A359012 Numbers k that are a substring of xPy where k=concatenation(x,y) and xPy is the number of permutations A008279(x,y).

318, 557, 692, 729, 2226, 2437, 2776, 3209, 4436, 5336, 5549, 5718, 5956, 6068, 6141, 6353, 6958, 7045, 7046, 7338, 7345, 7643, 7865, 8261, 8409, 9153, 9178, 9242, 9544, 9569, 9664, 9894, 9999, 10174, 10889, 12389, 12434, 13497, 13516, 16308, 18695, 19707, 21940, 21954, 22535

Offset: 1

Author

John Samuel, Dec 11 2022

Keywords

• nonn
• base

Comments

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)!.

Examples

			318 is present in 31P8 (= 318073392000 = A008279(31, 8)).
557 is present in 55P7 (= 1022755734000 = A008279(55, 7)).
692 is present in 69P2 (= 4692 = A008279(69, 2)).
		

Crossrefs

Cf. A008279.

Programs

PARI

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

Python

import math
def is_valid_sequence_number(n):
    num_str = str(n)
    length = len(num_str)
    for count in range(math.ceil(length / 2), length):
        if num_str in str(
            math.perm(int(num_str[:count]), int(num_str[-(length - count) :]))
        ):
            return True
    return False
A359012 = []
for num in range(10, 10**4):
    if is_valid_sequence_number(num):
        A359012.append(num)