Vladimir Letsko - cp's OEIS frontend

A370601 a(n) is the number of integer values of the function F_2n(x) = Product_{i=0..2n-1} (x + i) / Sum_{i=0..2n-1} (x + i).

0, 1, 3, 23, 37, 156, 371, 1207, 2826, 8738, 14839, 60738, 80177, 148702, 466545, 1673039, 2077633, 4771287, 10665251, 26790730, 72170979, 212182718, 248771227, 1074691776, 1488526850, 2533202074, 8444006973, 13950326222, 18313391221, 74263958970, 164820521219

Offset: 1

Vladimir Letsko, Feb 23 2024

nonn

If k is an odd prime then F_k(x) = Product_{i=0..k} (x + i)/ Sum_{i=0..k} (x + i) is not an integer if and only if x==(k+1)/2 (mod k). If k is odd but not prime, then F_k(x) is an integer for all positive integers x. On the other hand, for every even k, there is only a finite number of integer values of F_k(x).

			a(3) = 3 because F_6(x) has exactly 3 integer values: F_6(5) = 3360, F_6(10) = 48048, and F_6(35) = 12282816.

Vladimir Letsko, Math. vertical, YouTube video, 2023 (in Russian).

Cf. A000005, A001818, A126695.

a := proc(n) local d; d := doublefactorial(2*n-1)^2; numtheory[tau](d/igcd(n, d)) - n end: seq(a(n), n = 1..31);

Table[DivisorSigma[0, (2*n - 1)!!^2/GCD[(2*n - 1)!!^2, n]] - n, {n, 1, 30}] (* Vaclav Kotesovec, Feb 23 2024 *)

a(n) = tau(d/gcd(d, n)) - n, where d = ((2*n-1)!!)^2 and tau(k) is the number of divisors of the positive integer k.

A338160 Number of ways to represent n as a product of the greatest number of distinct factors.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 2, 2, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 1, 1, 3, 1, 2, 1, 1, 1, 4

Offset: 1

Vladimir Letsko, Oct 14 2020

nonn

a(n) = A058060(n) for 1 < n < 60; a(60) = 3, A058060(60) = 1.
a(n) is the number of factorizations of n into A086435(n) distinct factors > 1.
a(n) depends only on the prime signature of n.

			a(72) = 3 because 72 = 2*3*12 = 2*4*9 = 3*4*6 and 72 cannot be represented as a product of 4 distinct factors each greater than 1 (adding the factor 1 to each product doesn't change anything).

Cf. A086435, A140207, A338159.

a(n)={my(d=divisors(n)); my(F(r,k)=if(r==1, [0,1], my(b=-1,c=0); for(k=2, k, if(r%d[k]==0, my([tb,tc]=self()(r/d[k], k-1)); if(tb>b, b=tb; c=0); if(tb==b, c+=tc))); [b+1, c])); F(n, #d)[2]} \\ Andrew Howroyd, Oct 14 2020

More terms from Andrew Howroyd, Oct 14 2020

A338159 The least number which can be represented as a product of the greatest number of distinct positive integers in exactly n ways.

Original entry on oeis.org

1, 12, 60, 96, 360, 576, 480, 15120, 864, 2880, 3360, 6912, 25200, 7680, 20160, 36960, 4320, 93312, 46080, 82944, 221760, 34560, 2494800, 311040, 53760, 88200, 15966720, 30240, 3880800, 1995840, 43200, 322560, 388800, 345600, 970200, 241920, 414720, 5832000, 529200, 5598720

Offset: 1

Vladimir Letsko, Oct 14 2020

nonn

k = p_1^2*p_2*...*p_n obviously has exactly n required representations. Hence a(n) exists for any n.
a(n) is the least k such that A338160(k) = n.
All terms are in A025487.

			a(60) = 3 because 60 = 2*3*10 = 2*5*6 = 3*4*5 and each number less than 60 does not have exactly 3 such representations (adding the factor 1 to each product doesn't change anything).

David A. Corneth, Table of n, a(n) for n = 1..1283 (first 150 terms from Dmitry Khomovsky)
David A. Corneth, More terms

Cf. A338160.

a(A338160(n)) = n.

A338160(k) <> n for k < a(n).

a(23)-a(40) from Andrew Howroyd, Oct 14 2020

A336486 The least prime s for which there exist primes p, q, r such that phi(pqs^n) = phi(rs^(2n+1)) and sigma(pqs^n) = sigma(rs^(2n+1)).

Original entry on oeis.org

2, 2, 2, 2, 5, 37, 13, 7, 2, 19, 7, 7, 2, 19, 4447, 2, 2, 2, 2, 5, 73, 23, 37, 2, 2, 19, 19, 2, 2

Offset: 1

Vladimir Letsko, Jul 23 2020

This sequence is an inversion of A336485.
Let "a and b are similar" mean that for positive integers a and b we have phi(a) = phi(b), tau(a) = tau(b) and sigma(a) = sigma(b).
Conjecture: For each positive integer n there are infinitely many primes s such that numbers p*q*s^n and r*s^(2n+1) are similar for some primes p, q, r.

			a(2) = 2 because for n = 2 and prime s = 2, a = 19*89*s^n and b = 199*s^(2n+1), we have phi(a) = phi(b) and sigma(a) = sigma(b).

Cf. A000005, A000010, A000203, A134922, A336485.

with(NumberTheory):
KS := []; for k to 29 do tf := false;
for ii do s := ithprime(ii); c := 2*s^(k+1)+1; cc := (c^2-1)*(1/2); Q := Divisors(cc);
for d in Q do q := d+c; if isprime(q) then p := c+cc/(q-c); if p < q then break end if;
if isprime(p) then r := 2*(p+q)-c; if isprime(r) then print([k, [p, q], r], s); KS := [op(KS), s]; tf := true; break end if end if end if end do;
if tf then break end if end do end do; KS

is(t, u, x, y) = ispseudoprime(t*x+1) && ispseudoprime(u*y/t+1) && ispseudoprime(x*y+1);
a(n) = {my(s=1, t, u); while(s=nextprime(s+1), for(i=0, 1+n\2, t=s^i; fordiv(2*(1+u=s^(n+1)), d, if(is(t, u, 2*u/t+d, 2*t+(2*u+2)/d) || is(t, u, 2*u/t-d, 2*t-(2*u+2)/d), return(s))))); } \\ Jinyuan Wang, Sep 30 2020

A336485 The least positive integer k for which there exist primes p, q, r such that phi(pqs^k) = phi(rs^(2k+1)) and sigma(pqs^k) = sigma(rs^(2k+1)), where s is the n-th prime.

Original entry on oeis.org

1, 1, 1, 1, 17, 3, 29, 4, 4, 4, 1, 5, 4, 1, 20, 32, 2, 38, 12, 29, 9, 4, 26, 20, 8, 14, 2, 14, 8, 41, 4

Offset: 1

Vladimir Letsko, Jul 23 2020

This sequence is an inversion of A336486.
A236255 contains the primes s = prime(m) for which a(m) = 1.
Let "a and b are similar" mean that for positive integers a and b we have phi(a) = phi(b), tau(a) = tau(b) and sigma(a) = sigma(b).
Conjecture: For each prime s there are infinitely many positive integers k such that numbers p*q*s^k and r*s^(2k+1) are similar for some primes p, q, r.

			a(6) = 3 because:
1. For the 6th prime, s = 13, k = 3 and with primes p = 62807837, q = 57149, r = 125672849 we have phi(p*q*s^k) = phi(r*s^(2k+1)) and sigma(p*q*s^k) = sigma(r*s^(2k+1)).
2. There is no such equality for s = 13 and k less than 3.

Cf. A000005, A000010, A000203, A134922, A336486.

with(NumberTheory):
SK := []; for ii to 31 do s := ithprime(ii); tf := false;
for k do c := 2*s^(k+1)+1; cc := (c^2-1)*(1/2); Q := Divisors(cc);
for d in Q do q := d+c; if isprime(q) then p := c+cc/(q-c); if p < q then break end if;
if isprime(p) then r := 2*(p+q)-c; if isprime(r) then print([s, [p, q], r], k); SK := [op(SK), [s, k]]; tf := true; break end if end if end if end do;
if tf then break end if end do end do; SK

is(t, u, x, y) = ispseudoprime(t*x+1) && ispseudoprime(u*y/t+1) && ispseudoprime(x*y+1);
a(n) = {my(s=prime(n), t, u); for(k=1, oo, for(i=0, 1+k\2, t=s^i; fordiv(2*(1+u=s^(k+1)), d, if(is(t, u, 2*u/t+d, 2*t+(2*u+2)/d) || is(t, u, 2*u/t-d, 2*t-(2*u+2)/d), return(k))))); } \\ Jinyuan Wang, Sep 30 2020

A336297 Prime numbers p such that equation x = p*sopf(x) (where sopf(x) is the sum of distinct prime factors of x) has exactly 1 solution in positive integers.

Original entry on oeis.org

2, 61, 97, 113, 151, 173, 241, 277, 317, 353, 389, 449, 457, 593, 601, 607, 653, 673, 683, 727, 733, 797, 907, 929, 941, 947, 953, 977, 997, 1021, 1051, 1087, 1153, 1181, 1193, 1217, 1249, 1307, 1321, 1361, 1373, 1409, 1433, 1489, 1493, 1523, 1553, 1579, 1597, 1609, 1627

Offset: 1

Vladimir Letsko, Jul 16 2020

nonn

			4 is the unique integer x such that x = 2*sopf(x), a prime, so 2 is a term.

Vladimir Letsko, Mathematical Marathon, Problem 227 (in Russian).

Cf. A008472, A336098, A336099, A336296.

A336296 The least prime p such that equation x = p*sopf(x) (where sopf(x) is the sum of distinct prime factors of x) has exactly n solutions in positive integers.

Original entry on oeis.org

2, 3, 7, 19, 71, 431, 1259, 4679, 9719, 23399, 7559, 42839, 134399, 181439, 477359, 241919, 262079, 453599

Offset: 1

Vladimir Letsko, Jul 16 2020

It seems that a(n) is the least number for which equation x = p*sopf(x) has exactly n solutions in positive integers not only for prime numbers.

			a(3) = 7 because there are 3 solutions of the equation x = 7*sopf(x), which are {49, 84, 105}, and this is the smallest prime that gives 3 solutions.

Vladimir Letsko, Mathematical Marathon, Problem 227 (in Russian)

Cf. A008472, A089352, A336098, A336099, A336297, A157190 (note overlap of values).

A336027 The least k such that i*k + 1 is a product of i different primes for i = 1..n.

Original entry on oeis.org

1, 10, 268, 7576, 652726, 913180816

Offset: 1

Vladimir Letsko, Jul 05 2020

Since for k = 5000000000420503488, i*k+1 is a product of i different primes for i = 1..7, a(7) <= 5000000000420503488.

a(n) is the least parameter k such that equation tau(x^k) = x has at least A005425(n) solutions in positive integers.

			268 is in the sequence because 268 + 1 is prime, 2*268 + 1 is a product of 2 different primes, 3*268 + 1 is a product of 3 different primes, and 268 is the least number with such properties.

Vladimir Letsko, Problem K7 (in Russian).

Cf. A000005, A005425, A336028.

nn := 1; for kk to 6 do
n := nn; do n := nextprime(n); f := true;
for k from 2 to kk do a := k*(n-1)+1; if not IsSquareFree(a) or nops(ifactors(a)[2]) <> k then f := false; break end if end do;
if f then nn := n-1; print(nn); break end if end do end do

isok(k,n) = {for (i=1, n, if ((omega(i*k+1) != i) || (bigomega(i*k+1) != i), return (0));); return (1);}
a(n) = {my(k=1); while(!isok(k,n), k++); k;} \\ Michel Marcus, Jul 15 2020

A336099 Number of solutions of the equation k = n*sopf(k) in positive integers where sopf(k) is the sum of distinct prime factors of k.

Original entry on oeis.org

1, 2, 1, 2, 1, 3, 1, 2, 1, 3, 1, 3, 1, 1, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 2, 2, 1, 2, 1, 4, 1, 2, 2, 2, 1, 3, 2, 1, 1, 2, 1, 4, 1, 1, 0, 3, 1, 3, 1, 1, 2, 2, 1, 0, 1, 2, 2, 4, 1, 1, 2, 2, 1, 1, 1, 4, 2, 1, 1, 5, 1, 2, 2, 1, 2, 1, 1, 2, 1

Offset: 2

Vladimir Letsko, Jul 08 2020

nonn

Offset is 2 because a(1) cannot be defined since there are infinitely many solutions for n = 1, the primes.
If n = p^s then p^(s+1) is solution of k = n*sopf(k). Hence a(p^s) > 0. On the other hand there are infinitely many 0's in the sequence. For example a(5^s*11^t) = 0 for all positive integers s, t.
Records appear to occur only at prime n. These are seen in A336296, although note that A336296 is not monotonic, so it includes other terms. - Bill McEachen, Dec 02 2023

			a(3) = 2 because there are exactly 2 solutions of the equation k = 3*sopf(k) in positive integers (9 and 30).

Vladimir Letsko, Mathematical Marathon, Problem 227 (in Russian).

Cf. A008472, A336098, A336296.

Cf. A158804 (all possible k's).

A336098 Numbers k such that equation x = k*sopf(x) has no solutions in positive integers.

Original entry on oeis.org

46, 55, 85, 87, 92, 110, 123, 138, 141, 145, 155, 158, 183, 184, 187, 190, 194, 203, 205, 217, 219, 220, 230, 238, 247, 253, 259, 261, 265, 275, 276, 282, 287, 290, 291, 295, 302, 305, 310, 316, 319, 327, 334, 339, 366, 368, 369, 380, 388, 391, 395, 403, 406, 407, 410, 414, 415, 423, 425, 426, 427, 434

Offset: 1

Vladimir Letsko, Jul 08 2020

nonn

If k = p^s then p^(s+1) is solution of x = k*sopf(x). Hence powers of primes are not in the sequence.

Let p_1*...*p_t is in the sequence. Then p_1^a_1*...*p_t^a_t is in the sequence for all positive integers a_1, ..., a_t. It means that the sequence is infinite.

Vladimir Letsko, Mathematical Marathon, Problem 227 (in Russian).

Cf. A008472, A336099.

sopf(n) = vecsum(factor(n)[, 1]); \\ A008472
pp(n) = prod(k=1, n, prime(k)); \\ A002110
sp(n) = sum(k=1, n, prime(k)); \\ A007504
ip(n) = {my(k=1); while (pp(k)/sp(k) <= n, k++); k+1;}
listako(nn) = {my(lim = pp(ip(nn))); my(v = vector(lim, k, k++; k/sopf(k))); my(w = vector(nn-1, k, #select(x->(x==k+1), v))); apply(x->(x+1), Vec(select(x->(x==0), w, 1)));} \\ Michel Marcus, Jul 16 2020

cp's OEIS Frontend

User: Vladimir Letsko

A370601 a(n) is the number of integer values of the function F_2n(x) = Product_{i=0..2n-1} (x + i) / Sum_{i=0..2n-1} (x + i).

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A338160 Number of ways to represent n as a product of the greatest number of distinct factors.

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A338159 The least number which can be represented as a product of the greatest number of distinct positive integers in exactly n ways.

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A336486 The least prime s for which there exist primes p, q, r such that phi(p*q*s^n) = phi(r*s^(2n+1)) and sigma(p*q*s^n) = sigma(r*s^(2n+1)).

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A336485 The least positive integer k for which there exist primes p, q, r such that phi(p*q*s^k) = phi(r*s^(2k+1)) and sigma(p*q*s^k) = sigma(r*s^(2k+1)), where s is the n-th prime.

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A336297 Prime numbers p such that equation x = p*sopf(x) (where sopf(x) is the sum of distinct prime factors of x) has exactly 1 solution in positive integers.

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A336296 The least prime p such that equation x = p*sopf(x) (where sopf(x) is the sum of distinct prime factors of x) has exactly n solutions in positive integers.

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A336027 The least k such that i*k + 1 is a product of i different primes for i = 1..n.

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A336486 The least prime s for which there exist primes p, q, r such that phi(pqs^n) = phi(rs^(2n+1)) and sigma(pqs^n) = sigma(rs^(2n+1)).

A336485 The least positive integer k for which there exist primes p, q, r such that phi(pqs^k) = phi(rs^(2k+1)) and sigma(pqs^k) = sigma(rs^(2k+1)), where s is the n-th prime.