A260258 T(n,k) is the array read by rows, n>0 and k=1..q (with q = number of prime distinct divisors of n^2+1) giving the number of occurrences of the k-th prime divisor of n^2+1 counted from the prime divisors of m^2+1 for m=1..n.
1, 1, 2, 2, 1, 3, 1, 1, 4, 3, 4, 2, 5, 1, 1, 6, 1, 5, 1, 7, 6, 2, 1, 8, 1, 1, 9, 7, 2, 8, 3, 10, 1, 1, 11, 4, 3, 9, 1, 12, 10, 1, 1, 13, 1, 1, 14, 11, 1, 12, 1, 15, 1, 4, 2, 16, 5, 2, 13, 2, 17, 14, 1, 6, 1, 18, 1, 1, 19, 15, 1, 16, 5, 20, 1, 1, 21, 3, 17, 1
Offset: 1
Examples
T(13,k) = [7,6,2] for k = 1,2,3 because 13^2+1 = 2*5*17 => The number of occurrences of the prime divisor 2 is 7: 1^2+1=2, 3^2+1=2*5, 5^2+1=2*13, 7^2+1=2*5^2, 9^2+1=2*41, 11^2+1=2*61 and 13^2+1=2*5*17; The number of occurrences of the prime divisor 5 is 6: 2^2+1=5, 3^2+1=2*5, 7^2+1=2*5^2, 8^2+1=5*13, 12^2+1=5*29; The number of occurrences of the prime divisor 17 is 2: 4^2+1=17 and 13^2+1=2*5*17. The array begins: [1] [1] [2,2] [1] [3,1] [1] [4,3] [4,2] [5,2] [1] ...
Links
- Michel Lagneau, Table of n, a(n) for n = 1..5000
Programs
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Maple
with(numtheory):lst:={2}:nn:=1000:T:=array(1..270,[0$270]): for j from 1 to nn do: p:=4*j+1: if isprime(p) then lst:=lst union {p}: fi: od: nn0:=nops(lst): for n from 1 to 60 do: q:=factorset(n^2+1):n0:=nops(q): for k from 1 to n0 do: for m from 1 to 270 do: if q[k]=lst[m] then T[m]:=T[m]+1:printf(`%d, `, T[m]): fi: od: od: od:
Comments