A358236 Number of factorizations of n where the sum of the factors is carryfree when the addition is done in the primorial base.
1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 3, 1, 1, 1, 1, 1, 5, 1, 2, 1, 4, 1, 3, 1, 1, 1, 1, 1, 5, 1, 2, 1, 2, 1, 4, 1, 1, 1, 1, 1, 5, 1, 2, 1, 4, 1, 4, 1, 2, 1, 1, 1, 5, 1, 2, 1, 3, 1, 3, 1, 1, 1, 2, 1, 9, 1, 2, 1, 4, 1, 4, 1, 1, 1, 1, 1, 5, 1, 2, 1, 2, 1, 5, 1, 1, 1, 1, 1, 8, 1, 3, 1, 4, 1, 3, 1, 2, 1
Offset: 1
Examples
36 has in total 9 = A001055(36) factorizations: factors in decimal in primorial base Do they generate carries when summed? [3, 3, 2, 2] [11, 11, 10, 10] Yes, as A049345(3+3+2+2) = "120". [4, 3, 3] [20, 11, 11] Yes, in the least significant place. [6, 3, 2] [100, 11, 10] No, 6+3+2 = 11 = "121". [6, 6] [100, 100] No, 6+6 = 12 = "200". [9, 2, 2] [111, 10, 10] Yes, in the second place from right. [9, 4] [111, 20] Ditto. [12, 3] [200, 11] No, 12+3 = 15 = "211". [18, 2] [300, 10] No, 18+2 = 20 = "310". [36] [1100] No, as a single factor never does. Thus only five of the sums are carryfree, and a(36) = 5.
Links
Programs
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PARI
A276086(n) = { my(m=1, p=2); while(n, m *= (p^(n%p)); n = n\p; p = nextprime(1+p)); (m); }; A327936(n) = { my(f = factor(n)); for(k=1, #f~, f[k,2] = (f[k,2]>=f[k,1])); factorback(f); }; A358236(n, m=n, facs=List([])) = if(1==n, 1==A327936(factorback(apply(A276086,Vec(facs)))), my(s=0, newfacs); fordiv(n, d, if((d>1)&&(d<=m), newfacs = List(facs); listput(newfacs,d); s += A358236(n/d, d, newfacs))); (s));