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a(n) = Sum over i and j with j<=i and i=1.. sopf(n) of binomial(d(i), d(j)), where d(i) are the prime divisors of n and sopf(n) = sum of the distinct primes dividing n (A008472).
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%I #24 Feb 24 2023 03:09:32

%S 0,1,1,1,1,5,1,1,1,12,1,5,1,23,12,1,1,5,1,12,37,57,1,5,1,80,1,23,1,26,

%T 1,1,167,138,23,5,1,173,288,12,1,62,1,57,12,255,1,5,1,12,682,80,1,5,

%U 464,23,971,408,1,26,1,467,37,1,1289,226,1,138,1773,55

%N a(n) = Sum over i and j with j<=i and i=1.. sopf(n) of binomial(d(i), d(j)), where d(i) are the prime divisors of n and sopf(n) = sum of the distinct primes dividing n (A008472).

%C This sequence is very rich in properties. We cite a few.

%C If n = p^m where p is prime, then a(n) = 1 => a(A025475(n)) = 1.

%C If n = 2^i*3^j, i and j >= 1, then a(n) = 5 => a(A033845(n)) = 5.

%C We observe a very interesting property with a fractal structure if a(n) = 23.

%C a(n) = 23 if n belongs to the set E = {n} = {14, 28, 35, 56, 98, 112, 175, 196, 224, 245, 392, 448, 686, ...} and we observe that E1 = {n/7} = {2, 4, 5, 8, 14, 16, 25, 28, 32, 35, 56, 64, 98, 112, 125, 128, 175, 196, 224, 245, 256, ...} = E2 union E3 union E4 where:

%C E2 = {2, 4, 8, 16, 32, 64, 128, ...} are powers of 2 (A000079),

%C E3 = {5, 25, 125, 625, 3125, ...} are powers of 5 (A000351),

%C E4 = {14, 28, 35, 56, 98, 112, 175, 196, 224, 245, ...} = E, a surprising result: E1 contains E => this shows a fractal structure.

%C But you can continue with other values of a(n) in order to find similar properties. For example, a(n) = 55 if n belongs to the set F = {n} = {70, 140, 280, 350, 490, 560, 700, 980, 1120, ...} => F1 = {n/70} = { 1, 2, 4, 5, 7, 8, 10, 14, 16, 20, 25, 28, 32, 35, 40, 49, 50, 56, 64, 70, 80, ...} = F2 union F3 union F4 union F5 where:

%C F2 = {2, 4, 8, 16, 32, 64, 128, ...} are powers of 2 (A000079),

%C F3 = {5, 25, 125, 625, 3125, ...} are powers of 5 (A000351),

%C F4 = {1, 7, 49, 343, 2401, ...} are powers of 7 (A000420)

%C and F5 = {10, 14, 20, 28, 35, 40, 50, 56, 70, 80, 98, 100, 112, 140, ...} contains F but also E.

%H Michel Lagneau, <a href="/A242404/b242404.txt">Table of n, a(n) for n = 1..10000</a>

%e a(10) = 12 because 12 = 2^2*3 => sopf(12) = 2+3=5 and a(10) = sum{j<=i=1..5} binomial(d(i),d(j)) = binomial(2,2)+ binomial(5,2)+ binomial(5,5) = 12.

%p with(numtheory):for n from 1 to 200 do:x:=factorset(n):n1:=nops(x): r:=sum('sum('binomial(x[j],x[i])','j'=i..n1)','i'=1..n1):printf(`%d, `,r):od:

%Y Cf. A000079, A000351, A000420, A008472.

%K nonn

%O 1,6

%A _Michel Lagneau_, May 13 2014