A385189 -id:A385189 - cp's OEIS frontend

A385415 Products of three consecutive integers whose prime divisors are consecutive primes starting at 2.

6, 24, 60, 120, 210, 720, 3360, 9240, 117600, 166320, 970200, 43243200, 85765680

Offset: 1

Ken Clements, Jun 28 2025

The final term is 440*441*442 = 85765680.

Let tau(a(n)) be the number of divisors of a(n), sigma(a(n)) be the sum of divisors of a(n), S be the strictly increasing sequence of divisors of a(n), and S(i) be the i-th element of S. Since sigma(a(n)) is even and for every i in the interval [1, tau(a(n))-1], 2*S(i) >= S(i+1), a(n) is a Zumkeller number (see Proposition 17 in Rao/Peng JNT paper at A083207). - Ivan N. Ianakiev, Jul 09 2025

			a(1) = 6 = 1*2*3 = 2^1 * 3^1.
a(2) = 24 = 2*3*4 = 2^3 * 3^1.
a(3) = 60 = 3*4*5 = 2^2 * 3^1 * 5^1.
a(4) = 120 = 4*5*6 = 2^3 * 3^1 * 5^1.
a(5) = 210 = 5*6*7 = 2^1 * 3^1 * 5^1 * 7^1.
a(6) = 720 = 8*9*10 = 2^4 * 3^2 * 5^1.
...
a(13) = 85765680 = 440*441*442 = 2^4 * 3^2 * 5^1 * 7^2 * 11^1 * 13^1 * 17^1.

Intersection of A007531 and A055932.

Cf. A385189, A083207 (supersequence).

Select[(#*(# + 1)*(# + 2)) & /@ Range[500], PrimePi[(f = FactorInteger[#1])[[-1, 1]]] == Length[f] &] (* Amiram Eldar, Jun 28 2025 *)

from sympy import prime, primefactors
def is_pi_complete(n): # Check for complete set of
    factors = primefactors(n) # prime factors
    return factors[-1] == prime(len(factors))
def aupto(limit):
    result = []
    for i in range(1, limit+1):
        n = i * (i+1) * (i+2)
        if is_pi_complete(n):
            result.append(n)
    return result
print(aupto(100_000))

A385956 Intersection of A025487 and A002378.

Original entry on oeis.org

2, 6, 12, 30, 72, 210, 240, 420, 1260, 6480, 50400, 147840, 510510, 4324320

Offset: 1

Ken Clements, Aug 10 2025

These numbers are the products of two consecutive integers that are also Hardy-Ramanujan integers; that is, of the form 2^k1*3^k2*...*p_n^k_n, where k1 >= k2 >= ... >= k_n. This sequence is finite with last term a(14) = 2079*2080 = 4324320.

			a(1) = 2 = 1*2 = 2^1.
a(2) = 6 = 2*3 = 2^1 * 3^1.
a(3) = 12 = 3*4 = 2^2 * 3^1.
a(4) = 30 = 5*6 = 2^1 * 3^1 * 5^1.
a(5) = 72 = 8*9 = 2^3 * 3^2.
a(6) = 210 = 14*15 = 2^1 * 3^1 * 5^1 * 7^1.

Cf. A055932, A025487, A002378, A385189, A386951.

Select[FactorialPower[Range[0, 3000], 2], (Max@Differences[(f = FactorInteger[#])[[;; , 2]]] < 1 && f[[-1, 1]] == Prime[Length[f]]) &] (* Amiram Eldar, Aug 10 2025 *)

from sympy import prime, factorint
def is_Hardy_Ramanujan(n):
    factors = factorint(n)
    p_idx = len(factors)
    if list(factors.keys())[-1] != prime(p_idx):
        return False
    expos = list(factors.values())
    e = expos[0]
    for i in range(1, p_idx):
        if expos[i] > e:
            return False
        e = expos[i]
    return True
print([ n*(n+1) for n in range(1, 10_000) if is_Hardy_Ramanujan(n*(n+1))])

A386951 Intersection of A025487 and A007531.

Original entry on oeis.org

6, 24, 60, 120, 210, 720, 3360, 9240, 166320, 970200, 43243200

Offset: 1

Ken Clements, Aug 10 2025

These numbers are the products of three consecutive integers that are also Hardy-Ramanujan integers; that is, of the form 2^k1*3^k2*...*p_n^k_n, where k1 >= k2 >= ... >= k_n. This sequence is finite with last term a(11) = 350*351*352 = 43243200.

			a(1) = 6 = 1*2*3 = 2^1 * 3^1.
a(2) = 24 = 2*3*4 = 2^3 * 3^1.
a(3) = 60 = 3*4*5 = 2^2 * 3^1 * 5^1.
a(4) = 120 = 4*5*6 = 2^3 * 3^1 * 5^1.
a(5) = 210 = 5*6*7 = 2^1 * 3^1 * 5^1 * 7^1.
a(6) = 720 = 8*9*10 = 2^4 * 3^2 * 5^1.

Cf. A055932, A025487, A007531, A385189, A385956.

Select[FactorialPower[Range[0, 1000], 3], (Max@ Differences[(f = FactorInteger[#])[[;; , 2]]] < 1 && f[[-1, 1]] == Prime[Length[f]]) &] (* Amiram Eldar, Aug 10 2025 *)

from sympy import prime, factorint
def is_Hardy_Ramanujan(n):
    factors = factorint(n)
    p_idx = len(factors)
    if list(factors.keys())[-1] != prime(p_idx):
        return False
    expos = list(factors.values())
    e = expos[0]
    for i in range(1, p_idx):
        if expos[i] > e:
            return False
        e = expos[i]
    return True
print([ n*(n+1)*(n+2) for n in range(1, 1000) if is_Hardy_Ramanujan(n*(n+1)*(n+2))])

A385631 Products of five consecutive integers whose prime divisors are consecutive primes starting at 2.

Original entry on oeis.org

120, 720, 2520, 6720, 15120, 30240, 55440, 240240, 360360

Offset: 1

Ken Clements, Jul 05 2025

			a(1) = 120 = 1*2*3*4*5 = 2^3 * 3^1 * 5^1.
a(2) = 720 = 2*3*4*5*6 = 2^4 * 3^2 * 5^1.
a(3) = 2520 = 3*4*5*6*7 = 2^3 * 3^2 * 5^1 * 7^1.
a(4) = 6720 = 4*5*6*7*8 = 2^6 * 3^1 * 5^1 * 7^1.
a(5) = 15120 = 5*6*7*8*9 = 2^4 * 3^3 * 5^1 * 7^1.
a(6) = 30240 = 6*7*8*9*10 = 2^5 * 3^3 * 5^1 * 7^1.
a(7) = 55440 = 7*8*9*10*11 = 2^4 * 3^2 * 5^1 * 7^1 * 11^1.
a(8) = 240240 = 10*11*12*13*14 = 2^4 * 3^1 * 5^1 * 7^1 * 11^1 * 13^1.
a(9) = 360360 = 11*12*13*14*15 = 2^3 * 3^2 * 5^1 * 7^1 * 11^1 * 13^1.

Intersection of A052787 and A055932.

Cf. A217056, A385189, A385415.

Select[(#*(# + 1)*(# + 2)*(# + 3)*(# + 4)) & /@ Range[12], PrimePi[(f = FactorInteger[#1])[[-1, 1]]] == Length[f] &] (* Amiram Eldar, Jul 05 2025 *)

from sympy import prime, primefactors
def is_pi_complete(n): # Check for complete set of
    factors = primefactors(n) # prime factors
    return factors[-1] == prime(len(factors))
def aupto(limit):
    result = []
    for i in range(1, limit+1):
        n = i * (i+1) * (i+2) * (i+3) * (i+4)
        if is_pi_complete(n):
            result.append(n)
    return result
print(aupto(1000))

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A385415 Products of three consecutive integers whose prime divisors are consecutive primes starting at 2.

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A385956 Intersection of A025487 and A002378.

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A386951 Intersection of A025487 and A007531.

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A385631 Products of five consecutive integers whose prime divisors are consecutive primes starting at 2.

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Mathematica

Python