proposed

approved

editing

proposed

......for i in range(nlen(st)):

proposed

editing

editing

proposed

(PARI) a(n)=for(k=1, 10^9, st=23^k; c=0; if(#Str(st)>n, for(i=1, n, if(((st-(st%10^(i-1)))/10^(i-1))%10==((st-(st%10^i))/10^i)%10+-1, c++)); if(c==n, return(k))))

....st = str(23**k)

........if int(st[len(st)-1-i]) == int(st[len(st)-2-i])+-1:

Cf. A244613.

nonn,changed,base,fini,full

allocated for Derek OrrLeast number k such that 3^k ends in exactly n consecutive decreasing digits.

1, 5, 27, 496, 4996, 49996, 499996, 4999996, 49999996

1,2

The n consecutive decreasing digits, given by 3^a(n)%10^n, are {3, 43, 987, 4321, 54321, 654321, 7654321, 87654321, 987654321}, respectively.

3^27 ends in '987' (3 digits) and is the smallest power of 3 to end in '987', '876', '765', '654', '543', '432', '321', or '210'. So a(3) = 27.

(PARI) a(n)=for(k=1, 10^9, st=2^k; c=0; if(#Str(st)>n, for(i=1, n, if(((st-(st%10^(i-1)))/10^(i-1))%10==((st-(st%10^i))/10^i)%10+1, c++)); if(c==n, return(k))))

n=0; while(n<10, print1(a(n), ", "); n++)

(Python)

def a(n):

..for k in range(1, 10**9):

....st = str(2**k)

....if len(st) > n:

......count = 0

......for i in range(n):

........if int(st[len(st)-1-i]) == int(st[len(st)-2-i])+1:

..........count += 1

........else:

..........break

......if count == n:

........return k

n = 0

while n < 10:

..print(a(n), end=', ')

..n += 1

allocated

nonn

Derek Orr, Jul 03 2014

approved

editing

allocated for Derek Orr

allocated

approved