A377109 -id:A377109 - cp's OEIS frontend

A377119 a(n) = coefficient of 2^(2/3) in the expansion of (2^(1/3) + 2^(2/3))^n.

0, 1, 1, 6, 12, 42, 108, 324, 900, 2592, 7344, 20952, 59616, 169776, 483408, 1376352, 3919104, 11158560, 31772736, 90465984, 257587776, 733432320, 2088322560, 5946120576, 16930529280, 48206658816, 137259899136, 390823128576, 1112799347712, 3168498166272

Offset: 0

Clark Kimberling, Oct 26 2024

Conjecture: every prime divides a(n) for infinitely many n, and if K(p) = (k(1), k(2),...) is the maximal subsequence of indices n such that p divides a(n), then the difference sequence of K(p) is eventually periodic; indeed, K(p) is purely periodic for the first 6 primes, with respective period lengths 2,2,4,5,2,5 and these periods:
p = 2: (2,1)
p = 3: (7, 1)
p = 5: (21, 5, 13, 1)
p = 7: (15, 17, 20, 43, 1)
p = 11: (9, 1)
p = 13: (102, 5, 17, 15, 1)
See A377109 for a guide to related sequences.

			((2^(1/3) + 2^(2/3)))^3 = 4 + 2*2^(1/3) + 2^(2/3), so a(3) = 1.

Index entries for linear recurrences with constant coefficients, signature (0,6,6).

Cf. A377109, A377117, A377118.

(* Program 1 generates sequences A377117-A377119. *)
tbl = Table[Expand[(2^(1/3) + 2^(2/3))^n], {n, 0, 30}];
u = MapApply[{#1/#2, #2} /. {1, #} -> {{1}, {#}} &,
   Map[({#1, #1 /. ^ -> 1} &), Map[(Apply[List, #1] &), tbl]]];
{s1, s2, s3} = Transpose[(PadRight[#1, 3] &) /@ Last /@ u][[1 ;; 3]];
s3   (* Peter J. C. Moses, Oct 16 2024 *)
(* Program 2 generates (a(n)) for n>=1. *)
LinearRecurrence[{0,6,6}, {0,1,1}, 30]

a(n) = 6*a(n-2) + 6*a(n-3) for n>=1, with a(0)=0, a(1)=1, a(3)=1.

G.f.: x*(1 + x)/(1 - 6*x^2 - 6*x^3).

A377118 a(n) = coefficient of 2^(1/3) in the expansion of (2^(1/3) + 2^(2/3))^n.

Original entry on oeis.org

0, 1, 2, 6, 18, 48, 144, 396, 1152, 3240, 9288, 26352, 75168, 213840, 609120, 1734048, 4937760, 14059008, 40030848, 113980608, 324539136, 924068736, 2631118464, 7491647232, 21331123200, 60736594176, 172936622592, 492406304256, 1402039300608, 3992057561088

Offset: 0

Clark Kimberling, Oct 26 2024

Conjecture: every prime divides a(n) for infinitely many n, and if K(p) = (k(1), k(2),...) is the maximal subsequence of indices n such that p divides a(n), then the difference sequence of K(p) is eventually periodic; indeed, K(p) is purely periodic for the first 6 primes, with respective period lengths 1,5,4,17,1,13 and these periods:
p = 2: (1)
p = 3: (9, 4, 1, 2, 8)
p = 5: (7, 6, 5, 22)
p = 7: (60, 23, 5, 9, 16, 8, 42, 7, 19, 1, 2, 10, 31, 4, 11, 6, 34)
p = 11: (30)
p = 13: (119, 13, 9, 25, 15, 51, 45, 1, 2, 17, 41, 28, 54)
See A377109 for a guide to related sequences.

			((2^(1/3) + 2^(2/3)))^3 = 4 + 2*2^(1/3) + 2^(2/3), so a(3) = 2.

Index entries for linear recurrences with constant coefficients, signature (0,6,6).

Cf. A377109, A377117, A377119.

(* Program 1 generates sequences A377117-A377119. *)
tbl = Table[Expand[(2^(1/3) + 2^(2/3))^n], {n, 0, 30}];
u = MapApply[{#1/#2, #2} /. {1, #} -> {{1}, {#}} &,
   Map[({#1, #1 /. ^ -> 1} &), Map[(Apply[List, #1] &), tbl]]];
{s1, s2, s3} = Transpose[(PadRight[#1, 3] &) /@ Last /@ u][[1 ;; 3]];
s2   (* Peter J. C. Moses, Oct 16 2024 *)
(* Program 2 generates (a(n)) for n>=1. *)
LinearRecurrence[{0,6,6}, {0,1,2}, 30]

a(n) = 6*a(n-2) + 6*a(n-3) for n>=1, with a(0)=0, a(1)=1, a(3)=2.

G.f.: x*(1 + 2*x)/(1 - 6*x^2 - 6*x^3).

A377314 a(n) = coefficient of the term that is independent of 2^(1/3) and 2^(2/3) in the expansion of (1 + 2^(1/3) + 2^(2/3))^n.

Original entry on oeis.org

1, 1, 5, 19, 73, 281, 1081, 4159, 16001, 61561, 236845, 911219, 3505753, 13487761, 51891761, 199644319, 768096001, 2955112721, 11369270485, 43741245619, 168286661033, 647452990441, 2490960200041, 9583526232479, 36870912288001, 141854275761481

Offset: 0

Clark Kimberling, Oct 26 2024

nonn

See A377109 for a guide to related sequences.

			((1 + 2^(1/3) + 2^(2/3)))^3 = 19 + 15 2^(1/3) + 12 2^(2/3), so a(3) = 19.

Michael De Vlieger, Table of n, a(n) for n = 0..1709
Sela Fried, Even-up words and their variants, arXiv:2505.14196 [math.CO], 2025. See p. 4.
Index entries for linear recurrences with constant coefficients, signature (3,3,1).

Cf. A377109, A377117, A377315, A108368 (coefficients of 2^(2/3)).

(* Program 1 generates sequences A377314-A377315 and A108368. *)
tbl = Table[Expand[(1 + 2^(1/3) + 2^(2/3))^n], {n, 0, 24}];
Take[tbl, 6]
u = MapApply[{#1/#2, #2} /. {1, #} -> {{1}, {#}} &,
   Map[({#1, #1 /. ^ -> 1} &), Map[(Apply[List, #1] &), tbl]]];
{s1, s2, s3} = Transpose[(PadRight[#1, 3] &) /@ Last /@ u][[1 ;; 3]];
s1  (* Peter J. C. Moses, Oct 16 2024 *)
(* Program 2 generates (a(n)) for n>=1. *)
LinearRecurrence[{3,3,1}, {1, 1, 5}, 15]

a(n) = 3*a(n-1) + 3*a(n-2) + a(n-3), with a(0)=1, a(1)=1, a(3)=5. [Corrected by Jianing Song, Oct 31 2024]

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

A377315 a(n) = coefficient of 2^(1/3) in the expansion of (1 + 2^(1/3) + 2^(2/3))^n.

Original entry on oeis.org

0, 1, 4, 15, 58, 223, 858, 3301, 12700, 48861, 187984, 723235, 2782518, 10705243, 41186518, 158457801, 609638200, 2345474521, 9023795964, 34717449655, 133569211378, 513883779063, 1977076420978, 7606449811501, 29264462476500, 112589813284981, 433169277095944

Offset: 0

Clark Kimberling, Oct 26 2024

See A377109 for a guide to related sequences.

			(1 + 2^(1/3) + 2^(2/3))^3 = 19 + 15 2^(1/3) + 12 2^(2/3), so a(3) = 15.

Index entries for linear recurrences with constant coefficients, signature (3,3,1).

Cf. A377109, A377117, A377314, A108368 (coefficients of 2^(2/3)).

(* Program 1 generates sequences A377314-A377315 and A108368. *)
tbl = Table[Expand[(1 + 2^(1/3) + 2^(2/3))^n], {n, 0, 24}];
Take[tbl, 6]
u = MapApply[{#1/#2, #2} /. {1, #} -> {{1}, {#}} &,
   Map[({#1, #1 /. ^ -> 1} &), Map[(Apply[List, #1] &), tbl]]];
{s1, s2, s3} = Transpose[(PadRight[#1, 3] &) /@ Last /@ u][[1 ;; 3]];
s2  (* Peter J. C. Moses, Oct 16 2024 *)
(* Program 2 generates (a(n)) for n>=1. *)
LinearRecurrence[{3,3,1}, {0, 1, 4}, 15]

a(n) = 3*a(n-1) + 3*a(n-2) + a(n-3), with a(0)=0, a(1)=1, a(3)=4.

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

A377110 a(n) = coefficient of sqrt(2) in the expansion of (2 + sqrt(2) + sqrt(3))^n.

Original entry on oeis.org

0, 1, 4, 23, 120, 629, 3260, 16843, 86832, 447241, 2302516, 11851487, 60995880, 313912637, 1615504748, 8313878227, 42785563488, 220186165393, 1133137017700, 5831424517415, 30010056528600, 154439694647429, 794787521046812, 4090186754982235, 21049182488180880

Offset: 0

Clark Kimberling, Oct 20 2024

Conjecture: every prime divides a(n) for infinitely many n, and if K(p) = (k(1), k(2),...) is the maximal subsequence of indices n such that p divides a(n), then the difference sequence of K(p) is eventually periodic; indeed, K(p) is purely periodic for the first 7 primes, with respective period lengths 1,8,18,17,3,11,50 and these periods:
 p = 2: (4)
 p = 3: (4,2,6,6,1,1,3,1)
 p = 5: (12,2,10,7,3,2,3,9,12,1,8,2,1,12,12,5,7,12)
 p = 7: (5, 16, 26, 22, 13, 2, 1, 24, 16, 57, 2, 30, 9, 19, 6, 28, 12)
 p = 11: (29, 70, 61)
 p = 13: (21, 3, 24, 15, 9, 24, 9, 15, 24, 3, 21)
 p = 17: (19, 11, 30, 4, 21, 2, 3, 22, 8, 7, 23, 24, 6, 30, 30, 11, 10, 9, 26, 4, 30, 18, 12, 30, 30, 15, 15, 30, 30, 8, 4, 11, 7, 30, 30, 9, 7, 14, 30, 5, 9, 16, 6, 24, 30, 30, 3, 27, 30, 30)
See A377109 for a guide to related sequences.

			(2 + sqrt(2) + sqrt(3))^3 = 9 + 4*sqrt(2) + 4*sqrt(3) + 2*sqrt(6), so a(3) = 4.

Index entries for linear recurrences with constant coefficients, signature (8,-14,-8,23).

Cf. A377109.

(* Program 1 generates sequences A377109-A377112. *)
tbl = Table[Expand[(2 + Sqrt[2] + Sqrt[3])^n], {n, 0, 24}];
u = MapApply[{#1/#2, #2} /. {1, #} -> {{1}, {#}} &,
   Map[({#1, #1 /. ^ -> 1} &), Map[(Apply[List, #1] &), tbl]]];
{s1,s2,s3,s4}=Transpose[(PadRight[#1,4]&)/@Last/@u][[1;;4]];
s2  (* Peter J. C. Moses, Oct 16 2024 *)
(* Program 2 generates this sequence. *)
LinearRecurrence[{8, -14, -8, 23}, {0, 1, 4, 23}, 25]

a(n) = 8*a(n-1) - 14*a(n-2) - 8*a(n-3) + 23*a(n-4), with a(0)=0, a(1)=1, a(3)=4, a(4)=23.
G.f.: x*(-1 + 4*x - 5*x^2)/(-1 + 8*x - 14*x^2 - 8*x^3 + 23*x^4).
a(n) = ((2 + sqrt(2) + sqrt(3))^n - (2 - sqrt(2) - sqrt(3))^n + (2 + sqrt(2) - sqrt(3))^n - (2 - sqrt(2) + sqrt(3))^n) / 2^(5/2). - Vaclav Kotesovec, Oct 21 2024

A377111 a(n) = coefficient of sqrt(3) in the expansion of (2 + sqrt(2) + sqrt(3))^n.

Original entry on oeis.org

0, 1, 4, 21, 104, 529, 2700, 13845, 71120, 365697, 1881236, 9679605, 49809720, 256324433, 1319090972, 6788338869, 34934465440, 179781713537, 925203573540, 4761340669269, 24503114321416, 126099496024593, 648941324534188, 3339623572751061, 17186585699725680

Offset: 0

Clark Kimberling, Oct 20 2024

Conjecture: every prime divides a(n) for infinitely many n, and if K(p) = (k(1), k(2),...) is the maximal subsequence of indices n such that p divides a(n), then the difference sequence of K(p) is eventually periodic; indeed, K(p) is purely periodic for the first 7 primes, with respective period lengths 6,8,7,28,16,14,50:
 p = 2: (3, 3, 1, 2, 2, 1)
 p = 3: (6, 1, 1, 3, 1, 4, 2, 6)
 p = 5: (32, 21, 20, 10, 20, 9, 8)
 p = 7: (18, 18, 8, 10, 18, 18, 11, 7, 18, 18, 1, 3, 13, 1, 5, 5, 8, 18, 14, 4, 18, 13, 5, 2, 16, 7, 9, 2)
 p = 11: (16, 16, 16, 16, 12, 4, 1, 15, 7, 8, 1, 4, 12, 16, 9, 7)
 p = 13: (24, 18, 6, 24, 12, 1, 11, 24, 5, 1, 17, 1, 7, 17)
 p = 17: (15, 30, 30, 8, 4, 11, 7, 30, 30, 9, 7, 14, 30, 5, 9, 16, 6, 24, 30, 30, 3, 27, 30, 30, 19, 11, 30, 4, 21, 2, 3, 22, 8, 7, 23, 24, 6, 30, 30, 11, 10, 9, 26, 4, 30, 18, 12, 30, 30, 15).
See A377109 for a guide to related sequences.

			(2 + sqrt(2) + sqrt(3))^3 = 9 + 4*sqrt(2) + 4*sqrt(3) + 2*sqrt(6), so a(3) = 4.

Index entries for linear recurrences with constant coefficients, signature (8,-14,-8,23).

Cf. A377109.

(* Program 1 generates sequences A377109-A377112. *)
tbl = Table[Expand[(2 + Sqrt[2] + Sqrt[3])^n], {n, 0, 24}];
u = MapApply[{#1/#2, #2} /. {1, #} -> {{1}, {#}} &,
   Map[({#1, #1 /. ^ -> 1} &), Map[(Apply[List, #1] &), tbl]]];
{s1,s2,s3,s4}=Transpose[(PadRight[#1,4]&)/@Last/@u][[1;;4]];
s3  (* Peter J. C. Moses, Oct 16 2024 *)
(* Program 2 generates this sequence. *)
LinearRecurrence[{8, -14, -8, 23}, {0, 1, 4, 21}, 25]

a(n) = 8*a(n-1) - 14*a(n-2) - 8*a(n-3) + 23*a(n-4), with a(0)=0, a(1)=1, a(3)=4, a(4)=21.
G.f.: x*(-1 + 4*x - 3*x^2)/(-1 + 8*x - 14*x^2 - 8*x^3 + 23*x^4).
a(n) = ((2 - sqrt(2) + sqrt(3))^n + (2 + sqrt(2) + sqrt(3))^n - (2 - sqrt(2) - sqrt(3))^n - (2 + sqrt(2) - sqrt(3))^n) / (4*sqrt(3)). - Vaclav Kotesovec, Oct 21 2024

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A377119 a(n) = coefficient of 2^(2/3) in the expansion of (2^(1/3) + 2^(2/3))^n.

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A377118 a(n) = coefficient of 2^(1/3) in the expansion of (2^(1/3) + 2^(2/3))^n.

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A377314 a(n) = coefficient of the term that is independent of 2^(1/3) and 2^(2/3) in the expansion of (1 + 2^(1/3) + 2^(2/3))^n.

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A377315 a(n) = coefficient of 2^(1/3) in the expansion of (1 + 2^(1/3) + 2^(2/3))^n.

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A377110 a(n) = coefficient of sqrt(2) in the expansion of (2 + sqrt(2) + sqrt(3))^n.

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A377111 a(n) = coefficient of sqrt(3) in the expansion of (2 + sqrt(2) + sqrt(3))^n.

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