A172246 - OEIS

%I #31 Jan 17 2024 11:24:59

%S 0,0,0,0,0,2,0,0,0,2,0,2,0,2,0,2,0,3,0,2,0,4,0,3,0,0,0,2,0,4,0,0,0,3,

%T 0,3,0,2,2,3,0,3,0,0,0,5,0,4,0,0,0,4,0,4,0,2,2,3,0,5,0,2,0,4,0,4,0,0,

%U 0,7,0,7,0,0,0,5,0,6,0,0,0,4,0,3,0,0,0,5,0,5,0,3,0,4,0,5,0,3,0,5,0,6,0,0,0

%N Consider all ways of writing n = a + b where a<=b, gcd(a,b,n)=1, and having the same value of the function N(a,b,n) defined as product of the distinct prime divisors of a*b*n. Let w be the list of frequencies of distinct terms N(a, b, n) found among these partitions. If no frequency is more than 1 then a(n) = 0, otherwise a(n) = [the sum of elements of w] - [the number of elements of w] + 1.

%e a(7)=2 because we have two partitions 7=1+6 and 7=3+4 with different values of N(a,b,n) respectively 1*2*3*7=42 and 2*3*7=42.

%e From _David A. Corneth_, Aug 25 2020: (Start)

%e a(23) = 4 as the terms occurring more than once (so omitting those that occur just once) listed with repetitions is (690, 690, 690, 966, 966) and so the list of frequencies is (3, 2) as 690 occurs thrice and 966 occurs twice.

%e As at least one of these multiplicities is larger than 1, a(n) = [sum of numbers in (3, 2)] - [number of numbers in (3, 2)] + 1 = (3 + 2) - 2 + 1 = 4. (End)

%o (PARI) a(n) = {l = List(); m = Map(); my(res = 0); for(i = 1, n\2, if(gcd([i,(n-i),n]) == 1, c = factorback(factor(i*(n-i)*n)[,1]); listput(l, c); if(mapisdefined(m, c), mapput(m, c, mapget(m, c) + 1); , mapput(m, c, 1) ) ) ); l = Set(l); for(i = 1, #l, if(mapget(m, l[i]) > 1, res+=mapget(m, l[i]); ) ); w = vector(#l, i, mapget(m, l[i])); w = select(x -> x > 1, w); if(#w == 0, return(0)); w = vecsort(Vec(w)); return(vecsum(w) - #w + 1); } \\ _David A. Corneth_, Aug 25 2020

%Y Cf. A023022, A172245, A172246, A172247, A172248.

%K nonn,less

%O 2,6

%A _Artur Jasinski_, Jan 29 2010

%E Edited by _David A. Corneth_, Aug 25 2020, and _N. J. A. Sloane_, Sep 05 2020