A130312 -id:A130312 - cp's OEIS frontend

A348366 a(n) = 2*a(n - A130312(n)) + (A072649(n) - A072649(n - A130312(n))) mod 2 for n > 1 with a(0) = 0, a(1) = 1.

0, 1, 3, 2, 7, 6, 5, 15, 4, 14, 13, 11, 31, 12, 10, 30, 9, 29, 27, 23, 63, 8, 28, 26, 22, 62, 25, 21, 61, 19, 59, 55, 47, 127, 24, 20, 60, 18, 58, 54, 46, 126, 17, 57, 53, 45, 125, 51, 43, 123, 39, 119, 111, 95, 255, 16, 56, 52, 44, 124, 50, 42, 122, 38, 118

Offset: 0

Mikhail Kurkov, Oct 15 2021

nonn

Cf. A072649, A130312.

an(n) = my(m=0); until(fibonacci(m)>n, m++); m-2; \\ A072649
af(n) = my(m=0); until(fibonacci(m)>n, m++); fibonacci(m-2); \\ A130312
a(n) = if (n <= 1, n, 2*a(n - af(n)) + (an(n) - an(n - af(n))) % 2); \\ Michel Marcus, Nov 26 2022

Name changed by Mikhail Kurkov, Nov 08 2024

A355429 Square array T(n, k), n >= 0, k > 0, read by antidiagonals, where T(0, k) = A001906(k) for k > 0 and where T(n, k) = n - A130312(n) + A000045(2k + A072649(n)) for n > 0, k > 0.

Original entry on oeis.org

1, 2, 3, 4, 5, 8, 6, 9, 13, 21, 7, 14, 22, 34, 55, 10, 15, 35, 56, 89, 144, 11, 23, 36, 90, 145, 233, 377, 12, 24, 57, 91, 234, 378, 610, 987, 16, 25, 58, 146, 235, 611, 988, 1597, 2584, 17, 37, 59, 147, 379, 612, 1598, 2585, 4181, 6765, 18, 38, 92, 148, 380, 989

Offset: 1

Mikhail Kurkov, Jul 20 2022 [verification needed]

Each positive integer occurs exactly once, so this sequence is a permutation of the natural numbers.

			Square array begins:
   1,  3,  8,  21,  55,  144,  377,   987, ...
   2,  5, 13,  34,  89,  233,  610,  1597, ...
   4,  9, 22,  56, 145,  378,  988,  2585, ...
   6, 14, 35,  90, 234,  611, 1598,  4182, ...
   7, 15, 36,  91, 235,  612, 1599,  4183, ...
  10, 23, 57, 146, 379,  989, 2586,  6767, ...
  11, 24, 58, 147, 380,  990, 2587,  6768, ...
  12, 25, 59, 148, 381,  991, 2588,  6769, ...
  16, 37, 92, 236, 613, 1600, 4184, 10949, ...

Index entries for sequences that are permutations of the natural numbers

Cf. A000045, A001906, A072649, A130312.

b1(n)=local(m); if(n<1, 0, m=0; until(fibonacci(m)>n, m++); m-2) \\ A072649
T(n, k)=if(n==0, fibonacci(2*k), n - fibonacci(b1(n)) + fibonacci(2*k + b1(n)))

T(0, k) = A001906(k) for k > 0.

T(n, k) = n - A130312(n) + A000045(2k + A072649(n)) for n > 0, k > 0.

A357589 a(n) = n - A130312(n).

Original entry on oeis.org

0, 1, 1, 2, 2, 3, 4, 3, 4, 5, 6, 7, 5, 6, 7, 8, 9, 10, 11, 12, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36

Offset: 1

Mikhail Kurkov, Oct 05 2022

Cf. A130312, A183544.

b1(n)=local(m); if(n<1, 0, m=0; until(fibonacci(m)>n, m++); m-2) \\ A072649
a(n)=n - fibonacci(b1(n))

from itertools import islice
def A357589_gen(): # generator of terms
    a, b, i = 1, 1, 1
    while True:
        yield from (j-a for j in range(i,i+a))
        i += a
        a, b = b, a+b
A357589_list = list(islice(A357589_gen(),30)) # Chai Wah Wu, Oct 13 2022

Conjecture: a(n) = 2*n - 1 - A183544(n).

A200649 Number of 1's in the Stolarsky representation of n.

Original entry on oeis.org

0, 1, 2, 1, 3, 2, 2, 4, 1, 3, 3, 3, 5, 2, 2, 4, 2, 4, 4, 4, 6, 1, 3, 3, 3, 5, 3, 3, 5, 3, 5, 5, 5, 7, 2, 2, 4, 2, 4, 4, 4, 6, 2, 4, 4, 4, 6, 4, 4, 6, 4, 6, 6, 6, 8, 1, 3, 3, 3, 5, 3, 3, 5, 3, 5, 5, 5, 7, 3, 3, 5, 3, 5, 5, 5, 7, 3, 5, 5, 5, 7, 5, 5, 7, 5, 7, 7

Offset: 1

Casey Mongoven, Nov 19 2011

For the Stolarsky representation of n, see the C. Mongoven link.

Conjecture: a(n) is the length of row n-1 of A385886. To obtain it, first take maximal anti-run lengths of binary indices of each nonnegative integer (giving A384877), then remove all duplicate rows (giving A385886), and finally take the length of each remaining row. For sum instead of length we appear to have A200648. For runs minus 1 instead of anti-runs see A200650. - Gus Wiseman, Jul 21 2025

			The Stolarsky representation of 19 is 11101. This has 4 1's. So a(19) = 4.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Casey Mongoven, Description of Stolarsky Representations.

Cf. A072649, A130312, A135818, A200651.
For length instead of number of 1's we have A200648.
For 0's instead of 1's we have A200650.
Stolarsky representation is listed by A385888, ranks A200714.
A000120 counts 1's in binary expansion.
A384877 lists anti-run lengths of binary indices, duplicates removed A385886.
A384890 counts maximal anti-runs of binary indices, ranked by A385816.
Cf. A023758, A044813, A048793, A069010, A164707, A245562, A245563, A328592, A384893.

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}]]];
a[n_] := Count[stol[n], 1]; Array[a, 100] (* Amiram Eldar, Jul 07 2023 *)

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])));}
a(n) = vecsum(stol(n)); \\ Amiram Eldar, Jul 07 2023

a(n) = a(n - A130312(n-1)) + (A072649(n-1) - A072649(n - A130312(n-1) - 1)) mod 2 for n > 2 with a(1) = 0, a(2) = 1. - Mikhail Kurkov, Oct 19 2021 [verification needed]

a(n) = A200648(n) - A200650(n). - Amiram Eldar, Jul 07 2023

More terms from Amiram Eldar, Jul 07 2023

A353654 Numbers whose binary expansion has the same number of trailing 0 bits as other 0 bits.

Original entry on oeis.org

1, 3, 7, 10, 15, 22, 26, 31, 36, 46, 54, 58, 63, 76, 84, 94, 100, 110, 118, 122, 127, 136, 156, 172, 180, 190, 204, 212, 222, 228, 238, 246, 250, 255, 280, 296, 316, 328, 348, 364, 372, 382, 392, 412, 428, 436, 446, 460, 468, 478, 484, 494, 502, 506, 511, 528, 568

Offset: 1

Mikhail Kurkov, Jul 15 2022

Numbers k such that A007814(k) = A086784(k).
To reproduce the sequence through itself, use the following rule: if binary 1xyz is a term then so are 11xyz and 10xyz0 (except for 1 alone where 100 is not a term).
The number of terms with bit length k is equal to Fibonacci(k-1) for k > 1.
Conjecture: 2*A247648(n-1) + 1 with rewrite 1 -> 1, 01 -> 0 applied to binary expansion is the same as a(n) without trailing 0 bits in binary.
Odd terms are positive Mersenne numbers (A000225), because there is no 0 in their binary expansion. - Bernard Schott, Oct 12 2022

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A000045, A000120, A000523, A007814, A010056, A025480, A030109, A054429, A060142, A072649, A084471, A086784, A091892, A132665, A133512, A200650, A213911, A232559, A280514, A343152, A348366.
Cf. A356385 (first differences).
Subsequences with same number k of trailing 0 bits and other 0 bits: A000225 (k=0), 2*A190620 (k=1), 4*A357773 (k=2), 8*A360573 (k=3).

N:= 10: # for terms <= 2^N
S:= {1};
for d from 1 to N do
  for k from 0 to d/2-1 do
    B:= combinat:-choose([$k+1..d-2],k);
    S:= S union convert(map(proc(t) local s; 2^d - 2^k - add(2^(s),s=t) end proc,B),set);
od od:
sort(convert(S,list)); # Robert Israel, Sep 21 2023

Join[{1}, Select[Range[2, 600], IntegerExponent[#, 2] == Floor[Log2[# - 1]] - DigitCount[# - 1, 2, 1] &]] (* Amiram Eldar, Jul 16 2022 *)

isok(k) = if (k==1, 1, (logint(k-1, 2)-hammingweight(k-1) == valuation(k, 2))); \\ Michel Marcus, Jul 16 2022

from itertools import islice, count
def A353654_gen(startvalue=1): # generator of terms >= startvalue
    return filter(lambda n:(m:=(~n & n-1).bit_length()) == bin(n>>m)[2:].count('0'),count(max(startvalue,1)))
A353654_list = list(islice(A353654_gen(),30)) # Chai Wah Wu, Oct 14 2022

a(n) = a(n-1) + A356385(n-1) for n > 1 with a(1) = 1.
Conjectured formulas: (Start)
a(n) = 2^g(n-1)*(h(n-1) + 2^A000523(h(n-1))*(2 - g(n-1))) for n > 2 with a(1) = 1, a(2) = 3 where f(n) = n - A130312(n), g(n) = [n > 2*f(n)] and where h(n) = a(f(n) + 1).
a(n) = 1 + 2^r(n-1) + Sum_{k=1..r(n-1)} (1 - g(s(n-1, k)))*2^(r(n-1) - k) for n > 1 with a(1) = 1 where r(n) = A072649(n) and where s(n, k) = f(s(n, k-1)) for n > 0, k > 1 with s(n, 1) = n.
a(n) = 2*(2 + Sum_{k=1..n-2} 2^(A213911(A280514(k)-1) + 1)) - 2^A200650(n) for n > 1 with a(1) = 1.
A025480(a(n)-1) = A348366(A343152(n-1)) for n > 1.
a(A000045(n)) = 2^(n-1) - 1 for n > 1. (End)

A361989 a(n) is the sum of the Fibonacci numbers missing from the dual Zeckendorf representation of n; a(0) = 0, and for n > 0, a(n) = A022290(A035327(A003754(n+1))).

Original entry on oeis.org

0, 0, 1, 0, 2, 1, 0, 4, 3, 2, 1, 0, 7, 6, 5, 4, 3, 2, 1, 0, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 33, 32, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0

Offset: 0

Rémy Sigrist, Apr 02 2023

We consider that a Fibonacci number is missing from the dual Zeckendorf representation of a number if it does not appear in this representation and a larger Fibonacci number appears in it.
The dual Zeckendorf representation is also known as the lazy Fibonacci representation (see A356771 for further details).
This sequence can also be seen as an irregular table T(n, k), n > 0, k = 1..A000045(n), where T(n, k) = A000045(n) - k.
a(n-1) for n>=1 is the starting position of the first occurrence of one of the longest words w in the Fibonacci word A003849 such that no length-n factor of w is repeated. The length of such words is 2n. (See links) - Gandhar Joshi, Mar 19 2024

			For n = 42:
- using F(k) = A000045(k),
- the dual Zeckendorf representation of 42 is F(8) + F(7) + F(5) + F(3) + F(2),
- the numbers F(6) and F(4) are missing,
- so a(42) = F(6) + F(4) = 8 + 3 = 11.
.
As an irregular triangle the sequence begins:
     0;
     0;
     1,  0;
     2,  1,  0;
     4,  3,  2, 1, 0;
     7,  6,  5, 4, 3, 2, 1, 0;
    12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0;
    ...

Gandhar Joshi, Walnut code and details
Index entries for sequences related to Zeckendorf expansion of n

Cf. A000045, A003754, A022290, A035327, A072649, A130312, A132665, A194029, A356771.

for (n = 1, 9, for (k = 1, f = fibonacci(n), print1 (f-k", ")))

a(n) = A000045(A072649(n)) - A194029(n) for n > 0.

a(n) = A130312(n) - A194029(n) for n > 0.

A375430 The maximum exponent in the unique factorization of n in terms of distinct terms of A115975 using the dual Zeckendorf representation of the exponents in the prime factorization of n; a(1) = 0.

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1, 3, 1, 2, 1, 2, 1, 1, 1, 2, 2, 1, 2, 2, 1, 1, 1, 3, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 3, 2, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 3, 3, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 3, 1, 2, 2, 2, 1, 1, 1, 2, 1

Offset: 1

Amiram Eldar, Aug 15 2024

First differs from A299090 at n = 128. Differs from A046951 and A159631 at n = 1, 36, 64, 72, ... .

When the exponents in the prime factorization of n are expanded as sums of distinct Fibonacci numbers using the dual Zeckendorf representation (A104326), we get a unique factorization of n in terms of distinct terms of A115975, i.e., n is represented as a product of prime powers (A246655) whose exponents are Fibonacci numbers. a(n) is the maximum exponent of these prime powers. Thus all the terms are Fibonacci numbers.

			For n = 8 = 2^3, the dual Zeckendorf representation of 3 is 11, i.e., 3 = Fibonacci(2) + Fibonacci(3) = 1 + 2. Therefore 8 = 2^(1+2) = 2^1 * 2^2, and a(8) = 2.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000045, A095791, A104326, A115975, A246655, A331109, A368781, A375428, A375431.

Cf. A046951, A159631, A299090.

A130312[n_] := Module[{k = 0}, While[Fibonacci[k] <= n, k++]; Fibonacci[k-2]]; a[n_] := A130312[1 + Max[FactorInteger[n][[;;, 2]]]]; a[1] = 0; Array[a, 100]

A130312(n) = {my(k = 0); while(fibonacci(k) <= n, k++); fibonacci(k-2);}
a(n) = if(n == 1, 0, A130312(1 + vecmax(factor(n)[,2])));

a(n) = A130312(1 + A051903(n)).
a(n) = A000045(A375431(n)).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1 + Sum_{k>=4} Fibonacci(k) * (1 - 1/zeta(Fibonacci(k)-1)) = 1.48543763231328442311... .

A345067 Consider the "Quilt Tiling"; T(n, k) is the area of the tile containing the unit square whose upper right corner has coordinates (n, k); square array T(n, k) read by antidiagonals upwards, n, k > 0.

Original entry on oeis.org

1, 2, 2, 2, 4, 2, 6, 4, 4, 6, 6, 6, 4, 6, 6, 6, 6, 1, 1, 6, 6, 15, 6, 2, 9, 2, 6, 15, 15, 15, 2, 9, 9, 2, 15, 15, 15, 15, 15, 9, 9, 9, 15, 15, 15, 15, 15, 15, 2, 9, 9, 2, 15, 15, 15, 15, 15, 15, 2, 4, 9, 4, 2, 15, 15, 15, 40, 15, 15, 6, 4, 4, 4, 4, 6, 15, 15, 40

Offset: 1

Rémy Sigrist, Jun 06 2021

The "Quilt Tiling" is described in Shectman's paper (see Links section).

All terms belong to A006498.

			Array T(n, k) begins:
  n\k|  1   2   3   4   5   6   7   8   9  10  11
  ---+---+-------+-----------+-------------------+
   1 |  1|  2   2|  6   6   6| 15  15  15  15  15|
     +-----------+           |                   |
   2 |  2|  4   4|  6   6   6| 15  15  15  15  15|
     |   |       +---+-------+                   |
   3 |  2|  4   4|  1|  2   2| 15  15  15  15  15|
     +---+---+---+---+-------+-------+-----------+
   4 |  6   6|  1|  9   9   9|  2   2|  6   6   6|
     |       +---+           +-------+           |
   5 |  6   6|  2|  9   9   9|  4   4|  6   6   6|
     |       |   |           |       +---+-------+
   6 |  6   6|  2|  9   9   9|  4   4|  1|  2   2|
     +-------+---+---+-------+-------+---+-------+
   7 | 15  15  15|  2|  4   4| 25  25  25  25  25|
     |           |   |       |                   |
   8 | 15  15  15|  2|  4   4| 25  25  25  25  25|
     |           +---+---+---+                   |
   9 | 15  15  15|  6   6|  1| 25  25  25  25  25|
     |           |       +---+                   |
  10 | 15  15  15|  6   6|  2| 25  25  25  25  25|
     |           |       |   |                   |
  11 | 15  15  15|  6   6|  2| 25  25  25  25  25|
     +-----------+-------+---+-------------------+

Rémy Sigrist, Table of n, a(n) for n = 1..10153
J. Parker Shectman, A Quilt after Fibonacci, Part 2 of 3: Cohorts, Free Monoids, and Numeration
Rémy Sigrist, Illustration of the connection between the "Quilt Tiling" and the sequences A000201 and A005206
Rémy Sigrist, PARI program for A345067

Cf. A000045, A001654, A005206, A006498, A095791, A005206.

PARI
```
See Links section.
```

T(n, k) = T(k, n).
T(n, n) = A130312(n+1)^2.
T(n, 1) = A001654(A095791(n)+1).
T(n, k) is the square of a Fibonacci number for n = 1+A005206(k+1)..A000201(k).

A347188 a(n) = A346422(4*A003754(n-1) + 3) for n > 1 with a(1) = 1.

Original entry on oeis.org

1, 2, 6, 4, 24, 18, 12, 120, 8, 96, 72, 48, 720, 54, 36, 600, 24, 480, 360, 240, 5040, 16, 384, 288, 192, 4320, 216, 144, 3600, 96, 2880, 2160, 1440, 40320, 162, 108, 3000, 72, 2400, 1800, 1200, 35280, 48, 1920, 1440, 960, 30240, 1080, 720, 25200, 480, 20160

Offset: 1

Mikhail Kurkov, Aug 21 2021 [verification needed]

nonn

To get the distinct terms of A346422 in the order of their appearance up to A346422(2^n - 1), just take the first A000045(n+1) terms of this sequence and remove the duplicates.

Cf. A003754, A130312, A200649, A346422.

a(n) = (1 + A200649(n))*a(n - A130312(n-1)) for n > 1 with a(1) = 1.

a(n) = A346422(4*A003754(n-1) + 3) for n > 1 with a(1) = 1.

cp's OEIS Frontend

A348366 a(n) = 2*a(n - A130312(n)) + (A072649(n) - A072649(n - A130312(n))) mod 2 for n > 1 with a(0) = 0, a(1) = 1.

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A355429 Square array T(n, k), n >= 0, k > 0, read by antidiagonals, where T(0, k) = A001906(k) for k > 0 and where T(n, k) = n - A130312(n) + A000045(2k + A072649(n)) for n > 0, k > 0.

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A357589 a(n) = n - A130312(n).

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A200649 Number of 1's in the Stolarsky representation of n.

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A353654 Numbers whose binary expansion has the same number of trailing 0 bits as other 0 bits.

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A361989 a(n) is the sum of the Fibonacci numbers missing from the dual Zeckendorf representation of n; a(0) = 0, and for n > 0, a(n) = A022290(A035327(A003754(n+1))).

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A375430 The maximum exponent in the unique factorization of n in terms of distinct terms of A115975 using the dual Zeckendorf representation of the exponents in the prime factorization of n; a(1) = 0.

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A345067 Consider the "Quilt Tiling"; T(n, k) is the area of the tile containing the unit square whose upper right corner has coordinates (n, k); square array T(n, k) read by antidiagonals upwards, n, k > 0.

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