A100933 -id:A100933 - cp's OEIS frontend

A141757 Even terms in A100933.

50, 98, 150, 242, 250, 294, 338, 350, 490, 550, 578, 650, 686, 722, 726, 750, 850, 950, 1014, 1050, 1058, 1078, 1150, 1210, 1274, 1450, 1470, 1550, 1650, 1666, 1682, 1690, 1694, 1734, 1750, 1850, 1862, 1922, 1950, 2050, 2058, 2150, 2166, 2254, 2350, 2366

Offset: 1

David Applegate and N. J. A. Sloane, Sep 15 2008

nonn

with(numtheory):
# For A100549: if n = prod_p p^e_p, then pp = largest prime <= 1 + max e_p
pp := proc(n) local f,m; option remember;
if (n = 1) then
return 1;
end if;
m := 1:
for f in op(2..-1,ifactors(n)) do
if (f[2] > m) then
m := f[2]:
end if;
end do;
prevprime(m+2);
end proc;
# For A100762: B = prod_{p <= pp(n)} p^e_p
B := proc(n) local v,f,pv; global pp; option remember;
pv := pp(n);
v := 1:
for f in op(2..-1,ifactors(n)) while f[1] <= pv do
v := v * f[1]^f[2];
end do;
return v;
end proc;
# For A100417: Bgood = (is pp(n) = pp(B(n))), that is, is B(n) enough to establish pp(n)?
Bgood := proc(n) global pp;
`if`(pp(B(n))=pp(n),true,false);
end proc;
# For A100933 and A141757:
t0:=select(not Bgood, [$1..3000]);
t1:=[];
for n from 1 to nops(t0) do
if t0[n] mod 2 = 0 then t1:=[op(t1),t0[n]]; fi; od: t1;

A141756 Intersection of A141586 and A100933.

Original entry on oeis.org

115325637083043831900183479190311008528007516613207384396965600343647846400000, 3575094749574358788905687854899641264368233015009428916305933610653083238400000, 16347325453573190128511213579615837273221059920118876332835279076344751718400000

Offset: 1

David Applegate and N. J. A. Sloane, Sep 15 2008

For the prime factorizations of the first four terms (only three are shown above) see the Maple code.

Since all terms > 1 in A141586 are even, this is also the intersection of A141586 and A141757.

B0:=2^23*3^14*5^5*7^3*11^3*13^3*17^2*19^2*23^2*29;
a1:=B0*31^28; a2:=B0*31^29; a3:=B0*37^28; a4:=B0*350*31^28;
[seq(a1,a2,a3,a4)];

A064224 Numbers having more than one representation as the product of consecutive integers > 1.

Original entry on oeis.org

120, 210, 720, 5040, 175560, 17297280, 19958400, 259459200, 20274183401472000, 25852016738884976640000, 368406749739154248105984000000

Offset: 1

Jon Perry, Sep 22 2001

Solutions to the equations: x(x+1)...(x+m)=y(y+1)...(y+n) with x>1, y>1.
Some patterns are impossible, e.g. x(x+1)(x+2)(x+3)=y(y+1) has been proved impossible.
The early terms in this sequence each have two representations. Is two the maximum possible? The sequence is infinite: for any n, the number n*(n+1)*...*(n^2+n-1) is in this sequence. The next number of this form is 20274183401472000, which is obtained when n=4. - T. D. Noe, Nov 22 2004
Using an improved algorithm I have performed an exhaustive search up to 2.15 * 10^33 and can confirm the terms shown above are all that exist up to that point. For all k = A002378(n) > 2 we can construct a member of this sequence by equating n(n+1)(n+2)...(k-1) to (n+2)(n+3)...(k-1)k. Also, as demonstrated in my examples below, 5040 is related to 720 as 259459200 is to 210. So we also know that 36055954861352887137197787308347629783163600896000000000 and 6244042313569035223343873483125151604764341428027427022254596874567680000000000000 are terms. - Robert Munafo, Aug 17 2007 [edited by Peter Munn, Aug 20 2023]
MacLeod and Barrodale prove that the equation x(x+1)...(x+m-1) = y(y+1)...(y+n-1) has no solutions x>1 and y>1 for the following pairs of (m,n): (2,4), (2,6), (2,8), (2,12), (4,8), (5,10). They also show that (2,3) has two solutions and (3,6) has one solution. They conjecture that (2,k) has no solution for k>3. [T. D. Noe, Jul 29 2009]

			120 is here because 120 = 2*3*4*5 = 4*5*6.
a(2)=210 because we can write 210=5*6*7 or 14*15. The term a(8) = 259459200 = 5*6*7*8*9*10*11*12*13 = 8*9*10*11*12*13*14*15 is related to 210 by adding the intervening integers (8 through 13) to both products.

H. L. Abbott, P. Erdos and D. Hanson, On the numbers of times an integer occurs as a binomial coefficient, Amer. Math. Monthly, (March 1974), 256-261.
R. A. MacLeod and I. Barrodale, On equal products of consecutive integers, Canad. Math. Bull., 13 (1970) 255-259. [_T. D. Noe_, Jul 29 2009]
Robert Munafo, Page dealing with this sequence

Cf. A003015 (numbers occurring 5 or more times in Pascal's triangle).
Cf. A002378, A160642.
Subsequence of A045619, A100934.
Cf. A163263 (non-overlapping case). [T. D. Noe, Jul 29 2009]

nn=10^10; t3={}; Do[m=0; p=n; While[m++; p=p(n+m); p<=nn, t3={t3, p}], {n, 2, Sqrt[nn]}]; t3=Sort[Flatten[t3]]; lst={}; Do[If[t3[[i]]==t3[[i+1]], AppendTo[lst, t3[[i]]]], {i, Length[t3]-1}]; Union[lst]

import heapq
def aupton(terms, verbose=False):
    p = 2*3; h = [(p, 2, 3)]; nextcount = 4; alst = []; oldv = None
    while len(alst) < terms:
        (v, s, l) = heapq.heappop(h)
        if v == oldv and v not in alst:
            alst.append(v)
            if verbose: print(f"{v}, [= Prod_{{i = {s}..{l}}} i = Prod_{{i = {olds}..{oldl}}} i]")
        if v >= p:
            p *= nextcount
            heapq.heappush(h, (p, 2, nextcount))
            nextcount += 1
        oldv, olds, oldl = v, s, l
        v //= s; s += 1; l += 1; v *= l
        heapq.heappush(h, (v, s, l))
    return alst
print(aupton(8, verbose=True)) # Michael S. Branicky, Jun 24 2021

a(1), a(7) and a(8) from T. D. Noe, Nov 22 2004
a(9) and a(10) from Robert Munafo, Aug 13 2007
a(11) from Robert Munafo, Aug 17 2007
Edited by N. J. A. Sloane, Sep 14 2008 at the suggestion of R. J. Mathar

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A141757 Even terms in A100933.

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A141756 Intersection of A141586 and A100933.

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A064224 Numbers having more than one representation as the product of consecutive integers > 1.

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