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User: Thomas Nordhaus

Thomas Nordhaus has authored 3 sequences.

A209407 Primes p(i) such that p(i+1)/p(i) > p(k+1)/p(k) for all k>i, where p(i) is the i-th prime.

3, 7, 13, 23, 31, 47, 113, 139, 199, 211, 293, 317, 523, 1327, 1669, 1951, 2179, 2477, 2971, 3271, 4297, 4831, 5591, 5749, 5953, 6491, 6917, 7253, 8467, 9551, 9973, 10799, 11743, 15683, 19609, 31397, 34061, 35617, 35677, 43331, 44293, 45893, 48679, 58831

Offset: 1

Thomas Nordhaus, Mar 08 2012

nonn

p(i) belongs to the sequence if p(i+1)/p(i) > p(k+1)/p(k) for all k>i.
It follows from the prime number theorem that p(i+1)/p(i) converges to 1 as i tends to infinity. a(n) is an infinite sequence therefore. The a(n) constitute "record holders" for the relative size of the prime number gaps.
The values a(n) given above were obtained by comparing p(i+1)/p(i) with p(k+1)/p(k) for 1<=i<=5949 and k ranging from i+1 to 200000 for given i.
In order to show that these values are correct one has to analyze the error terms in the formula p(k) ~ k*log(k) and extend the "test range" if needed. Using Dusart's bound: n*(log(n)+loglog(n)-1) < p(n) < n*(log(n)+loglog(n)) for n>=6 one gets
p(k+1)/p(k) < f(k):=(1+1/k)*(log(k+1)+loglog(k+1))/(log(k)+loglog(k)-1) for all k>=6. However this bound tends to 1 like 1+1/log(k) as k->oo. In order to verify, for example, that the term a(9)=199=p(46) is correct one must make sure that p(k+1)/p(k) < p(47)/p(46) = 211/199 =~ 1.0603 for all k>47. However f(10^7)~=1.06666 still, so k <= 10^7 is not sufficient to validate a(9). a(8) however is validated by checking the range k<=10^7.
In order to validate terms up to a(n)=31397 for example one even needs k<=10^20 roughly which needs considerable computational power.
This can be improved with another of Dusart's bounds: there is always a prime in (x, x + x/(25log^2 x)) for x > 396738. Hence it suffices to check up to the higher of exp(1/(25 (prime(i+1)/prime(i)-1))) and 396738. - Charles R Greathouse IV, Mar 06 2013

			The smallest prime belonging to the sequence is p(2)=3 because p(3)/p(2) = 5/3 > 7/5, 11/7, 13/11, 17/13,... p(1)=2 does not belong to the sequence since p(2)/p(1) = 3/2 <5/3 = p(3)/p(2).

Charles R Greathouse IV, Table of n, a(n) for n = 1..67

Cf. A002386, A144104.

{np=200000;a=vector(44);q=vector(np,k,prime(k+1)/prime(k));m=n=0;
while(n<=44,if(q[m++]>vecmax(vector(np-m,j,q[m+j])),a[n++]=prime(m)))} \\ computes the first 44 terms of sequence.

list(lim)=my(v=List([3]),u=List([2/3]),mn=.04/log(lim)^2,p=7,t);forprime(q=11,nextprime(lim+1),t=(q-p)/p;if(t>mn,if(t>u[#v],v[#v]=p;u[#u]=t,listput(v,p);listput(u,t)));p=q);t=u[#u];forstep(i=#u-1,6,-1,if(u[i]>t,t=u[i],v[i]=3));Set(v) \\ valid for lim > 396738; Charles R Greathouse IV, Jun 25 2014

A186775 Numbers k such that digitsum(2^k) > digitsum(2^(k+1)).

Original entry on oeis.org

3, 4, 8, 9, 15, 16, 20, 21, 23, 24, 26, 28, 29, 33, 34, 36, 39, 40, 41, 46, 48, 51, 52, 55, 56, 57, 60, 63, 64, 67, 68, 69, 74, 75, 76, 77, 80, 82, 83, 85, 86, 88, 91, 92, 94, 95, 97, 98, 100, 102, 106, 108, 112, 113, 116, 118, 121, 124, 126

Offset: 1

Thomas Nordhaus, Feb 26 2011

If 2^k and 2^(k+1) acted like random numbers of their size, the probability that k would be in the sequence would be 1/2 + O(1/k). So very possibly a(n) ~ 2n. - Charles R Greathouse IV, Aug 08 2022

			3 is in the sequence because digitsum(2^3) = 8 > 7 = digitsum(2^4).

J.W.L. (Jan) Eerland, Table of n, a(n) for n = 1..49688

Cf. A001370.

DeleteCases[Table[If[Total[Total[IntegerDigits[2^n]]]>Total[IntegerDigits[2^(n+1)]],n,k],{n,0, 10^5}],k] (* J.W.L. (Jan) Eerland, Aug 08 2022 *)

from itertools import count, islice, pairwise
def ds2(n): return sum(map(int, str(1< t[1])
print(list(islice(agen(), 60))) # Michael S. Branicky, Aug 08 2022

def is_A186775(n): return sum((2^n).digits()) > sum((2^(n+1)).digits()) # D. S. McNeil, Feb 27 2011

A079296 Primes ordered by decreasing value of the function p -> sqrt(q) - sqrt(p) where q is the next prime after p.

Original entry on oeis.org

7, 113, 23, 13, 31, 3, 1327, 19, 47, 199, 139, 89, 5, 211, 293, 53, 523, 317, 61, 181, 73, 887, 1129, 83, 37, 241, 2, 43, 283, 1669, 11, 467, 1069, 337, 509, 2477, 131, 2179, 2971, 1259, 773, 1951, 1637, 409, 3271, 421, 151, 1381, 67, 839, 619, 863, 157, 17, 661, 3137

Offset: 1

Thomas Nordhaus, Feb 09 2003

I computed a couple of thousand primes with EXCEL and ordered them accordingly. There is a very small chance that very large prime numbers will change the order of the given terms above.
This sequence only makes sense if the sequence n -> sqrt(p_(n+1)) - sqrt(p_n) is a zero-sequence which is a hard unsolved problem. See also Andrica's conjecture.
For each consecutive prime pair p < q, the number d = sqrt(q) - sqrt(p) is unique. Place d in order from greatest to least and specify p. See Table II in Wolf. A rearrangement of the primes. - Robert G. Wilson v, Oct 18 2012

Donovan Johnson, Table of n, a(n) for n = 1..10000
C. K. Caldwell, Gaps between primes.
Eric W. Weisstein, Andrica's Conjecture
Wikipedia, Andrica's conjecture
Marek Wolf, A note on the Andrica conjecture, arXiv:1010.3945 [math.NT], 2010.

Cf. A078692, A002386, A084974 (records).

lim = 1/5; lst = {}; p = 2; q = 3; While[p < 50000, If[ Sqrt[q] - Sqrt[p] > lim, AppendTo[lst, {p, Sqrt[q] - Sqrt[p]}]]; p = q; q = NextPrime[q]]; First@ Transpose@ Sort[lst, #1[[2]] > #2[[2]] &] (* Robert G. Wilson v, Oct 18 2012 *)

More terms from Robert G. Wilson v, Oct 18 2012

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User: Thomas Nordhaus

A209407 Primes p(i) such that p(i+1)/p(i) > p(k+1)/p(k) for all k>i, where p(i) is the i-th prime.

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A186775 Numbers k such that digitsum(2^k) > digitsum(2^(k+1)).

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A079296 Primes ordered by decreasing value of the function p -> sqrt(q) - sqrt(p) where q is the next prime after p.

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