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%I M2219 #148 Feb 17 2024 12:46:54
%S 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,
%T 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,
%U 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
%N Sprague-Grundy (or Nim) values for the game of Maundy cake on an n X 1 sheet.
%C There are three equivalent formulas for a(n). Suppose n >= 2, and let p1 <= p2 <= ... <= pk be the prime factors of n, with repetition.
%C Theorem 1: a(1) = 0. For n >= 2, a(n) = n*s(n), where
%C s(n) = 1/p1 + 1/(p1*p2) + 1/(p1*p2*p3) + ... + 1/(p1*p2*...*pk).
%C This is implicit in Berlekamp, Conway and Guy, Winning Ways, 2 vols., 1982, pp. 28, 53.
%C Note that s(n) = A322034(n) / A322035(n).
%C _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):
%C Theorem 2: a(1) = 0. For n >= 2, a(n) = p*a(n/p) + 1, where p is the largest prime factor of n.
%C 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.
%C (Th. 2 implies Th. 1) The reverse implication is equally easy.
%C Theorem 2 is equivalent to the following more complicated recurrence:
%C Theorem 3: a(1) = 0. For n >= 2, a(n) = max_{p|n, p prime} (p*a(n/p)+1).
%D E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways, Academic Press, NY, 2 vols., 1982, see p. 28, 53.
%D E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways, Second Edition, Vol. 1, A K Peters, 2001, pp. 27, 51.
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H Reinhard Zumkeller, <a href="/A006022/b006022.txt">Table of n, a(n) for n = 1..10000</a>
%H Jonathan Blanchette and Robert Laganière, <a href="https://arxiv.org/abs/1910.11749">A Curious Link Between Prime Numbers, the Maundy Cake Problem and Parallel Sorting</a>, arXiv:1910.11749 [cs.DS], 2019.
%F 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
%F From _Antti Karttunen_, Apr 12 2020: (Start)
%F a(n) = A322382(n) + A333791(n).
%F a(n) = A332993(n) - n = A001065(n) - A333783(n). (End)
%F 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
%e For n=24, s(24) = 1/2 + 1/4 + 1/8 + 1/24 = 11/12, so a(24) = 24*11/12 = 22.
%p P:=proc(n) local FM: FM:=ifactors(n)[2]: seq(seq(FM[j][1], k=1..FM[j][2]), j=1..nops(FM)) end: # A027746
%p s:=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/A322035
%p A006022 := n -> if n = 1 then 0 else n*s(n); fi;
%p # _N. J. A. Sloane_, Nov 28 2018
%t 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 *)
%o (Haskell)
%o a006022 1 = 0
%o a006022 n = (+ 1) $ sum $ takeWhile (> 1) $
%o iterate (\x -> x `div` a020639 x) (a032742 n)
%o -- _Reinhard Zumkeller_, Jun 03 2012
%o (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
%Y Cf. A020639, A032742, A001348, A002182, A008864.
%Y Cf. also A027746, A306264, A306363, A322034, A322035, A322036, A322382, A328898, A333783, A332993, A333791.
%Y Row sums of A328969.
%K nonn
%O 1,4
%A _N. J. A. Sloane_
%E Edited and extended by _Christian G. Bower_, Oct 18 2002
%E Entry revised by _N. J. A. Sloane_, Nov 28 2018
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