A248630 -id:A248630 - cp's OEIS frontend

A248632 Numbers k such that A248631(k+1) = A248631(k).

6, 10, 14, 17, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 77, 80, 83, 86, 89, 92, 95, 97, 100, 103, 106, 109, 111, 114, 117, 120, 123, 126, 128, 131, 134, 137, 140, 142, 145, 148, 151, 154, 156, 159, 162, 165, 168, 170, 173

Offset: 1

Clark Kimberling, Oct 11 2014

			(A248631(k+1) = A248631(k)) = (1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1,... ), so that A248632 = (6, 10, 14, ... ).

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

Cf. A248631, A248630.

z = 200; p[k_] := p[k] = Sum[(h^2/2^h), {h, 1, k}];
d = N[Table[6 - p[k], {k, 1, z/5}], 12];
f[n_] := f[n] = Select[Range[z], 6 - p[#] < 1/3^n &, 1];
u = Flatten[Table[f[n], {n, 1, z}]]; (* A248629 *)
d = Differences[u];
Flatten[Position[d, 1]];  (* A248630 *)

A248629 Least k such that 6 - sum{(h^2)/2^h, h = 1..k} < 1/3^n.

Original entry on oeis.org

9, 11, 13, 15, 17, 19, 21, 22, 24, 26, 28, 29, 31, 33, 35, 36, 38, 40, 41, 43, 45, 47, 48, 50, 52, 53, 55, 57, 58, 60, 62, 63, 65, 67, 68, 70, 72, 73, 75, 76, 78, 80, 81, 83, 85, 86, 88, 90, 91, 93, 95, 96, 98, 99, 101, 103, 104, 106, 108, 109, 111, 112, 114

Offset: 1

Clark Kimberling, Oct 10 2014

This sequence provides insight into the manner of convergence of sum{(h^2)/2^h, h = 1..k} to 6.

			Let s(n) = 6 - sum{(h^2)/2^h, h = 1..n}.  Approximations follow:
n ... s(n) ........ 1/3^n
1 ... 5.50000 ... 0.333333
2 ... 4.50000 ... 0.111111
3 ... 3.37500 ... 0.037037
4 ... 2.37500 ... 0.012345
5 ... 1.59375 ... 0.004115
6 ... 1.03125 ... 0.001371
7 ... 0.64843 ... 0.000457
8 ... 0.39843 ... 0.000152
9 ... 0.24023 ... 0.000050
10 .. 0.14257 ... 0.000018
11 .. 0.08349 ... 0.000006
a(2) = 11 because s(11) < 1/9 < s(10).

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

Cf. A248630, A248631.

z = 300; p[k_] := p[k] = Sum[(h^2/2^h), {h, 1, k}]
d = N[Table[6 - p[k], {k, 1, z/5}], 12]
f[n_] := f[n] = Select[Range[z], 6 - p[#] < 1/3^n &, 1]
u = Flatten[Table[f[n], {n, 1, z}]]  (* A248629 *)
d = Differences[u]
v = Flatten[Position[d, 1]]  (* A248630 *)

A248635 Numbers k such that A248633(k+1) = A248633(k) + 2.

Original entry on oeis.org

1, 2, 3, 4, 6, 7, 9, 10, 12, 14, 15, 17, 19, 21, 22, 24, 26, 28, 30, 32, 33, 35, 37, 39, 41, 43, 45, 47, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99, 100, 102, 104, 106, 108, 110, 112, 114, 116

Offset: 1

Clark Kimberling, Oct 11 2014

			(A248633(k+1) = A248633(k)) = (2, 2, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1,... ), so that A248634 = (5, 8, 11, 13, ... ) and A248635 = (1, 2, 3, 4, 6, 7, 9, 10, 12, ...).

Clark Kimberling, Table of n, a(n) for n = 1..500

Cf. A248633, A248634, A248630.

z= 300; p[k_] := p[k] = Sum[(h^2/4^h), {h, 1, k}];
d = N[Table[20/27 - p[k], {k, 1, z/5}], 12];
f[n_] := f[n] = Select[Range[z], 20/27 - p[#] < 1/8^n &, 1];
u = Flatten[Table[f[n], {n, 1, z}]]  (* A248633 *)
d = Differences[u]
Flatten[Position[d, 1]]  (* A248634 *)
Flatten[Position[d, 2]]  (* A248635 *)

A248631 Least k such that 3/2 - sum{(h^2)/3^h, h = 1..k} < 1/2^n.

Original entry on oeis.org

3, 4, 5, 6, 7, 8, 8, 9, 10, 11, 11, 12, 13, 14, 14, 15, 16, 16, 17, 18, 19, 19, 20, 21, 21, 22, 23, 23, 24, 25, 25, 26, 27, 27, 28, 29, 29, 30, 31, 31, 32, 33, 33, 34, 35, 35, 36, 37, 37, 38, 39, 39, 40, 41, 41, 42, 43, 43, 44, 45, 45, 46, 47, 47, 48, 49, 49

Offset: 1

Clark Kimberling, Oct 11 2014

This sequence provides insight into the manner of convergence of sum{(h^2)/3^h, h = 1..k} to 3/2.

			Let s(n) = 3/2 - sum{(h^2)/3^h, h = 1..n}.  Approximations follow:
n ... s(n) ...... 1/2^n
1 ... 1.16666 ... 0.500000
2 ... 0.72222 ... 0.250000
3 ... 0.38888 ... 0.125000
4 ... 0.03909 ... 0.062500
5 ... 0.08847 ... 0.031250
6 ... 0.03909 ... 0.015625
7 ... 0.01668 ... 0.007812
a(5) = 7 because s(7) < 1/32 < s(6).

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

Cf. A248632, A248630.

z = 200; p[k_] := p[k] = Sum[(h^2/2^h), {h, 1, k}];
d = N[Table[6 - p[k], {k, 1, z/5}], 12];
f[n_] := f[n] = Select[Range[z], 6 - p[#] < 1/3^n &, 1];
u = Flatten[Table[f[n], {n, 1, z}]]; (* A248629 *)
d = Differences[u];
Flatten[Position[d, 1]];  (* A248630 *)

A248633 Least k such that 20/27- sum{(h^2)/4^h, h = 1..k} < 1/8^n.

Original entry on oeis.org

3, 5, 7, 9, 11, 12, 14, 16, 17, 19, 21, 22, 24, 25, 27, 29, 30, 32, 33, 35, 36, 38, 40, 41, 43, 44, 46, 47, 49, 50, 52, 53, 55, 57, 58, 60, 61, 63, 64, 66, 67, 69, 70, 72, 73, 75, 76, 78, 80, 81, 83, 84, 86, 87, 89, 90, 92, 93, 95, 96, 98, 99, 101, 102, 104

Offset: 1

Clark Kimberling, Oct 11 2014

This sequence provides insight into the manner of convergence of sum{(h^2)/4^h, h = 1..k} to 20/27. The difference sequence of A248633 entirely of 1s and 2s, so that A248634 and A248635 partition the positive integers.

			Let s(n) = 20/27 - sum{(h^2)/4^h, h = 1..n}.  Approximations follow:
n ... s(n) ........ 1/8^n
1 ... 0.49074 ... 0.125000
2 ... 0.24074 ... 0.015625
3 ... 0.10011 ... 0.001953
4 ... 0.03761 ... 0.000244
5 ... 0.01320 ... 0.000030
a(2) = 5 because s(5) < 1/8^2 < s(2).

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

Cf. A248632, A248630.

z= 300; p[k_] := p[k] = Sum[(h^2/4^h), {h, 1, k}];
d = N[Table[20/27 - p[k], {k, 1, z/5}], 12];
f[n_] := f[n] = Select[Range[z], 20/27 - p[#] < 1/8^n &, 1];
u = Flatten[Table[f[n], {n, 1, z}]]  (* A248633 *)
d = Differences[u]
Flatten[Position[d, 1]]  (* A248634 *)
Flatten[Position[d, 2]]  (* A248635 *)

A248634 Numbers k such that A248633(k+1) = A248633(k) + 1.

Original entry on oeis.org

5, 8, 11, 13, 16, 18, 20, 23, 25, 27, 29, 31, 34, 36, 38, 40, 42, 44, 46, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 101, 103, 105, 107, 109, 111, 113, 115, 117, 119, 121, 123, 125, 127, 129, 131, 133

Offset: 1

Clark Kimberling, Oct 11 2014

			(A248633(k+1) = A248633(k)) = (2, 2, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1,... ), so that A248634 = (5, 8, 11, 13, ... ) and A248635 = (1, 2, 3, 4, 6, 7, 9, 10, 12, ...).

Clark Kimberling, Table of n, a(n) for n = 1..500

Cf. A248633, A248635, A248630.

z= 300; p[k_] := p[k] = Sum[(h^2/4^h), {h, 1, k}];
d = N[Table[20/27 - p[k], {k, 1, z/5}], 12];
f[n_] := f[n] = Select[Range[z], 20/27 - p[#] < 1/8^n &, 1];
u = Flatten[Table[f[n], {n, 1, z}]]  (* A248633 *)
d = Differences[u]
Flatten[Position[d, 1]]  (* A248634 *)
Flatten[Position[d, 2]]  (* A248635 *)

A248636 Least k such that 33/8- sum{(h^3)/3^h, h = 1..k} < 1/4^n.

Original entry on oeis.org

7, 9, 10, 12, 13, 15, 17, 18, 19, 21, 22, 24, 25, 27, 28, 29, 31, 32, 34, 35, 36, 38, 39, 40, 42, 43, 44, 46, 47, 48, 50, 51, 52, 54, 55, 56, 58, 59, 60, 62, 63, 64, 66, 67, 68, 70, 71, 72, 73, 75, 76, 77, 79, 80, 81, 83, 84, 85, 87, 88, 89, 90, 92, 93, 94

Offset: 1

Clark Kimberling, Oct 11 2014

This sequence provides insight into the manner of convergence of sum{(h^3)/3^h, h = 1..k} to 33/8. The difference sequence of A248636 entirely of 1s and 2s, so that A248637 and A248638 partition the positive integers.

			Let s(n) = 33/8- sum{(h^3)/3^h, h = 1..n}.  Approximations follow:
n ... s(n) ........ 1/4^n
1 ... 3.79167 ... 0.250000
2 ... 2.90278 ... 0.062500
3 ... 1.90278 ... 0.015625
4 ... 1.11265 ... 0.003906
5 ... 0.59825 ... 0.000976
6 ... 0.30195 ... 0.000244
7 ... 0.14511 ... 0.000061
8 ... 0.06798 ... 0.000015
9 ... 0.03004 ... 0.000004
a(2) = 9 because s(9) < 1/16 < s(8).

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

Cf. A248632, A248630.

z = 300; p[k_] := p[k] = Sum[(h^3/3^h), {h, 1, k}];
d = N[Table[33/8 - p[k], {k, 1, z/5}], 12]
f[n_] := f[n] = Select[Range[z], 33/8 - p[#] < 1/4^n &, 1];
u = Flatten[Table[f[n], {n, 1, z}]]  (* A248636 *)
d = Differences[u]
v = Flatten[Position[d, 1]] (* A248637 *)
w = Flatten[Position[d, 2]] (* A248638 *)

A248637 Numbers k such that A248636(k+1) = A248636(k) + 1.

Original entry on oeis.org

2, 4, 7, 8, 10, 12, 14, 15, 17, 19, 20, 22, 23, 25, 26, 28, 29, 31, 32, 34, 35, 37, 38, 40, 41, 43, 44, 46, 47, 48, 50, 51, 53, 54, 56, 57, 59, 60, 61, 63, 64, 66, 67, 69, 70, 71, 73, 74, 76, 77, 78, 80, 81, 83, 84, 86, 87, 88, 90, 91, 93, 94, 95, 97, 98

Offset: 1

Clark Kimberling, Oct 11 2014

			(A248636(k+1) = A248636(k)) = (2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 2, 1, 2, 1, 1, 2,... ), so that A248637 = (2, 4, 7, 8, 10, 12, 14, ... ) and A248638 = (1, 3, 5, 6, 9, 11, 13, ...).

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

Cf. A248635, A248636, A248630.

z = 300; p[k_] := p[k] = Sum[(h^3/3^h), {h, 1, k}];
d = N[Table[33/8 - p[k], {k, 1, z/5}], 12]
f[n_] := f[n] = Select[Range[z], 33/8 - p[#] < 1/4^n &, 1];
u = Flatten[Table[f[n], {n, 1, z}]]  (* A248636 *)
d = Differences[u]
v = Flatten[Position[d, 1]] (* A248637 *)
w = Flatten[Position[d, 2]] (* A248638 *)

cp's OEIS Frontend

A248632 Numbers k such that A248631(k+1) = A248631(k).

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A248629 Least k such that 6 - sum{(h^2)/2^h, h = 1..k} < 1/3^n.

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A248635 Numbers k such that A248633(k+1) = A248633(k) + 2.

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A248631 Least k such that 3/2 - sum{(h^2)/3^h, h = 1..k} < 1/2^n.

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A248633 Least k such that 20/27- sum{(h^2)/4^h, h = 1..k} < 1/8^n.

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A248634 Numbers k such that A248633(k+1) = A248633(k) + 1.

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A248636 Least k such that 33/8- sum{(h^3)/3^h, h = 1..k} < 1/4^n.

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A248637 Numbers k such that A248636(k+1) = A248636(k) + 1.

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