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.

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A243625 a(n) is the smallest positive integer not already in the sequence for which a(n)+a(n-1) is a semiprime, with a(1)=1.

Original entry on oeis.org

1, 3, 6, 4, 2, 7, 8, 13, 9, 5, 10, 11, 14, 12, 21, 17, 16, 18, 15, 19, 20, 26, 23, 28, 27, 22, 24, 25, 30, 32, 33, 29, 36, 38, 31, 34, 35, 39, 43, 42, 40, 37, 45, 41, 44, 47, 46, 48, 58, 53, 62, 49, 57, 54, 52, 59, 56, 50, 61, 60, 51, 55, 63, 66, 67, 74, 68
Offset: 1

Views

Author

Michel Lagneau, Jun 08 2014

Keywords

Comments

It is probable that every positive integer occurs, and that this is a permutation of natural numbers.
a(n) = n for n = 1, 4, 9, 18, 23, 48, 54, 60, 63, 77, 91, 92, .... (375 cases for first 3000 terms). - Zak Seidov, Feb 22 2017

Examples

			a(3)=6 because 1 and 3 have already been used in the sequence and 3+2=5, 3+4=7 and 3+5=8 are not semiprime while 3+6=9 is semiprime.
		

Crossrefs

Cf. A055265.

Programs

  • Maple
    N:= 1000; # to get all terms up to a(N)
    issp:= proc(n) local F; F:= ifactors(n)[2]; add(f[2],f=F)=2 end proc:
    S:= {1}; m:= 1; R:= {}; a[1]:= 1;
    for n from 2 to N do
      found:= false;
      for k in R do
        if issp(a[n-1]+k) then
          a[n]:= k;
          S:= S union {k};
          R:= R minus {k};
          found:= true;
          break
        fi;
      od;
      if not found then
        for k from m+1 do
          if issp(a[n-1]+k) then
            a[n]:= k;
            S:= S union {k};
            R:= R union {$(m+1)..(k-1)};
            m:= k;
            break
          fi
        od
      fi
    od:
    seq(a(n),n=1..N); # Robert Israel, Jun 08 2014
  • Mathematica
    f[s_List] := Block[{k = 1, a = s[[ -1]]}, While[ MemberQ[s, k] || ! Plus@@Last/@FactorInteger[a+k] == 2, k++ ]; Append[s, k]]; Nest[f, {1}, 71]

A329563 For all n >= 1, exactly five sums are prime among a(n+i) + a(n+j), 0 <= i < j < 5; lexicographically earliest such sequence of distinct positive numbers.

Original entry on oeis.org

1, 2, 3, 4, 5, 8, 9, 14, 6, 23, 17, 7, 12, 24, 10, 13, 19, 16, 18, 25, 22, 15, 28, 21, 26, 32, 75, 20, 11, 27, 56, 30, 41, 53, 29, 38, 60, 44, 35, 113, 36, 31, 48, 61, 37, 42, 46, 33, 34, 55, 39, 40, 49, 58, 45, 43, 52, 51, 106, 57, 62, 50, 87, 47, 54, 59, 80, 66, 83, 68
Offset: 1

Views

Author

M. F. Hasler, Feb 09 2020

Keywords

Comments

That is, there are 5 primes, counted with multiplicity, among the 10 pairwise sums of any 5 consecutive terms.
Conjectured to be a permutation of the positive integers.
This sequence is quite different from the restriction of the "nonnegative" variant A329564 to positive indices: it seems that the two have no common terms beyond a(6) = 8, except for the accidental a(22) = 15 and maybe some later coincidences of this type. There also appears to be no other simple relation between the terms of these sequences, in contrast to, e.g., A055265 vs. A128280.

Examples

			For n = 1, we consider pairwise sums among the first 5 terms chosen as small as possible, a(1..5) = (1, 2, 3, 4, 5). We see that we have indeed 5 primes among the sums 1+2, 1+3, 1+4, 1+5, 2+3, 2+4, 2+5, 3+4, 3+5, 4+5.
Then, to get a(6), consider first the pairwise sums among terms a(2..5), (2+3, 2+4, 2+5; 3+4, 3+5; 4+5), among which there are 3 primes, counted with multiplicity (i.e., the prime 7 is there two times). So the new term a(6) must give exactly two more prime sums with the terms a(2..5). We find that 6 or 7 would give just one more (5+6 resp. 4+7), but a(6) = 8 gives exactly two more, 3+8 and 5+8.
		

Crossrefs

Cf. A329425 (6 primes using 5 consecutive terms), A329566 (6 primes using 6 consecutive terms).
Cf. A329449 (4 primes using 4 consecutive terms), A329456 (4 primes using 5 consecutive terms).
Cf. A329454 (3 primes using 4 consecutive terms), A329455 (3 primes using 5 consecutive terms).
Cf. A329411 (2 primes using 3 consecutive terms), A329452 (2 primes using 4 consecutive terms), A329453 (2 primes using 5 consecutive terms).
Cf. A329333 (1 (odd) prime using 3 terms), A128280 & A055265 (1 prime using 2 terms); A055266 & A253074 (0 primes using 2 terms), A329405 & A329450 (0 primes using 3 terms), A329406 ff: other variants.

Programs

  • PARI
    {A329563(n,show=1,o=1,N=5,M=4,p=[],u=o,U)=for(n=o,n-1, show>0&& print1(o","); show<0&& listput(L,o); U+=1<<(o-u); U>>=-u+u+=valuation(U+1,2); p=concat(if(#p>=M,p[^1],p),o); my(c=N-sum(i=2,#p, sum(j=1,i-1, isprime(p[i]+p[j])))); if(#p
    				

A308334 Lexicographically earliest sequence of distinct positive numbers such that for any n > 0, a(n) OR a(n+1) is a prime number (where OR denotes the bitwise OR operator).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 16, 13, 8, 11, 9, 10, 21, 12, 17, 14, 19, 15, 18, 23, 20, 25, 22, 27, 28, 29, 24, 31, 26, 33, 36, 37, 32, 41, 34, 43, 35, 40, 39, 42, 45, 38, 47, 44, 49, 52, 53, 48, 59, 50, 57, 51, 56, 61, 60, 67, 62, 65, 63, 64, 71, 58, 69, 66, 77, 54
Offset: 1

Views

Author

Rémy Sigrist, May 20 2019

Keywords

Comments

By Dirichlet's theorem on arithmetic progressions, we can always extend the sequence: say a(n) < 2^k, then a(n) OR 1 and 2^k are coprime and there are infinitely many prime numbers of the form (a(n) OR 1) + m*2^k = a(n) OR (1 + m*2^k) and we can extend the sequence.
Will every integer appear in this sequence?
Numerous sequences are based on the same model: the sequence is the lexicographically earliest sequence of distinct positive terms such that some function in two variables yields prime numbers when applied to consecutive terms:
f(u,v) Analog sequence
------- -----------------
u OR v a (this sequence)
u + v A055265
u*v + 1 A073666
u*v - 1 A081943
abs(u-v) A065186
max(u,v) A282649
u^2 + v^2 A100208
The appearance of numbers much earlier or later than their corresponding index is flagged strikingly in the plot2 graph of a(n)/n (see links). - Peter Munn, Sep 10 2022

Examples

			The first terms, alongside a(n) OR a(n+1), are:
  n   a(n)  a(n) OR a(n+1)
  --  ----  --------------
   1     1               3
   2     2               3
   3     3               7
   4     4               5
   5     5               7
   6     6               7
   7     7              23
   8    16              29
   9    13              13
  10     8              11
  11    11              11
  12     9              11
		

Crossrefs

See A308340 for the corresponding prime numbers.
See A055265, A065186, A073666, A081943, A100208, A282649 for similar sequences.

Programs

  • PARI
    s=0; v=1; for (n=1, 67, s+=2^v; print1 (v ", "); for (w=1, oo, if (!bittest(s,w) && isprime(o=bitor(v,w)), v=w; break)))
    
  • Python
    from sympy import isprime
    from itertools import count, islice
    def agen():
        aset, k, mink = {1}, 1, 2
        for n in count(1):
            an = k; yield an; aset.add(an)
            s, k = set(str(an)), mink
            while k in aset or not isprime(an|k): k += 1
            while mink in aset: mink += 1
    print(list(islice(agen(), 67))) # Michael S. Branicky, Sep 10 2022

A329568 For all n >= 1, exactly 9 sums are prime among a(n+i) + a(n+j), 0 <= i < j < 6: lexicographically earliest such sequence of distinct positive numbers.

Original entry on oeis.org

1, 2, 3, 9, 4, 10, 27, 14, 33, 57, 26, 40, 87, 50, 21, 63, 16, 20, 51, 8, 81, 93, 46, 56, 15, 58, 135, 183, 28, 44, 39, 88, 69, 123, 34, 68, 105, 128, 45, 129, 22, 52, 141, 38, 75, 159, 32, 82, 99, 64, 117, 147, 80, 94, 177, 116, 237, 273, 74, 100, 387, 76, 207, 357, 62, 104, 165, 86, 77, 95
Offset: 1

Views

Author

M. F. Hasler, Feb 10 2020

Keywords

Comments

That is, there are nine primes, counted with multiplicity, among the 15 pairwise sums of any six consecutive terms. This is the maximum number of possible prime sums for any set of 6 numbers > 1, see wiki page for details.
Conjectured to be a permutation of the positive integers. See A329569 = (0, 1, 2, 5, 6, 11, 12, 17, ...) for the quite different variant for nonnegative integers.
For n > 6, a(n) is the smallest number not used earlier such that the set a(n) + {a(n-5), ..., a(n-1)} has the same number of primes as a(n-6) + {a(n-5), ..., a(n-1)}. Such a number always exists, by definition of the sequence. (If it would not exist for a given n, the term a(n-1) (or earlier) "was wrong and must be corrected", so to say.) See the wiki page for further considerations about existence and surjectivity.
For a(4), one must exclude the values {4, ..., 8} to get an infinite sequence, but for all other (at least several hundred) terms, the greedy choice gives the correct solution.

Crossrefs

Cf. A055265, A128280 (1 prime from 2 terms), A329333 (1 prime from 3 terms), A329405, ..., A329417 (N primes from M terms >= 1), A329425, A329449, ..., A329581 (N primes from M terms >= 0).

Programs

  • PARI
    {A329568(n,show=0,o=1,N=9,M=5,X=[[4,x]|x<-[4..8]],p=[],u=o,U)=for(n=o+1,n, show>0&& print1(o","); show<0&& listput(L,o); U+=1<<(o-u); U>>=-u+u+=valuation(U+1,2); p=concat(if(#p>=M,p[^1],p),o); my(c=N-sum(i=2,#p, sum(j=1,i-1, isprime(p[i]+p[j])))); for(k=u,oo,bittest(U,k-u)|| min(c-#[0|x<-p,isprime(x+k)],#p>=M)|| setsearch(X,[n,k])|| [o=k,break])); show&&print([u]);o} \\ optional args: show=1: print a(o..n-1), show=-1: append them on global list L, in both cases print [least unused number] at the end. Parameters N,M,o,... allow getting other variants, see the wiki page for more.

A329569 For all n >= 0, exactly 9 sums are prime among a(n+i) + a(n+j), 0 <= i < j < 6: lexicographically earliest such sequence of distinct nonnegative numbers.

Original entry on oeis.org

0, 1, 2, 5, 6, 11, 12, 17, 26, 35, 36, 47, 24, 77, 32, 65, 62, 149, 74, 9, 8, 39, 14, 15, 4, 3, 28, 33, 38, 69, 10, 51, 20, 21, 58, 93, 16, 81, 46, 13, 70, 27, 76, 37, 34, 97, 52, 7, 30, 49, 40, 31, 22, 67, 82, 19, 42, 25, 64, 85, 18, 109, 54, 43, 88, 139, 84, 145, 94, 79, 112, 55, 48, 289, 144
Offset: 0

Views

Author

M. F. Hasler, Feb 10 2020

Keywords

Comments

That is, there are nine primes, counted with multiplicity, among the 15 pairwise sums of any six consecutive terms. This is the maximum: there can't be more than 9 primes among the pairwise sums of any 6 numbers > 1, cf. wiki page in LINKS.
Conjectured to be a permutation of the nonnegative integers. The restriction to [1,oo) is then a permutation of the positive integers with similar properties, but different from the lexico-smallest one, A329568 = (1, 2, 3, 9, 4, 10, 27, ...).
For n > 5, a(n) is the smallest number not used earlier such that the set a(n) + {a(n-5), ..., a(n-1)} has the same number of primes as a(n-6) + {a(n-5), ..., a(n-1)}. Such a number always exists, by definition of the sequence. (If it would not exist for a given n, the term a(n-1) (or earlier) "is wrong and must be corrected", so to say.) See the wiki page for further considerations about existence and surjectivity.
For a(3) and a(4), one must exclude values 3 & 4 to be able to continue the sequence indefinitely, but in all other cases (at least for several hundred terms), the greedy choice gives the correct solution.
The values 3, 4 and 7 appear quite late at indices 25, 24 resp. 47.

Crossrefs

Cf. A055265, A128280 (1 prime from 2 terms), A329333 (1 prime from 3 terms), A329405, ..., A329416 (N primes from M terms >= 1), A329425, A329449, ..., A329581 (N primes from M terms >= 0).

Programs

  • PARI
    {A329569(n,show=0,o=0,N=9,M=5,X=[[3,3],[3,4],[4,3],[4,4]],p=[],u=o,U)=for(n=o+1,n, show>0&& print1(o","); show<0&& listput(L,o); U+=1<<(o-u); U>>=-u+u+=valuation(U+1,2); p=concat(if(#p>=M,p[^1],p),o); my(c=N-sum(i=2,#p, sum(j=1,i-1, isprime(p[i]+p[j])))); for(k=u,oo,bittest(U,k-u)|| min(c-#[0|x<-p,isprime(x+k)],#p>=M)|| setsearch(X,[n,k])|| [o=k,break])); show&&print([u]);o} \\ optional args: show=1: print a(o..n-1), show=-1: append them on global list L, in both cases print [least unused number] at the end. Parameters N,M,o,... allow getting other variants, see the wiki page for more.

A284049 a(n) is the smallest positive integer not already in the sequence such that a(n) + a(n-1) is a prime power, with a(1) = 1.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 9, 8, 11, 12, 13, 10, 15, 14, 17, 20, 21, 16, 25, 18, 19, 22, 27, 26, 23, 24, 29, 30, 31, 28, 33, 34, 37, 36, 35, 32, 39, 40, 41, 38, 43, 46, 51, 50, 47, 42, 55, 48, 49, 52, 45, 44, 53, 54, 59, 62, 63, 58, 67, 60, 61, 64, 57, 56, 65, 66, 71, 68, 69, 70, 79, 72, 77, 74, 75, 76, 73, 78, 85, 82
Offset: 1

Views

Author

Ilya Gutkovskiy, Mar 19 2017

Keywords

Comments

Conjectured to be a permutation of the natural numbers.

Examples

			a(8) = 9 because 1, 2, 3, 4, 5, 6 and 7 have already been used in the sequence, 7 + 8 = 15 is not prime power while 7 + 9 = 16 is a prime power.
		

Crossrefs

Programs

  • Maple
    N:= 100: # to get all terms before the first term > N
    S:= [$2..N]:
    a[1]:= 1: found:= true:
    for n from 2 while found do
      found:= false;
      for j from 1 to nops(S) do
        if ispp(a[n-1]+S[j]) then
          found:= true;
          a[n]:= S[j];
          S:= subsop(j=NULL,S);
          break
        fi
      od;
    od:
    seq(a[i],i=1..n-2); # Robert Israel, Apr 16 2017
  • Mathematica
    f[s_List] := Block[{k = 1, a = s[[-1]]}, While[MemberQ[s, k] || ! PrimePowerQ[a + k], k++]; Append[s, k]]; Nest[f, {1}, 80]

A329572 For all n >= 0, exactly 12 sums are prime among a(n+i) + a(n+j), 0 <= i < j < 7; lexicographically earliest such sequence of distinct nonnegative numbers.

Original entry on oeis.org

0, 1, 2, 5, 6, 11, 12, 17, 26, 35, 36, 47, 24, 54, 77, 7, 43, 60, 13, 30, 96, 4, 67, 97, 16, 133, 34, 3, 40, 27, 63, 100, 10, 20, 171, 9, 8, 51, 21, 22, 52, 15, 32, 38, 75, 141, 56, 41, 71, 122, 152, 45, 68, 29, 59, 14, 39, 44, 50, 23, 53, 57, 74, 107, 170, 176, 93, 134, 137, 86, 177, 65, 476, 62, 87, 92, 101
Offset: 0

Views

Author

M. F. Hasler, Feb 09 2020

Keywords

Comments

That is, there are 12 primes, counted with multiplicity, among the 21 pairwise sums of any 7 consecutive terms.
This is the theoretical maximum: there can't be more than 12 primes in pairwise sums of 7 distinct numbers > 1. See the wiki page for more details.
Conjectured to be a permutation of the nonnegative integers. See A329573 for the "positive" variant: same definition but with offset 1 and positive terms, leading to a quite different sequence.
For a(3) and a(4) resp. a(5) one must forbid the values < 5 resp. < 11 which would be the greedy choices, in order to get a solution for a(7), but from then on, the greedy choice gives the correct solution, at least for several hundred terms.

Crossrefs

Cf. A055273 (analog starting with a(1) = 1), A055265 & A128280 (1 prime using 2 terms), A055266 & A253074 (0 primes using 2 terms), A329405 - A329416, A329425, A329333, A329449 - A329456, A329563 - A329581.

Programs

  • PARI
    {A329572(n,show=0,o=0,N=12,M=6,D=[3,5,4,6,5,11],p=[],u=o,U)=for(n=o+1,n, show>0&& print1(o","); show<0&& listput(L,o); U+=1<<(o-u); U>>=-u+u+=valuation(U+1,2); p=concat(if(#p>=M,p[^1],p),o); D&& D[1]==n&& [o=D[2],D=D[3..-1]]&& next; my(c=N-sum(i=2,#p, sum(j=1,i-1, isprime(p[i]+p[j])))); for(k=u,oo,bittest(U,k-u)|| min(c-#[0|p<-p,isprime(p+k)],#p>=M)|| [o=k,break]));show&&print([u]);o} \\ optional args: show=1: print a(o..n-1), show=-1: append them on global list L, in both cases print [least unused number] at the end. See the wiki page for more.

A329573 For all n >= 1, exactly 12 sums are prime among a(n+i) + a(n+j), 0 <= i < j < 7; lexicographically earliest such sequence of distinct positive numbers.

Original entry on oeis.org

1, 2, 3, 4, 9, 10, 27, 14, 20, 33, 34, 69, 39, 28, 40, 13, 19, 70, 31, 43, 180, 220, 61, 36, 66, 91, 127, 7, 12, 5, 102, 186, 11, 6, 25, 18, 55, 41, 42, 48, 65, 72, 59, 38, 125, 24, 29, 35, 54, 32, 47, 77, 164, 26, 407, 15, 116, 63, 75, 404, 416, 8, 215, 45, 56, 183, 23, 134, 206, 17, 44, 50
Offset: 1

Views

Author

M. F. Hasler, Feb 09 2020

Keywords

Comments

That is, there are 12 primes, counted with multiplicity, among the 21 pairwise sums of any 7 consecutive terms.
This is the theoretical maximum: there can't be more than 12 primes in pairwise sums of 7 distinct numbers > 1. See the wiki page for more details.
Conjectured to be a permutation of the positive integers. See A329572 for the nonnegative variant (same definition but with n >= 0 and terms >= 0), leading to a quite different sequence.
For a(5) and a(6) one must forbid values up to 8 in order to be able to find a solution for a(7), but from then on, the greedy choice gives the correct solution, at least for several hundred terms. Small values appearing late are a(30) = 5, a(34) = 6, a(28) = 7, a(62) = 8.

Examples

			Up to and including the 6th term, there is no constraint other than not using a term more than once, since it is impossible to have more than 12 primes as pairwise sums of 6 numbers. So one would first try to use the lexicographically smallest possible choice a(1..6) =?= (1, 2, ..., 6). But then one would have only 7 pairs (i,j) such that a(i) + a(j) is prime, 1 <= i < j <= 6. So one would need 12 - 7 = 5 more primes in {1, 2, ..., 6} + a(7), which is impossible. One can check that even a(1..5) =?= (1,...,5) does not allow one to find a(6) and a(7) in order to have 12 prime sums a(i) + a(j), 1 <= i < j <= 7. Nor is it possible to find a solution with a(5) equal to 6 or 7 or 8. One finds that a(5) = 9, and a(6) = 10, are the smallest possible choices for which a(7) can be found as to satisfy the requirement. In that case, a(7) = 27 is the smallest possible solution, which yields the 12 prime sums 1+2, 2+3, 1+4, 3+4, 2+9, 4+9, 1+10, 3+10, 9+10, 2+27, 4+27, 10+27.
Now, to satisfy the definition of the sequence for n = 2, we drop the initial 1 from the set of consecutive terms, and search for a(8) producing the same number of additional primes together with {2, 3, 4, 9, 10, 27} as did a(1) = 1, namely 3. We see that a(8) = 14 is the smallest possibility. And so on.
It seems that once a(5) and a(6) are chosen, one may always take the smallest possible choice for the next term without ever again running into difficulty. This is in strong contrast to the (exceptional) case of the variant where we require 10 prime sums among 7 consecutive terms, cf. sequence A329574.
		

Crossrefs

Cf. A055272 (analog starting with a(0)=0), A055265 & A128280 (1 prime using 2 terms), A055266 & A253074 (0 primes using 2 terms), A329405 - A329416, A329425, A329333, A329449 - A329456, A329563 - A329581.

Programs

  • PARI
    {A329573(n,show=0,o=1,N=12,M=6,D=[5,9,6,10],p=[],u=o,U)=for(n=o+1,n, show>0&& print1(o","); show<0&& listput(L,o); U+=1<<(o-u); U>>=-u+u+=valuation(U+1,2); p=concat(if(#p>=M,p[^1],p),o); D&& D[1]==n&& [o=D[2],D=D[3..-1]]&& next; my(c=N-sum(i=2,#p, sum(j=1,i-1, isprime(p[i]+p[j])))); for(k=u,oo,bittest(U,k-u)|| min(c-#[0|p<-p,isprime(p+k)],#p>=M)|| [o=k,break]));show&&print([u]);o} \\ optional args: show=1: print a(o..n-1), show=-1: append them on global list L, in both cases print [least unused number] at the end. See the wiki page for more.

A086526 Beginning with 2, the smallest even number not included earlier such that the arithmetic mean of a pair of successive terms is a prime.

Original entry on oeis.org

2, 4, 6, 8, 14, 12, 10, 16, 18, 20, 26, 32, 30, 28, 34, 24, 22, 36, 38, 44, 42, 40, 46, 48, 58, 60, 62, 56, 50, 68, 54, 52, 66, 76, 70, 64, 78, 80, 86, 72, 74, 84, 82, 96, 98, 104, 90, 88, 106, 100, 94, 108, 110, 92, 102, 112, 114, 140, 122, 132, 130, 124, 138, 116, 146, 128
Offset: 1

Views

Author

Amarnath Murthy, Jul 31 2003

Keywords

Comments

A rearrangement of even numbers.

Crossrefs

Cf. A086527.

Formula

a(n) = 2 * A055265(n).

Extensions

More terms from David Wasserman, Mar 14 2005

A284048 a(n) is the smallest positive integer not already in the sequence such that a(n) + a(n-1) is a proper prime power (A246547), with a(1) = 1.

Original entry on oeis.org

1, 3, 5, 4, 12, 13, 14, 2, 6, 10, 15, 17, 8, 19, 30, 34, 47, 74, 7, 9, 16, 11, 21, 28, 36, 45, 76, 49, 32, 89, 39, 25, 24, 40, 41, 23, 26, 38, 43, 78, 50, 31, 18, 46, 35, 29, 20, 44, 37, 27, 22, 42, 79, 90, 153, 103, 66, 55, 70, 51, 77, 48, 33, 88, 81, 162, 94, 75, 53, 68, 57, 64, 61, 60, 65, 56, 69, 52, 73, 96, 147
Offset: 1

Views

Author

Ilya Gutkovskiy, Mar 19 2017

Keywords

Examples

			a(5) = 12 because 1, 3, 4 and 5 have already been used in the sequence, 4 + 2 = 6, 4 + 6 = 10, 4 + 7 = 11, 4 + 8 = 12, 4 + 9 = 13, 4 + 10 = 14 and 4 + 11 = 15 are not proper prime powers while 4 + 12 = 16 is a proper prime power.
		

Crossrefs

Programs

  • Maple
    N:= 2000: # to get all terms before the first where a(n)+a(n-1)>N
    PP:= {seq(seq(p^j, j =2..floor(log[p](N))),p = select(isprime,[2,seq(i,i=3..floor(sqrt(N)),2)]))}:
    PP:= sort(convert(PP,list)):
    V:= Vector(N,datatype=integer[1],1):
    A[1]:= 1; V[1]:= 0;
    for n from 2 do
      for pp in PP do
        t:= pp - A[n-1];
        if t > 0 and V[t] = 1 then
          A[n]:= t; V[t]:= 0; break
        fi;
      od;
      if not assigned(A[n]) then break fi
    od:
    seq(A[i],i=1..n-1); # Robert Israel, Apr 24 2017
  • Mathematica
    f[s_List] := Block[{k = 1, a = s[[-1]]}, While[MemberQ[s, k] || ! (PrimePowerQ[a + k] && PrimeOmega[a + k] > 1), k++];Append[s, k]]; Nest[f, {1}, 80]
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