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PROG
(PARI)
A276086(n) = { my(m=1, p=2); while(n, m *= (p^(n%p)); n = n\p; p = nextprime(1+p)); (m); };
CROSSREFS
Cf. A060735 (gives the positions of other zeros after the initial a(0)=0).
Sprague-Grundy (or Nim) values for the game of Maundy cake on an n X 1 sheet.
(Formerly M2219)
+10
12
0, 1, 1, 3, 1, 4, 1, 7, 4, 6, 1, 10, 1, 8, 6, 15, 1, 13, 1, 16, 8, 12, 1, 22, 6, 14, 13, 22, 1, 21, 1, 31, 12, 18, 8, 31, 1, 20, 14, 36, 1, 29, 1, 34, 21, 24, 1, 46, 8, 31, 18, 40, 1, 40, 12, 50, 20, 30, 1, 51, 1, 32, 29, 63, 14, 45, 1, 52, 24, 43, 1, 67, 1, 38, 31, 58, 12, 53, 1
COMMENTS
There are three equivalent formulas for a(n). Suppose n >= 2, and let p1 <= p2 <= ... <= pk be the prime factors of n, with repetition.
Theorem 1: a(1) = 0. For n >= 2, a(n) = n*s(n), where
s(n) = 1/p1 + 1/(p1*p2) + 1/(p1*p2*p3) + ... + 1/(p1*p2*...*pk).
This is implicit in Berlekamp, Conway and Guy, Winning Ways, 2 vols., 1982, pp. 28, 53.
David James Sycamore observed on Nov 24 2018 that Theorem 1 implies a(n) < n for all n (see comments in A322034), and also leads to a simple recurrence for a(n):
Theorem 2: a(1) = 0. For n >= 2, a(n) = p*a(n/p) + 1, where p is the largest prime factor of n.
Proof. (Th. 1 implies Th. 2) If n is a prime, Theorem 1 gives a(n) = 1 = n*a(1)+1. For a nonprime n, let n = m*p where p is the largest prime factor of n and m >= 2. From Theorem 1, a(m) = m*s(m), a(n) = q*m*(s(m) + 1/n) = q*a(m) + 1.
(Th. 2 implies Th. 1) The reverse implication is equally easy.
Theorem 2 is equivalent to the following more complicated recurrence:
Theorem 3: a(1) = 0. For n >= 2, a(n) = max_{p|n, p prime} (p*a(n/p)+1).
REFERENCES
E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways, Academic Press, NY, 2 vols., 1982, see p. 28, 53.
E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways, Second Edition, Vol. 1, A K Peters, 2001, pp. 27, 51.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
FORMULA
a(n) = n * Sum_{k=1..N} (1/(p1^m1*p2^m2*...*pk^mk)) * (pk^mk-1)/(pk-1) for n>=2, where pk is the k-th distinct prime factor of n, N is the number of distinct prime factors of n, and mk is the multiplicity of pk occurring in n. To prove this, expand the factors in Theorem 1 and use the geometrical series identity. - Jonathan Blanchette, Nov 01 2019 a(n) = Sum_{k=1..bigomega(n)} F^k(n), where F^k(n) is the k-th iterate of F(n) = A032742(n). - Ridouane Oudra, Jan 26 2024
EXAMPLE
For n=24, s(24) = 1/2 + 1/4 + 1/8 + 1/24 = 11/12, so a(24) = 24*11/12 = 22.
MAPLE
P:=proc(n) local FM: FM:=ifactors(n)[2]: seq(seq(FM[j][1], k=1..FM[j][2]), j=1..nops(FM)) end: # A027746s:=proc(n) local i, t, b; global P; t:=0; b:=1; for i in [P(n)] do b:=b*i; t:=t+1/b; od; t; end; # A322034/ A322035A006022 := n -> if n = 1 then 0 else n*s(n); fi;
MATHEMATICA
Nest[Function[{a, n}, Append[a, Max@ Map[# a[[n/#]] + 1 &, Rest@ Divisors@ n]]] @@ {#, Length@ # + 1} &, {0, 1}, 77] (* Michael De Vlieger, Nov 23 2018 *)
PROG
(Haskell)
a006022 1 = 0
a006022 n = (+ 1) $ sum $ takeWhile (> 1) $
iterate (\x -> x `div` a020639 x) (a032742 n)
(PARI) lista(nn) = {my(v = vector(nn)); for (n=1, nn, if (n>1, my(m = 0); fordiv (n, d, if (d>1, m = max(m, d*v[n/d]+1))); v[n] = m; ); print1(v[n], ", "); ); } \\ Michel Marcus, Nov 25 2018
CROSSREFS
Cf. also A027746, A306264, A306363, A322034, A322035, A322036, A322382, A328898, A333783, A332993, A333791.
a(1) = 1, for n > 1, a(n) = n + a( A032742(n)).
+10
11
1, 3, 4, 7, 6, 10, 8, 15, 13, 16, 12, 22, 14, 22, 21, 31, 18, 31, 20, 36, 29, 34, 24, 46, 31, 40, 40, 50, 30, 51, 32, 63, 45, 52, 43, 67, 38, 58, 53, 76, 42, 71, 44, 78, 66, 70, 48, 94, 57, 81, 69, 92, 54, 94, 67, 106, 77, 88, 60, 111, 62, 94, 92, 127, 79, 111, 68, 120, 93, 113, 72, 139, 74, 112, 106, 134, 89, 131, 80, 156, 121
COMMENTS
Sum of those divisors of n that can be obtained by repeatedly taking the largest proper divisor (of previous such divisor, starting from n, which is included in the sum), up to and including the terminal 1.
FORMULA
a(1) = 1; and for n > 1, a(n) = n + a( A032742(n)).
EXAMPLE
a(18) = 18 + 18/2 + 9/3 + 3/3 = 18 + 9 + 3 + 1 = 31.
PROG
(PARI) A332993(n) = if(1==n, n, n + A332993(n/vecmin(factor(n)[, 1])));
a(1) = 1, for n > 1, a(n) = n + a( A052126(n)).
+10
10
1, 3, 4, 7, 6, 9, 8, 15, 13, 13, 12, 19, 14, 17, 19, 31, 18, 27, 20, 27, 25, 25, 24, 39, 31, 29, 40, 35, 30, 39, 32, 63, 37, 37, 41, 55, 38, 41, 43, 55, 42, 51, 44, 51, 58, 49, 48, 79, 57, 63, 55, 59, 54, 81, 61, 71, 61, 61, 60, 79, 62, 65, 76, 127, 71, 75, 68, 75, 73, 83, 72, 111, 74, 77, 94, 83, 85, 87, 80, 111, 121
FORMULA
a(1) = 1; and for n > 1, a(n) = n + a( A052126(n)).
PROG
(PARI) A332994(n) = if(1==n, n, n + A332994(n/vecmax(factor(n)[, 1])));
0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 6, 0, 2, 3, 0, 0, 8, 0, 6, 3, 2, 0, 14, 0, 2, 0, 6, 0, 21, 0, 0, 3, 2, 5, 24, 0, 2, 3, 14, 0, 25, 0, 6, 12, 2, 0, 30, 0, 12, 3, 6, 0, 26, 5, 14, 3, 2, 0, 57, 0, 2, 12, 0, 5, 33, 0, 6, 3, 31, 0, 56, 0, 2, 18, 6, 7, 37, 0, 30, 0, 2, 0, 69, 5, 2, 3, 14, 0, 78, 7, 6, 3, 2, 5, 62, 0, 16, 12
COMMENTS
Sum of all other divisors of n, except those divisors that can be obtained by repeatedly taking the largest proper divisor (of previous such divisor, starting from n), up to and including the terminal 1.
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