A251656
4-step Fibonacci sequence starting with 1,0,1,0.
Original entry on oeis.org
1, 0, 1, 0, 2, 3, 6, 11, 22, 42, 81, 156, 301, 580, 1118, 2155, 4154, 8007, 15434, 29750, 57345, 110536, 213065, 410696, 791642, 1525939, 2941342, 5669619, 10928542, 21065442, 40604945, 78268548, 150867477, 290806412, 560547382, 1080489819
Offset: 0
Other 4-step Fibonacci sequences are
A000078,
A000288,
A001630,
A001631,
A001648,
A073817,
A100532,
A251654,
A251655,
A251703,
A251704,
A251705.
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NB. see A251655 for the program and apply it to 1,0,1,0.
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LinearRecurrence[Table[1, {4}], {1, 0, 1, 0}, 36] (* Michael De Vlieger, Dec 09 2014 *)
A251654
4-step Fibonacci sequence starting with 0, 1, 1, 0.
Original entry on oeis.org
0, 1, 1, 0, 2, 4, 7, 13, 26, 50, 96, 185, 357, 688, 1326, 2556, 4927, 9497, 18306, 35286, 68016, 131105, 252713, 487120, 938954, 1809892, 3488679, 6724645, 12962170, 24985386, 48160880, 92833081, 178941517, 344920864, 664856342, 1281551804
Offset: 0
Other 4-step Fibonacci sequences are
A000078,
A000288,
A001630,
A001631,
A001648,
A073817,
A100532,
A251655,
A251656,
A251672,
A251703,
A251704,
A251705.
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NB. see A251655 for the program and apply it to 0,1,1,0.
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LinearRecurrence[Table[1, {4}], {0, 1, 1, 0}, 36] (* Michael De Vlieger, Dec 09 2014 *)
A251655
4-step Fibonacci sequence starting with 0, 1, 1, 1.
Original entry on oeis.org
0, 1, 1, 1, 3, 6, 11, 21, 41, 79, 152, 293, 565, 1089, 2099, 4046, 7799, 15033, 28977, 55855, 107664, 207529, 400025, 771073, 1486291, 2864918, 5522307, 10644589, 20518105, 39549919, 76234920, 146947533, 283250477, 545982849, 1052415779, 2028596638
Offset: 0
Other 4-step Fibonacci sequences are
A000078,
A000288,
A001630,
A001631,
A001648,
A073817,
A100532,
A251654,
A251656,
A251672,
A251703,
A251704,
A251705.
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(see www.jsoftware.com) First construct the generating matrix
[M=: (#.@}: + {:)\"1&.|: <:/~i.4
1 1 1 1
1 2 2 2
2 3 4 4
4 6 7 8
Given that matrix, one can produce the first 4*250 numbers with
, M(+/ . *)^:(i.250) 0 1 1 1x
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LinearRecurrence[Table[1, {4}], {0, 1, 1, 1}, 36] (* Michael De Vlieger, Dec 09 2014 *)
A251704
4-step Fibonacci sequence starting with 1, 1, 0, 1.
Original entry on oeis.org
1, 1, 0, 1, 3, 5, 9, 18, 35, 67, 129, 249, 480, 925, 1783, 3437, 6625, 12770, 24615, 47447, 91457, 176289, 339808, 655001, 1262555, 2433653, 4691017, 9042226, 17429451, 33596347, 64759041, 124827065, 240611904, 463794357, 893992367, 1723225693
Offset: 0
Other 4-step Fibonacci sequences are
A000078,
A000288,
A001630,
A001631,
A001648,
A073817,
A100532,
A251654,
A251655,
A251656,
A251703,
A251705.
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NB. see A251655 for the program and apply it to 1,1,0,1.
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LinearRecurrence[Table[1, {4}], {1, 1, 0, 1}, 36] (* Michael De Vlieger, Dec 09 2014 *)
A251705
4-step Fibonacci sequence starting with 1, 1, 1, 0.
Original entry on oeis.org
1, 1, 1, 0, 3, 5, 9, 17, 34, 65, 125, 241, 465, 896, 1727, 3329, 6417, 12369, 23842, 45957, 88585, 170753, 329137, 634432, 1222907, 2357229, 4543705, 8758273, 16882114, 32541321, 62725413, 120907121, 233055969, 449229824, 865918327, 1669111241
Offset: 0
Other 4-step Fibonacci sequences are
A000078,
A000288,
A001630,
A001631,
A001648,
A073817,
A100532,
A251654,
A251655,
A251656,
A251703,
A251704.
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NB. see A251655 for the program and apply it to 1,1,1,0.
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LinearRecurrence[Table[1, {4}], {1, 1, 1, 0}, 36] (* Michael De Vlieger, Dec 09 2014 *)
A320122
Numbers that are not Keith numbers in any base.
Original entry on oeis.org
12, 30, 390, 1170, 1200, 1560, 2340, 2760, 3120, 3900, 4680, 6120, 6240, 7680, 7800, 8460, 10020, 10140, 10950, 11580, 15090, 15480, 17160, 17580, 18360, 19140, 20280, 20700, 20940, 21480, 23040, 23280, 24060, 24210, 24960, 26550, 28740, 29250, 29520, 29670, 30060, 31080, 32400
Offset: 1
a(1) = 12 because 12 is not a Keith number in any base from 2 to 12, while all previous numbers are in some base.
For example, with b = 2, the sequence is : 1, 1, 0, 0, 2, 3, 5, 10, 20, ...; it doesn't contain 12. See A251703.
Cf.
A007629 (Keith numbers in base 10).
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fibo:=proc(n, b) local L,m,M,k:
L:=convert(n,base,b):m:=nops(L):M:=seq(L[m+1-k],k=1..m):
while M[m]
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iskb(n, b) = if(nA007629
isok(n) = if (n<=2, 0, for(b=2, n-1, if (iskb(n, b), return(0))); return (1)); \\ Michel Marcus, Oct 08 2018
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def digits(n, b):
r = []
m = n
while m > 0:
r = [m % b] + r
m = m // b
return r
def fibo(n, b):
L = digits(n, b)
m = len(L) - 1
while L[m] < n:
L.append(sum(k for k in L))
L.pop(0)
return L[m] == n
def test(n):
for b in range(2, n + 1):
if fibo(n, b):
return True
return False
print([n for n in range(2, 2001) if not test(n)])
Showing 1-6 of 6 results.
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