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

1,3

Comments

A002313(n) are the numbers such that T(n,k)>1 for all k=1..q.

T(2n-1,1)=n and T(m,1)=1 if m =1, 2, 4, 6, 10, 14, ... = A005574(n)(numbers n such that n^2 + 1 is prime). The length of row n is A128428(n).

Links

Michel Lagneau, Table of n, a(n) for n = 1..5000

Example

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]
  ...

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:

Crossrefs

Cf. A005574, A002312, A002313, A128428.

Sequence in context: A241150 A051135 A325541 * A283196 A238882 A279287

Adjacent sequences: A260255 A260256 A260257 * A260259 A260260 A260261

Keyword

nonn,tabf

Author

Michel Lagneau, Jul 21 2015

Status

approved