login
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).
2

%I #18 Oct 30 2017 23:20:58

%S 1,1,3,2,5,15,8,14,35,105,48,88,154,385,1155,480,624,1144,2002,5005,

%T 15015,5760,8160,10608,19448,34034,85085,255255,92160,109440,155040,

%U 201552,369512,646646,1616615,4849845,1658880,2119680,2517120,3565920

%N 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).

%C 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.

%e 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.

%e From _Bob Selcoe_, Oct 12 2017: (Start)

%e Triangle starts:

%e n/i 1 2 3 4 5 6

%e 1 1

%e 2 1 3

%e 3 2 5 15

%e 4 8 14 35 105

%e 5 48 88 154 385 1155

%e 6 480 624 1144 2002 5005 15015

%e (End)

%t 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]

%t (* Second program: *)

%t 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 *)

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

%Y Cf. A293558 (transpose).

%K nonn,tabl

%O 1,3

%A _T. D. Noe_, Apr 01 2010