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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.

Showing 1-4 of 4 results.

A265745 a(n) is the number of Jacobsthal numbers (A001045) needed to sum to n using the greedy algorithm.

Original entry on oeis.org

0, 1, 2, 1, 2, 1, 2, 3, 2, 3, 2, 1, 2, 3, 2, 3, 2, 3, 4, 3, 4, 1, 2, 3, 2, 3, 2, 3, 4, 3, 4, 3, 2, 3, 4, 3, 4, 3, 4, 5, 4, 5, 2, 1, 2, 3, 2, 3, 2, 3, 4, 3, 4, 3, 2, 3, 4, 3, 4, 3, 4, 5, 4, 5, 2, 3, 4, 3, 4, 3, 4, 5, 4, 5, 4, 3, 4, 5, 4, 5, 4, 5, 6, 5, 6, 1, 2, 3, 2, 3, 2, 3, 4, 3, 4, 3, 2, 3, 4, 3, 4, 3, 4, 5, 4, 5, 2
Offset: 0

Views

Author

Antti Karttunen, Dec 17 2015

Keywords

Comments

Sum of digits in "Jacobsthal greedy base", A265747.
It would be nice to know for sure whether this sequence gives also the least number of Jacobsthal numbers that add to n, i.e., that there cannot be even better nongreedy solutions.
The integer 63=21+21+21 has 3 for its 'non-greedy' solution, and a(63) = 5 for its greedy solution 63=43+11+5+3+1. - Yuriko Suwa, Jul 11 2021
Positions where a(n) is different from A372555(n) are n=63, 84, 148, 169, 191, 212, 234, 255, etc. See A372557. - Antti Karttunen, May 07 2024

Examples

			a(0) = 0, because no numbers are needed to form an empty sum, which is zero.
For n=1 we need just A001045(2) = 1, thus a(1) = 1.
For n=2 we need A001045(2) + A001045(2) = 1 + 1, thus a(2) = 2.
For n=4 we need A001045(3) + A001045(2) = 3 + 1, thus a(4) = 2.
For n=6 we form the greedy sum as A001045(4) + A001045(2) = 5 + 1, thus a(6) = 2. Alternatively, we could form the sum as A001045(3) + A001045(3) = 3 + 3, but the number of summands in that case is no less.
For n=7 we need A001045(4) + A001045(2) + A001045(2) = 5 + 1 + 1, thus a(7) = 3.
For n=8 we need A001045(4) + A001045(3) = 5 + 3, thus a(8) = 2.
For n=10 we need A001045(4) + A001045(4) = 5 + 5, thus a(10) = 2.

Links

Crossrefs

Cf. A001045, A130249, A197911, A003158, A265746, A265747, A364377, A364379, A364380, A372288, A372555, A372557.
Cf. A054111 (apparently the positions of the first occurrence of each n > 0).
Cf. also A007895, A014420, A053610, A265404, A265743, A265744.

Programs

  • Mathematica
    jacob[n_] := (2^n - (-1)^n)/3; maxInd[n_] := Floor[Log2[3*n + 1]]; A265745[n_] := A265745[n] = 1 + A265745[n - jacob[maxInd[n]]]; A265745[0] = 0; Array[A265745, 100, 0] (* Amiram Eldar, Jul 21 2023 *)
  • PARI
    A130249(n) = floor(log(3*n + 1)/log(2));
A001045(n) = (2^n - (-1)^n) / 3;
A265745(n) = {if(n == 0, 0, my(d = n - A001045(A130249(n))); if(d == 0, 1, 1 + A265745(d)));} \\ Amiram Eldar, Jul 21 2023
  • Python
    def greedyJ(n): n1 = (3*n+1).bit_length() - 1; return (2**n1 - (-1)**n1)//3
def a(n): return 0 if n == 0 else 1 + a(n - greedyJ(n))
print([a(n) for n in range(107)]) # Michael S. Branicky, Jul 11 2021

Formula

a(0) = 0; for n >= 1, a(n) = 1 + a(n - A001045(A130249(n))). [This formula uses a simple greedy algorithm.]

A078510 Spiro-Fibonacci numbers, a(n) = sum of two previous terms that are nearest when terms arranged in a spiral.

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 18, 21, 24, 27, 31, 36, 42, 48, 54, 61, 69, 78, 88, 98, 108, 119, 131, 144, 158, 172, 186, 201, 217, 235, 256, 280, 304, 328, 355, 386, 422, 464, 512, 560, 608, 662, 723, 792, 870, 958, 1056
Offset: 0

Views

Author

Neil Fernandez, Jan 05 2003

Keywords

Comments

Or "Spironacci numbers" for short. See also Spironacci polynomials, A265408. This sequence has an interesting growth rate, see A265370 and A265404. - Antti Karttunen, Dec 13 2015

Examples

			Terms are written in square boxes radiating spirally (cf. Ulam prime spiral). a(0)=0 and a(1)=1, so write 0 and then 1 to its right. a(2) goes in the box below a(1). The nearest two filled boxes contain a(0) and a(1), so a(2)=a(0)+a(1)=0+1=1. a(3) goes in the box to the left of a(2). The nearest two filled boxes contain a(0) and a(2), so a(3)=a(0)+a(2)=0+1=1.
From _Antti Karttunen_, Dec 17 2015: (Start)
The above description places cells in clockwise direction. However, for the computation of this sequence the actual orientation of the spiral is irrelevant. Following the convention used at A265409, we draw this spiral counterclockwise:
+--------+--------+--------+--------+
|a(15)   |a(14)   |a(13)   |a(12)   |
| = a(14)| = a(13)| = a(12)| = a(11)|
| + a(4) | + a(3) | + a(2) | + a(2) |
| = 9    | = 8    | = 7    | = 6    |
+--------+--------+--------+--------+
|a(4)    |a(3)    |a(2)    |a(11)   |
| = a(3) | = a(2) | = a(1) | = a(10)|
| + a(0) | + a(0) | + a(0) | + a(2) |
| = 1    | = 1    | = 1    | = 5    |
+--------+--------+--------+--------+
|a(5)    | START  |   ^    |a(10)   |
| = a(4) | a(0)=0 | a(1)=1 | = a(9) |
| + a(0) |   -->  |        | + a(1) |
| = 1    |        |        | = 4    |
+--------+--------+--------+--------+
|a(6)    |a(7)    |a(8)    |a(9)    |
| = a(5) | = a(6) | = a(7) | = a(8) |
| + a(0) | + a(0) | + a(1) | + a(1) |
| = 1    | = 1    | = 2    | = 3    |
+--------+--------+--------+--------+
(End)

Links

Crossrefs

Cf. A000045, A001222, A033951, A063826, A265407, A265408, A265409, A265359, A265370, A265404.

Formula

From Antti Karttunen, Dec 13 2015: (Start)
a(0) = 0, a(1) = 1; for n > 1, a(n) = a(n-1) + a(A265409(n)).
equally, for n > 1, a(n) = a(n-1) + a(n - A265359(n)).
a(n) = A001222(A265408(n)).
(End)

A265744 a(n) is the number of Pell numbers (A000129) needed to sum to n using the greedy algorithm (A317204).

Original entry on oeis.org

0, 1, 1, 2, 2, 1, 2, 2, 3, 3, 2, 3, 1, 2, 2, 3, 3, 2, 3, 3, 4, 4, 3, 4, 2, 3, 3, 4, 4, 1, 2, 2, 3, 3, 2, 3, 3, 4, 4, 3, 4, 2, 3, 3, 4, 4, 3, 4, 4, 5, 5, 4, 5, 3, 4, 4, 5, 5, 2, 3, 3, 4, 4, 3, 4, 4, 5, 5, 4, 5, 1, 2, 2, 3, 3, 2, 3, 3, 4, 4, 3, 4, 2, 3, 3, 4, 4, 3, 4, 4, 5, 5, 4, 5, 3, 4, 4, 5, 5, 2, 3, 3, 4, 4, 3, 4, 4, 5, 5, 4, 5, 3
Offset: 0

Views

Author

Antti Karttunen, Dec 17 2015

Keywords

Comments

a(0) = 0, because no numbers are needed to form an empty sum, which is zero.
It would be nice to know for sure whether this sequence also gives the least number of Pell numbers that add to n, i.e., that there cannot be even better nongreedy solutions.

References

  • A. F. Horadam, Zeckendorf representations of positive and negative integers by Pell numbers, Applications of Fibonacci Numbers, Springer, Dordrecht, 1993, pp. 305-316.

Links

Crossrefs

Cf. A000129, A007953, A317204.
Cf. also A014420, A053610, A265404, A265743, A265745.
Similar sequences: A007895, A116543, A278043.

Programs

  • Mathematica
    pell[1] = 1; pell[2] = 2; pell[n_] := pell[n] = 2*pell[n - 1] + pell[n - 2]; a[n_] := Module[{s = {}, m = n, k}, While[m > 0, k = 1; While[pell[k] <= m, k++]; k--; AppendTo[s, k]; m -= pell[k]; k = 1]; Plus @@ IntegerDigits[Total[3^(s - 1)], 3]]; Array[a, 100, 0] (* Amiram Eldar, Mar 12 2022 *)

Formula

a(n) = A007953(A317204(n)). - Amiram Eldar, Mar 12 2022

A265743 a(n) = number of terms of A005187 needed to sum to n using the greedy algorithm.

Original entry on oeis.org

0, 1, 2, 1, 1, 2, 3, 1, 1, 2, 1, 1, 2, 3, 2, 1, 1, 2, 1, 1, 2, 3, 1, 1, 2, 1, 1, 2, 3, 2, 2, 1, 1, 2, 1, 1, 2, 3, 1, 1, 2, 1, 1, 2, 3, 2, 1, 1, 2, 1, 1, 2, 3, 1, 1, 2, 1, 1, 2, 3, 2, 2, 3, 1, 1, 2, 1, 1, 2, 3, 1, 1, 2, 1, 1, 2, 3, 2, 1, 1, 2, 1, 1, 2, 3, 1, 1, 2, 1, 1, 2, 3, 2, 2, 1, 1, 2, 1, 1, 2, 3, 1, 1, 2, 1, 1, 2, 3, 2, 1, 1, 2, 1, 1, 2, 3, 1, 1, 2, 1, 1, 2, 3, 2, 2, 3, 4, 1, 1
Offset: 0

Views

Author

Antti Karttunen, Dec 17 2015

Keywords

Comments

a(0) = 0, because no numbers are needed to form an empty sum, which is zero.

Links

Crossrefs

Cf. A005187, A055938.
Cf. also A000120, A007895, A053610, A265404, A265744, A265745.

Formula

Other identities. For all n >= 1:
a(A005187(n)) = 1 and a(A055938(n)) > 1.
Showing 1-4 of 4 results.

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