A174909 Triangle T(n,i) whose n-th row gives the number of numbers in any prime(n)# consecutive numbers whose smallest prime factor is prime(n-i+1).

1, 1, 3, 2, 5, 15, 8, 14, 35, 105, 48, 88, 154, 385, 1155, 480, 624, 1144, 2002, 5005, 15015, 5760, 8160, 10608, 19448, 34034, 85085, 255255, 92160, 109440, 155040, 201552, 369512, 646646, 1616615, 4849845, 1658880, 2119680, 2517120, 3565920

Offset: 1

Author

T. D. Noe, Apr 01 2010

Keywords

• nonn
• tabl

Comments

Here prime(n)# denotes the product of the first n primes. Row n begins with A005867(n-1). The other n-1 terms in row n are prime(n) times the previous row. The sum of the terms in row n is cototient(prime(n)#), which is A053144(n), and which equals prime(n)#-A005867(n). This sequence is a generalization of a comment in A005867 by Dennis Martin.

Examples

			For n=3, we have prime(n)=5 and any range of 2*3*5=30 consecutive numbers has 2 numbers whose smallest prime factor is 5, 5 numbers whose smallest prime factor is 3, and 15 numbers whose smallest prime factor is 2.
From _Bob Selcoe_, Oct 12 2017: (Start)
Triangle starts:
n/i  1   2   3    4    5     6
1    1
2    1   3
3    2   5   15
4    8  14   35  105
5   48  88  154  385  1155
6  480 624 1144 2002  5005 15015
(End)
		

Crossrefs

Cf. A002110, A005867 (first column), A020639, A053144, A070826 (main diagonal).

Cf. A293558 (transpose).

Programs

Mathematica

t={{1}}; q=2; Do[p=Prime[n]; t=AppendTo[t, Join[{(q-1)*t[[ -1,1]]}, p*t[[ -1]]]]; q=p, {n,2,9}]; Flatten[t]
(* Second program: *)
Block[{nn = 8, s}, s = Array[FactorInteger[#][[1, 1]] &, Product[Prime@i, {i, nn}]]; Table[With[{P = Product[Prime@ k, {k, n}]}, Count[Take[s, P], ?(# == Prime[n - i + 1] &)]], {n, nn}, {i, n}]] (* _Michael De Vlieger, Oct 14 2017 *)