1,3

Table of n, a(n) for n=1..9.

a(n) = primepi(A003173(n)).

Cf. A000720, A003173.

nonn,fini,full,new

Alexander R. Povolotsky, Sep 16 2022

approved

1,2

To find a(n), start by placing n stones labeled 1 on an infinite chessboard.

Set k=2. At step k, you must place a stone labeled k in a square where the sum of its neighbors is x*k for some k >= 1. That is, the sum of the neighbors must be an integral multiple of k.

If there is no way to do that, this game ends, and you win k-1 dollars. If you can do it, you increment k and repeat.

Then a(n) = maximum number of dollars you can win with optimal play for any initial placement of the n starting stones.

The sequence was suggested by Skylark Xentha Murphy on Sep 16 2022.  She found lower bounds for a(2), a(3), and a(4).  On Sep 17 2022, Hugo van der Sanden showed that her lower bound for a(2) was the correct value, and found the values of a(3) and a(4).

In the original version of the problem (see A337663) each stone that you place must equal the sum of of its 8 neighbors.

Table of n, a(n) for n=1..4.

Skylark Xentha Murphy and Hugo van der Sanden, Illustration for a(2) = 27 {Discovered by Skylark Xentha Murphy and proved optimal by Hugo van der Sanden) [There are two choices for the 27 stone, indicated by dashed lines. The 27 in the top left corner is not part of the arrangement.]

N. J. A. Sloane, Brady Haran and Pete McPartlan, Stones on an Infinite Chessboard, Numberphile video (2022).

Hugo van der Sanden, Maximal example for a(4) = 67.

Illustration for a(2) = 27 {Discovered by Skylark Xentha Murphy and proved optimal by Hugo van der Sanden)

    .  .  .  .  .  .  .  .

    . 21 20  .  .  .  .  .

   23 13  9 18  .  .  .  .

    . 12  5  4  . 14 25  .

    . 27 10  1  3 11 26  .

    .  .  .  .  2  6  . 22

    .  .  . 19  .  1  7 15

    .  .  .  . 17 16  8  .

    .  .  .  .  .  . 24  .

    .  .  .  .  .  .  .  .

Illustration for a(3) = 41 {Discovered and proved optimal by Hugo van der Sanden)

     .  .  .  .  .  .  .  .  .  .

     .  . 28 34 39 32  .  .  .  .

     . 33  9 19 12 20  .  .  .  .

    41 24  5  4  8  . 38  .  .  .

     . 25 10  1  3 26 30  .  .  .

     .  . 11 31  2  6  7 21  .  .

     .  .  . 13 22  1 14 40  .  .

     .  .  . 35  . 29 23 15  . 37

     .  .  .  .  .  . 27 16  1 36

     .  .  .  .  .  .  .  . 17 18

     .  .  .  .  .  .  .  .  .  .

Cf. A337663.

nonn,bref,more,new

N. J. A. Sloane, Sep 17 2022, based on emails from Skylark Xentha Murphy and Hugo van der Sanden

approved

0,2

Table of n, a(n) for n=0..70.

Michael De Vlieger, Plot of a(n), n = 1..2^10 in gold, with A357082(n) = b(n) in black, records in b(n) in red, local minima in b(n) in blue.

Rémy Sigrist, PARI program

nn = 2^10; c[_] = False; j = a[0] = 0; u = 1; w = "0"; Do[k = u; While[Or[c[k], StringContainsQ[w, Set[v, IntegerString[j + k, 2]]]], k++]; Set[{a[n], c[k], b[n]}, {k, True, u}]; Set[{j, w}, {k, w <> IntegerString[k, 2]}]; If[k == u, While[c[u], u++]], {n, nn}]; Array[b, nn] (* Michael De Vlieger, Sep 15 2022 *)

(PARI) See Links section.

Cf. A357082.

nonn,new

Michael De Vlieger, Sep 15 2022

approved

0,1

Equivalently, least k > 0 such that either 10^n + k + {0, 2, 6} or 10^n + k + {0, 4, 6} are primes.

The initial term, index n = 0, is the only even term and the only case where the last member of the triplet has one digit more than the first member. The value a(0) = 4 correspond to the prime triplet (5, 7, 11). We do not consider the triplets (2, 3, 5) or (3, 5, 7) which come earlier but do not follow the standard pattern.

Table of n, a(n) for n=0..45.

Norman Luhn, smallest 7000 digit prime triplet, primenumberstheory mailing list at groups.io, Sep 14 2022

a(n) = min{ k>0 | 10^n + k + [0, 6] contains 3 primes }.

a(n) = min A007529 ∩ [10^n, oo) for n > 0.

(11, 13, 17) and (101, 103, 107) are the smallest 2-digit and 3-digit prime triplets, at distance a(1) = a(2) = 1 from 10^1 and 10^2, respectively.

(1087, 1091, 1093) is the smallest 4-digit prime triplet, at distance a(3) = 87 from 10^3.

a(6999) = 1141791245437 is the distance from 10^6999 to the smallest 7000 digit prime triplet, of the form (p, p+2, p+6).

f:= proc(n) local p;

   for p from 10^n + 1 by 2 do

     if p mod 3 = 1 then if isprime(p) and isprime(p+4) and isprime(p+6) then return p-10^n fi

     elif p mod 3 = 2 and isprime(p) and isprime(p+2) and isprime(p+6) then

return p-10^n

     fi

   od;

end proc:

f(0):= 4:

map(f, [$0..45]); # Robert Israel, Sep 15 2022

A357052 := proc(n) local p, q, r; p, q, r := 10^n, 0, 0; while p-r <> 6 do p, q, r := nextprime(p+1), p, q; od; r-10^n; end; # M. F. Hasler, Sep 15 2022

(PARI) A357052(n, q=-9, r=-9)=forprime(p=10^n, , p-r<7 && return(r-10^n); [q, r]=[p, q])

Cf. A007529 (start of prime triplets), A022004 and A022005 (start of prime triples {0,2,6} resp. {0,4,6}), A343635 (same for quintuplets).

nonn,base,new

M. F. Hasler, Sep 14 2022

approved

1,1

There are no runs of 7 consecutive odd numbers with this property, since in every run of 7 consecutive odd numbers one is divisible by 7 which is not a prime-indexed prime.

Are there such runs of 5 consecutive odd numbers? There are none below 6.6*10^7.

Table of n, a(n) for n=1..29.

121 = 11^2 is a term since 123 = 3 * 41, 125 = 5^3, and 3 = prime(2), 5 = prime(3), 11 = prime(5), 41 = prime(13) and 127 = prime(31) are all prime-indexed primes.

q[n_] := AllTrue[FactorInteger[n][[;; , 1]], PrimeQ[PrimePi[#]] &]; q[1] = True; v = q /@ {1, 3, 5, 7}; seq = {}; Do[If[And @@ v, AppendTo[seq, k - 8]]; v = Join[Rest[v], {q[k]}], {k, 9, 10^6, 2}]; seq

(PARI) is(n) = {my(p = factor(n)[, 1]); for(i = 1, #p, if(!isprime(primepi(p[i])), return(0))); return(1)};

v = vector(4); forstep(k = 3, 9, 2, v[(k-1)/2] = is(k));

forstep(k=11, 1e8, 2, q = is(k); v = concat(vecextract(v, "^1"), q); if(v[1]&&v[2]&&v[3]&&v[4], print1(k-6, ", ")))

Cf. A006450, A076610.

Subsequence of A357167 and A357168.

nonn,new

Amiram Eldar, Sep 16 2022

approved

1,2

Table of n, a(n) for n=1..41.

81 is a term since 81 = 3^4, 85 = 5 * 17, and 3 = prime(2), 5 = prime(3), 17 = prime(7) and 83 = prime(23) are all prime-indexed primes.

q[n_] := AllTrue[FactorInteger[n][[;; , 1]], PrimeQ[PrimePi[#]] &]; q[1] = True; v = q /@ {1, 3, 5}; seq = {}; Do[If[And @@ v, AppendTo[seq, k - 6]]; v = Join[Rest[v], {q[k]}], {k, 7, 10^5, 2}]; seq

(PARI) isokf(k) = my(f = factor(k)[, 1]); sum(i=1, #f, isprime(primepi(f[i]))) == #f; \\ A076610

isok(k) = (k % 2) && isokf(k) && isokf(k+2) && isokf(k+4); \\ Michel Marcus, Sep 16 2022

Cf. A006450, A076610.

Subsequence of A357167.

A357169 is a subsequence.

nonn,new

Amiram Eldar, Sep 16 2022

approved

1,2

Numbers k such that both k and k+2 are in A076610.

Since 2 is not a prime-indexed prime, all the terms of A076610 are odd, so there are no 2 consecutive integers in A076610.

Table of n, a(n) for n=1..51.

3 is a term since 3 = prime(2) and 5 = prime(3) are both prime-indexed primes.

15 is a term since 15 = 3 * 5, 15 + 2 = 17, and 3 = prime(2), 5 = prime (3) and 17 = prime(7) are all prime-indexed primes.

q[n_] := AllTrue[FactorInteger[n][[;; , 1]], PrimeQ[PrimePi[#]] &]; q[1] = True; Select[Range[1, 2500, 2], q[#] && q[# + 2] &]

(PARI) isokf(k) = my(f = factor(k)[, 1]); sum(i=1, #f, isprime(primepi(f[i]))) == #f; \\ A076610

isok(k) = (k % 2) && isokf(k) && isokf(k+2); \\ Michel Marcus, Sep 16 2022

Cf. A006450, A076610.

Subsequences: A357168, A357169.

nonn,new

Amiram Eldar, Sep 16 2022

approved

1,2

As A009993 is finite with 512 terms, a(n) is bounded with a(n) <= 511 and not 512, since A009993(1) = 0.

Table of n, a(n) for n=1..100.

G.f.: Sum_{n in A009993} x^n/(1-x^n). - Robert Israel, Sep 16 2022

22 has 4 divisors {1, 2, 11, 22} of which two have decimal digits that are not in strictly increasing order: {11, 22}, hence a(22) = 4-2 = 2.

52 has divisors {1, 2, 4, 13, 26, 52} and a(52) = 5 of them have decimal digits that are in strictly increasing order (all except 52 itself).

f:= proc(n) local d, L, i, t;

  t:= 0;

  for d in numtheory:-divisors(n) do

    L:= convert(d, base, 10);

    if `and`(seq(L[i]>L[i+1], i=1..nops(L)-1)) then t:= t+1 fi

  od;

  t

end proc:

map(f, [$1..100]); # Robert Israel, Sep 16 2022

a[n_] := DivisorSum[n, 1 &, Less @@ IntegerDigits[#] &]; Array[a, 100] (* Amiram Eldar, Sep 16 2022 *)

(PARI) isok(d) = Set(d=digits(d)) == d; \\ A009993

a(n) = sumdiv(n, d, isok(d)); \\ Michel Marcus, Sep 16 2022

(Python)

from sympy import divisors

def c(n): s = str(n); return s == "".join(sorted(set(s)))

def a(n): return sum(1 for d in divisors(n, generator=True) if c(d))

print([a(n) for n in range(1, 101)]) # Michael S. Branicky, Sep 16 2022

Cf. A009993, A357172, A357173, A160218.

Similar: A087990 (palindromic), A355302 (undulating), A355593 (alternating).

nonn,base,new

Bernard Schott, Sep 16 2022

approved

1,2

In the definition, "exactly" means the run is not part of a longer run.

Table of n, a(n) for n=1..15.

2 and 3 are 2 consecutive numbers and have the same number of prime factors, and 2 is the smallest such number, hence a(2) = 2.

(PARI) card(m)=my(c=0, k=bigomega(m)); if(bigomega(m-1)!=k, while(bigomega(m)==k, c++; m++)); c

a(n)=if(n==1, return(1)); for(m=2, +oo, if(card(m)==n, return(m)))

Cf. A001222, A077657, A356855.

Cf. A045920, A115186, A113752, A356893.

nonn,more,new

Jean-Marc Rebert, Sep 06 2022

approved

1,2

This sequence is finite since A009993 is finite with 511 nonzero terms, hence the last term is a(511) = lcm of the 511 nonzero terms of A009993.

a(511) = 8222356410...6120992000 and has 1036 digits. - Michael S. Branicky, Sep 16 2022

Table of n, a(n) for n=1..44.

For n=7, the divisors of 54 are {1, 2, 3, 6, 9, 18, 27, 54} of which 7 have their digits in strictly increasing order (all except 54). No integer < 54 has 7 such divisors, so a(7) = 54.

s[n_] := DivisorSum[n, 1 &, Less @@ IntegerDigits[#] &]; seq[len_, nmax_] := Module[{v = Table[0, {len}], n = 1, c = 0, i}, While[c < len && n < nmax, i = s[n]; If[i <= len && v[[i]] == 0, v[[i]] = n; c++]; n++]; v]; seq[25, 10^4] (* Amiram Eldar, Sep 16 2022 *)

(PARI) isok(d) = Set(d=digits(d)) == d; \\ A009993

f(n) = sumdiv(n, d, isok(d)); \\ A357171

a(n) = my(k=1); while (f(k) !=n, k++); k; \\ Michel Marcus, Sep 16 2022

(Python)

from sympy import divisors

from itertools import count, islice

def c(n): s = str(n); return s == "".join(sorted(set(s)))

def f(n): return sum(1 for d in divisors(n, generator=True) if c(d))

def agen():

    n, adict = 1, dict()

    for k in count(1):

        fk = f(k)

        if fk not in adict: adict[fk] = k

        while n in adict: yield adict[n]; n += 1

print(list(islice(agen(), 37))) # Michael S. Branicky, Sep 16 2022

Cf. A009993, A357171, A357173, A160218.

Similar: A087997 (palindromic), A355303 (undulating), A355594 (alternating).

nonn,base,fini,new

Bernard Schott, Sep 16 2022

More terms from Amiram Eldar, Sep 16 2022

approved