A002390 -id:A002390 - cp's OEIS frontend

A351794 Decimal expansion of Sum_{k>=1} AH(k)*L(k)/2^k, where AH(k) = A058313(k)/A058312(k) is the k-th alternating harmonic number and L(k) = A000032(k) is the k-th Lucas number.

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

Offset: 1

Amiram Eldar, Feb 19 2022

			2.82147535876264946174605143568253063724665666934546...

Cf. A000032, A001622, A002390, A058312, A058313, A349850, A349851, A351789.

RealDigits[3*Log[5/4] + 10*Log[GoldenRatio]/Sqrt[5], 10, 100][[1]]

Equals 3*log(5/4) + 10*log(phi)/sqrt(5), where phi is the golden ratio (A001622).

Equals (4/3)*log(5/4) + (5/3)*A351789.

A370742 Decimal expansion of Sum_{k>=2} H(k-1) * F(k) / (k*2^k), where H(k) = A001008(k)/A002805(k) is the k-th harmonic number and F(k) = A000045(k) is the k-th Fibonacci number.

Original entry on oeis.org

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

Offset: 0

Amiram Eldar, Feb 29 2024

			0.59667348783398269737770682436833083924687967042183...

Kenny B. Davenport, Problem B-1222, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 56, No. 1 (2018), p. 81; The Generating Function for Harmonic Numbers, Solution to Problem B-1222 by Amanda M. Andrews and Samantha L. Zimmerman, ibid., Vol. 57, No. 1 (2019), pp. 83-84.

Cf. A000045, A001008, A001622, A002162, A002163, A002390, A002805, A349850, A370743.

RealDigits[4 * Log[2] * Log[GoldenRatio] / Sqrt[5], 10, 120][[1]]

PARI
```
4 * log(2) * log(quadgen(5)) / sqrt(5)
```

Equals 4 * log(2) * log(phi) / sqrt(5), where phi is the golden ratio (A001622) (Davenport, 2018).

A370743 Decimal expansion of Sum_{k>=2} H(k-1) * L(k) / (k*2^k), where H(k) = A001008(k)/A002805(k) is the k-th harmonic number and L(k) = A000032(k) is the k-th Lucas number.

Original entry on oeis.org

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

Offset: 1

Amiram Eldar, Feb 29 2024

			1.40671229622697899465481881125279601179617835179174...

Kenny B. Davenport, Problem B-1222, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 56, No. 1 (2018), p. 81; The Generating Function for Harmonic Numbers, Solution to Problem B-1222 by Amanda M. Andrews and Samantha L. Zimmerman, ibid., Vol. 57, No. 1 (2019), pp. 83-84.

Cf. A000032, A001008, A001622, A002162, A002390, A002805, A349851, A370742.

RealDigits[Log[2]^2 + 4*Log[GoldenRatio]^2, 10, 120][[1]]

PARI
```
log(2)^2 + 4*log(quadgen(5))^2
```

Equals log(2)^2 + 4*log(phi)^2, where phi is the golden ratio (A001622) (Davenport, 2018).

A119579 a(n) = (n + n^2)(binomial(2n, n)).

Original entry on oeis.org

0, 4, 36, 240, 1400, 7560, 38808, 192192, 926640, 4375800, 20323160, 93117024, 421848336, 1892909200, 8424486000, 37228204800, 163493866080, 714083503320, 3103696272600, 13431200244000, 57895542104400, 248675137991280

Offset: 0

Zerinvary Lajos, May 31 2006

Cf. A000984, A002378, A002390.

[seq ((n+n^2)*(binomial(2*n,n)),n=0..29)];

Table[n*(n + 1)*Binomial[2*n, n], {n, 0, 20}] (* Amiram Eldar, Feb 20 2021 *)

a(n) = (n + n^2)*(binomial(2*n, n)); \\ Michel Marcus, Feb 20 2021

From Amiram Eldar, Feb 20 2021: (Start)
a(n) = A002378(n)*A000984(n).
Sum_{n>=1} (-1)^(n+1)/a(n) = -1 + 2*sqrt(5)*log(phi) - 4*log(phi)^2, where log(phi) = A002390. (End)
Sum_{n>=1} 1/a(n) = Pi^2/9 - Pi/sqrt(3) + 1. - Amiram Eldar, Jan 24 2022

A145424 Decimal expansion of Sum_{n>=1} 1/(n*(25n^2-1)).

Original entry on oeis.org

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

Offset: 0

R. J. Mathar, Feb 08 2009

			0.049808566747630516862...

Alexander Apelblat, Tables of Integrals and Series, Harri Deutsch, 1996, 4.1.26.

Cf. A001622, A002390.

evalf(5*log(5)/4-5/2+sqrt(5)/2*log((1+sqrt(5))/2)) ;

RealDigits[Log[GoldenRatio]*Sqrt[5]/2 - 5*(2 - Log[5])/4, 10, 120, -1][[1]] (* Amiram Eldar, Aug 23 2024 *)

Equals log(phi)*sqrt(5)/2 - 5*(2 - log(5))/4, where phi is the golden ratio (A001622). - Amiram Eldar, Aug 23 2024

A157701 Decimal expansion of 2(14sigma+5)/625 where sigma = sqrt(5)*log(golden ratio).

Original entry on oeis.org

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

Offset: 0

R. J. Mathar, Mar 04 2009

The factor 28 in the Lehmer paper has been corrected to 14.

Equals sum_{n=1..infinity} (-1)^n*n^3/binomial(2n,n).

			0.064205801387968452355..

D. H. Lehmer, Interesting series involving the Central Binomial Coefficient, Am. Math. Monthly 92, no 7 (1985) 449-457.
R. J. Mathar, Corrigenda to "Interesting series involving...", arXiv:0905.0215
Index entries for transcendental numbers

Cf. A145434, A145433.

2/625*(14*sqrt(5)*log((1+sqrt(5))/2)+5) ;

Join[{0},RealDigits[2*(14*Sqrt[5]*Log[GoldenRatio]+5)/625,10,120][[1]]] (* Harvey P. Dale, Mar 13 2015 *)

2*(14*sqrt(5)*log((sqrt(5)+1)/2)+5)/625 \\ Charles R Greathouse IV, May 15 2019

Equals 2*(14*A002163*A002390+5)/625 .

A245800 a(n) is the least number k such that Sum_{j=S(n)+1..S(n)+k} 1/j >= 1/2, where S(n) = Sum_{i=1..n-1} a(i) and S(1) = 0.

Original entry on oeis.org

1, 1, 2, 3, 5, 9, 14, 24, 39, 64, 106, 175, 288, 475, 783, 1291, 2129, 3510, 5787, 9541, 15730, 25935, 42759, 70498, 116232, 191634, 315951, 520915, 858844, 1415994, 2334579, 3849070, 6346044, 10462858, 17250336, 28440996, 46891275, 77310643, 127463701, 210152115, 346482262

Offset: 1

Rob van den Tillaart, Aug 22 2014

nonn

The limiting ratio of consecutive terms is sqrt(e) ~ 1.648721270700128.
Conjecture 1: If the summation threshold (which, for this sequence, is defined as 1/2) is set to any positive number R, then the limiting ratio of consecutive terms in the resulting sequence is e^R.
Conjecture 2: If the summation threshold is set to log(phi) = 0.4812118250..., then the Fibonacci sequence is generated.
Partition the harmonic series into the smallest-sized groups that sum to 1/2 or greater. You get 1, 1/2, 1/3 + 1/4, 1/5 + 1/6 + 1/7, 1/8 + 1/9 + 1/10 + 1/11 + 1/12, 1/13 + 1/14 + ... + 1/21, 1/22 + ... ; a(n) gives the number of terms in the n-th group. - Derek Orr, Nov 08 2014
A partition sums to exactly 1/2 for only n=2. - Jon Perry, Nov 09 2014

			Start with 1/1; 1/1 >= 1/2, so a(1) = 1.
1/1 has been used, so start a new sum with 1/2; 1/2 >= 1/2, so a(2) = 1.
1/1 and 1/2 have been used, so start a new sum with 1/3; 1/3 + 1/4 >= 1/2, so a(3) = 2.
1/1 through 1/4 have been used, so start a new sum with 1/5; 1/5 + 1/6 + 1/7 >= 1/2, so a(4) = 3, etc.

Cf. A226161.

Cf. A019774, A002390.

lista(nn) = {n = 1; while (n < nn, k = 1; while (sum(x=n, n+k-1, 1/x) < 1/2, k++); print1(k, ", "); n = n+k;);} \\ Michel Marcus, Sep 14 2014

a(n)=if(n==1,return(1));p=sum(i=1,n-1,a(i));k=1;while(sum(x=p+1,p+k,1/x)<1/2,k++);k
n=1;while(n<100,print1(a(n),", ");n++) \\ Derek Orr, Oct 16 2014

import math
total = 0
prev = 1
count = 0
for n in range(1, 10**7):
    total = total + 1/n
    count = count + 1
    if (total >= 0.5):
        print(count,end=', ')
        prev = count
        total = 0
        count = 0
 
print("\ndone")
print(math.sqrt(math.e))
# Rob van den Tillaart, Aug 22 2014 [Corrected by Derek Orr, Oct 16 2014]

def a(n):
  if n == 1:
    return 1
  tot,c,k = 0,0,0
  for x in range(1,10**7):
    tot += 1/x
    if tot >= 0.5:
      k += 1
      tot = 0
    if k == n-1:
      c += 1
    if k == n:
      return c
n = 1
while n < 100:
  print(a(n),end=', ')
  n += 1
# Derek Orr, Oct 16 2014

Edited by Derek Orr, Nov 08 2014

A278765 Engel expansion of natural logarithm of golden ratio.

Original entry on oeis.org

3, 3, 4, 4, 4, 6, 15, 48, 118, 147, 671, 1026, 3075, 44641, 48364, 1868380, 75080506, 96848501, 911582093, 2511879981, 8700005050, 15888441652, 108526838262, 446779835336, 632466801279, 1084794852728, 1184722346307, 1657692322844, 12376968750845, 17341469111712, 27996895637798, 38935285631573

Offset: 1

Ilya Gutkovskiy, Nov 28 2016

			log(phi)  = 0.4812118250596... = 1/3 + 1/(3*3) + 1/(3*3*4) + 1/(3*3*4*4) + 1/(3*3*4*4*4) + 1/(3*3*4*4*4*6) + ...

Eric Weisstein's World of Mathematics, Engel Expansion
Eric Weisstein's World of Mathematics, Golden Ratio
Index entries for sequences related to Engel expansions

Cf. A006784 (for definition of Engel expansion).

Cf. A001622, A002390, A028259, A059195.

EngelExp[A_, n_]:=Join[Array[1&, Floor[A]], First@Transpose@NestList[{Ceiling[1/Expand[ #[[1]]#[[2]]-1]], Expand[ #[[1]]#[[2]]-1]}&, {Ceiling[1/(A-Floor[A])], A-Floor[A]}, n-1]]; EngelExp[N[Log[GoldenRatio], 7! ], 40]

A309408 Start with X = F(n) = A000045(n). Repeatedly replace X with X - ceiling(X/n); a(n) is the number of steps to reach 0.

Original entry on oeis.org

0, 1, 1, 2, 3, 5, 7, 10, 13, 18, 23, 29, 36, 44, 53, 63, 74, 85, 98, 111, 125, 140, 157, 174, 192, 211, 231, 251, 273, 296, 319, 343, 369, 395, 423, 451, 481, 510, 541, 573, 606, 640, 675, 710, 747, 785, 823, 863, 903, 944, 987, 1030, 1074, 1119, 1165, 1212, 1260, 1309, 1359, 1409, 1462, 1514, 1568, 1622, 1678, 1734, 1791

Offset: 0

A.H.M. Smeets, Jul 29 2019

Inspired by A278586.

Cf. A000045, A002390, A278586.

f[n_] := Length[NestWhileList[# - Ceiling[#/n] &, Fibonacci[n], # > 0 &]] - 1; f /@ Range[0,70] (* Amiram Eldar, Aug 08 2019 *)

f(x, n) = x - ceil(x/n);
a(n) = my(nb=0, x=fibonacci(n)); while(x, x = f(x, n); nb++); nb; \\ Michel Marcus, Aug 03 2019

n, f1, f0 = 0, 0, 1
while n <= 20000:
    fn, a = f1, 0
    while fn > 0:
        fn, a = fn - (fn+n-1)//n, a+1
    print(n,a)
    n, f1, f0 = n+1, f0, f1+f0

Lim_{n -> inf} (a(n)/(n^2)) = log(phi) = A002390.
a(n) = n^2*log(phi) - n*log(n) + O(n), phi = (1+sqrt(5))/2.
Lim_{n -> inf} (a(n) - n^2*log(phi) + n*log(n))/ n = -0.4681... .

A335605 Decimal expansion of arctan(log(phi)/(Pi/2)), the polar slope angle (in radians) of the golden spiral.

Original entry on oeis.org

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

Offset: 0

Stefano Occhetti, Jun 15 2020

In the polar equation for a logarithmic spiral: r = a*e^(b*theta), b represents the tangent of angle alpha, where alpha is the polar slope of the curve. In a golden spiral equation, b = log(phi)/(Pi/2) (being: log the natural logarithm; phi the golden ratio: 1.61803398...), therefore, alpha = arctan(log(phi)/(Pi/2)).

			0.29727130470530517362994810317214622995424798032442...

Wikipedia, Golden spiral.

Cf. A000796, A001622, A002390, A212225.

RealDigits[ArcTan[2 * Log[GoldenRatio]/Pi], 10, 100][[1]] (* Amiram Eldar, Jun 15 2020 *)

PARI
```
atan(log(((1+sqrt(5))/2))/(Pi/2))
```

alpha = arctan(log(phi)/(Pi/2)).

Equals arctan(A212225). - Amiram Eldar, Jun 15 2020

cp's OEIS Frontend

A351794 Decimal expansion of Sum_{k>=1} AH(k)*L(k)/2^k, where AH(k) = A058313(k)/A058312(k) is the k-th alternating harmonic number and L(k) = A000032(k) is the k-th Lucas number.

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A370742 Decimal expansion of Sum_{k>=2} H(k-1) * F(k) / (k*2^k), where H(k) = A001008(k)/A002805(k) is the k-th harmonic number and F(k) = A000045(k) is the k-th Fibonacci number.

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A370743 Decimal expansion of Sum_{k>=2} H(k-1) * L(k) / (k*2^k), where H(k) = A001008(k)/A002805(k) is the k-th harmonic number and L(k) = A000032(k) is the k-th Lucas number.

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A119579 a(n) = (n + n^2)*(binomial(2*n, n)).

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A145424 Decimal expansion of Sum_{n>=1} 1/(n*(25n^2-1)).

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A157701 Decimal expansion of 2*(14*sigma+5)/625 where sigma = sqrt(5)*log(golden ratio).

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A245800 a(n) is the least number k such that Sum_{j=S(n)+1..S(n)+k} 1/j >= 1/2, where S(n) = Sum_{i=1..n-1} a(i) and S(1) = 0.

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A278765 Engel expansion of natural logarithm of golden ratio.

A119579 a(n) = (n + n^2)(binomial(2n, n)).

A157701 Decimal expansion of 2(14sigma+5)/625 where sigma = sqrt(5)*log(golden ratio).