A364377 -id:A364377 - cp's OEIS frontend

A324477 k appears t+1 times, where t = A364377(k) is the number of trailing zeros in the greedy Jacobsthal representation of k, A265747(k).

1, 2, 3, 3, 4, 5, 5, 5, 6, 7, 8, 8, 9, 10, 10, 10, 11, 11, 11, 11, 12, 13, 14, 14, 15, 16, 16, 16, 17, 18, 19, 19, 20, 21, 21, 21, 21, 21, 22, 23, 24, 24, 25, 26, 26, 26, 27, 28, 29, 29, 30, 31, 31, 31, 32, 32, 32, 32, 33, 34, 35, 35, 36, 37, 37, 37, 38, 39, 40, 40

Offset: 1

Nathan Fox and N. J. A. Sloane, Mar 09 2019

Amiram Eldar, Table of n, a(n) for n = 1..10000
Nathan Fox, Trees, Fibonacci Numbers, and Nested Recurrences, Rutgers University Experimental Math Seminar, Mar 07, 2019.

Cf. A001045, A265747, A364377.

A046699, A316628, A324473 and A324475 have similar definitions.

Table[Table[k, {IntegerExponent[A265747[k], 10] + 1}], {k, 1, 40}] // Flatten (* Amiram Eldar, Jul 21 2023 using A265747[n] *)

More terms from Amiram Eldar, Jul 21 2023

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

Antti Karttunen, Dec 17 2015

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

			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.

Antti Karttunen, Table of n, a(n) for n = 0..10923

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.

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 *)

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

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

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

cp's OEIS Frontend

A324477 k appears t+1 times, where t = A364377(k) is the number of trailing zeros in the greedy Jacobsthal representation of k, A265747(k).

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

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