A297174 An auxiliary sequence for computing A300250. See comments and examples.
0, 1, 1, 5, 1, 19, 1, 69, 5, 19, 1, 2123, 1, 19, 19, 4165, 1, 2131, 1, 2125, 19, 19, 1, 4228171, 5, 19, 69, 2125, 1, 526631, 1, 2101317, 19, 19, 19, 268706123, 1, 19, 19, 4228237, 1, 526643, 1, 2125, 2123, 19, 1, 550026380363, 5, 2131, 19, 2125, 1, 4229203, 19, 4228237, 19, 19, 1, 8798249190555, 1, 19, 2123, 17181970501, 19, 526643, 1, 2125
Offset: 1
Keywords
Examples
a(1) = 0 by convention (as 1 has no prime divisors). a(p) = 1 for any prime p. For any n > 1, the least significant 1-bit is at rightmost position (bit-0), signifying the smallest prime factor of n, which is always the least divisor > 1. For n = 4 = 2*2, the next divisor of 4 after 2 is 4, for which A101296(4) = 3, thus the second least significant 1-bit comes 3-1 = 2 positions left of the rightmost 1, thus a(4) = 2^0 + 2^(3-1) = 1+4 = 5. For n = 6 with divisors d = 2, 3 and 6 larger than one, for which A101296(d)-1 gives 1, 1 and 3, thus a(6) = 2^(1-1) + 2^(1-1+1) + 2^(1-1+1+3) = 2^0 + 2^1 + 2^4 = 19. For n = 12 with divisors d = 2, 3, 2*2, 2*3, 2*2*3 larger than one, A101296(d)-1 gives 1, 1, 2, 3 and 5 thus a(12) = 2^0 + 2^(0+1) + 2^(0+1+2) + 2^(0+1+2+3) + 2^(0+1+2+3+5) = 2123. For n = 18 with divisors d = 2, 3, 2*3, 3*3, 2*3*3 larger than one, A101296(d)-1 gives 1, 1, 3, 2, and 5 thus a(18) = 2^0 + 2^(0+1) + 2^(0+1+3) + 2^(0+1+3+2) + 2^(0+1+3+2+5) = 2131.
Links
Crossrefs
Programs
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PARI
up_to = 4096; rgs_transform(invec) = { my(om = Map(), outvec = vector(length(invec)), u=1); for(i=1, length(invec), if(mapisdefined(om,invec[i]), my(pp = mapget(om, invec[i])); outvec[i] = outvec[pp] , mapput(om,invec[i],i); outvec[i] = u; u++ )); outvec; }; A046523(n) = { my(f=vecsort(factor(n)[, 2], , 4), p); prod(i=1, #f, (p=nextprime(p+1))^f[i]); }; \\ From A046523. v101296 = rgs_transform(vector(up_to, n, A046523(n))); A101296(n) = v101296[n]; A297174(n) = { my(s=0,i=-1); fordiv(n, d, if(d>1, i += (A101296(d)-1); s += 2^i)); (s); };
Comments