A250477 Number of times prime(n) (the n-th prime) occurs as the least prime factor among numbers 1 .. (prime(n)^2 * prime(n+1)): a(n) = A078898(A251720(n)).

6, 8, 12, 21, 33, 45, 63, 80, 116, 148, 182, 232, 265, 296, 356, 433, 490, 548, 625, 674, 740, 829, 919, 1055, 1187, 1252, 1313, 1376, 1446, 1657, 1897, 2029, 2134, 2301, 2484, 2605, 2785, 2946, 3110, 3301, 3439, 3654, 3869, 3978, 4086, 4349, 4811, 5147, 5273, 5395, 5604, 5787, 6049, 6403, 6684, 6954, 7153

Offset: 1

Author

Antti Karttunen, Dec 14 2014

Keywords

• nonn

Comments

a(n) = Position of 6 on row n of array A249821. This is always larger than A250474(n), the position of 4 on row n, as 4 is guaranteed to be the first composite term on each row of A249821.

From Antti Karttunen, Mar 29 2015: (Start)

a(n) = 1 + number of positive integers <= (prime(n)*prime(n+1)) whose smallest prime factor is at least prime(n).

That a(n) > A250474(n) can also be seen by realizing that prime(n) must occur at least as many times as the smallest prime factor for the numbers in range 1 .. (prime(n)^2 * prime(n+1)) than for numbers in (smaller) range 1 .. (prime(n)^3), and also by realizing that a(n) cannot be equal to A250474(n) because each row of A249822 is a permutation of natural numbers.

Or more simply, by considering the comment given in A256447 which follows from the new interpretation given above.

(End)

Links

• Antti Karttunen, Table of n, a(n) for n = 1..564
• Antti Karttunen, Ratio a(n)/A250474(n+1), plotted up to n=65 with OEIS Plot2-utility

Crossrefs

Column 6 of A249822. Cf. also A250474 (column 4), A250478 (column 8).

First differences: A256446. Cf. also A256447, A256448.

Cf. A002110, A006094, A020639, A078898, A249821, A251720, A281890.

Programs

Mathematica

f[n_] := Count[Range[Prime[n]^2*Prime[n + 1]], x_ /; Min[First /@ FactorInteger[x]] == Prime@ n]; Array[f, 20] (* Michael De Vlieger, Mar 30 2015 *)

PARI

allocatemem(234567890);
A002110(n) = prod(i=1, n, prime(i));
A250477(n) = { my(m); m = (prime(n) * prime(n+1)); sumdiv(A002110(n-1), d, (moebius(d)*(m\d))); };
for(n=1, 23, print1(A250477(n),", "));
\\ A more practical program:

PARI

allocatemem(234567890);
vecsize = (2^24)-4;
v020639 = vector(vecsize);
v020639[1] = 1; for(n=2,vecsize, v020639[n] = vecmin(factor(n)[, 1]));
A020639(n) = v020639[n];
A250477(n) = { my(p=prime(n),q=prime(n+1),u=p*q,k=1,s=1); while(k <= u, if(A020639(k) >= p, s++); k++); s; };
for(n=1, 564, write("b250477.txt", n, " ", A250477(n)));
\\ Antti Karttunen, Mar 29 2015

Formula

a(n) = A078898(A251720(n)).

a(1) = 1, a(n) = Sum_{d | A002110(n-1)} moebius(d) * floor(A006094(n) / d). [Follows when A251720, (p_n)^2 * p_{n+1} is substituted to the similar formula given for A078898. Here p_n is the n-th prime (A000040(n)), A006094(n) gives the product p_n * p{n+1} and A002110(n) gives the product of the first n primes. Because the latter is always squarefree, one could use here also Liouville's lambda (A008836) instead of Moebius mu (A008683)].

a(n) = A250474(n) + A256447(n).