A224820 -id:A224820 - cp's OEIS frontend

A227728 a(1) = greatest k such that H(k) - H(2) < 1/1 + 1/2; a(2) = greatest k such that H(k) - H(a(1)) < H(a(1)) - H(2); and for n>2, a(n) = greatest k such that H(k) - H(a(n-1)) < H(a(n-1)) - H(a(n-2)), where H = harmonic number.

10, 43, 179, 740, 3054, 12599, 51971, 214376, 884278, 3647546, 15045706, 62061794, 255997704, 1055960840, 4355715996, 17966823308, 74111062350, 305699536774, 1260975134078, 5201376179830, 21455073484758, 88499689759294, 365050956038686, 1505792854949114

Offset: 1

Clark Kimberling, Jul 29 2013

Suppose that x and y are positive integers and that x <= y. Let c(1) = y and c(2) = greatest k such that H(k) - H(y) < H(y) - H(x); for n > 2, let c(n) = greatest such that H(k) - H(c(n-1)) < H(c(n-1)) - H(c(n-2)). Then 1/x + ... + 1/c(1) > 1/(c(1)+1) + ... + 1/(c(2)) > 1/(c(2)+1) + ... + 1/(c(3)) > ... The decreasing sequences H(c(n)) - H(c(n-1)) and c(n)/c(n-1) converge. For what choices of (x,y) is the sequence c(n) linearly recurrent?

For A224868, (x,y) = (1,2) and a(n) = c(n+1) for n >= 1. It appears that the sequence a(n) is linearly recurrent with signature (5,-4,2,-2,2,-2,2,-2,2,-2). See A227729 and A225815 for limits of the sequences H(c(n)) - H(c(n-1)) and c(n)/c(n-1).

			The first three values (a(1),a(2),a(3)) = (10,43,179) match the beginning of the following inequality chain (and partition of the harmonic numbers): 1/1 + 1/2 > 1/3 + 1/4 + ... + 1/10 > 1/11 + ... + 1/43 > 1/44 + ... + 1/179 > ...

Clark Kimberling, Table of n, a(n) for n = 1..1000

Cf. A224820, A224868, A225815, A227729.

z = 100; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 1; y = 2; a[1] = -1 + Ceiling[w /. FindRoot[h[w] == 2 h[y] - h[x - 1], {w, 1}, WorkingPrecision -> 400]]; a[2] = -1 + Ceiling[w /. FindRoot[h[w] == 2 h[a[1]] - h[y], {w, a[1]}, WorkingPrecision -> 400]]; Do[s = 0; a[t] = -1 + Ceiling[w /. FindRoot[h[w] == 2 h[a[t - 1]] - h[a[t - 2]], {w, a[t - 1]}, WorkingPrecision -> 400]], {t, 3, z}]; m = Map[a, Range[z]] (* A227728 *)
N[Table[h[a[t]] - h[a[t - 1]], {t, 2, z, 25}], 50]  (* A227729 *)
N[Table[a[n]/a[n - 1], {n, 2, z, 25}], 50]  (* A225815 *)

a(n) = 5*a(n-1) - 4*a(n-2) + 2*a(n-3) - 2*a(n-4) + 2*a(n-5) - 2*a(n-6) + 2*a(n-7) - 2*a(n-8) + 2*a(n-9) - 2*a(n-10) (conjectured).

Empirical g.f.: -x*(4*x^9-4*x^8+3*x^7-4*x^6+3*x^5-4*x^4+3*x^3-4*x^2+7*x-10) / ((x-1)*(2*x^9+2*x^7+2*x^5+2*x^3+4*x-1)). - Colin Barker, Mar 23 2015

Definition corrected by Clark Kimberling, Jan 14 2017

A224868 a(1) = greatest k such that H(k) - H(4) < 1/3 + 1/4; a(2) = greatest k such that H(k) - H(a(1)) < H(a(1)) - H(4); and for n > 2, a(n) = greatest k such that H(k) - H(a(n-1)) > H(a(n-1)) - H(a(n-2)), where H = harmonic number.

Original entry on oeis.org

7, 11, 17, 26, 39, 58, 86, 127, 187, 275, 404, 593, 870, 1276, 1871, 2743, 4021, 5894, 8639, 12662, 18558, 27199, 39863, 58423, 85624, 125489, 183914, 269540, 395031, 578947, 848489, 1243522, 1822471, 2670962, 3914486, 5736959, 8407923, 12322411, 18059372

Offset: 1

Clark Kimberling, Jul 23 2013

Suppose that x and y are positive integers and that x <=y. Let c(1) = y and c(2) = greatest k such that H(k) - H(y) < H(y) - H(x); for n > 2, let c(n) = greatest such that H(k) - H(c(n-1)) < H(c(n-1)) - H(c(n-2)). Then 1/x + ... + 1/c(1) > 1/(c(1)+1) + ... + 1/(c(2)) > 1/(c(2)+1) + ... + 1/(c(3)) > ... The decreasing sequences H(c(n)) - H(c(n-1)) and c(n)/c(n-1) converge. For what choices of (x,y) is the sequence c(n) linearly recurrent?

For A224868, (x,y) = (3,4); it appears that the sequence a(n) is linearly recurrent with signature (2,-1,1,-1). Possibly the constant at A202537 is the limit of the sequences H(c(n))-H(c(n-1)). Possibly the constant at A092526 is the limit of c(n)/c(n-1).

			The first three values (a(1),a(2),a(3)) = (7,11,17) match the beginning of the following inequality chain (and partition of {1/m: m>=3}):
1/3+1/4 > 1/5+1/6+1/7 > 1/8+1/9+1/10+1/11 > 1/12+ ... +1/17 > ...

Clark Kimberling, Table of n, a(n) for n = 1..1000

Cf. A224820.

z = 100; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 3; y = 4; a[1] = -1 + Ceiling[w /. FindRoot[h[w] == 2 h[y] - h[x - 1], {w, 1}, WorkingPrecision -> 400]]; a[2] = -1 + Ceiling[w /. FindRoot[h[w] == 2 h[a[1]] - h[y], {w, a[1]}, WorkingPrecision -> 400]]; Do[s = 0; a[t] = -1 + Ceiling[w /. FindRoot[h[w] == 2 h[a[t - 1]] - h[a[t - 2]], {w, a[t - 1]}, WorkingPrecision -> 400]], {t, 3, z}]; m = Map[a, Range[z]] (* A224868 *)
N[Table[h[a[t]] - h[a[t - 1]], {t, 2, z, 25}], 5]  (* A202537? *)
N[Table[a[n]/a[n - 1], {n, 2, z, 25}], 5]  (* A092526? *)
(* Peter J. C. Moses, Jul 23 2013 *)

a(n) = 2*a(n-1) - a(n-2) + a(n-3) - a(n-4) (conjectured).

G.f.: (7 - 3 x + 2 x^2 - 4 x^3)/(1 - 2 x + x^2 - x^3 + x^4) (conjectured).

A227965 a(1) = least k such that 1 + 1/2 < H(k) - H(2); a(2) = least k such that H(a(1)) - 1/2 < H(k) -H(a(1)), and for n > 2, a(n) = least k such that H(a(n-1)) - H(a(n-2)) > H(k) - H(a(n-1)), where H = harmonic number.

Original entry on oeis.org

11, 53, 249, 1164, 5435, 25371, 118428, 552798, 2580343, 12044484, 56221045, 262427666, 1224955522, 5717827134, 26689578960, 124581175389, 581517950673, 2714399875409, 12670230858892, 59141894115145, 276061555506087, 1288595564424512, 6014885070144844

Offset: 1

Clark Kimberling, Aug 01 2013

Suppose that x and y are positive integers and that x <=y. Let a(1) = least k such that H(y) - H(x-1) < H(k) - H(y); let a(2) = least k such that H(a(1)) - H(y) < H(k) - H(a(1)); and for n > 2, let a(n) = least k such that greatest such H(a(n-1)) - H(a(n-2)) < H(k) - H(a(n-1)). The increasing sequences H(a(n)) - H(a(n-1)) and a(n)/a(n-1) converge. For what choices of (x,y) is the sequence a(n) linearly recurrent?

For A227965, (x,y) = (1,2); H(a(n)) - H(a(n-1)) approaches a limit 1.540684... given by A227966, and a(n)/a(n-1) approaches a limit 4.6677834... given by A227967. It is unknown whether the sequence a(n) is linearly recurrent.

			The first two values (a(1),a(2)) = (11,53) match the beginning of the following inequality chain (and partition of the harmonic numbers):  1/1 + 1/2 < 1/3 + ... + 1/11 < 1/12 + ... + 1/53 < ...

Clark Kimberling, Table of n, a(n) for n = 1..100

Cf. A224820, A224868, A227728, A227966, A227967.

z = 300; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 1; y = 2;
a[1] = Ceiling[w /. FindRoot[h[w] == 2 h[y] - h[x - 1], {w, 1}, WorkingPrecision -> 400]]; a[2] = Ceiling[w /. FindRoot[h[w] == 2 h[a[1]] - h[y], {w, a[1]}, WorkingPrecision -> 400]]; Do[s = 0; a[t] = Ceiling[w /. FindRoot[h[w] == 2 h[a[t - 1]] - h[a[t - 2]], {w, a[t - 1]}, WorkingPrecision -> 400]], {t, 3, z}];
m = Map[a, Range[z]] (* A227965 *)
t = N[Table[h[a[t]] - h[a[t - 1]], {t, 2, z, 25}], 60]
Last[RealDigits[t, 10]]  (* A227966 *)
t = N[Table[a[t]/a[t - 1], {t, 2, z, 50}], 60]
Last[RealDigits[t, 10]]  (* A227967 *)
(* A227965,  Peter J. C. Moses, Jul 12 2013*)

A227653 a(1) = least k such that 1/2 + 1/3 < H(k) - H(3); a(2) = least k such that H(a(1)) - H(3) < H(k) -H(a(1)), and for n > 2, a(n) = least k such that H(a(n-1)) - H(a(n-2)) > H(k) - H(a(n-1)), where H = harmonic number.

Original entry on oeis.org

8, 21, 54, 138, 352, 897, 2285, 5820, 14823, 37752, 96148, 244872, 623645, 1588311, 4045140, 10302237, 26237926, 66823230, 170186624, 433434405, 1103878665, 2811378360, 7160069791, 18235396608, 46442241368, 118279949136, 301237536249, 767197263003

Offset: 1

Clark Kimberling, Aug 02 2013

Suppose that x and y are positive integers and that x <=y. Let a(1) = least k such that H(y) - H(x-1) < H(k) - H(y); let a(2) = least k such that H(a(1)) - H(y) < H(k) - H(a(1)); and for n > 2, let a(n) = least k such that greatest such H(a(n-1)) - H(a(n-2)) < H(k) - H(a(n-1)). The increasing sequences H(a(n)) - H(a(n-1)) and a(n)/a(n-1) converge. For what choices of (x,y) is the sequence a(n) linearly recurrent?

For A227965, (x,y) = (2,3); H(a(n)) - H(a(n-1)) approaches a limit 0.9348448455..., and a(n)/a(n-1) approaches a limit 2.546818276...

			The first two values (a(1),a(2)) = (8,21) match the beginning of the following inequality chain:  1/2 + 1/3 < 1/4 + ... + 1/8 < 1/9 + ... + 1/21 < ...

Clark Kimberling, Table of n, a(n) for n = 1..100

Cf. A224820, A224965.

z = 100; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 2; y = 3;
a[1] = Ceiling[w /. FindRoot[h[w] == 2 h[y] - h[x - 1], {w, 1}, WorkingPrecision -> 400]]; a[2] = Ceiling[w /. FindRoot[h[w] == 2 h[a[1]] - h[y], {w, a[1]}, WorkingPrecision -> 400]]; Do[s = 0; a[t] = Ceiling[w /. FindRoot[h[w] == 2 h[a[t - 1]] - h[a[t - 2]], {w, a[t - 1]}, WorkingPrecision -> 400]], {t, 3, z}];
m = Map[a, Range[z]] (* A227653, Peter J. C. Moses, Jul 12 2013 *)

a(n) = A077849(n+1) (conjectured).

a(n) = 3*a(n-1) - a(n-2) - a(n-4) (conjectured).G.f.: (8 - 3 x - x^2 - 3 x^3)/(1 - 3 x + x^2 + x^4) (conjectured).

A227816 a(1) = greatest k such that H(k) - H(6) < H(6) - H(3); a(2) = greatest k such that H(k) - H(a(1)) < H(a(1)) - H(6), and for n > 2, a(n) = greatest k such that H(k) - H(a(n-1)) > H(a(n-1)) - H(a(n-2)), where H = harmonic number.

Original entry on oeis.org

16, 41, 103, 257, 640, 1592, 3958, 9839, 24457, 60792, 151107, 375596, 933591, 2320556, 5768028, 14337143, 35636731, 88579473, 220175161, 547272407, 1360312788, 3381224518, 8404448844, 20890289891, 51925381404, 129066913288, 320811665802, 797416799492

Offset: 1

Clark Kimberling, Jul 31 2013

Suppose that x and y are positive integers and that x <=y. Let c(1) = y and c(2) = greatest k such that H(k) - H(y) < H(y) - H(x); for n > 2, let c(n) = greatest such that H(k) - H(c(n-1)) < H(c(n-1)) - H(c(n-2)). Then 1/x + ... + 1/c(1) > 1/(c(1)+1) + ... + 1/(c(2)) > 1/(c(2)+1) + ... + 1/(c(3)) > ... The decreasing sequences H(c(n)) - H(c(n-1)) and c(n)/c(n-1) converge. It appears that for many choices of (x,y), the sequence c(n) is linearly recurrent with signature of length less than 10; (3,6) is not one of them. H(c(n)) - H(c(n-1)) and c(n)/c(n-1) approach limits given by A227817 and A227818.

			The first two values (a(1),a(2)) = (16,41) match the beginning of the following inequality chain (and partition of the harmonic numbers H(n) for n >= 3 ):
1/3 + 1/4 + 1/5 + 1/6 > 1/7 + ... + 1/16 < 1/17 + ... + 1/41 <  ...

Clark Kimberling, Table of n, a(n) for n = 1..1000

Cf. A001008, A002805 (numerator and denominator of harmonic numbers).

Cf. A224820, A224868, A227728, A227817, A227818.

z = 100; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 3; y = 6; a[1] = -1 + Ceiling[w /. FindRoot[h[w] == 2 h[y] - h[x - 1], {w, 1}, WorkingPrecision -> 400]]; a[2] = -1 + Ceiling[w /. FindRoot[h[w] == 2 h[a[1]] - h[y], {w, a[1]}, WorkingPrecision -> 400]]; Do[s = 0; a[t] = -1 + Ceiling[w /. FindRoot[h[w] == 2 h[a[t - 1]] - h[a[t - 2]], {w, a[t - 1]}, WorkingPrecision -> 400]], {t, 3, z}]; m = Map[a, Range[z]] (* (A227804) Peter J. C. Moses, Jul 23 2013 *)

A228016 a(1) = least k such that 1/1+1/2+1/3+1/4+1/5 < H(k) - H(5); a(2) = least k such that H(a(1)) - H(5) < H(k) -H(a(1)), and for n > 2, a(n) = least k such that H(a(n-1)) - H(a(n-2)) > H(k) - H(a(n-1)), where H = harmonic number.

Original entry on oeis.org

54, 539, 5340, 52865, 523314, 5180279, 51279480, 507614525, 5024865774, 49741043219, 492385566420, 4874114620985, 48248760643434, 477613491813359, 4727886157490160, 46801248083088245, 463284594673392294, 4586044698650834699, 45397162391834954700

Offset: 1

Clark Kimberling, Aug 02 2013

Suppose that x and y are positive integers and that x <=y. Let a(1) = least k such that H(y) - H(x-1) < H(k) - H(y); let a(2) = least k such that H(a(1)) - H(y) < H(k) - H(a(1)); and for n > 2, let a(n) = least k such that greatest such H(a(n-1)) - H(a(n-2)) < H(k) - H(a(n-1)). The increasing sequences H(a(n)) - H(a(n-1)) and a(n)/a(n-1) converge. For what choices of (x,y) is the sequence a(n) linearly recurrent?

For A227965, (x,y) = (1,5); H(a(n)) - H(a(n-1)) approaches a limit 2.29243166956117768780078... and a(n)/a(n-1) approaches a limit 0.8989794855663561963945681494117

			The first two values (a(1),a(2)) = (54,539) match the beginning of the following inequality chain:  1/1+1/2+1/3+1/4+1/5 < 1/6+...+1/54 < 1/55+...+1/539 < ...

Clark Kimberling, Table of n, a(n) for n = 1..100

Cf. A224820, A224965.

z = 100; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 1; y = 5;
a[1] = Ceiling[w /. FindRoot[h[w] == 2 h[y] - h[x - 1], {w, 1}, WorkingPrecision -> 400]]; a[2] = Ceiling[w /. FindRoot[h[w] == 2 h[a[1]] - h[y], {w, a[1]}, WorkingPrecision -> 400]]; Do[s = 0; a[t] = Ceiling[w /. FindRoot[h[w] == 2 h[a[t - 1]] - h[a[t - 2]], {w, a[t - 1]}, WorkingPrecision -> 400]], {t, 3, z}];
m = Map[a, Range[z]] (* A227653, Peter J. C. Moses, Jul 12 2013 *)

a(n) = 11*a(n-1) - 11*a(n-2) + a(n-3) (conjectured).

G.f.: (-54 + 55 x - 5 x^2)/(-1 + 11 x - 11 x^2 + x^3) (conjectured)

A227804 a(1) = greatest k such that H(k) - H(8) < H(8) - H(4); a(2) = greatest k such that H(k) - H(a(1)) < H(a(1)) - H(8), and for n > 2, a(n) = greatest k such that H(k) - H(a(n-1)) > H(a(n-1)) - H(a(n-2)), where H = harmonic number.

Original entry on oeis.org

15, 27, 48, 85, 150, 264, 464, 815, 1431, 2512, 4409, 7738, 13580, 23832, 41823, 73395, 128800, 226029, 396654, 696080, 1221536, 2143647, 3761839, 6601568, 11584945, 20330162, 35676948, 62608680, 109870575, 192809419, 338356944, 593775045, 1042002566

Offset: 1

Clark Kimberling, Jul 31 2013

Suppose that x and y are positive integers and that x <=y. Let c(1) = y and c(2) = greatest k such that H(k) - H(y) < H(y) - H(x); for n > 2, let c(n) = greatest such that H(k) - H(c(n-1)) < H(c(n-1)) - H(c(n-2)). Then 1/x + ... + 1/c(1) > 1/(c(1)+1) + ... + 1/(c(2)) > 1/(c(2)+1) + ... + 1/(c(3)) > ... The decreasing sequences H(c(n)) - H(c(n-1)) and c(n)/c(n-1) converge. For what choices of (x,y) is the sequence c(n) linearly recurrent?

For A227804, (x,y) = (5,8); it appears that the sequence a(n) is linearly recurrent with signature (3,-3,2,-1), that H(c(n)) - H(c(n-1)) approaches a limit 0.56239..., and that c(n)/c(n-1) approaches the constant 1.75487... given at A109134.

			The first three values (a(1),a(2),a(3)) = (10,43,179) match the beginning of the following inequality chain (and partition of the harmonic numbers H(n) for n >= 5 ):  1/5 + 1/6 + 1/7 + 1/8 > 1/9 + ... + 1/15 < 1/16 + ... + 1/27 < 1/28 + ... + 1/48 > ...

Clark Kimberling, Table of n, a(n) for n = 1..1000

Cf. A001008, A002805 (numerator and denominator of harmonic numbers).

Cf. A224820, A224868, A227728.

z = 100; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 3; y = 5; a[1] = -1 + Ceiling[w /. FindRoot[h[w] == 2 h[y] - h[x - 1], {w, 1}, WorkingPrecision -> 400]]; a[2] = -1 + Ceiling[w /. FindRoot[h[w] == 2 h[a[1]] - h[y], {w, a[1]}, WorkingPrecision -> 400]]; Do[s = 0; a[t] = -1 + Ceiling[w /. FindRoot[h[w] == 2 h[a[t - 1]] - h[a[t - 2]], {w, a[t - 1]}, WorkingPrecision -> 400]], {t, 3, z}]; m = Map[a, Range[z]] (* A227804, Peter J. C. Moses, Jul 23 2013 *)

a(n) = 3*a(n-1) - 3*a(n-2) + 2*a(n-3) - a(n-4) (conjectured).

G.f.: (15 - 18 x + 12 x^2 - 8 x^3)/(1 - 3 x + 3 x^2 - 2 x^3 + x^4) (conjectured).

A225605 a(1) = least k such that 1/3 < H(k) - 1/3; a(2) = least k such that H(a(1)) - H(3) < H(k) - H(a(1)), and for n > 2, a(n) = least k such that H(a(n-1)) - H(a(n-2)) > H(k) - H(a(n-1)), where H = harmonic number.

Original entry on oeis.org

5, 9, 16, 29, 53, 97, 178, 327, 601, 1105, 2032, 3737, 6873, 12641, 23250, 42763, 78653, 144665, 266080, 489397, 900141, 1655617, 3045154, 5600911, 10301681, 18947745, 34850336, 64099761, 117897841, 216847937, 398845538, 733591315, 1349284789, 2481721641

Offset: 1

Clark Kimberling, Aug 03 2013

Suppose that x and y are positive integers and that x <=y. Let a(1) = least k such that H(y) - H(x-1) < H(k) - H(y); let a(2) = least k such that H(a(1)) - H(y) < H(k) - H(a(1)); and for n > 2, let a(n) = least k such that greatest such H(a(n-1)) - H(a(n-2)) < H(k) - H(a(n-1)). The increasing sequences H(a(n)) - H(a(n-1)) and a(n)/a(n-1) converge. For what choices of (x,y) is the sequence a(n) linearly recurrent?

For A225605, (x,y) = (3,3); it appears that H(a(n)) - H(a(n-1)) approaches 0.60937786343... ; it is conjectured a(n)/a(n-1) approaches the constant given at A058265..

			The first two values (a(1),a(2)) = (5,9) match the beginning of the following inequality chain:
  1/3 < 1/4 + 1/5 < 1/6 + 1/7 + 1/8 + 1/9 < ...

Clark Kimberling, Table of n, a(n) for n = 1..300

Cf. A224820, A224965, A228016.

z = 100; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 3; y = 3;
a[1] = Ceiling[w /. FindRoot[h[w] == 2 h[y] - h[x - 1], {w, 1}, WorkingPrecision -> 400]]; a[2] = Ceiling[w /. FindRoot[h[w] == 2 h[a[1]] - h[y], {w, a[1]}, WorkingPrecision -> 400]]; Do[s = 0; a[t] = Ceiling[w /. FindRoot[h[w] == 2 h[a[t - 1]] - h[a[t - 2]], {w, a[t - 1]}, WorkingPrecision -> 400]], {t, 3, z}];
m = Map[a, Range[z]] (* A225605, Peter J. C. Moses, Jul 12 2013 *)

a(n) = A192804(n+4)  (conjectured).
a(n) = 2*a(n-1) - a(n-4)  (conjectured).
G.f.: (5 - x - 2 x^2 - 3 x^3)/(1 - 2 x + x^4)  (conjectured)

A228025 a(1) = least k such that 1/2+1/3+1/4+1/5 < H(k) - H(5); a(2) = least k such that H(a(1)) - H(5) < H(k) -H(a(1)), and for n > 2, a(n) = least k such that H(a(n-1)) - H(a(n-2)) > H(k) - H(a(n-1)), where H = harmonic number.

Original entry on oeis.org

20, 76, 285, 1065, 3976, 14840, 55385, 206701, 771420, 2878980, 10744501, 40099025, 149651600, 558507376, 2084377905, 7779004245, 29031639076, 108347552060, 404358569165, 1509086724601, 5631988329240, 21018866592360, 78443478040201, 292755045568445

Offset: 1

Clark Kimberling, Aug 03 2013

Suppose that x and y are positive integers and that x <=y. Let a(1) = least k such that H(y) - H(x-1) < H(k) - H(y); let a(2) = least k such that H(a(1)) - H(y) < H(k) - H(a(1)); and for n > 2, let a(n) = least k such that greatest such H(a(n-1)) - H(a(n-2)) < H(k) - H(a(n-1)). The increasing sequences H(a(n)) - H(a(n-1)) and a(n)/a(n-1) converge. For what choices of (x,y) is the sequence a(n) linearly recurrent?

For A228025, (x,y) = (2,5); it appears that H(a(n)) - H(a(n-1)) approaches log(2 + sqrt(3)) and that and a(n)/a(n-1) approaches sqrt(3).

			The first two values (a(1),a(2)) = (20,76) match the beginning of the following inequality chain:  1/2+1/3+1/4+1/5 < 1/6+...+1/20 < 1/21+...+1/76 < ...

Clark Kimberling, Table of n, a(n) for n = 1..100

Cf. A224820, A224965, A228016.

z = 100; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 2; y = 5;
a[1] = Ceiling[w /. FindRoot[h[w] == 2 h[y] - h[x - 1], {w, 1}, WorkingPrecision -> 400]]; a[2] = Ceiling[w /. FindRoot[h[w] == 2 h[a[1]] - h[y], {w, a[1]},      WorkingPrecision -> 400]]; Do[s = 0; a[t] = Ceiling[w /. FindRoot[h[w] == 2 h[a[t - 1]] - h[a[t - 2]], {w, a[t - 1]}, WorkingPrecision -> 400]], {t, 3, z}];
m = Map[a, Range[z]] (* A227653, Peter J. C. Moses, Jul 12 2013 *)

a(n) = A061278(n+1) (conjectured).
a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3) (conjectured).
G.f.: (-20 + 24 x - 5 x^2)/(-1 + 5 x - 5 x^2 + x^3) (conjectured).

A364200 Minimal number of terms of mixed-sign Egyptian fraction f such that H(n) + f is an integer, where H(n) is the n-th harmonic number.

Original entry on oeis.org

0, 1, 1, 1, 2, 2, 3, 3, 3, 2, 2, 3, 4, 3, 4, 4, 4, 4, 4, 5, 5, 4, 5, 5, 5, 4, 5, 5, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6

Offset: 1

Denis Ivanov, Jul 13 2023

For H(n) - floor(H(n)) and ceiling(H(n)) - H(n), the shortest mixed-sign Egyptian fractions are calculated, and the smaller length of fractions is selected.
Similar to A106394 and A224820. But those sequences use the greedy algorithm, which does not guarantee the shortest length of expansion.
For 1 < n < 41, a(n) < A363937(n) only for n = 10 and n = 22.

			For n=10: H(10) = 7381/2520 = 2.928...; H(10) - floor(H(10)) = 7381/2520 - 2 = 2341/2520 = 1/2 + 1/7 + 1/8 + 1/9 + 1/20, which cannot be expressed as the sum of fewer than 5 reciprocals, and ceiling(H(10)) - H(10) = 3 - 7381/2520 = 179/2520 = 1/30 + 1/42 + 1/72, which cannot be expressed as the sum of fewer than 3 reciprocals, so A363937(10) = 3.
But 179/2520 = 1/14 - 1/2520 (a "mixed-sign Egyptian fraction"), so a(10) = 2.

Ron Knott, Egyptian Fractions

Cf. A363937.

check[f_, k_] := (If[Numerator@f == 1, Return@True];
   If[k == 1, Return@False];
   Catch[Do[If[check[f - 1/i, k - 1], Throw@True],
     {i, Range[Ceiling[1/f], Floor[k/f]]}];
    Throw@False]);
checkMixed[f_, k_, m_] := If[m == 1,
   Catch[Do[If[check[1/i - f, k], Throw@True],
     {i, Range[2, Floor[1/f]]}];
    Throw@False],
   checkMixed[f, k, m - 1]];
a[n_] := (h = HarmonicNumber[n];
  d = Min[h - Floor@h, Ceiling@h - h];
  j = 1;
  While[Not@check[d, j], j++];
  res = j;
  Do[
   If[checkMixed[d, i - m, m], res = i],
   {i, 2, j - 1}, {m, 1, i - 1}];
  res);

a(n) <= A363937(n).

cp's OEIS Frontend

A227728 a(1) = greatest k such that H(k) - H(2) < 1/1 + 1/2; a(2) = greatest k such that H(k) - H(a(1)) < H(a(1)) - H(2); and for n>2, a(n) = greatest k such that H(k) - H(a(n-1)) < H(a(n-1)) - H(a(n-2)), where H = harmonic number.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A224868 a(1) = greatest k such that H(k) - H(4) < 1/3 + 1/4; a(2) = greatest k such that H(k) - H(a(1)) < H(a(1)) - H(4); and for n > 2, a(n) = greatest k such that H(k) - H(a(n-1)) > H(a(n-1)) - H(a(n-2)), where H = harmonic number.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A227965 a(1) = least k such that 1 + 1/2 < H(k) - H(2); a(2) = least k such that H(a(1)) - 1/2 < H(k) -H(a(1)), and for n > 2, a(n) = least k such that H(a(n-1)) - H(a(n-2)) > H(k) - H(a(n-1)), where H = harmonic number.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A227653 a(1) = least k such that 1/2 + 1/3 < H(k) - H(3); a(2) = least k such that H(a(1)) - H(3) < H(k) -H(a(1)), and for n > 2, a(n) = least k such that H(a(n-1)) - H(a(n-2)) > H(k) - H(a(n-1)), where H = harmonic number.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A227816 a(1) = greatest k such that H(k) - H(6) < H(6) - H(3); a(2) = greatest k such that H(k) - H(a(1)) < H(a(1)) - H(6), and for n > 2, a(n) = greatest k such that H(k) - H(a(n-1)) > H(a(n-1)) - H(a(n-2)), where H = harmonic number.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A228016 a(1) = least k such that 1/1+1/2+1/3+1/4+1/5 < H(k) - H(5); a(2) = least k such that H(a(1)) - H(5) < H(k) -H(a(1)), and for n > 2, a(n) = least k such that H(a(n-1)) - H(a(n-2)) > H(k) - H(a(n-1)), where H = harmonic number.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A227804 a(1) = greatest k such that H(k) - H(8) < H(8) - H(4); a(2) = greatest k such that H(k) - H(a(1)) < H(a(1)) - H(8), and for n > 2, a(n) = greatest k such that H(k) - H(a(n-1)) > H(a(n-1)) - H(a(n-2)), where H = harmonic number.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A225605 a(1) = least k such that 1/3 < H(k) - 1/3; a(2) = least k such that H(a(1)) - H(3) < H(k) - H(a(1)), and for n > 2, a(n) = least k such that H(a(n-1)) - H(a(n-2)) > H(k) - H(a(n-1)), where H = harmonic number.

Views

Author