cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A358233 Number of ways n can be expressed as an unordered product of two natural numbers that do not generate any carries when added together in the primorial base.

Original entry on oeis.org

0, 1, 0, 2, 0, 2, 0, 1, 0, 1, 0, 2, 0, 2, 0, 2, 0, 3, 0, 1, 0, 1, 0, 4, 0, 2, 0, 3, 0, 3, 0, 1, 0, 1, 0, 4, 0, 2, 0, 2, 0, 4, 0, 1, 0, 1, 0, 4, 0, 2, 0, 3, 0, 4, 0, 2, 0, 1, 0, 5, 0, 2, 0, 3, 0, 3, 0, 1, 0, 2, 0, 6, 0, 2, 0, 3, 0, 4, 0, 1, 0, 1, 0, 4, 0, 2, 0, 2, 0, 5, 0, 1, 0, 1, 0, 6, 0, 3, 0, 3, 0, 3, 0, 2, 0
Offset: 1

Views

Author

Antti Karttunen, Nov 26 2022

Keywords

Examples

			a(6) = 2, because 6 has only two factor pairs, {1, 6} and {2, 3}, and for both of those pairs the criterion is satisfied, as we have A329041(1, 6) = 1 and A329041(2, 3) = 1. In the latter case the primorial base expansions of 2 and 3 are "10" and "11" (see A049345), which can be added together cleanly (i.e., without carries) to obtain "21" = A049345(2+3).
a(8) = 1, because while there are two ways to factor 8 into two factors, as 1*8 and 2*4, only 1+8 yields a carry-free sum ("1" and "110" added together gives "111" = 9 in primorial base, A049345), while 2+4 (= "10" + "20") is not carry-free, as 2 is the max. allowed digit in the second rightmost place.
		

Crossrefs

Cf. A038548, A049345, A100484, A276086, A329041, A358234 (even bisection), A358671.
Cf. also A358235, A358236.

Programs

  • 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); };
    A329041sq(row,col) = A327936(A276086(row)*A276086(col));
    A358233(n) = sumdiv(n, d, ((d <= (n/d)) && 1==A329041sq(d,n/d)));

Formula

a(n) = Sum_{d|n} [d <= (n/d) and A329041(d,n/d) == 1], where [ ] is the Iverson bracket, and the dyadic function A329041 returns 1 only when its two arguments do not generate any carries when added together in the primorial base.
For all n >= 1, a(n) <= A038548(n) [see A358671 for the indices where the equality is attained] and a(n) <= A358236(n).
For all n >= 1, a(A100484(n)) = A358235(A100484(n)).
For all n >= 1, a(2n-1) = 0, a(4n-2) = A358236(4n-2).

A359778 Number of factorizations of n into factors not divisible by p^p for any prime p (terms of A048103).

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 2, 2, 1, 1, 4, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 5, 1, 1, 2, 2, 2, 5, 1, 2, 2, 2, 1, 5, 1, 2, 4, 2, 1, 2, 2, 4, 2, 2, 1, 5, 2, 2, 2, 2, 1, 6, 1, 2, 4, 1, 2, 5, 1, 2, 2, 5, 1, 5, 1, 2, 4, 2, 2, 5, 1, 2, 3, 2, 1, 6, 2, 2, 2, 2, 1, 11, 2, 2, 2, 2, 2, 2, 1, 4, 4, 5, 1, 5, 1, 2, 5, 2, 1, 7
Offset: 1

Views

Author

Antti Karttunen, Jan 16 2023

Keywords

Examples

			108 has in total 16 = A001055(108) factorizations:
  Factors           Are there any factors that are divisible by p^p,
                    where p is any prime?
  -------------------------------------------------------------------
  [3, 3, 3, 2, 2]   No
  [4, 3, 3, 3]      Yes (4, divisible by 2^2)
  [6, 3, 3, 2]      No
  [6, 6, 3]         No
  [9, 3, 2, 2]      No
  [9, 4, 3]         Yes (4)
  [9, 6, 2]         No
  [12, 3, 3]        Yes (12, divisible by 2^2)
  [12, 9]           Yes (12)
  [18, 3, 2]        No
  [18, 6]           No
  [27, 2, 2]        Yes (27, divisible by 3^3)
  [27, 4]           Yes (both 27 and 4)
  [36, 3]           Yes (36)
  [54, 2]           Yes (54, divisible by 3^3)
  [108]             Yes (108 = 2^2 * 3^3)
Thus only seven of the factorizations satisfy the criterion, and a(108) = 7.
		

Crossrefs

Cf. A001055, A048103, A276086, A317836, A359550, A359779 (Dirichlet inverse).
Cf. also A358236.

Programs

  • PARI
    A359550(n) = { my(f = factor(n)); prod(k=1, #f~, (f[k, 1]>f[k, 2])); };
    A359778(n, m=n) = if(1==n, 1, my(s=0); fordiv(n, d, if((d>1) && (d<=m) && A359550(d), s += A359778(n/d, d))); (s));

Formula

a(n) <= A001055(n).
For all n >= 0, a(A276086(n)) = A317836(n).
Showing 1-2 of 2 results.