A233394 -id:A233394 - cp's OEIS frontend

A233416 c-perfect numbers.

11, 71, 226, 3676, 16911, 1143267, 4721203, 8906035

Offset: 1

Vladimir Shevelev and Peter J. C. Moses, Dec 09 2013

A number k is called a c-perfect number if the sum of its proper c-divisors equals k.
For the definition of a c-divisor of an integer, see comment in A124771.
From Charlie Neder, Jan 17 2019: (Start)
Sequence in binary: 1011, 1000111, 11100010, 111001011100, 100001000001111, 100010111000111100011, 10010000000101000110011, 100001111110010100110011...
Next term > 10^7. (End)

			For n=11 which is a concatenation of binary parts (10)(1)(1); we have proper positive c-divisors 1, 2, 3, and 5, the sum of which is 11, so 11 is in the sequence.

Index entries for sequences related to binary expansion of n

Cf. A124771, A233394.

A233394(a(n))=2*a(n).

a(6)-a(8) from Charlie Neder, Jan 17 2019

A233655 Sum of parts power divisors of canonical representation of n (A233569).

Original entry on oeis.org

1, 2, 4, 4, 9, 9, 11, 8, 17, 12, 26, 17, 26, 26, 26, 16, 33, 26, 48, 26, 45, 45, 63, 33, 48, 45, 63, 48, 63, 63, 57, 32, 65, 50, 92, 40, 97, 97, 115, 50, 97, 54, 120, 97, 120, 120, 140, 65, 92, 97, 115, 97, 120, 120, 140, 92, 115, 120, 140, 115, 140, 140, 120, 64

Offset: 1

Vladimir Shevelev, Dec 14 2013

nonn

If the canonical representation of n is A233569(n)=(1)^k_1[*](10)^k_2[*]...[*](10...0)^k_t, where [*] means concatenation, then we say that a number (1)^r_1[*](10)^r_2[*]...[*](10...0)^r_t is a parts power divisor of canonical representation of n, iff all r_i<=k_i.

Note that, by agreement, (10...0)^0 means the absence of the corresponding part.

			Since A233569(5)=6, then the canonical representation of 5 is (1)^1[*](10)^1 which has parts power divisors 0, (1)^1, (10)^1, (1)^1[*](10)^1. Converting to decimal, they are 0,1,2,6 with sum 9. So a(5)=9. Note that 6 is a parts power divisor of 5, but not a c-divisors of 5 (see comment in A124771).
Analogously, 12 = (1)^1[*](10)^0[*](100)^1 is a parts power divisor of 52 = (1)^1[*](10)^1[*](100)^1, but not a c-divisor of 52.

Cf. A233394, A233569, A124771.

a((10...0[m zeros])^k) = 2^m/(2^(m+1)-1)^2 * (2^((m+1)*(k+1)) - 1) - (k+1)*2^m/(2^(m+1)-1). For example, a(101010)[here m=1,k=3] = 2/9*(2^8-1) - 4*2/3 = 54.

Thus a(42)=54. What is a general formula for a(n)?

cp's OEIS Frontend

A233416 c-perfect numbers.

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