A179244 -id:A179244 - cp's OEIS frontend

A179242 Numbers that have two terms in their Zeckendorf representation.

4, 6, 7, 9, 10, 11, 14, 15, 16, 18, 22, 23, 24, 26, 29, 35, 36, 37, 39, 42, 47, 56, 57, 58, 60, 63, 68, 76, 90, 91, 92, 94, 97, 102, 110, 123, 145, 146, 147, 149, 152, 157, 165, 178, 199, 234, 235, 236, 238, 241, 246, 254, 267, 288, 322, 378, 379, 380, 382, 385, 390

Offset: 1

Emeric Deutsch, Jul 05 2010

nonn

A007895(a(n)) = 2. - Reinhard Zumkeller, Mar 10 2013

			4 = 1+3;
6 = 1+5;
7 = 2+5;
9 = 1+8;
10 = 2+8;

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A035517, A007895, A179243, A179244, A179245, A179246, A179247, A179248, A179249, A179250, A179251, A179252, A179253.

Cf. A000045.

import Data.List (inits)
a179242 n = a179242_list !! (n-1)
a179242_list = concatMap h $ drop 3 $ inits $ drop 2 a000045_list where
   h is = reverse $ map (+ f) fs where
          (f:_:fs) = reverse is
-- Reinhard Zumkeller, Mar 10 2013

with(combinat): B := proc (n) local A, ct, m, j: A := proc (n) local i; for i while fibonacci(i) <= n do n-fibonacci(i) end do end proc: ct := 0: m := n: for j while 0 < A(m) do ct := ct+1: m := A(m) end do: ct+1 end proc: Q := {}: for i from fibonacci(5)-1 to 400 do if B(i) = 2 then Q := `union`(Q, {i}) else end if end do: Q;

f[n_] := (k = 1; ff = {}; While[(fi = Fibonacci[k]) <= n, AppendTo[ff, fi]; k++]; Drop[ff, 1]); z[n_] := If[n == 0, 0, r = n; s = {}; fr = f[n]; While[r > 0, lf = Last[fr]; If[lf <= r, r = r - lf; PrependTo[s, lf]]; fr = Drop[fr, -1]]; s]; Select[ Range[400], Length[z[#]] == 2 &] (* Jean-François Alcover, Sep 27 2011 *)
zeck = DigitCount[Select[Range[5000], BitAnd[#, 2*#] == 0&], 2, 1];
Position[zeck, 2] // Flatten (* Jean-François Alcover, Jan 25 2018 *)

from math import comb, isqrt
from sympy import fibonacci
def A179242(n): return fibonacci((m:=isqrt(n<<3)+1>>1)+3)+fibonacci(n+1-comb(m, 2)) # Chai Wah Wu, Apr 09 2025

A179253 Numbers k that have 13 terms in their Zeckendorf representation.

Original entry on oeis.org

196417, 271442, 300099, 311045, 315226, 316823, 317433, 317666, 317755, 317789, 317802, 317807, 317809, 317810, 392835, 421492, 432438, 436619, 438216, 438826, 439059, 439148, 439182, 439195, 439200, 439202, 439203, 467860, 478806, 482987

Offset: 1

Emeric Deutsch, Jul 05 2010

nonn

A007895(a(n)) = 13. - Reinhard Zumkeller, Mar 10 2013

			196417 = 1 + 3 + 8 + 21 + 55 + 144 + 377 + 987 + 2584 + 6765 + 17711 + 46368 + 121393;
271442 = 1 + 3 + 8 + 21 + 55 + 144 + 377 + 987 + 2584 + 6765 + 17711 + 46368 + 196418;

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000

Cf. A035517, A007895, A179242, A179243, A179244, A179245, A179246, A179247, A179248, A179249, A179250, A179251, A179252.

a179253 n = a179253_list !! (n-1)
a179253_list = filter ((== 13) . a007895) [1..]
-- Reinhard Zumkeller, Mar 10 2013

with(combinat): seq(add(fibonacci(2*k), k = 1 .. 13-m)+add(fibonacci(27-2*k+2), k = 1 .. m), m = 0 .. 13); # this program yields only the first 14 terms of the sequence
From R. J. Mathar, Jul 23 2010: (Start)
Lzto10 := proc(L) local i ; add( op(i,L)*combinat[fibonacci](i+1),i=1..nops(L) ) ; end proc:
zbits := proc(numbits,toset,upbits) local L,hibi ; if 2*toset-1 > numbits then return ; end if; if toset = 0 then L := [(seq(0,i=1..numbits)),op(upbits)] ; Lzto10(L); print(%) ; else for hibi from toset-1 to numbits -1 do if toset = 1 then procname(hibi,toset-1,[1,seq(0,i=1..numbits-hibi-1),op(upbits)]) ; else procname(hibi-1,toset-1,[0,1,seq(0,i=1..numbits-hibi-1),op(upbits)]) ; end if; end do; end if; return ; end proc:
ztot := 13 : for numbits from 2*ztot -1 to 50 do zbits(numbits-2,ztot-1,[0,1]) ; end do: (End)

Reap[For[m = 0; k = 1, k <= 10^8, k++, If[BitAnd[k, 2 k] == 0, m++; If[DigitCount[k, 2, 1] == 13, Print[m]; Sow[m]]]]][[2, 1]] (* Jean-François Alcover, Aug 20 2023 *)

More terms from R. J. Mathar, Jul 23 2010

A179243 Numbers that have three terms in their Zeckendorf representation.

Original entry on oeis.org

12, 17, 19, 20, 25, 27, 28, 30, 31, 32, 38, 40, 41, 43, 44, 45, 48, 49, 50, 52, 59, 61, 62, 64, 65, 66, 69, 70, 71, 73, 77, 78, 79, 81, 84, 93, 95, 96, 98, 99, 100, 103, 104, 105, 107, 111, 112, 113, 115, 118, 124, 125, 126, 128, 131, 136, 148, 150, 151, 153, 154, 155

Offset: 1

Emeric Deutsch, Jul 05 2010

			12 = 1+3+8;
17 = 1+3+13;
19 = 1+5+13;
20 = 2+5+13;
25 = 21+3+1;

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A035517, A007895, A179242, A179244, A179245, A179246, A179247, A179248, A179249, A179250, A179251, A179252, A179253.

a179243 n = a179243_list !! (n-1)
a179243_list = filter ((== 3) . a007895) [1..]
-- Reinhard Zumkeller, Mar 10 2013

with(combinat): B := proc (n) local A, ct, m, j: A := proc (n) local i: for i while fibonacci(i) <= n do n-fibonacci(i) end do end proc: ct := 0: m := n: for j while 0 < A(m) do ct := ct+1: m := A(m) end do: ct+1 end proc: Q := {}: for i from fibonacci(7)-1 to 160 do if B(i) = 3 then Q := `union`(Q, {i}) else end if end do: Q

zeck = DigitCount[Select[Range[2000], BitAnd[#, 2*#] == 0 &], 2, 1];
Position[zeck, 3] // Flatten (* Jean-François Alcover, Jan 30 2018 *)

from math import comb, isqrt
from sympy import fibonacci, integer_nthroot
def A179243(n): return fibonacci(2+(r:=n-1-comb((m:=integer_nthroot(6*n, 3)[0])+(t:=(n>comb(m+2, 3)))+1, 3))-comb((k:=isqrt(b:=r+1<<1))+(b>k*(k+1)), 2))+fibonacci((a:=isqrt(s:=n-comb(m-(t^1)+2, 3)<<1))+((s<<2)>(a<<2)*(a+1)+1)+3)+fibonacci(m+t+5) # Chai Wah Wu, Apr 09 2025

A007895(a(n)) = 3. - Reinhard Zumkeller, Mar 10 2013

A179245 Numbers that have 5 terms in their Zeckendorf representation.

Original entry on oeis.org

88, 122, 135, 140, 142, 143, 177, 190, 195, 197, 198, 211, 216, 218, 219, 224, 226, 227, 229, 230, 231, 266, 279, 284, 286, 287, 300, 305, 307, 308, 313, 315, 316, 318, 319, 320, 334, 339, 341, 342, 347, 349, 350, 352, 353, 354, 360, 362, 363, 365, 366, 367

Offset: 1

Emeric Deutsch, Jul 05 2010

A007895(a(n)) = 5. - Reinhard Zumkeller, Mar 10 2013

Numbers that are the sum of five non-consecutive Fibonacci numbers. Their Zeckendorf representation thus consists of five 1's with at least one 0 between each pair of 1's; for example, 122 is represented as 1001010101. - Alonso del Arte, Nov 17 2013

			88  = 55 + 21 + 8 + 3 + 1.
122 = 89 + 21 + 8 + 3 + 1.
135 = 89 + 34 + 8 + 3 + 1.
140 = 89 + 34 + 13 + 3 + 1.
142 = 89 + 34 + 13 + 5 + 1.
81 is not in the sequence because, although it is the sum of five Fibonacci numbers (81 = 5 + 8 + 13 + 21 + 34), its Zeckendorf representation only has three terms: 81 = 55 + 21 + 5.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A035517, A007895. Numbers that have m terms in their Zeckendorf representations: A179242 (m = 2), A179243 (m = 3), A179244 (m = 4), A179246 (m = 6), A179247 (m = 7), A179248 (m = 8), A179249 (m = 9), A179250 (m = 10), A179251 (m = 11), A179252 (m = 12), A179253 (m = 13).

a179245 n = a179245_list !! (n-1)
a179245_list = filter ((== 5) . a007895) [1..]
-- Reinhard Zumkeller, Mar 10 2013

with(combinat): B := proc (n) local A, ct, m, j: A := proc (n) local i: for i while fibonacci(i) <= n do n-fibonacci(i) end do end proc: ct := 0: m := n: for j while 0 < A(m) do ct := ct+1: m := A(m) end do: ct+1 end proc: Q := {}: for i from fibonacci(11)-1 to 400 do if B(i) = 5 then Q := `union`(Q, {i}) else end if end do: Q;

zeck = DigitCount[Select[Range[3000], BitAnd[#, 2*#] == 0 &], 2, 1];
Position[zeck, 5] // Flatten (* Jean-François Alcover, Jan 30 2018 *)

A048680(n)=my(k=1,s);while(n,if(n%2,s+=fibonacci(k++)); k++; n>>=1); s
[A048680(n)|n<-[1..100],hammingweight(n)==5] \\ Charles R Greathouse IV, Nov 17 2013

a(n) = A048680(A014313(n)). - Charles R Greathouse IV, Nov 17 2013

A179246 Numbers that have 6 terms in their Zeckendorf representation.

Original entry on oeis.org

232, 321, 355, 368, 373, 375, 376, 465, 499, 512, 517, 519, 520, 554, 567, 572, 574, 575, 588, 593, 595, 596, 601, 603, 604, 606, 607, 608, 698, 732, 745, 750, 752, 753, 787, 800, 805, 807, 808, 821, 826, 828, 829, 834, 836, 837, 839, 840, 841, 876, 889

Offset: 1

Emeric Deutsch, Jul 05 2010

nonn

A007895(a(n)) = 6. - Reinhard Zumkeller, Mar 10 2013

			232 = 144 + 55 + 21 +  8 + 3 + 1;
321 = 233 + 55 + 21 +  8 + 3 + 1;
355 = 233 + 89 + 21 +  8 + 3 + 1;
368 = 233 + 89 + 34 +  8 + 3 + 1;
373 = 233 + 89 + 34 + 13 + 3 + 1.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A035517, A007895, A179242, A179243, A179244, A179245, A179247, A179248, A179249, A179250, A179251, A179252, A179253.

a179246 n = a179246_list !! (n-1)
a179246_list = filter ((== 6) . a007895) [1..]
-- Reinhard Zumkeller, Mar 10 2013

with(combinat): B := proc (n) local A, ct, m, j: A := proc (n) local i; for i while fibonacci(i) <= n do n-fibonacci(i) end do end proc: ct := 0: m := n: for j while 0 < A(m) do ct := ct+1: m := A(m) end do: ct+1 end proc: Q := {}: for i from fibonacci(13)-1 to 900 do if B(i) = 6 then Q := `union`(Q, {i}) else end if end do: Q;

zeck = DigitCount[Select[Range[12000], BitAnd[#, 2*#] == 0 &], 2, 1];
Position[zeck, 6] // Flatten (* Jean-François Alcover, Jan 30 2018 *)

A179247 Numbers that have 7 terms in their Zeckendorf representation.

Original entry on oeis.org

609, 842, 931, 965, 978, 983, 985, 986, 1219, 1308, 1342, 1355, 1360, 1362, 1363, 1452, 1486, 1499, 1504, 1506, 1507, 1541, 1554, 1559, 1561, 1562, 1575, 1580, 1582, 1583, 1588, 1590, 1591, 1593, 1594, 1595, 1829, 1918, 1952, 1965, 1970, 1972, 1973

Offset: 1

Emeric Deutsch, Jul 05 2010

nonn

A007895(a(n)) = 7. - Reinhard Zumkeller, Mar 10 2013

			609 = 377+144+55+21+8+3+1;
842 = 610+144+55+21+8+3+1;
931 = 610+233+55+21+8+3+1;
965 = 610+233+89+21+8+3+1;
978 = 610+233+89+34+8+3+1.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A035517, A007895, A179242, A179243, A179244, A179245, A179246, A179248, A179249, A179250, A179251, A179252, A179253.

a179247 n = a179247_list !! (n-1)
a179247_list = filter ((== 7) . a007895) [1..]
-- Reinhard Zumkeller, Mar 10 2013

with(combinat): B := proc (n) local A, ct, m, j: A := proc (n) local i: for i while fibonacci(i) <= n do n-fibonacci(i) end do end proc: ct := 0: m := n: for j while 0 < A(m) do ct := ct+1: m := A(m) end do: ct+1 end proc: Q := {}: for i from fibonacci(15)-1 to 2100 do if B(i) = 7 then Q := `union`(Q, {i}) else end if end do: Q;

zeck = DigitCount[Select[Range[4*10^4], BitAnd[#, 2*#] == 0 &], 2, 1];
Position[zeck, 7] // Flatten (* Jean-François Alcover, Jan 30 2018 *)

A179248 Numbers that have 8 terms in their Zeckendorf representation.

Original entry on oeis.org

1596, 2206, 2439, 2528, 2562, 2575, 2580, 2582, 2583, 3193, 3426, 3515, 3549, 3562, 3567, 3569, 3570, 3803, 3892, 3926, 3939, 3944, 3946, 3947, 4036, 4070, 4083, 4088, 4090, 4091, 4125, 4138, 4143, 4145, 4146, 4159, 4164, 4166, 4167, 4172, 4174, 4175

Offset: 1

Emeric Deutsch, Jul 05 2010

nonn

A007895(a(n)) = 8. - Reinhard Zumkeller, Mar 10 2013

			1596 =  987+377+144+55+21+8+3+1;
2206 = 1597+377+144+55+21+8+3+1;
2439 = 1597+610+144+55+21+8+3+1;
2528 = 1597+610+233+55+21+8+3+1;
2562 = 1597+610+233+89+21+8+3+1.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A035517, A007895, A179242, A179243, A179244, A179245, A179246, A179247, A179249, A179250, A179251, A179252, A179253.

a179249 n = a179249_list !! (n-1)
a179249_list = filter ((== 9) . a007895) [1..]
-- Reinhard Zumkeller, Mar 10 2013

with(combinat): B := proc (n) local A, ct, m, j: A := proc (n) local i: for i while fibonacci(i) <= n do n-fibonacci(i) end do end proc: ct := 0: m := n: for j while 0 < A(m) do ct := ct+1: m := A(m) end do: ct+1 end proc: Q := {}: for i from fibonacci(17)-1 to 5000 do if B(i) = 8 then Q := `union`(Q, {i}) else end if end do: Q;

zeck = DigitCount[Select[Range[10^5], BitAnd[#, 2*#] == 0 &], 2, 1];
Position[zeck, 8] // Flatten (* Jean-François Alcover, Jan 30 2018 *)

A179249 Numbers that have 9 terms in their Zeckendorf representation.

Original entry on oeis.org

4180, 5777, 6387, 6620, 6709, 6743, 6756, 6761, 6763, 6764, 8361, 8971, 9204, 9293, 9327, 9340, 9345, 9347, 9348, 9958, 10191, 10280, 10314, 10327, 10332, 10334, 10335, 10568, 10657, 10691, 10704, 10709, 10711, 10712, 10801, 10835, 10848

Offset: 1

Emeric Deutsch, Jul 05 2010

nonn

A007895(a(n)) = 9. - Reinhard Zumkeller, Mar 10 2013

			4180 = 2584 +987+377+144+55+21+8+3+1;
5777 = 4181 +987+377+144+55+21+8+3+1;
6387 = 4181+1597+377+144+55+21+8+3+1;
6620 = 4181+1597+610+144+55+21+8+3+1;
6709 = 4181+1597+610+233+55+21+8+3+1.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A035517, A007895, A179242, A179243, A179244, A179245, A179246, A179247, A179248, A179250, A179251, A179252, A179253.

a179249 n = a179249_list !! (n-1)
a179249_list = filter ((== 9) . a007895) [1..]
-- Reinhard Zumkeller, Mar 10 2013

with(combinat): B := proc (n) local A, ct, m, j: A := proc (n) local i: for i while fibonacci(i) <= n do n-fibonacci(i) end do end proc: ct := 0: m := n: for j while 0 < A(m) do ct := ct+1: m := A(m) end do: ct+1 end proc: Q := {}: for i from fibonacci(19)-1 to 10900 do if B(i) = 9 then Q := `union`(Q, {i}) else end if end do: Q;

zeck = DigitCount[Select[Range[4*10^5], BitAnd[#, 2*#] == 0 &], 2, 1];
Position[zeck, 9] // Flatten (* Jean-François Alcover, Jan 30 2018 *)

A179250 Numbers that have 10 terms in their Zeckendorf representation.

Original entry on oeis.org

10945, 15126, 16723, 17333, 17566, 17655, 17689, 17702, 17707, 17709, 17710, 21891, 23488, 24098, 24331, 24420, 24454, 24467, 24472, 24474, 24475, 26072, 26682, 26915, 27004, 27038, 27051, 27056, 27058, 27059, 27669, 27902, 27991

Offset: 1

Emeric Deutsch, Jul 05 2010

A007895(a(n)) = 10. - Reinhard Zumkeller, Mar 10 2013

			10945=6765+2584+987+377+144+55+21+8+3+1;
15126=10946+2584+987+377+144+55+21+8+3+1;
16723=10946+4181+987+377+144+55+21+8+3+1;

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A035517, A007895, A179242, A179243, A179244, A179245, A179246, A179247, A179248, A179249, A179251, A179252, A179253.

a179250 n = a179250_list !! (n-1)
a179250_list = filter ((== 10) . a007895) [1..]
-- Reinhard Zumkeller, Mar 10 2013

with(combinat): B := proc (n) local A, ct, m, j: A := proc (n) local i: for i while fibonacci(i) <= n do n-fibonacci(i) end do end proc: ct := 0: m := n: for j while 0 < A(m) do ct := ct+1: m := A(m) end do: ct+1 end proc: Q := {}: for i from fibonacci(21)-1 to 28500 do if B(i) = 10 then Q := `union`(Q, {i}) else end if end do: Q;

zeck = DigitCount[Select[Range[2*10^6], BitAnd[#, 2*#] == 0 &], 2, 1];
Position[zeck, 10] // Flatten (* Jean-François Alcover, Jan 30 2018 *)

A179251 Numbers that have 11 terms in their Zeckendorf representation.

Original entry on oeis.org

28656, 39602, 43783, 45380, 45990, 46223, 46312, 46346, 46359, 46364, 46366, 46367, 57313, 61494, 63091, 63701, 63934, 64023, 64057, 64070, 64075, 64077, 64078, 68259, 69856, 70466, 70699, 70788, 70822, 70835, 70840, 70842, 70843

Offset: 1

Emeric Deutsch, Jul 05 2010

nonn

A007895(a(n)) = 11. - Reinhard Zumkeller, Mar 10 2013

			28656=17711+6765+2584+987+377+144+55+21+8+3+1;
39602=28657+6765+2584+987+377+144+55+21+8+3+1;

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A035517, A007895, A179242, A179243, A179244, A179245, A179246, A179247, A179248, A179249, A179250, A179252, A179253.

a179251 n = a179251_list !! (n-1)
a179251_list = filter ((== 11) . a007895) [1..]
-- Reinhard Zumkeller, Mar 10 2013

with(combinat): B := proc (n) local A, ct, m, j: A := proc (n) local i: for i while fibonacci(i) <= n do n-fibonacci(i) end do end proc: ct := 0: m := n: for j while 0 < A(m) do ct := ct+1: m := A(m) end do: ct+1 end proc: Q := {}: for i from fibonacci(23)-1 to 73000 do if B(i) = 11 then Q := `union`(Q, {i}) else end if end do: Q;

Select[Range[6*10^6], BitAnd[#, 2*#] == 0&] // DigitCount[#, 2, 1]& // Position[#, 11]& // Flatten (* Jean-François Alcover, Feb 15 2018 *)

cp's OEIS Frontend

A179242 Numbers that have two terms in their Zeckendorf representation.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

Python

A179253 Numbers k that have 13 terms in their Zeckendorf representation.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

Extensions

A179243 Numbers that have three terms in their Zeckendorf representation.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

Python

Formula

A179245 Numbers that have 5 terms in their Zeckendorf representation.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

PARI

Formula

A179246 Numbers that have 6 terms in their Zeckendorf representation.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

A179247 Numbers that have 7 terms in their Zeckendorf representation.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

A179248 Numbers that have 8 terms in their Zeckendorf representation.

Views