A344444 - OEIS

%I #17 Jun 14 2022 06:54:47

%S 0,12,19,24,28,31,34,36,38,40,42,43,45,46,47,48,49,50,51,52,53,54,55,

%T 55,56,57,57,58,59,59,60,60,61,61,62,62,63,63,64,64,65,65,66,66,66,67,

%U 67,67,68,68,68,69,69,69,70,70,70,71,71,71,72,72,72,72,73,73,73,73,74,74

%N Completely additive with a(2) = 12, a(3) = 19; for prime p > 3, a(p) = ceiling((a(p-1) + a(p+1))/2).

%C Monotonic until a(143) = 87 > 86 = a(144).

%C The only infinite monotonic completely additive integer sequence is the all 0's sequence (cf. A000004). The challenge taken up here is to specify one that is monotonic for a modestly long number of terms, using a comparatively short prescriptive definition.

%C To start we specify values for a(2) and a(3) so that a(3)/a(2) approximates log(3)/log(2). 19/12 is a good approximation relative to the size of denominator. This reflects 2^19 = 524288 having a similar magnitude to 3^12 = 531441. Equivalently, we can say 3 is approximately the 19th power of the 12th root of 2. This approximation is used to construct musical scales. (See the Enevoldsen link, also A143800.) There is no better approximation with a denominator smaller than 29. [Revised by _Peter Munn_, Jun 14 2022]

%C To find a good specification to use for a(p) for larger primes, p, we are guided by knowing that if 2*a(n) < a(n-1) + a(n+1) then a completely additive sequence is not monotonic after a(n^2-1) because a(n^2) < a((n-1)*(n+1)) = a(n^2-1). Considering n = p, we see we want a(p) >= (a(p-1) + a(p+1))/2; but the same consideration for n = p-1 shows we don't want a(p) larger than necessary. These considerations lead towards the choice of "a(p) = ceiling((a(p-1) + a(p+1))/2)" for use in the definition.

%H Encyclopedia of Mathematics, <a href="https://encyclopediaofmath.org/wiki/Additive_arithmetic_function">Additive arithmetic function</a>

%H Keith Enevoldsen, <a href="https://thinkzone.wlonk.com/Music/12Tone.htm">Twelve-Tone Musical Scale</a>

%H <a href="/wiki/Index_to_OEIS:_Section_Mu#music">Index entries for sequences related to music</a>

%F a(n*k) = a(n) + a(k).

%e a(4) = a(2*2) = a(2) + a(2) from the definition of completely additive. So a(4) = 12 + 12 = 24. Similarly, a(6) = a(2*3) = a(2) + a(3) = 12 + 19 = 31.

%e 5 is a prime number greater than 3, so a(5) = ceiling((a(5-1) + a(5+1))/2). Using the values a(4) = 24 and a(6) = 31 that we calculated earlier, we get a(5) = ceiling((24 + 31)/2) = ceiling(27.5) = 28.

%e The sequence is defined as completely additive, so a(1) = 0, the identity element for addition. (To see this, note that "completely additive" implies a(2) = a(2*1) = a(2)+a(1), and solve the equation for a(1).)

%t a[1] = 0; a[n_] := a[n] = Plus @@ ((Last[#] * Which[First[#] == 2, 12, First[#] == 3, 19, First[#] > 3, Ceiling[(a[First[#] - 1] + a[First[#] + 1])/2]]) & /@ FactorInteger[n]); Array[a, 100] (* _Amiram Eldar_, Jun 27 2021 *)

%Y Equivalent sequence with a(2)=5, a(3)=8: A344443.

%Y First 10 terms match A143800.

%Y Cf. row 23 of A352957.

%Y Cf. A000004, A020857, A060528, A061920.

%Y For other completely additive sequences see the references in A104244.

%K nonn,easy

%O 1,2

%A _Peter Munn_, May 20 2021