A199767 -id:A199767 - cp's OEIS frontend

A198391 Numbers m such that Sum_{i=1..k} (1-1/p_i) + Product_{i=1..k} (1-1/p_i) is an integer, where p_i are the k prime factors of m (with multiplicity).

2, 15, 20, 272, 476, 19024, 47425, 65792, 125172, 216900, 539280, 1222976, 1372736, 2770496, 3494336, 5321808, 5844528, 6177168, 7032528, 8885808, 20670768, 60727876, 69081344, 82724356, 95579136, 544382208, 907440192, 1657497600, 4295032832, 5048574976

Offset: 1

Paolo P. Lava, Oct 24 2011

nonn

The numbers of the sequence are solutions of the differential equation m’=(k-a)*m+b, which can be written as A003415(m)=(k-a)*m+A003958(m), where k is the number of prime factors of m, and a is the integer Sum_{i=1..k} (1-1/p_i) + Product_{i=1..k} (1-1/p_i).
If k = a we have m’ = b or A003415(m) = A003958(m). For instance 15 has prime factors 3, 5; its arithmetic derivative is 15’ = 8 and b = 3*5 - 3 - 5 + 1 = 8. The term 47425 has prime factors 5, 5, 7, 271. Its arithmetic derivative is 47425’ = 25920 and b = 5*5*7*271 - 5*5*7 - 5*5*271 - 5*7*271 - 5*7*271 + 5*5 + 5*7 + 5*271 + 5*7 + 5*271 + 7*271 - 5 - 5 - 7 - 271 + 1 = 25920.
The numbers of the sequence satisfy also Sum_{i=1..k} (1+1/p_i) - Product_{i=1..k} (1-1/p_i) = some integer.

			125172 has prime factors 2, 2, 3, 3, 3, 19, 61. 1 - 1/2 + 1 - 1/2 + 1 - 1/3 + 1 - 1/3 + 1 - 1/3 + 1 - 1/19 + 1 - 1/61 = 5715/1159 is the sum over the 1-1/p_i. (1-1/2) * (1-1/2) * (1-1/3) * (1-1/3) * (1-1/3) * (1-1/19) * (1-1/61) = 80/1159 is the product of the 1-1/p_i. The sum over sum and product is 5715/1159 + 80/1159 = 5, an integer.

Giovanni Resta, Table of n, a(n) for n = 1..48 (terms < 10^12)
J. M. Borwein and E. Wong, A survey of results relating to Giuga’s conjecture on primality, May 8, 1995.
Romeo Mestrovic, On a Congruence Modulo n^3 Involving Two Consecutive Sums of Powers, Journal of Integer Sequences, Vol. 17 (2014), 14.8.4.

Cf. A003415, A003958, A007850, A199767.

isA198391 := proc(n)
    p := ifactors(n)[2] ;
    add(op(2,d)-op(2,d)/op(1,d),d=p) + mul((1-1/op(1,d))^op(2,d),d=p) ;
    type(%,'integer') ;
end proc:
for n from 2 to 20000000 do
    if isA198391(n) then
        printf("%d,\n",n);
    end if;
end do: # R. J. Mathar, Nov 26 2011

Select[Range[2, 10^5], IntegerQ[(Plus @@ # + Times @@ #) &@ (1 - 1/ Flatten[ Table[#1, {#2}] & @@@ FactorInteger@#])] &] (* Giovanni Resta, May 23 2016 *)

Missing a(23) and a(26)-a(30) from Giovanni Resta, May 23 2016

A224912 Numbers m for which Sum_{i=1..k} (p(i)/(p(i)-1)) + Product_{i=1..k} (p(i)/(p(i)-1)) is an integer, where p(i) are the k prime factors of m (with multiplicity).

Original entry on oeis.org

2, 3, 4, 8, 16, 32, 36, 64, 72, 108, 128, 144, 200, 256, 288, 396, 512, 528, 576, 588, 1024, 1040, 1152, 1296, 2000, 2048, 2304, 2320, 2400, 2592, 3888, 4096, 4160, 4608, 4752, 4800, 5184, 5600, 6552, 7200, 8192, 8448, 9216, 9600, 9936, 10368, 11316, 12000

Offset: 1

Paolo P. Lava, Apr 19 2013

nonn

Apart from 3 all terms are even.

			Prime factors of 11316 are 2^2, 3, 23 and 41.
Sum_{i=1..5} (p(i)/(p(i)-1)) = 2*(2/(2-1)) + 3/(3-1) + 23/(23-1) + 41/(41-1) = 3331/440.
Sroduct_{i=1..5} (p(i)/(p(i)-1)) = 2*(2/(2-1)) * 3/(3-1) * 23/(23-1) * 41/(41-1) = 2829/440.
Their sum is an integer: 3331/440 + 2829/440 = 14.

Paolo P. Lava, Table of n, a(n) for n = 1..200

Cf. A199767, A198391.

with(numtheory);
A224912:=proc(i) local b,c,d,n,p;
for n from 2 to i do p:=ifactors(n)[2];
  b:=add(op(2,d)*op(1,d)/(op(1,d)-1),d=p)+mul((op(1,d)/(op(1,d)-1))^op(2,d),d=p);
  if trunc(b)=b then print(n); fi; od; end:
A224912(10^6);

A230110 Composite numbers m such that Sum_{i=1..k} (p_i/(p_i+1)) + Product_{i=1..k} (p_i/(p_i-1)) is an integer, where p_i are the k prime factors of m (with multiplicity).

Original entry on oeis.org

8, 10, 30, 63, 64, 512, 588, 720, 800, 1320, 3960, 4096, 5184, 5760, 6400, 7200, 21600, 27720, 27900, 32768, 35280, 41472, 46080, 51200, 70840, 84672, 92400, 95040, 105600, 151200, 188160, 262144, 331776, 368640, 376320, 409600, 504000, 518400, 576000, 640000

Offset: 1

Paolo P. Lava, Oct 09 2013

nonn

Includes 2^(3*a) * 3^(4*b) if 3*a >= 4*b. - Robert Israel, Mar 30 2023

			Prime factors of 3960 are 2^3, 3^2, 5 and 11.
Sum_{i=1..7} (p(i)/(p(i)+1)) = 3*(2/(2+1)) + 2*(3/(3+1)) + 5/(5+1) + 11/(11+1) = 21/4.
Product_{i=1..7} (p(i)/(p(i)-1)) = (2/(2+1))^3 * (3/(3-1))^2 * 5/(5-1) * 11/(11-1) = 99/4.
Their sum is an integer: 21/4 + 99/4 = 30.

Robert Israel, Table of n, a(n) for n = 1..155

Cf. A199767, A198391, A227034, A227248, A230111, A230112.

with(numtheory); P:=proc(i) local b,d,n,p;
for n from 2 to i do p:=ifactors(n)[2];
b:=add(op(2,d)*op(1,d)/(op(1,d)+1),d=p)+mul((op(1,d)/(op(1,d)-1))^op(2,d),d=p);
if trunc(b)=b then print(n); fi; od; end: P(10^7);

A226365 Composite numbers such that Sum_{i=1..k} (1 + 1/p_i) - Product_{i=1..k} (1 + 1/p_i) is an integer, where p_i are the k prime factors of n (with multiplicity).

Original entry on oeis.org

152, 432, 1620, 1728, 3456, 4752, 22464, 46656, 80892, 139968, 186624, 237168, 326592, 746496, 1651968, 2052864, 2426112, 2985984, 5971968, 10257408, 12177216, 12690432, 14048240, 14183424, 20155392, 20901888, 26127360, 38817792

Offset: 1

Paolo P. Lava, Jun 12 2013

nonn

The numbers of the sequence are the solution of the differential equation n' = (a-k)*n + b, which can also be written as A003415(n) = (a-k)*n + A003958(n), where k is the number of prime factors of n, and a is the integer Sum_{i=1..k} (1 + 1/p_i) - Product_{1=1..k} (1 + 1/p_i).

The numbers of the sequence satisfy also Sum_{i=1..k} (1 - 1/p_i) + Product_{i=1..k} (1 + 1/p_i) = some integer.

			237168 has prime factors 2, 2, 2, 2, 3, 3, 3, 3, 3, 61. 4*(1 + 1/2) + 5*(1 + 1/3) + (1 + 1/61) = 2504/183 is the sum over the 1 + 1/p_i. (1 + 1/2)^4 * (1 + 1/3)^5 * (1 + 1/61) = 3968/183 is the product of the 1 + 1/p_i. The difference over sum and product is 2504/183 - 3968/183 = -8, an integer.

J. M. Borwein and E. Wong, A survey of results relating to Giuga's conjecture on primality, May 8, 1995.
R. Mestrovic, On a Congruence Modulo n^3 Involving Two Consecutive Sums of Powers, Journal of Integer Sequences, Vol. 17 (2014), 14.8.4.

Cf. A003415, A003958, A198391, A199767, A224912.

with(numtheory); ListA226365:=proc(q) local a,d,n,p;
for n from 1 to q do if not isprime(n) then p:=ifactors(n)[2];
a:=add(op(2,d)+op(2,d)/op(1,d),d=p)-mul((1+1/op(1,d))^op(2,d),d=p);
if type(a,integer) then print(n); fi; fi;
od; end: ListA226365(10^9);

A230111 Composite numbers m such that Sum_{i=1..k} (p_i/(p_i+1)) - Product_{i=1..k} (p_i/(p_i-1)) is an integer, where p_i are the k prime factors of m (with multiplicity).

Original entry on oeis.org

8, 10, 64, 512, 720, 800, 1320, 1944, 4096, 5184, 5760, 6400, 7200, 8370, 23520, 32768, 41472, 44000, 46080, 47040, 51200, 69580, 74088, 76096, 84672, 93000, 95040, 105600, 129360, 235200, 240000, 262144, 331776, 368640, 409600, 518400, 546480, 576000, 640000

Offset: 1

Paolo P. Lava, Oct 09 2013

nonn

			Prime factors of 7200 are 2^5, 3^2 and 5^2.
Sum_{i=1..9} (p(i)/(p(i)+1)) = 5*(2/(2+1)) + 2*(3/(3+1)) + 2*(5/(5+1)) = 13/2.
Product_{i=1..9} (p(i)/(p(i)-1)) = (2/(2+1))^5 * (3/(3-1))^2 * (5/(5-1))^2 = 225/2.
Their sum is an integer: 13/2 - 225/2 = -106.

Cf. A199767, A198391, A227034, A227248, A230110, A230112.

with(numtheory); P:=proc(i) local b,d,n,p;
for n from 2 to i do p:=ifactors(n)[2];
b:=add(op(2,d)*op(1,d)/(op(1,d)+1),d=p)-mul((op(1,d)/(op(1,d)-1))^op(2,d),d=p);
if trunc(b)=b then print(n); fi; od; end: P(10^7);

A230112 Composite numbers m such that Product_{i=1..k} (p_i/(p_i-1)) / Sum_{i=1..k} (p_i/(p_i+1)) is an integer, where p_i are the k prime factors of m (with multiplicity).

Original entry on oeis.org

4, 8, 16, 64, 256, 720, 800, 2200, 4096, 25600, 33600, 36288, 41472, 46080, 65536, 92400, 104960, 235200, 282240, 338688, 376320, 403200, 419840, 535680, 556640, 576000, 580800, 640000, 844800, 979776, 1088640, 1244160, 1354752, 1382400, 1505280, 1689600, 1995840

Offset: 1

Paolo P. Lava, Oct 09 2013

nonn

			Prime factors of 2200 are 2^3, 5^2 and 11.
Sum_{i=1..6} (p(i)/(p(i)+1)) = 3*(2/(2+1)) + 2*(5/(5+1)) + 11/(11+1) = 55/12.
Product_{i=1..6} (p(i)/(p(i)-1)) = (2/(2-1))^3*(5/(5-1))^2*11/(11-1) = 55/4.
The ratio is an integer: (55/4) / (55/12) = 3.

Cf. A199767, A198391, A227034, A227248, A230110, A230111.

with(numtheory); P:=proc(q) local a, d, n, p;
for n from 2 to q do if not isprime(n) then p:=ifactors(n)[2];
a:=mul((op(1, d)/(op(1, d)-1))^op(2, d), d=p)/add((op(1, d)/(op(1, d)+1))*op(2, d), d=p); if type(a, integer) then print(n); fi; fi;
od; end: P(10^7);

cp's OEIS Frontend

A198391 Numbers m such that Sum_{i=1..k} (1-1/p_i) + Product_{i=1..k} (1-1/p_i) is an integer, where p_i are the k prime factors of m (with multiplicity).

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A224912 Numbers m for which Sum_{i=1..k} (p(i)/(p(i)-1)) + Product_{i=1..k} (p(i)/(p(i)-1)) is an integer, where p(i) are the k prime factors of m (with multiplicity).

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A230110 Composite numbers m such that Sum_{i=1..k} (p_i/(p_i+1)) + Product_{i=1..k} (p_i/(p_i-1)) is an integer, where p_i are the k prime factors of m (with multiplicity).

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A226365 Composite numbers such that Sum_{i=1..k} (1 + 1/p_i) - Product_{i=1..k} (1 + 1/p_i) is an integer, where p_i are the k prime factors of n (with multiplicity).

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A230111 Composite numbers m such that Sum_{i=1..k} (p_i/(p_i+1)) - Product_{i=1..k} (p_i/(p_i-1)) is an integer, where p_i are the k prime factors of m (with multiplicity).

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A230112 Composite numbers m such that Product_{i=1..k} (p_i/(p_i-1)) / Sum_{i=1..k} (p_i/(p_i+1)) is an integer, where p_i are the k prime factors of m (with multiplicity).

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