cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A184837 a(n) = n + floor(n*t) + floor(n*t^2) + floor(n/t) + floor(n/t^2), where t is the pentanacci constant.

Original entry on oeis.org

5, 13, 20, 29, 36, 44, 51, 59, 66, 74, 81, 90, 97, 105, 111, 120, 127, 135, 142, 151, 158, 166, 172, 181, 188, 196, 203, 212, 219, 225, 233, 241, 248, 256, 264, 272, 279, 286, 294, 302, 309, 317, 325, 333, 339, 347, 355, 363, 370, 378, 386, 393, 400, 408, 416, 424, 431, 440, 447, 453, 461, 469, 477, 484, 492, 500, 507, 514, 522, 530, 538, 545, 553, 561, 568, 575, 583, 591, 599, 606, 614, 621, 629, 636, 644, 652, 660, 667, 674, 681, 689, 696, 705, 712, 720, 727, 735, 742, 750, 757, 766, 773, 781, 787, 796, 803
Offset: 1

Views

Author

Paul D. Hanna, Jan 23 2011

Keywords

Comments

This is one of five sequences that partition the positive integers.
Given t is the pentanacci constant, then the following sequences are disjoint:
. A184835(n) = n + [n/t] + [n/t^2] + [n/t^3] + [n/t^4],
. A184836(n) = n + [n*t] + [n/t] + [n/t^2] + [n/t^3],
. A184837(n) = n + [n*t] + [n*t^2] + [n/t] + [n/t^2],
. A184838(n) = n + [n*t] + [n*t^2] + [n*t^3] + [n/t],
. A184839(n) = n + [n*t] + [n*t^2] + [n*t^3] + [n*t^4], where []=floor.
This is a special case of Clark Kimberling's results given in A184812.

Examples

			Given t = pentanacci constant, then t^3 = 1 + t + t^2 + 1/t + 1/t^2,
t = 1.965948236645..., t^2 = 3.864952469169..., t^3 = 7.598296491482..., t^4 = 14.93785758893..., t^5 = 29.36705478623...

Crossrefs

Cf. A184835, A184836, A184838, A184839; A184812, A103814.

Programs

  • PARI
    {a(n)=local(t=real(polroots(1+x+x^2+x^3+x^4-x^5)[1])); n+floor(n*t)+floor(n*t^2)+floor(n/t)+floor(n/t^2)}

Formula

Limit a(n)/n = t^3 = 7.5982964914823797216620775...
a(n) = n + floor(n*p/r) + floor(n*q/r) + floor(n*s/r) + floor(n*u/r), where p=t, q=t^2, r=t^3, s=t^4, u=t^5, and t is the pentanacci constant.

A184838 a(n) = n + floor(n*t) + floor(n*t^2) + floor(n*t^3) + floor(n/t), where t is the pentanacci constant.

Original entry on oeis.org

12, 28, 42, 58, 72, 88, 103, 117, 132, 147, 162, 178, 192, 208, 221, 237, 252, 267, 282, 297, 312, 328, 341, 357, 371, 387, 402, 417, 432, 445, 460, 476, 490, 506, 520, 536, 551, 565, 580, 595, 610, 626, 640, 656, 669, 685, 700, 715, 730, 745, 760, 775, 789, 805, 819, 835, 850, 865, 880, 893, 909, 924, 939, 954, 969, 984, 999, 1013, 1029, 1043, 1059, 1074, 1089, 1104, 1118, 1133, 1149, 1163, 1179, 1193, 1209, 1223, 1238, 1253, 1268, 1283, 1299, 1313, 1327, 1341, 1357, 1372, 1387, 1402, 1417, 1432, 1447
Offset: 1

Views

Author

Paul D. Hanna, Jan 23 2011

Keywords

Comments

This is one of five sequences that partition the positive integers.
Given t is the pentanacci constant, then the following sequences are disjoint:
. A184835(n) = n + [n/t] + [n/t^2] + [n/t^3] + [n/t^4],
. A184836(n) = n + [n*t] + [n/t] + [n/t^2] + [n/t^3],
. A184837(n) = n + [n*t] + [n*t^2] + [n/t] + [n/t^2],
. A184838(n) = n + [n*t] + [n*t^2] + [n*t^3] + [n/t],
. A184839(n) = n + [n*t] + [n*t^2] + [n*t^3] + [n*t^4], where []=floor.
This is a special case of Clark Kimberling's results given in A184812.

Examples

			Given t = pentanacci constant, then t^4 = 1 + t + t^2 + t^3 + 1/t,
t = 1.965948236645..., t^2 = 3.864952469169..., t^3 = 7.598296491482..., t^4 = 14.93785758893..., t^5 = 29.36705478623...

Crossrefs

Cf. A184835, A184836, A184837, A184839; A184812, A103814.

Programs

  • PARI
    {a(n)=local(t=real(polroots(1+x+x^2+x^3+x^4-x^5)[1])); n+floor(n*t)+floor(n*t^2)+floor(n*t^3)+floor(n/t)}

Formula

Limit a(n)/n = t^4 = 14.937857588939362411757354...
a(n) = n + floor(n*p/q) + floor(n*r/q) + floor(n*s/q) + floor(n*u/q), where p=t, q=t^2, r=t^3, s=t^4, u=t^5, and t is the pentanacci constant.

A184839 a(n) = n + floor(n*t) + floor(n*t^2) + floor(n*t^3) + floor(n*t^4), where t is the pentanacci constant.

Original entry on oeis.org

26, 56, 85, 115, 144, 174, 204, 232, 262, 291, 321, 351, 380, 410, 438, 468, 497, 526, 556, 585, 615, 645, 673, 703, 732, 762, 792, 821, 851, 878, 908, 938, 966, 996, 1025, 1055, 1085, 1113, 1143, 1172, 1202, 1232, 1261, 1291, 1319, 1349, 1379, 1408, 1437, 1466, 1496, 1525, 1554, 1584, 1613, 1643, 1673, 1702, 1731, 1759, 1789, 1819, 1848, 1878, 1906, 1936, 1965, 1994, 2024, 2053, 2083, 2113, 2142, 2172, 2200, 2230, 2260, 2289, 2319, 2348, 2377, 2406, 2435, 2465, 2494, 2524, 2554, 2583, 2611, 2640, 2670
Offset: 1

Views

Author

Paul D. Hanna, Jan 23 2011

Keywords

Comments

This is one of five sequences that partition the positive integers.
Given t is the pentanacci constant, then the following sequences are disjoint:
. A184835(n) = n + [n/t] + [n/t^2] + [n/t^3] + [n/t^4],
. A184836(n) = n + [n*t] + [n/t] + [n/t^2] + [n/t^3],
. A184837(n) = n + [n*t] + [n*t^2] + [n/t] + [n/t^2],
. A184838(n) = n + [n*t] + [n*t^2] + [n*t^3] + [n/t],
. A184839(n) = n + [n*t] + [n*t^2] + [n*t^3] + [n*t^4], where []=floor.
This is a special case of Clark Kimberling's results given in A184812.

Examples

			Given t = pentanacci constant, then t^5 = 1 + t + t^2 + t^3 + t^4,
t = 1.965948236645..., t^2 = 3.864952469169..., t^3 = 7.598296491482..., t^4 = 14.93785758893..., t^5 = 29.36705478623...

Links

Crossrefs

Cf. A184835, A184836, A184837, A184838; A184812, A103814.

Programs

  • Mathematica
    With[{t=Root[x^5-x^4-x^3-x^2-x-1,1]},Table[n+Total@@Through[ Floor[ n*t^Range[4]]],{n,100}]] (* Harvey P. Dale, Dec 12 2019 *)
  • PARI
    {a(n)=local(t=real(polroots(1+x+x^2+x^3+x^4-x^5)[1])); n+floor(n*t)+floor(n*t^2)+floor(n*t^3)+floor(n*t^4)}

Formula

Limit a(n)/n = t^5 = 29.367054786236720687050865...
a(n) = n + floor(n*q/p) + floor(n*r/p) + floor(n*s/p) + floor(n*u/p), where p=t, q=t^2, r=t^3, s=t^4, u=t^5, and t is the pentanacci constant.

A202805 a(n) is the largest k in an n_nacci(k) sequence (Fibonacci(k) for n=2, tribonacci(k) for n=3, etc.) such that n_nacci(k) >= 2^(k-n-1).

Original entry on oeis.org

6, 12, 25, 48, 94, 184, 363, 719, 1430, 2851, 5691, 11371, 22728, 45443, 90870, 181724, 363429, 726839, 1453658, 2907295, 5814566, 11629107
Offset: 2

Views

Author

Frank M Jackson, Dec 24 2011

Keywords

Comments

From Frank M Jackson, Jul 02 2023: (Start)
Define the n_nacci sequence, basically row n in A092921, with an offset of 0, n_nacci(k) = 0 for 0 <= k <= n-2 and n_nacci(n-1) = 1. Thereafter, n_nacci(k) for k >= n continues as the sum of its previous n terms.
This means that n_nacci(k) = 2^(k-n) for n <= k <= 2n-1. In the limit as n tends to infinity the n_nacci sequence after an initial large set of zeros followed by 1 has successive terms of ascending powers of 2.
As the n-acci constants, (A001622, A058265, A086088, A103814,...) are smaller than 2, for each n_nacci sequence there is a largest k such that n_nacci(k) >= 2^(k-n-1). (End)

Examples

			For n=3, the tribonacci sequence is 0,0,1,1,2,4,7,...,149,274,504,... and the 13th term is 504 < 512 so a(n)=12 because 274 is greatest term >= 2^(12-3-1) = 256.

Crossrefs

Cf. A000045, A000073, A000078.

Programs

  • Maple
    nAcci := proc(n,k)
    option remember ;
    if k <= n-2 then
        0;
    elif k = n-1 then
        1;
    else
        add( procname(n,i),i=k-n..k-1) ;
    end if;
end proc:
A202805 := proc(n)
    local k ;
    for k from n do
        if nAcci(n,k) < 2^(k-n-1) then
            return k-1;
        end if;
    end do:
end proc:
for n from 2 do
    print(n,A202805(n)) ;
end do: # R. J. Mathar, Mar 11 2024
  • Mathematica
    fib[n_, m_] := (Block[{nacci}, (Do[nacci[g]=0, {g, 0, m - 2}];
nacci[m-1]=1;nacci[p_] := (nacci[p]=Sum[nacci[h], {h, p-m, p-1}]);nacci[n])]);
crossover[q_] := (Block[{$RecursionLimit=Infinity}, (k=0;While[fib[k+q+1, q]>=2^k, k++];k+q)]);
Table[crossover[j], {j, 2, 12}]
  • Python
    def nacci(n): # generator of n_nacci terms
    window = [0]*(n-1) + [1]
    yield from window
    while True:
        an = sum(window)
        yield an
        window = window[1:] + [an]
def a(n):
    pow2 = 1
    for k, t in enumerate(nacci(n)):
        if k > n + 1: pow2 <<= 1
        if 0 < t < pow2: return k-1
print([a(n) for n in range(2, 12)]) # Michael S. Branicky, Jan 29 2025

Extensions

Edited by N. J. A. Sloane, May 20 2023
There seems to be an error in the Comment. See "History" tab. - N. J. A. Sloane, Jun 24 2023
Removed musing about what might define "complete" sequences. - R. J. Mathar, Mar 11 2024
a(17)-a(23) from Michael S. Branicky, Jan 29 2025

A382627 Decimal expansion of the smallest (in absolute value) root of 1-x-x^2-x^3-x^4-x^5.

Original entry on oeis.org

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

Views

Author

R. J. Mathar, Apr 01 2025

Keywords

Comments

The base of the asymptotic growth derived from Binet's formula for 5-nacci sequences.
The 5 roots are -1.0118..+-0.68395..i, 0.2575066..+-1.118790..i and this.

Examples

			0.5086603916420041364638429658984...

Crossrefs

Cf. A103814, A001622 (Fibonacci), A192918 (3-nacci), A382626 (4-nacci).

Formula

Equals 1/A103814.

A239640 a(n) is the smallest number such that for n-bonacci constant c_n satisfies round(c_n^prime(m)) == 1 (mod 2*p_m) for every m>=a(n).

Original entry on oeis.org

3, 3, 4, 5, 7, 7, 10, 13, 14, 14, 19, 23, 23, 31, 34, 34, 46, 50, 60, 65, 73, 79, 88, 92, 107, 113, 126, 139, 149, 168, 182, 198, 210, 227, 244, 265, 276, 292, 317, 340, 369, 384, 408, 436, 444, 480, 516, 540, 565, 606, 628, 669, 704, 735, 759, 810, 829, 895, 925
Offset: 2

Views

Author

Vladimir Shevelev and Peter J. C. Moses, Mar 23 2014

Keywords

Comments

The n-bonacci constant is a unique root x_1>1 of the equation x^n-x^(n-1)-...-x-1=0. So, for n=2 we have Fibonacci constant phi or golden ratio (A001622); for n=3 we have tribonacci constant (A058265); for n=4 we have tetranacci constant (A086088), for n=5 (A103814), for n=6 (A118427), etc.

Examples

			Let n=2, then c_2 = phi (Fibonacci constant). We have round(c_2^2)=3 is not == 1 (mod 4), round(c_2^3)=4 is not == 1 (mod 6), while round(c_2^5)=11 == 1 (mod 10) and one can prove that for p>=5, we have round(c_2^p) == 1 (mod 2*p). Since 5=prime(3), then a(2)=3.

Links

Crossrefs

Cf. A001622, A007619, A007663, A058265, A086088, A103814, A118427, A118428, A238693, A238697, A238698, A238700, A239502, A239544, A239564, A239565, A239566.
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