A359012 - OEIS

%I #35 Jan 06 2023 20:54:31

%S 318,557,692,729,2226,2437,2776,3209,4436,5336,5549,5718,5956,6068,

%T 6141,6353,6958,7045,7046,7338,7345,7643,7865,8261,8409,9153,9178,

%U 9242,9544,9569,9664,9894,9999,10174,10889,12389,12434,13497,13516,16308,18695,19707,21940,21954,22535

%N Numbers k that are a substring of xPy where k=concatenation(x,y) and xPy is the number of permutations A008279(x,y).

%C If n and d are two nonnegative integers, and d <= n, then the number of permutations is obtained by the formula nPd = n!/(n-d)!.

%e 318 is present in 31P8 (= 318073392000 = A008279(31, 8)).

%e 557 is present in 55P7 (= 1022755734000 = A008279(55, 7)).

%e 692 is present in 69P2 (= 4692 = A008279(69, 2)).

%o (Python)

%o import math

%o def is_valid_sequence_number(n):

%o     num_str = str(n)

%o     length = len(num_str)

%o     for count in range(math.ceil(length / 2), length):

%o         if num_str in str(

%o             math.perm(int(num_str[:count]), int(num_str[-(length - count) :]))

%o         ):

%o             return True

%o     return False

%o A359012 = []

%o for num in range(10, 10**4):

%o     if is_valid_sequence_number(num):

%o         A359012.append(num)

%o (PARI) T(n,k) = n!/(n-k)!; \\ A008279

%o isok(k) = my(d=digits(k), s=Str(k), d1, d2); for (i=1, #d-1, d1=fromdigits(Vec(d, i)); d2=fromdigits(vector(#d-i, k, d[i+k])); if ((d1 >= d2) && (#strsplit(Str(T(d1,d2)), s) > 1), return(1));); \\ _Michel Marcus_, Dec 12 2022

%Y Cf. A008279.

%K nonn,base

%O 1,1

%A _John Samuel_, Dec 11 2022