Pratik Koirala - cp's OEIS frontend

A260360 The absolute difference between the largest prime factors of prime(n)-1 and prime(n+1)-1.

0, 1, 2, 2, 1, 1, 8, 4, 2, 2, 2, 2, 16, 10, 16, 24, 6, 4, 4, 10, 28, 30, 8, 2, 12, 36, 50, 4, 0, 6, 4, 6, 14, 32, 8, 10, 80, 40, 46, 84, 14, 16, 4, 4, 4, 30, 76, 94, 10, 12, 12, 0, 3, 129, 64, 62, 18, 16, 40, 26, 56, 14, 18, 66, 68, 4, 166, 144, 18, 168, 118, 30, 24, 184, 94, 86, 6, 12, 2, 12, 36, 40, 70, 56, 10

Offset: 2

Pratik Koirala, Otis Tweneboah, Nathan Fox, Eugene Fiorini, Jul 23 2015

nonn

a(n)=0 if and only if n is in A105403.

It is an open question whether there are infinitely many zeros in this sequence. Are there infinitely many terms below some fixed upper bound?

			n=4: The prime factors of prime(4)-1 are 2,3 and the prime factors of prime(5)-1 are 2,5. The largest are 3 and 5, so a(4)=2.

Robert Israel, Table of n, a(n) for n = 2..10000

Cf. A023503, A105403.

B:= [seq(max(numtheory:-factorset(ithprime(i)-1)),i=2..101)]:
seq(abs(B[n+1]-B[n]),n=1..99); # Robert Israel, Aug 06 2015

Table[Abs[FactorInteger[Prime[n] - 1][[-1, 1]] - FactorInteger[Prime[n + 1] - 1][[-1, 1]]], {n, 2, 86}] (* Michael De Vlieger, Jul 24 2015 *)
Rest[Abs[Differences[Table[FactorInteger[p-1][[-1,1]],{p,Prime[ Range[ 90]]}]]]] (* Harvey P. Dale, Aug 08 2021 *)

gpf(n) = if(n>1, vecmax(factor(n)[, 1]), 1);
a(n) = gpf(prime(n)-1) - gpf(prime(n+1)-1); \\ Michel Marcus, Aug 05 2015

a(n) = abs(A023503(n+1) - A023503(n)). - Robert Israel, Aug 06 2015

A260375 Numbers k such that A260374(k) is a perfect square.

Original entry on oeis.org

0, 1, 2, 4, 5, 6, 7, 8, 10, 11, 14, 15, 16

Offset: 1

Otis Tweneboah, Pratik Koirala, Eugene Fiorini, Nathan Fox, Jul 23 2015

There are a surprising number of small terms in this sequence.
Heuristic: The square root of x has an average distance of 1/4 to an integer, so |x - round(sqrt(x))^2| is around |x - (sqrt(x) - 1/4)^2| or about sqrt(x)/2, hence A260374(n) is around sqrt(n!)/2. By Stirling's approximation this is around (n/e)^(n/2) which is a square with probability (n/e)^(-n/4). The integral of this function converges, so this sequence should be finite. This heuristic is crude, though, because it does not model the extreme values of A260374. - Charles R Greathouse IV, Jul 23 2015
There are no further terms up to 10^5, so probably the list is complete. - Charles R Greathouse IV, Jul 23 2015

			6! = 720. The nearest perfect square is 729. The difference is 9, which is itself a perfect square. So, 6 is in this sequence.

Cf. A260373, A260374.

is(n)=my(N=n!,s=sqrtint(N)); issquare(min(N-s^2, (s+1)^2-N)) \\ Charles R Greathouse IV, Jul 23 2015

from gmpy2 import isqrt, is_square
A260375_list, g = [0], 1
for i in range(1, 1001):
    g *= i
    s = isqrt(g)
    t = g-s**2
    if is_square(t if t-s <= 0 else 2*s+1-t):
        A260375_list.append(i) # Chai Wah Wu, Jul 23 2015

A260374 The distance between n! and the nearest perfect square.

Original entry on oeis.org

0, 0, 1, 2, 1, 1, 9, 1, 81, 476, 225, 324, 4604, 74879, 176400, 215296, 3444736, 11551671, 45680444, 255004929, 1158920361, 2657058876, 24923993276, 130518272975, 97216010329, 2430400258225, 1553580508516, 4666092737476, 538347188396016, 2137864362693921

Offset: 0

Otis Tweneboah, Pratik Koirala, Eugene Fiorini, Nathan Fox, Jul 23 2015

nonn

			6!=720. The nearest perfect square is 729. The difference between these is 9, so a(6)=9.

Cf. A260373, A260375.

a(n)=abs(n!-round(sqrt(n!))^2) \\ Charles R Greathouse IV, Jul 23 2015

from gmpy2 import isqrt
A260374_list, g = [0], 1
for i in range(1, 1001):
    g *= i
    s = isqrt(g)
    t = g-s**2
    A260374_list.append(int(t if t-s <= 0 else 2*s+1-t)) # Chai Wah Wu, Jul 23 2015

a(n) = abs(n!-A260373(n)).

A260373 The nearest perfect square to n!

Original entry on oeis.org

1, 1, 1, 4, 25, 121, 729, 5041, 40401, 362404, 3629025, 39917124, 478996996, 6226945921, 87178467600, 1307674583296, 20922793332736, 355687416544329, 6402373660047556, 121645100663836929, 2432902009335560361, 51090942169052381124, 1124000727752683686724

Offset: 0

Otis Tweneboah, Pratik Koirala, Eugene Fiorini, Nathan Fox, Jul 23 2015

a(n) is well defined as the squares are alternatingly odd and even and thus the average of two successive squares is not an integer and thus no integer is equidistant to two successive squares. - Chai Wah Wu, Jul 24 2015

			6! = 720. The nearest perfect square is 729.

Cf. A055227, A260374.

a(n)=round(sqrt(n!))^2 \\ Charles R Greathouse IV, Jul 23 2015

from gmpy2 import isqrt
A260373_list, g = [1], 1
for i in range(1, 101):
    g *= i
    s = isqrt(g)
    t = s**2
    A260373_list.append(int(t if g-t-s <= 0 else t+2*s+1)) # Chai Wah Wu, Jul 23 2015

a(n) = A055227(n)^2.

A259559 Numbers n such that prime(n)-1 and prime(n+1)-1 have the same number of prime factors, including repeats.

Original entry on oeis.org

3, 4, 10, 12, 19, 29, 34, 36, 45, 46, 50, 61, 85, 89, 91, 104, 112, 117, 118, 119, 129, 130, 137, 138, 143, 147, 148, 158, 178, 179, 181, 185, 200, 202, 206, 214, 220, 233, 238, 239, 244, 248, 249, 258, 262, 275, 299, 304, 314, 333, 338, 340

Offset: 1

Pratik Koirala, Otis Tweneboah, Nathan Fox, Eugene Fiorini, Jun 30 2015

nonn

Unlike A105403, this sequence appears to be infinite.

			The prime factors of prime(10)-1 are 2,2,7 and the prime factors of prime(11)-1 are 2,3,5 and so they have the same number of prime factors, including repeats.

Cf. A105403, A259558.

Select[Range@ 360, PrimeOmega[Prime[#] - 1] == PrimeOmega[Prime[# + 1] - 1] &] (* Michael De Vlieger, Jul 01 2015 *)
Transpose[SequencePosition[Table[PrimeOmega[Prime[n]-1],{n,400}],{x_,x_}]][[1]] (* The program uses the SequencePosition function from Mathematica version 10 *) (* Harvey P. Dale, Nov 29 2015 *)

lista(nn) = {forprime(p=2, nn, if (bigomega(p-1)==bigomega(nextprime(p+1)-1), print1(primepi(p), ", ")););} \\ Michel Marcus, Jul 01 2015

A259558 Numbers n such that prime(n)-1 and prime(n+1)-1 have the same number of distinct prime factors.

Original entry on oeis.org

2, 4, 5, 8, 9, 12, 15, 16, 18, 19, 23, 24, 25, 28, 29, 31, 36, 38, 39, 40, 42, 44, 52, 56, 58, 59, 60, 63, 64, 71, 73, 74, 76, 80, 85, 88, 91, 92, 94, 96, 98, 99, 102, 103, 106, 107, 109, 110, 111, 112, 113, 117, 126, 129, 130, 131, 132, 133, 134, 136, 139, 141, 142, 143, 144, 151, 152, 159, 160, 161, 165, 168, 169, 173

Offset: 1

Pratik Koirala, Otis Tweneboah, Nathan Fox, Eugene Fiorini, Jun 30 2015

nonn

Unlike A105403, this sequence appears to be infinite.

Dickson's conjecture would imply that there are infinitely many p such that p, p+6, 2*p+1 and 2*p+13 are prime and there are no primes between 2*p+1 and 2*p+13. Then n is in the sequence where 2*p+1=prime(n). - Robert Israel, Jun 30 2015

			The prime factors of prime(5)-1 are 2,5. The prime factors of prime(6)-1 are 2,3,3 and they have the same number of distinct prime factors.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A105403, A259559.

N:= 2000: # to use primes <= N
Primes:= select(isprime,[2,seq(2*i+1,i=1..floor((N-1)/2))]):
npf:= map(t -> nops(numtheory:-factorset(Primes[t]-1)), [$1..nops(Primes)]):
select(t -> npf[t+1]=npf[t],[$1..nops(Primes)-1]); # Robert Israel, Jun 30 2015

Select[Range@ 173, PrimeNu[Prime[#] - 1] == PrimeNu[Prime[# + 1] - 1] &] (* Michael De Vlieger, Jul 01 2015 *)

lista(nn) = {forprime(p=2, nn, if (omega(p-1)==omega(nextprime(p+1)-1), print1(primepi(p), ", ")););} \\ Michel Marcus, Jul 01 2015

A259564 Numbers n such that the sum of the prime factors (including repeats) of prime(n)-1 and prime(n+1)-1 are the same.

Original entry on oeis.org

5, 7, 11, 30, 133, 160, 415, 527, 883, 1257, 2025, 2771, 2775, 6650, 6932, 13793, 19091, 30695, 32341, 33722, 36372, 37944, 40532, 42141, 47230, 60986, 77956, 82165, 90564, 111414, 113106, 136036, 147573, 148357, 158279, 169137, 169604, 171549, 174540, 187679

Offset: 1

Otis Tweneboah, Pratik Koirala, Eugene Fiorini, Nathan Fox, Jun 30 2015

nonn

Although there are more terms than A105403 so far, these numbers are still fairly uncommon.

Is this sequence infinite?

			The prime factors of prime(30)-1 are 2,2,2,2,7 and the prime factors of prime(31)-1 are 2,3,3,7. The sum of entries in each of these lists is 15.

Cf. A001414, A105403, A259562.

SequencePosition[Table[Total[Flatten[Table[#[[1]],#[[2]]]&/@ FactorInteger[ p-1]]],{p,Prime[Range[200000]]}],{x_,x_}][[All,1]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Oct 12 2020 *)

spff(n) = {my(f=factor(n)); sum(k=1, #f~, f[k,1]*f[k,2]);}
lista(nn) = {forprime(p=2, nn, if (spff(p-1)==spff(nextprime(p+1)-1), print1(primepi(p), ", ")););} \\ Michel Marcus, Jun 30 2015

More terms from Alois P. Heinz, Jun 30 2015

Name edited by Zak Seidov, Jul 02 2015

A259562 Numbers n such that the sum of the distinct prime factors of prime(n)-1 and prime(n+1)-1 are the same.

Original entry on oeis.org

2, 414, 556, 3962, 4972, 6151, 6521, 8440, 8665, 13769, 13909, 15576, 16696, 17176, 19926, 20630, 21541, 27090, 30822, 62118, 65349, 74014, 94203, 98600, 101231, 103058, 108333, 112332, 136036, 142714, 145588, 147150, 160730, 162366, 169137, 194681, 200837

Offset: 1

Otis Tweneboah, Pratik Koirala, Eugene Fiorini, Nathan Fox, Jun 30 2015

nonn

Although there are more terms than A105403 so far, these numbers are still fairly uncommon.
Is this sequence infinite?
It would follow from the generalized Bunyakovsky conjecture that, e.g., there are infinitely many primes p such that p+2, p+12, p+14, 6*p^2+84*p+1 and 6*p^2+84*p+145 are all prime, and there are no primes between 6*p^2+84*p+1 and 6*p^2+84*p+145. If so, then the sequence is infinite, because it contains n where prime(n) = 6*p^2+84*p+1, with prime(n)-1 having distinct prime factors 2,3,p,p+14 and prime(n+1) having distinct prime factors 2,3,p+2,p+12. - Robert Israel, Jun 30 2015

			The prime factors of prime(414)-1 are 2,3,5,5,19 and the prime factors of prime(415)-1 are 2,2,2,3,7,17. The sum of the distinct entries in each of these lists is 29.

Cf. A008472, A105403, A259564.

Primes:= select(isprime,[2,seq(2*i+1,i=1..10^6)]):
spf:= map(p -> convert(numtheory:-factorset(p-1),`+`), Primes):
select(t -> spf[t+1]=spf[t], [$1..nops(Primes)-1]);

Select[Range@ 250000, Total[First /@ FactorInteger[Prime@ # - 1]] == Total[First /@ FactorInteger[Prime[# + 1] - 1]] &] (* Michael De Vlieger, Jul 01 2015 *)

spf(n) = {my(f=factor(n)); sum(k=1, #f~, f[k,1]);}
lista(nn) = {forprime(p=2, nn, if (spf(p-1)==spf(nextprime(p+1)-1), print1(primepi(p), ", ")););} \\ Michel Marcus, Jun 30 2015

More terms from Alois P. Heinz, Jun 30 2015

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User: Pratik Koirala

A260360 The absolute difference between the largest prime factors of prime(n)-1 and prime(n+1)-1.

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A260375 Numbers k such that A260374(k) is a perfect square.

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A260374 The distance between n! and the nearest perfect square.

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A260373 The nearest perfect square to n!

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A259559 Numbers n such that prime(n)-1 and prime(n+1)-1 have the same number of prime factors, including repeats.

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A259558 Numbers n such that prime(n)-1 and prime(n+1)-1 have the same number of distinct prime factors.

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A259564 Numbers n such that the sum of the prime factors (including repeats) of prime(n)-1 and prime(n+1)-1 are the same.

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A259562 Numbers n such that the sum of the distinct prime factors of prime(n)-1 and prime(n+1)-1 are the same.