0, 1, 2, 2, 3, 3, 4, 3, 4, 4, 5, 4, 5, 5, 5, 4, 5, 5, 6, 5, 6, 6, 7, 5, 6, 6, 6, 6, 7, 6, 7, 5, 6, 6, 7, 6, 7, 7, 7, 6, 7, 7, 8, 7, 7, 8, 9, 6, 7, 7, 7, 7, 8, 7, 8, 7, 8, 8, 9, 7, 8, 8, 8, 6, 7, 7, 8, 7, 8, 8, 9, 7, 8, 8, 8, 8, 9, 8, 9, 7, 8, 8, 9, 8, 8, 9, 9, 8, 9, 8, 9, 9, 9, 10, 9, 7, 8, 8, 8, 8, 9, 8, 9, 8, 9
Offset: 1
A117632
Number of 1's required to build n using {+,T} and parentheses, where T(i) = i*(i+1)/2.
Original entry on oeis.org
1, 2, 2, 3, 4, 2, 3, 4, 4, 3, 4, 4, 5, 6, 4, 5, 6, 6, 7, 6, 2, 3, 4, 4, 5, 6, 4, 3, 4, 5, 5, 6, 6, 5, 6, 4, 5, 6, 6, 7, 8, 4, 5, 6, 4, 5, 6, 6, 5, 6, 6, 7, 8, 8, 3, 4, 5, 5, 6, 7, 5, 6, 6, 7, 6, 4, 5, 6, 6, 7, 8, 6, 7, 8, 8, 5, 6, 4, 5, 6, 6, 7, 6, 6, 7, 8, 6, 7, 8, 8, 5, 6, 7, 7, 8, 9, 7, 8, 6, 7, 8, 8
Offset: 1
a(1) = 1 because "1" has a single 1.
a(2) = 2 because "1+1" has two 1's.
a(3) = 2 because 3 = T(1+1) has two 1's.
a(6) = 2 because 6 = T(T(1+1)).
a(10) = 3 because 10 = T(T(1+1)+1).
a(12) = 4 because 12 = T(T(1+1)) + T(T(1+1)).
a(15) = 4 because 15 = T(T(1+1)+1+1).
a(21) = 2 because 21 = T(T(T(1+1))).
a(28) = 3 because 28 = T(T(T(1+1))+1).
a(55) = 3 because 55 = T(T(T(1+1)+1)).
- W. A. Beyer, M. L. Stein and S. M. Ulam, The Notion of Complexity. Report LA-4822, Los Alamos Scientific Laboratory of the University of California, Los Alamos, NM, 1971.
- R. K. Guy, Unsolved Problems Number Theory, Sect. F26.
- Alois P. Heinz, Table of n, a(n) for n = 1..10000
- W. A. Beyer, M. L. Stein and S. M. Ulam, The Notion of Complexity. Report LA-4822, Los Alamos Scientific Laboratory of the University of California, Los Alamos, NM, December 1971. [Annotated scanned copy]
- R. K. Guy, Some suspiciously simple sequences, Amer. Math. Monthly 93 (1986), 186-190; 94 (1987), 965; 96 (1989), 905.
- Ed Pegg, Jr., Integer Complexity.
- Eric Weisstein's World of Mathematics, Integer Complexity.
- Wikipedia, Optimal substructure.
Cf.
A000217,
A005245,
A005520,
A003313,
A076142,
A076091,
A061373,
A005421,
A023361,
A053614,
A064097,
A025280,
A003037,
A099129,
A050536,
A050542,
A050548,
A050909,
A007501.
-
a:= proc(n) option remember; local m; m:= floor (sqrt (n*2));
if n<3 then n
elif n=m*(m+1)/2 then a(m)
else min (seq (a(i)+a(n-i), i=1..floor(n/2)))
fi
end:
seq (a(n), n=1..110); # Alois P. Heinz, Jan 05 2011
-
a[n_] := a[n] = Module[{m = Floor[Sqrt[n*2]]}, If[n < 3, n, If[n == m*(m + 1)/2, a[m], Min[Table[a[i] + a[n - i], {i, 1, Floor[n/2]}]]]]];
Array[a, 110] (* Jean-François Alcover, Jun 02 2018, from Maple *)
I do not know how many of these entries have been proved to be minimal. -
N. J. A. Sloane, Apr 15 2006
A118387
Record values in A079300 (number of shortest addition chains of n).
Original entry on oeis.org
1, 2, 5, 15, 33, 40, 132, 220, 352, 1258, 2661, 3903, 9787, 12989, 17961, 31373, 33076, 55262, 110624
Offset: 1
A186435
Number of evaluation schemes for x^n achieving the minimal number of multiplications.
Original entry on oeis.org
1, 1, 1, 1, 2, 2, 6, 1, 3, 4, 19, 3, 10, 16, 4, 1, 2, 7, 37, 6, 31, 48, 4, 4, 14, 24, 5, 26, 152, 12, 80, 1, 2, 4, 51, 12, 39, 100, 20, 8, 23, 90, 4, 81, 14, 8, 242, 5, 12, 36, 4, 38, 215, 16, 172, 36, 190, 395, 40, 24
Offset: 1
For n=7, we can evaluate x^7 using only 4 multiplications in 6 ways:
x^2 = x * x ; x^3 = x * x^2 ; x^4 = x * x^3 ; x^7 = x^3 * x^4
x^2 = x * x ; x^3 = x * x^2 ; x^4 = x^2 * x^2 ; x^7 = x^3 * x^4
x^2 = x * x ; x^3 = x * x^2 ; x^5 = x^2 * x^3 ; x^7 = x^2 * x^5
x^2 = x * x ; x^3 = x * x^2 ; x^6 = x^3 * x^3 ; x^7 = x * x^6
x^2 = x * x ; x^4 = x^2 * x^2 ; x^5 = x * x^4 ; x^7 = x^2 * x^5
x^2 = x * x ; x^4 = x^2 * x^2 ; x^6 = x^2 * x^4 ; x^7 = x * x^6
See
A003313 for the minimal number of multiplications to evaluate x^n.
See
A001190 for the total number of evaluation schemes for x^n (regardless of the number of effective multiplications).
A079300 gives the number of minimal chains (= sequences of powers of x) ending at x^n. This is actually a bit less than the number of evaluation schemes since two schemes may produce the same chain, like the first and second schemes in the example above, where the corresponding chain is (x^2, x^3, x^4, x^7).
A186437
Maximal number of squarings in an evaluation scheme for x^n achieving the minimal number of operations.
Original entry on oeis.org
0, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 3, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 4, 5, 5, 5, 4, 5, 5, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5
Offset: 1
For n=5, we can evaluate x^5 using only 3 operations in 2 ways:
x^2 = (x)^2; x^3 = x * x^2; x^5 = x^2 * x^3
x^2 = (x)^2; x^4 = (x^2)^2; x^5 = x * x^4
The second way achieves the maximal number of doublings, which is a(5) = 2.
A353058
Minimum number of iterations {add or subtract 1, or half if even} needed to reach 1, starting from n.
Original entry on oeis.org
0, 1, 2, 2, 3, 3, 4, 3, 4, 4, 5, 4, 5, 5, 5, 4, 5, 5, 6, 5, 6, 6, 6, 5, 6, 6, 7, 6, 7, 6, 6, 5, 6, 6, 7, 6, 7, 7, 7, 6, 7, 7, 8, 7, 8, 7, 7, 6, 7, 7, 8, 7, 8, 8, 8, 7, 8, 8, 8, 7, 8, 7, 7, 6, 7, 7, 8, 7, 8, 8, 8, 7, 8, 8, 9, 8, 9, 8, 8, 7, 8, 8, 9, 8, 9, 9, 9, 8
Offset: 1
For n = 1, the value 1 is already reached, so a(1) = 0 iterations are needed.
For n = 2, one can either subtract 1 or divide by 2 to reach 1, i.e., a(2) = 1 iterations are needed.
For n = 3, one must subtract 1 twice in order to reach the goal 1 in the minimum number of a(3) = 2 steps: Initially, one cannot divide by 2 since 3 is even, and if 1 is added, to get 3 + 1 = 4, at least two divisions by 2 (for a total of 3 steps) would be needed to reach 1.
For n = 7, the fastest is to add 1 (to get 7 + 1 = 8) and then divide three times by 2, to reach 1 in the minimum number of a(7) = 4 steps.
Cf.
A003313 (shortest addition chain),
A137813 (minimal set with n-topology),
A128998 (shortest addition-subtraction chain).
-
apply( {A353058(n,o=[n])=for(i=0,n,o[1]>1||return(i);o=Set(concat([if(n%2,[n-1,n+1],n\2)|n<-o])))}, [1..90])
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