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.

Previous Showing 31-39 of 39 results.

A364006 Wythoff-Niven numbers: numbers that are divisible by the number of 1's in their Wythoff representation.

Original entry on oeis.org

1, 3, 4, 6, 7, 8, 10, 12, 15, 18, 20, 21, 24, 26, 28, 32, 35, 39, 40, 42, 45, 47, 51, 52, 54, 55, 56, 60, 68, 72, 76, 80, 84, 86, 88, 90, 91, 98, 100, 102, 105, 117, 120, 123, 125, 135, 136, 138, 141, 143, 144, 156, 164, 168, 172, 174, 176, 178, 180, 188, 192
Offset: 1

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Author

Amiram Eldar, Jul 01 2023

Keywords

Comments

Numbers k such that A135818(k) | k.
Includes all the positive even-indexed Fibonacci numbers (A001906), since the Wythoff representation of Fibonacci(2*n), for n >= 1, is 1 followed by n-1 0's.

Crossrefs

Programs

  • Mathematica
    wnQ[n_] := (s = Total[w[n]]) > 0 && Divisible[n, s] (* using the function w[n] from A364005 *)

A364123 Stolarsky-Niven numbers: numbers that are divisible by the number of 1's in their Stolarsky representation (A364121).

Original entry on oeis.org

2, 4, 6, 8, 9, 12, 14, 16, 20, 22, 24, 27, 30, 36, 38, 40, 42, 44, 48, 54, 56, 57, 60, 65, 69, 72, 75, 80, 84, 85, 90, 92, 96, 98, 100, 102, 104, 108, 112, 116, 120, 124, 126, 132, 136, 138, 145, 147, 150, 153, 155, 159, 160, 175, 180, 185, 190, 195, 196, 205
Offset: 1

Views

Author

Amiram Eldar, Jul 07 2023

Keywords

Comments

Numbers k such that A200649(k) | k.
Fibonacci(k) + 1 is a term if k !== 3 (mod 6) (i.e., k is in A047263).

Examples

			4 is a term since its Stolarsky representation, A364121(4) = 10, has one 1 and 4 is divisible by 1.
6 is a term since its Stolarsky representation, A364121(6) = 101, has 2 1's and 6 is divisible by 2.
		

Crossrefs

Programs

  • Mathematica
    stol[n_] := stol[n] = If[n == 1, {}, If[n != Round[Round[n/GoldenRatio]*GoldenRatio], Join[stol[Floor[n/GoldenRatio^2] + 1], {0}], Join[stol[Round[n/GoldenRatio]], {1}]]];
    stolNivQ[n_] := n > 1 && Divisible[n, Total[stol[n]]];
    Select[Range[200], stolNivQ]
  • PARI
    stol(n) = {my(phi=quadgen(5)); if(n==1, [], if(n != round(round(n/phi)*phi), concat(stol(floor(n/phi^2) + 1), [0]), concat(stol(round(n/phi)), [1])));}
    isA364123(n) = n > 1 && !(n % vecsum(stol(n)));

A377271 Numbers k such that k and k+1 are both terms in A377209.

Original entry on oeis.org

1, 2, 3, 4, 5, 12, 89, 1824, 3024, 7024, 15084, 17184, 18935, 22624, 28657, 29424, 31464, 37024, 38835, 40032, 42679, 44975, 47375, 66744, 66815, 78219, 89495, 107456, 112175, 119744, 144599, 148519, 169883, 171941, 172025, 188208, 207935, 226624, 244404, 248255
Offset: 1

Views

Author

Amiram Eldar, Oct 22 2024

Keywords

Examples

			1824 is a term since both 1824 and 1825 are in A377209: 1824/A007895(1824) = 304 and 304/A007895(304) = 76 are integers, and 1825/A007895(1825) = 365 and 365/A007895(365) = 73 are integers.
		

Crossrefs

Cf. A007895, A376793 (binary analog).
Subsequence of A328208, A328209 and A377209.
Subsequences: A377272, A377273.

Programs

  • Mathematica
    zeck[n_] := Length[DeleteCases[NestWhileList[# - Fibonacci[Floor[Log[Sqrt[5]*# + 3/2]/Log[GoldenRatio]]] &, n, # > 1 &], 0]]; (* Alonso del Arte at A007895 *)
    q[k_] := q[k] = Module[{z = zeck[k]}, Divisible[k, z] && Divisible[k/z, zeck[k/z]]]; Select[Range[250000], q[#] && q[#+1] &]
  • PARI
    zeck(n) = if(n<4, n>0, my(k=2, s, t); while(fibonacci(k++)<=n, ); while(k && n, t=fibonacci(k); if(t<=n, n-=t; s++); k--); s); \\ Charles R Greathouse IV at A007895
    is1(k) = {my(z = zeck(k)); !(k % z) && !((k/z) % zeck(k/z)); }
    lista(kmax) = {my(q1 = is1(1), q2); for(k = 2, kmax, q2 = is1(k); if(q1 && q2, print1(k-1, ", ")); q1 = q2); }

A377208 a(n) is the number of iterations that n requires to reach a noninteger or a Fibonacci number under the map x -> x / z(x), where z(k) = A007895(k) is the number of terms in the Zeckendorf representation of k; a(n) = 0 if n is a Fibonacci number.

Original entry on oeis.org

0, 0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 2, 0, 2, 1, 1, 1, 2, 1, 1, 0, 2, 1, 3, 1, 1, 2, 1, 1, 2, 1, 1, 1, 0, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 0, 2, 1, 2, 1, 3, 1, 1, 1, 1, 1, 3, 1, 1, 2, 1, 1, 3, 1, 1, 1, 2, 1, 2, 1, 2, 3, 1, 1, 2, 1, 1, 1, 1, 0, 3, 1, 2, 2, 2, 1, 2, 1, 1, 2, 1
Offset: 1

Views

Author

Amiram Eldar, Oct 20 2024

Keywords

Comments

The Fibonacci numbers are the fixed points of the map, since z(Fibonacci(k)) = 1 for all k >= 1. Therefore they are arbitrarily assigned the value a(Fibonacci(k)) = 0.
Each number n starts a chain of a(n) integers: n, n/z(n), (n/z(n))/z(n/z(n)), ..., of them the first a(n)-1 integers are Zeckendorf-Niven numbers (A328208).

Examples

			a(12) = 2 since 12/z(12) = 4 and 4/z(4) = 2 is a Fibonacci number that is reached after 2 iterations.
a(36) = 3 since 36/z(36) = 18, 18/z(18) = 9 and 9/z(9) = 9/2 is a noninteger that is reached after 3 iterations.
		

Crossrefs

Cf. A000005, A000045, A007895, A328208, A376615 (binary analog), A377209, A377210.

Programs

  • Mathematica
    zeck[n_] := Length[DeleteCases[NestWhileList[# - Fibonacci[Floor[Log[Sqrt[5]*# + 3/2]/Log[GoldenRatio]]] &, n, # > 1 &], 0]]; (* Alonso del Arte at A007895 *)
    a[n_] := a[n] = Module[{z = zeck[n]}, If[z == 1, 0, If[!Divisible[n, z], 1, 1 + a[n/z]]]]; Array[a, 100]
  • PARI
    zeck(n) = if(n<4, n>0, my(k=2, s, t); while(fibonacci(k++)<=n, ); while(k && n, t=fibonacci(k); if(t<=n, n-=t; s++); k--); s) \\ Charles R Greathouse IV at A007895
    a(n) = {my(z = zeck(n)); if(z == 1, 0, if(n % z, 1, 1 + a(n/z)));}

Formula

a(n) = 0 if and only if n is in A000045 (by definition).
a(n) >= 2 if and only if n is in A328208 \ A000079 (i.e., n is a Zeckendorf-Niven number that is not a Fibonacci number).
a(n) >= 3 if and only if n is in A377209 \ A000079.
a(n) >= 4 if and only if n is in A377210 \ A000079.
a(n) < A000005(n).

A377272 Numbers k such that k and k+1 are both terms in A377210.

Original entry on oeis.org

1, 2, 3, 4, 5, 12, 47375, 2310399, 3525200, 6506367, 9388224, 17613504, 29373839, 41534800, 48191759, 48344120, 66927384, 68094999, 71982999, 92547279, 95497919, 110146959, 110395439, 126123920, 148865535, 152546030, 154451583, 171570069, 193628799, 232058519
Offset: 1

Views

Author

Amiram Eldar, Oct 22 2024

Keywords

Examples

			47375 is a term since both 47375 and 47376 are in A377210: 47375/A007895(47375) = 9475, 9475/A007895(9475) = 1895 and 1895/A007895(1895) = 379 are integers, and 47376/A007895(47376) = 15792, 15792/A007895(15792) = 3948 and 3948/A007895(3948) = 1316 are integers.
		

Crossrefs

Cf. A007895, A376795 (binary analog).
Subsequence of A328208, A328209, A377210 and A377271.

Programs

  • Mathematica
    zeck[n_] := Length[DeleteCases[NestWhileList[# - Fibonacci[Floor[Log[Sqrt[5]*# + 3/2]/Log[GoldenRatio]]] &, n, # > 1 &], 0]]; (* Alonso del Arte at A007895 *)
    q[k_] := q[k] = Module[{z = zeck[k], z2, m, n}, IntegerQ[m = k/z] && Divisible[m, z2 = zeck[m]] && Divisible[n = m/z2, zeck[n]]]; Select[Range[50000], q[#] && q[#+1] &]
  • PARI
    zeck(n) = if(n<4, n>0, my(k=2, s, t); while(fibonacci(k++)<=n, ); while(k && n, t=fibonacci(k); if(t<=n, n-=t; s++); k--); s); \\ Charles R Greathouse IV at A007895
    is1(k) = {my(z = zeck(k), z2, m); if(k % z, return(0)); m = k/z; z2 = zeck(m); !(m % z2) && !((m/z2) % zeck(m/z2)); }
    lista(kmax) = {my(q1 = is1(1), q2); for(k = 2, kmax, q2 = is1(k); if(q1 && q2, print1(k-1, ", ")); q1 = q2); }

A377273 Starts of runs of 3 consecutive integers that are terms in A377209.

Original entry on oeis.org

1, 2, 3, 4, 231700599, 1069467839, 1156703470, 1241186868, 2533742848, 2684864798, 3037193808, 5056780650, 7073145000, 7557047134, 9623855878, 12090760318, 12120887700, 13816479742, 14430478270, 15811947072, 16864260048, 20905152190, 22735441078, 23224253128, 23269229774, 23766221400, 25175490262
Offset: 1

Views

Author

Amiram Eldar, Oct 22 2024

Keywords

Examples

			231700599 is a term since 231700599, 231700600 and 231700601 are all terms in A377209: 231700599/A007895(231700599) = 17823123 and 17823123/A007895(17823123) = 1980347 are integers, 231700600/A007895(231700600) = 23170060 and 23170060/A007895(23170060) = 2317006 are integers, and 231700601/A007895(231700601) = 21063691 and 21063691/A007895(21063691) = 1914881 are integers.
		

Crossrefs

Cf. A007895, A376794 (binary analog).
Subsequence of A328208, A328209, A328210, A377209 and A377271.

Programs

  • PARI
    zeck(n) = if(n<4, n>0, my(k=2, s, t); while(fibonacci(k++)<=n, ); while(k && n, t=fibonacci(k); if(t<=n, n-=t; s++); k--); s); \\ Charles R Greathouse IV at A007895
    is1(k) = {my(z = zeck(k)); !(k % z) && !((k/z) % zeck(k/z)); }
    lista(kmax) = {my(q1 = is1(1), q2 = is1(2), q3); for(k = 3, kmax, q3 = is1(k); if(q1 && q2 && q3, print1(k-2, ", ")); q1 = q2; q2 = q3);}

A330712 Numbers k such that F(k) - 1 is divisible by floor((k - 1)/2), where F(k) is the k-th Fibonacci number (A000045).

Original entry on oeis.org

3, 4, 5, 7, 15, 22, 25, 26, 27, 35, 41, 47, 49, 50, 73, 74, 75, 87, 89, 95, 97, 98, 101, 107, 121, 122, 135, 145, 146, 147, 167, 193, 194, 195, 207, 215, 217, 218, 221, 227, 241, 242, 255, 275, 289, 290, 315, 327, 335, 337, 338, 347, 361, 362, 385, 386, 387, 395
Offset: 1

Views

Author

Amiram Eldar, Dec 27 2019

Keywords

Comments

Numbers of the form F(k) - 1 have the same Zeckendorf (A014417) and dual Zeckendorf (A104326) representations: alternating digits of 1 and 0 whose sum is floor((k - 1)/2). Thus, if k is in this sequence then F(k) - 1 is both a Zeckendorf-Niven number (A328208) and a lazy-Fibonacci-Niven number (A328212), i.e., A000071(a(n)) is in A330711.

Examples

			7 is in this sequence since F(7) - 1 = 13 - 1 = 12 is divisible by floor((7 - 1)/2) = 3. The Zeckendorf and dual Zeckendorf representations of 7 are both 1010, whose sum of digits, 2, divides 12. Thus 12 is both a Zeckendorf-Niven number and a lazy-Fibonacci-Niven number.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[3, 400], Divisible[Fibonacci[#] - 1, Floor[(# - 1)/2]] &]

A377211 a(n) is the least number k such that A377208(k) = n, or -1 if no such number exists.

Original entry on oeis.org

1, 4, 12, 24, 180, 1056, 2592, 15552, 46656, 544320, 20528640, 238085568, 3547348992, 46438023168, 599501979648
Offset: 0

Views

Author

Amiram Eldar, Oct 20 2024

Keywords

Comments

a(15) > 2.4*10^12, if it exists.
All the terms are Zeckendorf-Niven numbers (A328208).

Examples

			  n | The n iterations
  --+---------------------------------------------
  1 | 4 -> 2 = Fibonacci(3)
  2 | 12 -> 4 -> 2
  3 | 24 -> 12 -> 4 -> 2
  4 | 180 -> 60 -> 30 -> 10 -> 5 = Fibonacci(5)
  5 | 1056 -> 264 -> 66 -> 22 -> 11 -> 11/2
  6 | 2592 -> 1296 -> 324 -> 108 -> 27 -> 9 -> 9/2
		

Crossrefs

Cf. A000045, A376619 (binary analog), A377208.
Subsequence of A328208.

Programs

  • Mathematica
    zeck[n_] := Length[DeleteCases[NestWhileList[# - Fibonacci[Floor[Log[Sqrt[5]*# + 3/2]/Log[GoldenRatio]]] &, n, # > 1 &], 0]]; (* Alonso del Arte at A007895 *)
    s[n_] := s[n] = Module[{z = zeck[n]}, If[z == 1, 0, If[!Divisible[n, z], 1, 1 + s[n/z]]]];
    seq[len_] := Module[{v = Table[0, {len}], c = 0, k = 1, i}, While[c < len, i = s[k] + 1; If[v[[i]] == 0, c++; v[[i]] = k]; k++]; v]; seq[9]
  • PARI
    zeck(n) = if(n<4, n>0, my(k=2, s, t); while(fibonacci(k++)<=n, ); while(k && n, t=fibonacci(k); if(t<=n, n-=t; s++); k--); s) \\ Charles R Greathouse IV at A007895
    s(n) = {my(z = zeck(n)); if(z == 1, 0, if(n % z, 1, 1 + s(n/z)));}
    lista(len) = {my(v = vector(len), c = 0, k = 1, i); while(c < len, i = s(k) + 1; if(v[i] == 0, c++; v[i] = k); k++); v; }

A328216 Weak-Fibonacci-Niven numbers: numbers divisible by the number of terms in at least one representation as a sum of distinct Fibonacci numbers.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 13, 14, 15, 16, 18, 21, 22, 24, 26, 27, 28, 30, 32, 34, 35, 36, 39, 40, 42, 44, 45, 48, 50, 52, 55, 56, 57, 58, 60, 63, 64, 65, 66, 68, 69, 70, 72, 75, 76, 78, 80, 81, 84, 85, 89, 90, 92, 93, 94, 95, 96, 99, 100, 102, 104, 105, 108
Offset: 1

Views

Author

Amiram Eldar, Oct 07 2019

Keywords

Comments

The Fibonacci numbers F(1) = F(2) = 1 can be used at most once in the representation.
Grundman proved that there are infinitely many runs of 6 consecutive weak-Fibonacci-Niven numbers by showing that if m = F(240k) + F(14) + F(9) for k >= 1, then m, m+1, ... m+5 are 6 consecutive weak-Fibonacci-Niven numbers.

Examples

			6 is in the sequence since it can be represented as the sum of 2 Fibonacci numbers, 1 + 5, and 2 is a divisor of 6.
		

Crossrefs

Supersequence of A328208 and A328212.

Programs

  • Mathematica
    m = 10; v = Array[Fibonacci, m, 2]; vm = v[[-1]]; seq = {}; Do[s = Subsets[v, 2^m, {k}]; If[(sum = Total @@ s) <= vm && Divisible[sum, Length @@ s], AppendTo[seq, sum]] , {k, 2, 2^m}]; Union @ seq
Previous Showing 31-39 of 39 results.