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Search: a322036 -id:a322036
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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
OFFSET
1,4
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.
Note that s(n) = A322034(n) / A322035(n).
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).
LINKS
Jonathan Blanchette and Robert Laganière, A Curious Link Between Prime Numbers, the Maundy Cake Problem and Parallel Sorting, arXiv:1910.11749 [cs.DS], 2019.
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
From Antti Karttunen, Apr 12 2020: (Start)
a(n) = A322382(n) + A333791(n).
a(n) = A332993(n) - n = A001065(n) - A333783(n). (End)
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: # A027746
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
A006022 := n -> if n = 1 then 0 else n*s(n); fi;
# N. J. A. Sloane, Nov 28 2018
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)
-- Reinhard Zumkeller, Jun 03 2012
(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
KEYWORD
nonn
EXTENSIONS
Edited and extended by Christian G. Bower, Oct 18 2002
Entry revised by N. J. A. Sloane, Nov 28 2018
STATUS
approved
Let p1 <= p2 <= ... <= pk be the prime factors of n, with repetition; let s = 1/p1 + 1/(p1*p2) + 1/(p1*p2*p3) + ... + 1/(p1*p2*...*pk); a(n) = numerator of s. a(1)=0 by convention.
+10
5
0, 1, 1, 3, 1, 2, 1, 7, 4, 3, 1, 5, 1, 4, 2, 15, 1, 13, 1, 4, 8, 6, 1, 11, 6, 7, 13, 11, 1, 7, 1, 31, 4, 9, 8, 31, 1, 10, 14, 9, 1, 29, 1, 17, 7, 12, 1, 23, 8, 31, 6, 10, 1, 20, 12, 25, 20, 15, 1, 17, 1, 16, 29, 63, 14, 15, 1, 13, 8, 43, 1, 67
OFFSET
1,4
COMMENTS
Note that s < 1 for all n (compare A322036). This follows easily by induction, since when we increase n by multiplying it by a new (not-smaller) prime, we increase s by less than 1-s.
EXAMPLE
If n=12 we get the prime factors 2,2,3, and s = 1/2 + 1/4 + 1/12 = 5/6. So a(12) = 5.
The fractions s for n >= 2 are 1/2, 1/3, 3/4, 1/5, 2/3, 1/7, 7/8, 4/9, 3/5, 1/11, 5/6, 1/13, 4/7, 2/5, 15/16, 1/17, 13/18, 1/19, 4/5, 8/21, ...
MAPLE
# This generates the terms starting at n=2:
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
f0:=[]; f1:=[]; f2:=[];
for n from 2 to 120 do
a:=0; b:=1; t1:=[P(n)];
for i from 1 to nops(t1) do b:=b/t1[i]; a:=a+b; od;
f0:=[op(f0), a]; f1:=[op(f1), numer(a)]; f2:=[op(f2), denom(a)]; od:
f0; # s
f1; # A322034
f2; # A322035
f2-f1; # A322036
PROG
(PARI) A322034(n) = if(1==n, 0, my(f=factor(n), pm=1, s=0); for(i=1, #f~, while(f[i, 2], pm *= f[i, 1]; f[i, 2]--; s += 1/pm)); numerator(s)); \\ Antti Karttunen, Feb 28 2019
CROSSREFS
A017665/A017666 = sum of reciprocals of all divisors of n.
KEYWORD
nonn,frac
AUTHOR
STATUS
approved
Let p1 <= p2 <= ... <= pk be the prime factors of n, with repetition; let s = 1/p1 + 1/(p1*p2) + 1/(p1*p2*p3) + ... + 1/(p1*p2*...*pk); a(n) = denominator of s. a(1)=1 by convention.
+10
5
1, 2, 3, 4, 5, 3, 7, 8, 9, 5, 11, 6, 13, 7, 5, 16, 17, 18, 19, 5, 21, 11, 23, 12, 25, 13, 27, 14, 29, 10, 31, 32, 11, 17, 35, 36, 37, 19, 39, 10, 41, 42, 43, 22, 15, 23, 47, 24, 49, 50, 17, 13, 53, 27, 55, 28, 57, 29, 59, 20, 61, 31, 63, 64, 65, 22
OFFSET
1,2
LINKS
EXAMPLE
If n=12 we get the prime factors 2,2,3, and s = 1/2 + 1/4 + 1/12 = 5/6. So a(12) = 6.
The fractions s for n >= 2 are 1/2, 1/3, 3/4, 1/5, 2/3, 1/7, 7/8, 4/9, 3/5, 1/11, 5/6, 1/13, 4/7, 2/5, 15/16, 1/17, 13/18, 1/19, 4/5, 8/21, ...
MAPLE
# This generates the terms starting at n=2:
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
f0:=[]; f1:=[]; f2:=[];
for n from 2 to 120 do
a:=0; b:=1; t1:=[P(n)];
for i from 1 to nops(t1) do b:=b/t1[i]; a:=a+b; od;
f0:=[op(f0), a]; f1:=[op(f1), numer(a)]; f2:=[op(f2), denom(a)]; od:
f0; # s
f1; # A322034
f2; # A322035
f2-f1; # A322036
CROSSREFS
A017665/A017666 = sum of reciprocals of all divisors of n.
KEYWORD
nonn,frac
AUTHOR
STATUS
approved

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