A134269 Number of solutions to the equation p^k - p^(k-1) = n, where k is a positive integer and p is prime.
1, 2, 0, 2, 0, 2, 0, 1, 0, 1, 0, 1, 0, 0, 0, 2, 0, 2, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 2, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 2, 0, 1, 0, 0, 0
Offset: 1
Keywords
Examples
Notice that it is not possible to have more than 2 solutions, but say when n=4 there are two solutions, namely 5^1 - 5^0 and 2^3 - 2^2. a(2) = 2 refers to 2^2 - 2^1 = 2 and 3^1 - 3^0 = 2. a(6) = 2 as 6 = 3^2 - 3^1 = 7^1 - 7^0.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..65537
Programs
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Maple
A134269 := proc(n) local a,p,r ; a := 0 ; p :=2 ; while p <= n+1 do r := n/(p-1) ; if type(r,'integer') then if r = 1 then a := a+1 ; else r := ifactors(r)[2] ; if nops(r) = 1 then if op(1,op(1,r)) = p then a := a+1 ; end if; end if; end if; end if; p := nextprime(p) ; end do: return a; end proc: # R. J. Mathar, Aug 06 2013
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PARI
lista(N=100) = {tab = vector(N); for (i=1, N, p = prime(i); for (j=1, N, v = p^j-p^(j-1); if (v <= #tab, tab[v]++););); for (i=1, #tab, print1(tab[i], ", "));} \\ Michel Marcus, Aug 06 2013
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PARI
A134269list(up_to) = { my(v=vector(up_to)); forprime(p=2,1+up_to, for(j=1,oo,my(d = (p^j)-(p^(j-1))); if(d>up_to,break,v[d]++))); (v); }; v134269 = A134269list(up_to); A134269(n) = v134269[n]; \\ Antti Karttunen, Nov 09 2018
Extensions
a(2) corrected by Michel Marcus, Aug 06 2013
More terms from Antti Karttunen, Nov 09 2018
Comments