A225918 -id:A225918 - cp's OEIS frontend

A134931 a(n) = (5*3^n-3)/2.

1, 6, 21, 66, 201, 606, 1821, 5466, 16401, 49206, 147621, 442866, 1328601, 3985806, 11957421, 35872266, 107616801, 322850406, 968551221, 2905653666, 8716961001, 26150883006, 78452649021, 235357947066, 706073841201, 2118221523606

Offset: 0

Rolf Pleisch, Jan 29 2008

Numbers n where the recurrence s(0)=1, if s(n-1) >= n then s(n) = s(n-1) - n else s(n) = s(n-1) + n produces s(n)=0. - Hugo Pfoertner, Jan 05 2012
A046901(a(n)) = 1. - Reinhard Zumkeller, Jan 31 2013
Binomial transform of A146523: (1, 5, 10, 20, 40, ...) and double binomial transform of A010685: (1, 4, 1, 4, 1, 4, ...). - Gary W. Adamson, Aug 25 2016
Also the number of maximal cliques in the (n+1)-Hanoi graph. - Eric W. Weisstein, Dec 01 2017
a(n) is the least k such that f(a(n-1)+1) + ... + f(k) > f(a(n-2)+1) + ... + f(a(n-1)) for n > 1, where f(n) = 1/(n+1). Because Sum_{k=1..5*3^(n-1)} 1/(a(n)+3*k-1) + 1/(a(n)+3*k) + 1/(a(n)+3*k+1) - 1/((a(n)+1+5*3^n)*5*3^(n-1)) < Sum_{k=1..5*3^(n-1)} 1/(a(n-1)+k+1) < Sum_{k=1..5*3^(n-1)} 1/(a(n)+3*k-1) + 1/(a(n)+3*k) + 1/(a(n)+3*k+1), we have 1 < 1/3 + 1/4 + ... + 1/7 < 1/8 + 1/9 + ... + 1/22 < ... . - Jinyuan Wang, Jun 15 2020

Vincenzo Librandi, Table of n, a(n) for n = 0..500
Eric Weisstein's World of Mathematics, Hanoi Graph
Eric Weisstein's World of Mathematics, Maximal Clique
Index entries for linear recurrences with constant coefficients, signature (4,-3).

Cf. A003462, A007051, A034472, A024023, A067771, A029858.

Cf. A005030, A010685, A046901, A060816, A116952, A146523, A225918.

[(5*3^n-3)/2: n in [0..30]]; // Vincenzo Librandi, Jun 05 2011

seq((5*3^n-3)/2, n= 0..25); # Gary Detlefs, Jun 22 2010

a=1; lst={a}; Do[a=a*3+3; AppendTo[lst,a], {n,0,100}]; lst (* Vladimir Joseph Stephan Orlovsky, Dec 25 2008 *)
Table[(5 3^n - 9)/6, {n, 20}] (* Eric W. Weisstein, Dec 01 2017 *)
(5 3^Range[20] - 9)/6 (* Eric W. Weisstein, Dec 01 2017 *)
LinearRecurrence[{4, -3}, {1, 6}, 20] (* Eric W. Weisstein, Dec 01 2017 *)
CoefficientList[Series[(1 + 2 x)/(1 - 4 x + 3 x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Dec 01 2017 *)

a(n) = (5*3^n-3)/2; /* Joerg Arndt, Apr 14 2013 */

a(n) = 3*(a(n-1) + 1), with a(0)=1.
From R. J. Mathar, Jan 31 2008: (Start)
O.g.f.: (5/2)/(1-3*x) - (3/2)/(1-x).
a(n) = (A005030(n) - 3)/2. (End)
a(n) = A060816(n+1) - 1. - Philippe Deléham, Apr 14 2013
E.g.f.: exp(x)*(5*exp(2*x) - 3)/2. - Stefano Spezia, Aug 28 2023

More terms from Vladimir Joseph Stephan Orlovsky, Dec 25 2008

A224820 Array r(n,m), where r(n,1) = n; r(n,2) = least k such that H(k) - H(n) > 1/n; and for m > 2, r(n,m) = least k such that H(k)-H(r(n,m-1)) > H(r(n,m-1)) - H(r(n,m-2)), where H = harmonic number.

Original entry on oeis.org

1, 4, 2, 13, 4, 3, 40, 8, 5, 4, 121, 16, 9, 6, 5, 364, 32, 16, 9, 7, 6, 1093, 64, 29, 14, 10, 8, 7, 3280, 128, 53, 22, 15, 11, 9, 8, 9841, 256, 97, 35, 23, 16, 12, 10, 9, 29524, 512, 178, 56, 36, 24, 16, 13, 11, 10, 88573, 1024, 327, 90, 57, 36, 22, 17, 14, 12, 11

Offset: 1

Clark Kimberling, Jul 21 2013

For every n, the sequence H(r(n,m)) - H(r(n,m-1)) converges as m -> oo. Which row-sequences are linearly recurrent? Is r(4,m) = 1 + F(m+3), where F = A000045 (Fibonacci numbers)?

More generally, suppose that x and y are positive integers and that x <= y. Let c(1) = y and c(2) = least k such that H(k) - H(y) > H(y) - H(x); for n > 2, let c(n) = least k such that H(k) - H(c(n-1)) > H(c(n-1)) - H(c(n-2)). Thus the Egyptian fractions for m >= x are partitioned, and 1/x + ... + 1/c(1) < 1/(c(1)+1) + ... + 1/(c(2)) < 1/(c(2)+1) + ... + 1/(c(3)) < ... The 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?

			Northwest corner:
       m=1  m=2  m=3  m=4  m=5  m=6   m=7   m=8
  n=1:   1,   4,  13,  40, 121, 364, 1093, 3280
  n=2:   2,   4,   8,  16,  32,  64,  128,  256
  n=3:   3,   5,   9,  16,  29,  53,   97,  178
  n=4:   4,   6,   9,  14,  22,  35,   56,   90
  n=5:   5,   7,  10,  15,  23,  36,   57,   91
  n=6:   6,   8,  11,  16,  24,  36,   54,   81
  n=7:   7,   9,  12,  16,  22,  31,   44,   63
  n=8:   8,  10,  13,  17,  23,  32,   45,   64
The chain indicated by row n=4 is
1/4 < 1/5 + 1/6 < 1/7 + 1/8 + 1/9 < 1/10 + ... + 1/14 < ...

Clark Kimberling, Table of n, a(n) for n = 1..1830

Cf. A225918.

h[n_] := h[n] = HarmonicNumber[N[n, 300]]; z = 12; Table[s = 0; a[1] = NestWhile[# + 1 &, x + 1, ! (s += 1/#) >= h[x] - h[x - 1] &];   s = 0; a[2] = NestWhile[# + 1 &, a[1] + 1, ! (s += 1/#) >= h[a[1]] - h[x] &]; Do[test = h[a[t - 1]] - h[a[t - 2]] + h[a[t - 1]]; s = 0; a[t] = Floor[x /. FindRoot[h[x] == test, {x, a[t - 1]}, WorkingPrecision -> 100]] + 1, {t, 3, z}]; Flatten[{x, Map[a, Range[z]]}], {x, 1, z}] // TableForm (* A224820 array *)
t = Flatten[Table[%[[n - k + 1]][[k]], {n, z}, {k, n, 1, -1}]]; (* A224820 sequence *) (* Peter J. C. Moses, Jul 20 2013 *)

A225919 a(n) is the least k such that f(a(n-1)+1) + ... + f(k) > f(a(n-2)+1) + ... + f(a(n-1)) for n > 1, where f(n) = 1/(n+4) and a(1) = 1.

Original entry on oeis.org

1, 11, 40, 124, 367, 1070, 3104, 8989, 26016, 75280, 217815, 630210, 1823388, 5275597, 15263836, 44162700, 127775471, 369691398, 1069624136, 3094731965, 8953954568, 25906380024, 74954649447, 216865477466, 627454543012, 1815407450477, 5252498763364, 15196997925652

Offset: 1

Clark Kimberling, May 21 2013

nonn

Conjecture: a(n) is linearly recurrent. See A225918 for details.

			a(1) = 1 by decree; a(2) = 11 because 1/6 + ... + 1/14 < 1 < 1/6 + ... + 1/(11+4), so that a(3) = 40 because 1/16 + ... + 1/43 < 1/6 + ... + 1/15 < 1/16 + ... + 1/(40+4).
Successive values of a(n) yield a chain: 1 < 1/(1+5) + ... + 1/(11+4) < 1/(11+5) + ... + 1/(40+4) < 1/(40+5) + ... + 1/(124+4) < ...
Abbreviating this chain as b(1) = 1 < b(2) < b(3) < b(4) < ... < R = 2.8931..., it appears that lim_{n->infinity} b(n) = log(R) = 1.0623... .

Cf. A225918.

nn = 11; f[n_] := 1/(n + 4); a[1] = 1; g[n_] := g[n] = Sum[f[k], {k, 1, n}]; s = 0; a[2] = NestWhile[# + 1 &, 2, ! (s += f[#]) >= a[1] &]; s = 0; a[3] = NestWhile[# + 1 &, a[2] + 1, ! (s += f[#]) >= g[a[2]] - f[1] &]; Do[s = 0; a[z] = NestWhile[# + 1 &, a[z - 1] + 1, ! (s += f[#]) >= g[a[z - 1]] - g[a[z - 2]] &], {z, 4, nn}]; m = Map[a, Range[nn]]

For n>=3, a(n) = ceiling( (a(n-1)+4.5)^2 / (a(n-2)+4.5) - 4.5 ) unless the fractional part of the number inside ceiling() is very small (~ 1/a(n-2)). - Max Alekseyev, Jan 27 2022

a(11)-a(16) from Robert G. Wilson v, May 22 2013
a(17)-a(18) from Robert G. Wilson v, Jun 13 2013
a(19) from Jinyuan Wang, Jun 14 2020
Terms a(20) on from Max Alekseyev, Jan 27 2022

A225920 a(n) is the least k such that f(a(n-1)+1) + ... + f(k) > f(a(n-2)+1) + ... + f(a(n-1)) for n > 1, where f(n) = 1/(n+5) and a(1) = 1.

Original entry on oeis.org

1, 13, 48, 150, 447, 1312, 3831, 11167, 32531, 94748, 275938, 803605, 2340292, 6815476, 19848236, 57802615, 168334451, 490228448, 1427657419, 4157665074, 12108072013, 35261476137, 102689486632, 299055281267, 870917405325, 2536310757258, 7386317253546, 21510645891422

Offset: 1

Clark Kimberling, May 21 2013

nonn

Conjecture: a(n) is linearly recurrent. See A225918 for details.

The sequence does not satisfy any linear recurrence of order below 50, which suggests it's unlikely to exist. - Max Alekseyev, Jan 27 2022

			a(1) = 1 by decree; a(2) = 13 because 1/7 + ... + 1/17 < 1 < 1/7 + ... + 1/(13+5), so that a(3) = 48 because 1/19 + ... + 1/52 < 1/7 + ... + 1/18 < 1/19 + ... + 1/(48+5).
Successive values of a(n) yield a chain: 1 < 1/(1+6) + ... + 1/(13+5) < 1/(13+6) + ... + 1/(48+5) < 1/(48+6) + ... + 1/(150+5) < ...
Abbreviating this chain as b(1) = 1 < b(2) < b(3) < b(4) < ... < R = 2.912229..., it appears that lim_{n->infinity} b(n) = log(R) = 1.068918... .

Cf. A225918.

nn = 11; f[n_] := 1/(n + 5); a[1] = 1; g[n_] := g[n] = Sum[f[k], {k, 1, n}]; s = 0; a[2] = NestWhile[# + 1 &, 2, ! (s += f[#]) >= a[1] &]; s = 0; a[3] = NestWhile[# + 1 &, a[2] + 1, ! (s += f[#]) >= g[a[2]] - f[1] &]; Do[s = 0; a[z] = NestWhile[# + 1 &, a[z - 1] + 1, ! (s += f[#]) >= g[a[z - 1]] - g[a[z - 2]] &], {z, 4, nn}]; m = Map[a, Range[nn]]

For n>=3, a(n) = ceiling( (a(n-1)+5.5)^2 / (a(n-2)+5.5) - 5.5 ) unless the fractional part of the number inside ceiling() is very small (~ 1/a(n-2)). - Max Alekseyev, Jan 27 2022

a(12)-a(16) from Robert G. Wilson v, May 22 2013
a(17)-a(18) from Robert G. Wilson v, Jun 13 2013
a(18) corrected by and a(19) from Jinyuan Wang, Jun 14 2020
Terms a(20) on from Max Alekseyev, Jan 27 2022

A225921 a(n) is the least k such that f(a(n-1)+1) + ... + f(k) > f(a(n-2)+1) + ... + f(a(n-1)) for n > 1, where f(n) = 1/(n+6) and a(1) = 1.

Original entry on oeis.org

1, 14, 50, 150, 427, 1195, 3324, 9226, 25587, 70942, 196672, 545212, 1511411, 4189842, 11614806, 32197786, 89256522, 247430866, 685911016, 1901435842, 5271031028, 14611993445, 40506373648, 112289011899, 311279955644, 862909105217, 2392097886032, 6631210937220, 18382591594862

Offset: 1

Clark Kimberling, May 21 2013

nonn

Conjecture: a(n) is linearly recurrent. See A225918 for details.

The sequence does not satisfy any linear recurrence of order below 50, which suggests it's unlikely to exist. - Max Alekseyev, Jan 27 2022

			a(1) = 1 by decree; a(2) = 14 because 1/8 + ... + 1/19 < 1 < 1/8 + ... + 1/(14+6), so that a(3) = 50 because 1/21 + ... + 1/55 < 1/8 + ... + 1/20 < 1/21 + ... + 1/(50+6).
Successive values of a(n) yield a chain: 1 < 1/(1+7) + ... + 1/(14+6) < 1/(14+7) + ... + 1/(50+6) < 1/(50+7) + ... + 1/(150+6) < ...
Abbreviating this chain as b(1) = 1 < b(2) < b(3) < b(4) < ... < R = 2.7721..., it appears that lim_{n->infinity} b(n) = log(R) = 1.0196... .

Cf. A225918.

nn = 11; f[n_] := 1/(n + 6); a[1] = 1; g[n_] := g[n] = Sum[f[k], {k, 1, n}]; s = 0; a[2] = NestWhile[# + 1 &, 2, ! (s += f[#]) >= a[1] &];
s = 0; a[3] = NestWhile[# + 1 &, a[2] + 1, ! (s += f[#]) >= g[a[2]] - f[1] &]; Do[s = 0; a[z] = NestWhile[# + 1 &, a[z - 1] + 1, ! (s += f[#]) >= g[a[z - 1]] - g[a[z - 2]] &], {z, 4, nn}]; m = Map[a, Range[nn]]

For n>=3, a(n) = ceiling( (a(n-1)+6.5)^2 / (a(n-2)+6.5) - 6.5 ) unless the fractional part of the number inside ceiling() is very small (~ 1/a(n-2)). - Max Alekseyev, Jan 27 2022

a(13)-a(16) from Robert G. Wilson v, May 22 2013
a(17)-a(18) from Robert G. Wilson v, Jun 13 2013
a(19) from Jinyuan Wang, Jun 14 2020
Terms a(20) on from Max Alekseyev, Jan 27 2022

A225922 a(n) is the least k such that f(a(n-1)+1) + ... + f(k) > f(a(n-2)+1) + ... + f(a(n-1)) for n > 1, where f(n) = 1/(n+7) and a(1) = 1.

Original entry on oeis.org

1, 16, 58, 176, 507, 1436, 4043, 11359, 31890, 89506, 251193, 704933, 1978258, 5551574, 15579326, 43720081, 122691130, 344306598, 966223316, 2711500429, 7609249779, 21353742568, 59924740904, 168166051476, 471922288540, 1324349620283, 3716505787761, 10429583743512

Offset: 1

Clark Kimberling, May 21 2013

nonn

Conjecture: a(n) is linearly recurrent. See A225918 for details.

The sequence does not satisfy any linear recurrence of order below 50, which suggests it's unlikely to exist. - Max Alekseyev, Jan 27 2022

			a(1) = 1 by decree; a(2) = 15 because 1/9 + ... + 1/21 < 1 < 1/9 + ... + 1/(15+7), so that a(3) = 58 because 1/23 + ... + 1/57 < 1/9 + ... + 1/22 < 1/23 + ... + 1/(58+7).
Successive values of a(n) yield a chain: 1 < 1/(1+8) + ... + 1/(15+7) < 1/(15+8) + ... + 1/(58+7) < 1/(58+8) + ... + 1/(176+7) < ...
Abbreviating this chain as b(1) = 1 < b(2) < b(3) < b(4) < ... < R = 2.80628..., it appears that lim_{n->infinity} b(n) = log(R) = 1.03186... .

Cf. A225918.

nn = 11; f[n_] := 1/(n + 7); a[1] = 1; g[n_] := g[n] = Sum[f[k], {k, 1, n}]; s = 0; a[2] = NestWhile[# + 1 &, 2, ! (s += f[#]) >= a[1] &]; s = 0; a[3] = NestWhile[# + 1 &, a[2] + 1, ! (s += f[#]) >= g[a[2]] - f[1] &]; Do[s = 0; a[z] = NestWhile[# + 1 &, a[z - 1] + 1, ! (s += f[#]) >= g[a[z - 1]] - g[a[z - 2]] &], {z, 4, nn}]; m = Map[a, Range[nn]]

For n>=3, a(n) = ceiling( (a(n-1)+7.5)^2 / (a(n-2)+7.5) - 7.5 ) unless the fractional part of the number inside ceiling() is very small (~ 1/a(n-2)). - Max Alekseyev, Jan 27 2022

a(10)-a(17) from Robert G. Wilson v, May 22 2013
a(18) from Robert G. Wilson v, Jun 13 2013
a(19) from Jinyuan Wang, Jun 14 2020
Terms a(20) on from Max Alekseyev, Jan 27 2022

cp's OEIS Frontend

A134931 a(n) = (5*3^n-3)/2.

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A224820 Array r(n,m), where r(n,1) = n; r(n,2) = least k such that H(k) - H(n) > 1/n; and for m > 2, r(n,m) = least k such that H(k)-H(r(n,m-1)) > H(r(n,m-1)) - H(r(n,m-2)), where H = harmonic number.

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A225919 a(n) is the least k such that f(a(n-1)+1) + ... + f(k) > f(a(n-2)+1) + ... + f(a(n-1)) for n > 1, where f(n) = 1/(n+4) and a(1) = 1.

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A225920 a(n) is the least k such that f(a(n-1)+1) + ... + f(k) > f(a(n-2)+1) + ... + f(a(n-1)) for n > 1, where f(n) = 1/(n+5) and a(1) = 1.

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A225921 a(n) is the least k such that f(a(n-1)+1) + ... + f(k) > f(a(n-2)+1) + ... + f(a(n-1)) for n > 1, where f(n) = 1/(n+6) and a(1) = 1.

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A225922 a(n) is the least k such that f(a(n-1)+1) + ... + f(k) > f(a(n-2)+1) + ... + f(a(n-1)) for n > 1, where f(n) = 1/(n+7) and a(1) = 1.

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