1};
For[r=1,r<39,++r,p=Append[p,x/.NSolve[{Sum[1/(x-p[[k]]),{k,1,Length[p]}]==0,(x-p[[-1]])*(x-p[[-2]])<0},x,WorkingPrecision -> 30][[1,1]]];];
A365125 = RealDigits[Floor[10^10*p[[-1]]]][[1]]

from decimal import Decimal, getcontext, ROUND_UP, ROUND_DOWN
getcontext().prec = 100
def nm(f, df, x):
    for i in range(10):
        x -= f(x)/df(x)
    return x
def flip_rounding():
    if getcontext().rounding == ROUND_UP: getcontext().rounding = ROUND_DOWN
    else: getcontext().rounding = ROUND_UP
def get_zero(vs, rounding):
    getcontext().rounding = rounding
    def p(x,v):
        flip_rounding(); t = x-v
        flip_rounding(); return 1/t
    def dp(x,v):
        flip_rounding(); t = x-v; t = t**2
        flip_rounding(); return -1/t
    def f(x): return sum(p(x,vs[n]) for n in range(len(vs)))
    def df(x): return sum(dp(x,vs[n]) for n in range(len(vs)))
    sign = -1 if rounding == ROUND_DOWN else 1
    return nm(f, df, (vs[-1]+vs[-2])/2+sign*abs(vs[-1]-vs[-2])/3)
v_lo = [Decimal(0), Decimal(1)]
v_up = [Decimal(0), Decimal(1)]
for r in range(150):
    v_lo.append(get_zero(v_lo, ROUND_DOWN))
    v_up.append(get_zero(v_up, ROUND_UP))
lower_bounds = [v_lo[i] for i in range(0, len(v_lo), 2)]
upper_bounds = [v_up[i] for i in range(1, len(v_up), 2)]
right = True
A365125 = [int(l) for l, u in zip(str(lower_bounds[-1])[2:], str(upper_bounds[-1])[2:]) if right and (right := (l == u))]

A365350 Decimal expansion of 1/(Pi-3).

Original entry on oeis.org

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

Offset: 1

Rok Cestnik, Sep 02 2023

The continued fraction expansion is the same as Pi (A001203) with initial 3 omitted.

			7.062513305931045769793...

Index entries for transcendental numbers

Cf. A000796, A201775, A001203, A194807.

A365350 = RealDigits[N[1/(Pi-3), #+1]][[1]][[1;; -2]]&;

PARI
```
1/(Pi-3)
```

A365307 Decimal expansion of 1/(2*e-5).

Original entry on oeis.org

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

Offset: 1

Rok Cestnik, Aug 31 2023

The continued fraction expansion is A081750 with initial term 5 omitted.

			2.2906166927853624221...

Cf. A001113, A194807, A019762, A113011, A081750, A073333, A185108.

A365307 = RealDigits[N[1/(2*E-5),#+1]][[1]][[1;;-2]]&;

PARI
```
1/(2*exp(1)-5).
```

Equals 2 + 1/(3 + 2/(4 + 3/(5 + 4/(6 + 5/( ... /(n+1 + n/(n+2 + ... ))))))).
From Peter Bala, Oct 23 2023: (Start)
Define s(n) = Sum_{k = 3..n} 1/k!. Then 1/(2*e - 5) = 3 - (1/2)*Sum_{n >= 3 } 1/( (n+1)!*s(n)*s(n+1) ) is a rapidly converging series of rationals. Cf. A073333 and A194807.
Equivalently, 1/(2*e - 5) = 3 - (1/2)*(3!/(1*5) + 4!/(5*26) + 5!/(26*157) + 6!/(157*1100) + ...), where [1, 5, 26, 157, 1100, ... ] is A185108. (End)

A364835 a(1) = 1; a(n+1) = (number of times a(n)+1 has appeared) - (number of times a(n) has appeared).

Original entry on oeis.org

1, -1, -1, -2, 1, -2, 0, 1, -3, 1, -4, 0, 2, -1, -1, -2, 1, -4, -1, -3, 1, -5, 1, -6, 0, 4, -1, -3, 0, 3, 0, 2, -1, -2, 3, -1, -3, 0, 1, -6, -1, -3, -1, -4, 2, -1, -5, 1, -6, -1, -6, -2, 7, -1, -7, 3, -2, 7, -2, 6, 1, -7, 2, -1, -8, 1, -7, 1, -8, 1, -9, 1, -10, 0, 7, -3, 1, -11, 0

Offset: 1

Rok Cestnik, Aug 28 2023

Irregular until n=149695 at which point it follows a simple pattern (see formula).

			a(5) = 2 - 1 = 1 because a(4) + 1 = -1 has appeared twice before and a(4) = -2 has appeared once.
a(7) = 2 - 2 = 0 because a(6) + 1 = -1 and a(6) = -2 have both appeared twice before.

Rok Cestnik, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,-1).

Cf. A330004, A329934.

nmax=78; a={1}; For[n=1, n<=nmax, n++, AppendTo[a,Count[a,Part[a,n]+1]-Count[a,Part[a,n]]]]; a (* Stefano Spezia, Aug 29 2023 *)

a=[1]
for n in range(1000):
    a.append(a.count(a[n]+1)-a.count(a[n]))

from itertools import islice
from collections import Counter
def agen(): # generator of terms
    an = 1; c = Counter([1])
    while True: yield an; an = c[an+1] - c[an]; c[an] += 1
print(list(islice(agen(),1001))) # Michael S. Branicky, Aug 29 2023

For n >= 149695: a(n) = 49456 - n/3 if (n mod 3) = 0, otherwise a(n) = (n mod 3) - 1.

A365116 Greedy Egyptian fraction expansion of 1/(2+3/(4+5/(6+7/(...)))) = A113014.

Original entry on oeis.org

3, 22, 1060, 1471180, 4470565318951, 21387196871513452925199541, 508406155285302398938678134812723800438323137635884, 293206664431782985993415302840424364324000216460330294129310314248712028637333711113413230006858255456

Offset: 1

Rok Cestnik, Aug 22 2023

nonn

			1/3 + 1/22 + 1/1060 + 1/1471180 + ... = 1/(2+3/(4+5/(6+7/(...)))).

Cf. A113014, A365105.

A365105 Continued fraction expansion of 1/(2+3/(4+5/(6+7/(...)))) = A113014.

Original entry on oeis.org

0, 2, 1, 1, 1, 2, 1, 2, 10, 2, 2, 66, 1, 1, 13, 66, 9, 5, 8, 1, 9, 1, 1, 1, 5, 2, 2, 3, 1, 1, 1, 16, 99, 6, 1, 5, 1, 2, 1, 55, 2, 2, 1, 1, 6, 4, 1, 1, 1, 40, 1, 1, 1, 6, 14, 7, 9, 1, 1, 2, 3, 2, 2, 2, 1, 1, 2, 7, 12, 1, 2, 2, 1, 4, 2, 4, 2, 1, 3, 2, 1, 10, 7, 1, 4, 1, 119, 1, 1, 1, 3, 5, 2, 12, 1

Offset: 0

Rok Cestnik, Aug 21 2023

A113014 is defined by a generalized continued fraction and the expansion here is its simple continued fraction.

			1/(2+1/(1+1/(1+1/(1+1/(2+1/(...)))))) = 1/(2+3/(4+5/(6+7/(...)))).

Cf. A113014, A365116.

A365105 = ContinuedFraction[Sqrt[2*E/Pi]/Erfi[1/Sqrt[2]]-1,#]&;

A365052 Decimal expansion of continued fraction [1; 4, 9, 16, 25, ... n^2, ... ].

Original entry on oeis.org

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

Offset: 1

Rok Cestnik, Aug 18 2023

			1.243288478399715644...

Cf. A073824 (reciprocal), A036246/A036245 (convergents).

Cf. A226771, A060997, A052119.

A365052 = RealDigits[FromContinuedFraction[Range[1,50]^2],10,#][[1]]&;

p(N) = my(m=contfracpnqn(vector(N, i, i^2))); m[1,1]/m[2,1];
A365052(N) = {my(t=2); while(floor(10^N*p(t)) != floor(10^N*p(t+1)), t++); digits(floor(10^(N-1)*p(t)))};

Equals 1/A073824.

A364873 Decimal expansion of the lexicographically earliest continued fraction which equals its own sum of reciprocals.

Original entry on oeis.org

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

Offset: 1

Rok Cestnik, Aug 11 2023

This continued fraction (A364872) is the earliest infinite sequence {a0,a1,a2,a3,...} such that: a0+1/(a1+1/(a2+1/(a3+...))) = 1/a0 + 1/a1 + 1/a2 + 1/a3 + ....

There are infinitely many real numbers whose continued fraction is also their sum of reciprocals - they are dense on the interval (2,oo).

			2.71053359137351078733864566...

Cf. A364872.

cf(a) = my(m=contfracpnqn(a)); m[1, 1]/m[2, 1];
uf(a) = sum(i=1, #a, 1/a[i]);
A364872(N) = {a=[2]; for(i=2, N, a=concat(a, if(cf(a)==uf(a), a[i-1], ceil(1/(cf(a)-uf(a))))); while(cf(a)<=uf(a), a[i]++)); a};
A364873(N) = {t=2; while(floor(10^N*cf(A364872(t))) != floor(10^N*cf(A364872(t+1))), t++); digits(floor(10^(N-1)*cf(A364872(t))))};

cp's OEIS Frontend

User: Rok Cestnik

A366397 Decimal expansion of the number whose continued fraction terms are one larger than those of Pi.

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A367120 Decimal expansion of continued fraction 2+1/(4+3/(6+5/(8+7/(...)))).

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A365125 Put a positive charge at 0 and a negative charge at 1, then keep adding alternating charges at points of zero potential; this is the decimal expansion of the limit.

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A365350 Decimal expansion of 1/(Pi-3).

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A365307 Decimal expansion of 1/(2*e-5).

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A364835 a(1) = 1; a(n+1) = (number of times a(n)+1 has appeared) - (number of times a(n) has appeared).

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A365116 Greedy Egyptian fraction expansion of 1/(2+3/(4+5/(6+7/(...)))) = A113014.

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A365105 Continued fraction expansion of 1/(2+3/(4+5/(6+7/(...)))) = A113014.

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