cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A227248 Composite numbers 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 n (with multiplicity).

Original entry on oeis.org

4, 72, 102, 216, 260, 264, 270, 432, 462, 504, 612, 720, 1980, 2592, 3672, 5184, 5550, 8352, 10368, 15552, 20736, 22032, 25488, 29016, 30240, 31104, 36288, 38880, 41976, 46656, 49248, 57888, 62208, 93312, 103680, 132192, 186624, 238464, 286848, 373248, 410688
Offset: 1

Views

Author

Paolo P. Lava, Jul 04 2013

Keywords

Examples

			Prime factors of 270 are 2, 3^3, 5; therefore (2/(2+1)+3*3/(3+1)+5/(5+1))/(2/(2+1)*(3/(3+1))^3*5/(5+1)) = 16.

Links

Crossrefs

Cf. A224346, A224912, A226365, A227034.

Programs

  • Maple
    with(numtheory); ListA227248:=proc(q) local a, d, n, p;
for n from 2 to q do if not isprime(n) then p:=ifactors(n)[2];
a:=add((op(1, d)/(op(1, d)+1))*op(2, d), d=p)/mul((op(1, d)/(op(1, d)+1))^op(2, d), d=p);
if type(a, integer) then print(n); fi; fi;
od; end: ListA227248(10^10);

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

Views

Author

Paolo P. Lava, Oct 09 2013

Keywords

Comments

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

Examples

			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.

Links

Crossrefs

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

Programs

  • Maple
    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);

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

Views

Author

Paolo P. Lava, Oct 09 2013

Keywords

Examples

			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.

Crossrefs

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

Programs

  • Maple
    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

Views

Author

Paolo P. Lava, Oct 09 2013

Keywords

Examples

			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.

Crossrefs

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

Programs

  • Maple
    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);
Showing 1-4 of 4 results.

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